SL ARORA CLASS 12TH PHYSICS BY ROCKY TRICKSTER.pdf

SL ARORA CLASS 12 TH PHYSICS

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2. cHAP T E R
ELECTRIC C
HARGES
AND FIELD
1.1
FRICTIONAL ELECTRICITY
1. Whatisfrictionalelectricity? When is abodysaid
to be electrified orcharged ?
t"
Frictional electricity.If a glass rod is rubbed with a
silk cloth, or a fountain-pen with a coat-sleeve, it is able
to attract small pieces of paper, straw, lint, light feathers,
etc.Similarly, a plastic comb passed through dry hair
can attract such light objects. In all these examples, we
can say that the rubbed substance has becomeelectrified
orelectricallycharged. It is because of friction that the
substances get charged on rubbing.
Theproperty of rubbedsubstances duetowhich they
attract light objectsiscalled electricity. The electricity
developed by rubbing orfrictioniscalledfrictional or static
electricitu.The rubbed substances which show this property
of attraction are said to have become electrified or
electrically charged.
2.Give a historical view offrictionalelectricity. From
where did the term electricity get its origin?
Historical view of frictional electricity.In600B.C.,
Thales of Miletus, one of the founders of Greek science,
first noticed that if a piece of amber is rubbed with a
woollen cloth, it then acquires the property of attrac-
ting light feathers, dust, lint, pieces of leaves, etc.
In1600AD.,William Cillbert,the personal doctor to
Queen Elizabeth -I of England, made a systematic study
of the substances that behave like amber. In his book
De Magnete(on the magnet), he introduced the name
electrica for such substances. In fact, the Greek name for
amber iselektronwhich is the origin of all such words:
electricity, electric force, electric charge and electron.
For Your Knowledge
~Amber is a yellow resinous (gum like) substance
found on the shores of the Baltic sea.
~Both electric and magnetic phenomena can be derived
from charged particles. Magnetism arises from
charges in motion. The charged particles in motion
exert both electric and magnetic forces on each other.
Hence electricity and magnetism are studied together
aselectromagnetism.
1.2
ELECTRIC CHARGE
3. Whatiselectric charge ?Isita scalar or vector
quantity? Name its 51 unit.
Electric charge.Electric charge is an intrinsic property
of the elementary particles like electrons, protons, etc.,
of which all the objects are made up of. It is because of
these electric charges that various objects exert strong
electric forces of attraction or repulsion on each other.
Electric chargeisan intrinsic property of elementary
particles of matter which gives rise to electric force between
various objects.
Electric charge is a scalar quantity. Its51unit is
coulomb(C).A proton has a positive charge (+e)and an
electron has a negative charge(-e),where
e=1.6x10-19coulomb
(1.1){{{0rsxiwhvmzi0gsq

3. 1.2
Large-scale matter
that consists of equal number of
electrons and protons iselectrically neutral. If there is
an excess of electrons, the body has anegative charge
and an excess of protons results in a positive charge.
1.3ELECTROSTATICS
4. Whatiselectrostatics ?Mentionsome of its
important applications.
Electrostatics.Electrostatics isthe study of electric
charges at rest.Here westudy the forces, fields and
potentials associated with static charges.
Applications of electrostatics.The attraction and
repulsion between charged bodies have many indus-
trial applications. Some of these are as follows:
1. In electrostatic loudspeaker.
2. In electrostatic spraying of paints and powder
coating.
3. In flyash collection in chimneys.
4. In a Xerox copying machine.
5. In the design of a cathode-ray tube used in
television and radar.
1.4TWO KINDS OF ELECTRIC CHARGES
5. How will you show experimentally that (i)thereare
only two kinds ofelectric charges and (ii) like charges
repel and unlike charges attract each other ?
Two kinds of electric charges. About 100years ago,
Charles DuFayof France showed that electric charges
on various objects are of only two kinds. The following
simple experiments prove this fact.
EXPERIMENT1
(i)Rub a glass rod with silk and suspend it from a
rigid support by means of a silk thread. Bring
another similarly charged rod near it. The two
rods repel each other [Fig. l.1(a)].
Silk ~ilk
Glass + Glass
++)t ++JI
'JRepulslOn 1 tic ./
~ ~_p_a_5___ Attraction
~ ~
(a) (b)
~ilk
-_ Plastic
----JRepulsion
~
(c)
Fig. 1.1 Like charges repel and unlike charges
attract each other.
PHYSICS-XII
(ii)Bring a plastic rod rubbed withwool near the
chargedglassrod.Thetworods attract each
other [Fig. l.1(b)].
(iii)Now rub a plastic rod withwool and suspend it
from a rigid support. Bring another similarly
charged plastic rod near it.There will be a
repulsion between the two rods[Fig. 1.1(c)].
EXPERIMENT2.If a glass rod, rubbed withsilk, is
made totouch two small pith balls (orpolystyrene
balls) which are suspended by silkthreads,thenthe two
balls repel eachother, as shown in Fig. 1.2(a). Similarly,
two pith balls touched withaplasticrod rubbed with
fur are found to repel each other [Fig.1.2(b)]. Butitis
seen that a pith ball touched withglassrodattracts
another pith ball touched with a plastic rod [Fig. 1.2(c)].
(a)Repulsion (b)Repulsion
+
--(c)Attraction
Fig. 1. 2Like charges repel and unlike charges attract.
From the above experiments, we note that the
charge produced on a glass rod is different from the
charge produced onaplastic rod. Also the charge
produced ona pith balltouched with a glass rod is
different from the charge produced on pith ball
touched withaplastic rod. We can conclude that:
1. There are only two kinds of electric charges - positive
and negative.
2.Like chargesrepeland unlike chargesattract eachother.
The statement 2 is known asthefundamental law
of electrostatics.
Theaboveexperiments also demonstrate-that the
charges are transferred from the rods tothepith balls
on contact. We say that the pith balls havebeen
electrified or charged by contact. This property which
distinguishes the two kinds of chargesiscalled thepolarity
of charge.{{{0rsxiwhvmzi0gsq

4. ELECTR
IC CHARGES AND FIELD
6. What are vitreous andresinous charges ?What
was wrong with this nomenclature?
Vitreous and resinous charges.CharlesDu Fayusedthe
terms vitreous and resinous for thetwokinds ofcharges.
1.The charge developed on glass rod when rubbed with
silkwas calledvitreous charge (Latinvirtum=glass).
2.The charge developed on amber when rubbed with
woolwas called resinous charge (amber is a resin).
But later on, these terms were found to be
misleading. For example, a ground glass rod develops
resinous electricity while a highly polished ebonite rod
developsvitreous electricity.
7.What are positive and negative charges ?Whatis
thenature of chargeon anelectron inthis convention ?
Positive and negative charges. Benjamin Franklin
(1706-1790), an American pioneer of electrostatics
introduced the present-day convention by replacing
the terms vitreous and resinous by positive and
negative, respectively. According to this convention:
1.The charge developed on a glass rod when rubbed
withsilkiscalled positive charge.
2.Thecharge developed on a plastic rod when rubbed
with wooliscalled negative charge.
The above convention is consistent with the fact
that when two opposite kinds of charges are brought in
contact, theytend to cancel each other's effect. According
to this convention, thecharge on an electron isnegative.
Table 1.1 gives a list of the pairs of objects which get
charged on rubbing against each other. On rubbing, an
object of column I will acquire positive charge while
that of column IIwill acquire negative charge.
Tae 1 .1Two kinds of charges developed on rubing
ColumnI Column II
(Positive charge) (Negative charge)
Glass rod Silkcloth
Flannel or catskin Ebonite rod
Woollencloth Amber rod
Woollen coat Plasticseat
Woollen carpet Rubber shoes
Obviously, anytwo charged objects belonging to
the same column will repel each other while those of
twodifferent columns will attract each other.
For Your Knowledge
~Benjamine's choice of positive and negative charges is
purely conventional one.However, it isunfortunate
that the charge on an electron (which is so important
to physical andchemical properties of materials)
1.3
turns out to be negative in this convention. It would
have been more convenient if electrons were assigned
positive charge. But in science, sometimes we have to
live with the historical conventions.
~Different substances can be arranged in a series in
such a way that if any two of them are rubbed together,
then the one occurring earlier in the series acquires a
positive charge while the other occurring later acquires
a negative charge:
1.Fur 2.Flannel 3.Sealing wax
4.Glass 5.Cotton 6.Paper
7.Silk 8.Human body 9.Wood
10.Metals 11.Rubber 12.Resin
13.Amber 14.Sulphur 15.Ebonite
16.Gutaparcha
Thus glass acquires a positive charge when rubbed
with silk but it acquires negative charge when rubbed
with flannel. )
1.5
ELECTRONIC THEORY OF FRICTIONAL
ELECTRICITY
8.Describe the electronic theory of frictional
electricity. Are the frictional forces electric inorigin ?
Electronic theory of frictional electricity.All matter
is made of atoms. An atom consists of a small central
nucleus containing protons and neutrons, around
which revolve a number of electrons. In any piece of
matter, the positive proton charges and the negative
electron charges cancel each other and so the matter in
bulk is electrically neutral.
The electrons of the outershell of an atom are
loosely bound to the nucleus. The energy required to
remove an electron from thesurface of a material is
called its'work function'. When two different bodies
are rubbed against each other, electrons are transferred
from the material with lower work function to the
material with higherwork function. For example,
when a glass rod is rubbedwith asilkcloth, some
electrons are transferred from glass rod to silk. The
glass rod develops a positive charge due to deficiency
of electrons while the silk cloth develops an equal
negative charge due to excess of electrons. The
combined total charge of the glass rod and silk cloth is
still zero, as it was before rubbingi.e.,electric charge is
conserved during rubbing. ,
\ Electric origin of frictional forces.The only wayby
which an electron can be pulled away from'an atom is
to exert a strong electric force on it. As electrons are
actually transferred from one body to another during
rubbing, so frictional forces must have an electric origin.°°°2tu}kyjxo£k2ius

5. 1.4
Fo
r Your Knowledge
~The cause of charging is the actual transfer of elec-
trons from one material to another during rubbing.
Protons are not transferred during rubbing.
~The material with lower work function loses electrons
and becomes positively charged.
~As an electron has a finite mass, therefore, there always
occurs some change in mass during charging. The
mass of a positively charged body slightly decreases
due to loss of some electrons. The mass of a negatively
charged body slightly increases due to gain in some)
electrons. _
1.6 CONDUCTORS AND INSULATORS
9. How do the conductors difer from the insulators?
Why cannot we electrify a metal rod by rubbing it while
holding it in our hand?How can we charge it?
Conductors.The substances through which electric
charges can ow easily are called conductors.They contain
a large number of free electrons which make them
good conductor of electricity. Metals, human and animal
bodies, graphite, acids, alkalies, etc. are conductors.
Insulators.The substances through which electriccharges
cannot ow easily are called insulators.In the atoms of
such substances, electrons of the outer shell are tightly
bound to the nucleus. Due to the absence of free charge
carriers, these substances offer high resistance to the
flow of electricity through them. Most of the non-
metals like glass, diamond, porcelain, plastic, nylon,
wood, mica, etc. are insulators.
An important difference between conductors and
insulators is that when some charge is transferred to a
conductor, it readily gets distributed over its entire
surface. On the other hand, if some charge is put on an
insulator, it stays at the same place. We shall discuss
this distinguishing feature in the next chapter.
A metal rod held in hand and rubbed with wool
does not develop any charge. This is because the
human body is a good conductor of electricity, so any
charge developed on the metal rod is transferred to the
earth through the human body. We can electrify the
rodby providing it a plastic or a rubber handle and
rubbing it without touching its metal part.
10.What is meant by earthing or grounding in
household circuits?What is its importance?
Earthing and safety.When a charged body is
brought in contact with the earth (through a connecting
conductor), its entire charge passes to the ground in
the form of a momentary current. This processinwhich a
body sharesitscharges with the earthiscalledgrounding or
earthing.
PHYSICS-XII
+ +
++
+ + <J) <J)
-;:: ~
+ +
c cc
+ +~ln ~ll~
uw u
W
- -
(a) (b)
Fig. 1.3 (a)Positively charge(b)Negatively charge, earthed body.
The electricity from the mains is supplied to our
houses using a three-core wiring :live, neutraland
earth wires.The live wire red in colour brings in the
current. The black neutral wire is the return wire. The
green earth wire is connected to a thick metal plate
buried deep into the earth. The metallic bodies of the
electric appliances such as electric iron, refrigerator, TV,
etc. are connected to the earth wire. When any fault
occurs or live wire touches the metallic body, the charge
flows to the earth and the person who happens to touch
the body of the appliance does not receive any shock.
1.7 ELECTROSTATIC INDUCTION
11. Whatismeant by electrostatic induction?
Electrostatic induction.As shown in Fig. 1.4, hold a
conducting rodABover an insulating stand. Bring a
positively charged glass rod near its endA.The free
electrons of the conducting rod get attracted towards
the endAwhile the end Bbecomes electron deficient.
The closer endAacquires a negative charge while the
remote end B acquires an equal positive charge. As
soon as the glass rod is taken away, the charges at the
endsAand Bdisappear.
Conducting rod
Excess of
electrons
Deficiency of
electrons
Insulating
stand
Fig.1.4Electrostatic induction.
Electrostatic inductionisthe phenomenon of
temporary electrification of a conductor in which opposite
charges appear atitscloser end and similar charges appear at
itsfarther endinthe presence of a nearby charged body.
The positive and negative charges produced at the
ends of the conducting rod are calledinduced chargesand
the charge on the glass rod which induces these
charges on conducting rod is calledinducing charge.{{{0rsxiwhvmzi0gsq

6. ELECTRIC CHARGES
ANDFIELD
12. Describe how two metal spheres can be oppositely
charged by induction.
Charging of two spheres by induction. Figure 1.S
shows thevarious steps involved ininducing opposite
charges on two metal spheres.
rl~n
(a) (b)
~2222
(c) (d)
22
(e)
Fig. 1.5 Two metal spheres get oppositely charged by induction.
(a)Hold the two metal spheres on insulatingstands
and place them in contact, as shown in Fig. 1.S(a).
(b)Bring a positively charged glass rod near the left
sphere. The free electrons of the spheres get attracted
towards theglassrod.Theleftsurface of theleftsphere
develops an excess of negative charge while the right
side of the rightsphere develops an excess of positive
charge. However, all of the electrons of the spheres do
not collect at the left face. As the negative charge
begins to build up at the left face, itstartsrepelling the
new incoming electrons. Soon an equilibrium is
established under the actionof force ofattrac- tion of
the rod and the force of repulsion due to the
accumulated electrons. The equilibrium situation is
shown in Fig.1.S(b).
(c)Holding the glass rod near the left sphere, sepa-
rate the two spheres byasmall distance, asshown in
Fig.l.S(c). The two spheres now have opposite charges.
(d)Remove the glass rod. Thecharges on thespheres
get redistributed. Their positive and negative charges
face each other, asshown in Fig. 1.S(d). The two
spheres attracteach other.
(e)When the two spheres areseparated quite apart,
the charges on them get uniformly distributed, as
shown in Fig.1.S(e).
Thus the two metal spheres get charged by a
process calledchargingbyinduction. In contrast to the
process of charging by contact, here the glass rod does
notlose any of its charge.
1.5
13. How can you charge a metal sphere positively
without touching it?
Charging of a sphere by induction.Fig. 1.6 shows
thevarious steps involved in inducing a positive
charge on a metal sphere.
(a)Hold the metal sphere on an insulating stand.
Bring a negatively charged plastic rod near it.
The free electrons of the sphere are repelled to
the farther end. The near end becomes posi-
tively charged due to deficit of electrons.
(b)When the far end of the sphere is connected to
the ground by a connecting wire, its free
electrons flow to the ground.
(c)When the sphere is disconnected from the
ground, its positive charge atthe near end
remains held there due to the attractive force of
theexternal charge.
(d)When the plastic rod is removed, the positive
charge spreads uniformly on the sphere.
p:~2GPlld
(a) (b)
(e) (d)
Fig.1.6 Charging by induction.
Similarly, the metal sphere can be negatively charged
bybringing a positively charged glass rod near it.
For Your Knowledge
~Gold-leaf electroscope. Itis adevice used for detecting
an electric charge and identifying its polarity. It
consists of a vertical conducting rod passing through
a rubber stopper fitted in the mouthof a glass vessel.
Two thin gold leaves are attached to lower endof the
rod.When a charged object touches themetal knob at
the outer end of the rod, the charge flows down to the
leaves.The leaves Metal--+
diverge due to knob
repulsion of thelike
charges they have
received. The degree
of divergence ofthe
leaves gives a
measure ofthe
amount of charge.
Rubber
stopper
Glass
vessel
Gold lea
Tin foil°°°2tu}kyjxo£k2ius

7. 1.6
1.8BASIC PROPERTIES
OF ELECTRIC CHARGE
It is observed from experiments that electric charge
hasfollowingthreebasic properties :
1.Additivity 2. Quantization 3. Conservation.
We shall discuss these properties in detail in the
next few sections.
1.9ADDITIVITY OF ELECTRIC CHARGE
14.Whatdoyoumean by additive nature of electric
charges?
Additive nature of electric charges. Like mass,
electric charge is a scalar quantity. Just as the mass of
anextended body is thesum of the masses of its
individual particles, the total charge of an extended
body is the algebraic sum (i.e.,thesum taking into
account the positive and negative signs) of all the
charges located at different points inside it. Thus,the
electric charge isadditive innature.
Additivity of electric chargemeans that thetotal
charge of a system isthealgebraic sum of all the individual
charges located at diferent points inside thesystem.
If a system contains chargesql'q2'.....,qn'then its
total charge is
q=ql+q2+ .....+qn
The total charge of asystem containing four
charges 2 1lC,-31lC,4 IlC and-5 IlC is
q=2IlC-3IlC+4IlC-5 IlC= -2IlC
1.10QUANTIZATION OF ELECTRIC CHARGE
15.Whatismeant by quantization of a physical
quantity?
Quantization of a physical quantity.Thequanti-
zationof aphysical quantity means thatitcannot vary conti-
nuously to have any arbitraryvaluebut itcanchangedisconti-
nuously to take anyone of only a discrete set of values. For
example, a building can have different floors (ground,
first, second, etc.) from the ground floor upwards but it
cannot'have a floor of the value in-between. Thus the
energy of an electron in atom or the electric charge of a
system is quantized.The minimum amount by which a
physical quantity can changeiscalled its quantum.
16.Whatismeant by quantization of electric charge ?
Whatisthe cause of quantization ofelectric charge?
Quantization of electric charge. It is found
experimentally that the electric charge of any body,
large or small, is always an integral multiple of a
-certainminimum amount of charge. This basic charge
is the charge on an electron, which is denoted byeand
has magnitude 1.6 x10-19coulomb. Thus the charge on
an electron is -e,on a proton is+eand that on
a-particle is+2e.
PHYSICS-XII
Theexperimental fact thatelectric charges occurin
discrete amounts instead of continuous amounts iscalled
quantization of electric charge. Thequantization of electric
chargemeans that the total charge (q) of a bodyisalways an
integral multiple of a basic quantum ofcharge (e),i.e.,
q=ne,where n=0,±1,±2,±3, .
Cause of quantization.The basic cause of quanti-
zation of electriccharge is that during rubbing only an
integral numberof electrons can be transferred from
one body to another.
Quantization of electric charge is an experi-
mentally verified law:
1.The experimental laws of electrolysis discov-
ered by Faraday first suggested the quanti-
zation of electric charge.
2. Millikan's oil drop experiment in1912on the
measurement of electric charge further estab-
lished the quantization of electric charge.
17.Canweignore thequantization of electric
charge ?If yes,underwhat conditions ?
When can we ignore the quantization of electric
charge.While dealing with macroscopic charges(q=ne),
we can ignore the quantization of electric charge. This
is because the basic chargeeis very small andnis very
large in most practical situations, soqbehaves as if it
were continuousi.e.,as if a large amount of charge
were flowing. For example, when we switch on a 60 W
bulb, nearly 2x1018electrons pass through its filament
per second. Here the graininess or structure of charge
does not show up i.e., the bulb does not flicker with the
entry of each electron. Quantization of charge becomes
important at the microscopic level, where the charges
involved are of the order of a few tens or hundreds ofe.
/ ,
For Your Knowledge
~The smallest amount of charge or basic quantum of
charge is the charge on an electron or a proton. Its
exact magnitude ise=1.602192x10-19C
~Quantization of electric charge cannot be explained on
the basis of classical electrodynamics or even modem
physics. However, the physical and chemical properties
of atoms, molecules and bulk matter cannot be explained
without considering the quantization of electric charge.
~Recent discoveries in high energy physics have indi-
cated that the elementary particles like protons and
neutrons are themselves built out of more elementary
units, calledquarks,which have charges(2/3)eand
(- 1/3)eEven if quark-model is established in {tIture,
the quantization of charge will still hold. Only the
quantum of charge will reduce frometo e/3.
~Quantization is a universal law of nature. Like charge,
energy and angular momentum of an electron are also
quantized. However, quantization of mass is yet to be
established.{{{0rsxiwhvmzi0gsq

8. ELECTRIC C
HARGES AND FIELD
Examples based on
uanrisation of Electric Charge
Formulae Used
1.q=ne
2.Mass transferred during charging=mexn
Units Used
qandeare in coulomb,nis pure integer.
Constants Used
e=1.6x10-19 C,me=9.1x10-31 kg
Example1.Which isbigger -a coulomb or a charge on an
electron ?Howmanyelectronic charges formone coulomb of
charge ? [Haryana 01]
Solution. One coulomb of charge is bigger than the
charge on an electron.
Charge on one electron, e=1.6x10-19 C
:.Number of electroniccharges in1coulomb,
q 1 C 18
n=-= =6.25x10 .
e1.6x10-19C
Example 2. Acomb drawn through person's hair on a dry
daycauses1022electrons to leave theperson's hair and stick
to the comb. Calculate the charge carried by the comb.
Solution. Heren=1022, e=1.6x10-19 C
:.q=ne=1022 x1.6x10-19 =1.6 x103C
Asthe comb has excess of electrons,
:. Charge on comb =-1.6x103C.
Example 3. If a body gives out 109electrons every second,
how much time isrequired to get a totalcharge of1Cfrom
it? [NCERT]
Solution. Number of electrons given out by the
body in one second=109
Charge given out by the body in one second
=ne=109x1.6x10-19 C
=1.6x10-10C
Time required to get a charge of1.6x10-10 C
=ls
Time required to get a charge of 1 C
110S=6.25x109s
1.6x10-
6.25x109
------ years=198.18 years.
365x24x3600
Thus from a body emitting 109 electrons per
second, it will take nearly200years to get a charge of
1 C from that body. This shows how large is one
coulomb as the unit of charge.
1.7
Example 4.How much positive and negative chargeis
there in a cup of water? [NCERT]
Solution. Suppose the mass of water contained in a
cup is250g. The molecular mass of water is18g.
Number of molecules present in18g of water
=Avogadro's number=6.02x1023
:. Number of molecules present in a cup (or250g)
of water
23
n= 6.02x10x250 = 8.36 x1024
18
Each molecule of water(HzO) contains 2+8=10
electrons as well as10protons.
Total number of electrons or protons present in a
cup of water,
n'=nx10 =8.36x1025
Total negative charge carried by electrons or total
positive charge carried by protons in a cup of water,
q=n' e
= 8.36 x1025x1.6x10-19C=1.33x107C
rproemsFor Practice
1.Calculate the charge carried by 12.5 x108electrons.
[CBSE D 92]
(Ans.2x10-10C)
2.How many electrons would have to be removed
from a copper penny to leave it with a positive
charge of 10-7 C ? (Ans. 6.25xUy1electrons)
3.Calculate the charge on an alpha particle. Given
charge on a proton=1.6x10-19C.
(Ans. +3.2x10-19C)
4.Calculate the charge on ~ Fe nucleus. Given char~e on
a proton=1.6x10-19 C. (Ans.+4.16x10-8C)
5.Determine the total charge on 75.0 kg of electrons.
(Ans.-1.33xuPC)
6.How many mega coulombs of positive (or
negative) charge are present in 2.0 mole of neutral
hydrogen (H2) gas?
7.Estimate the total number of electrons present in
100 g of water. How much is the total negative
charge carried by these electrons ? Avogadro's
number=6.02x1023and molecular mass of water
=18. (Ans. 5.35x106C)
HINTS
3.An alpha particle contains 2 protons and 2
neutrons.
q=+2e.°°°2tu}kyjxo£k2ius

9. 1.8
4.~~Fe nuc
leus contains 26protons and30neutrons.
.. q=+26e
5.n= Total mass 75.0 =25x1031
Mass of an electron 9x10-31 3
q=-ne= - 25x1031x1.6x10-19=-1.33 x1013e.
3
6.Number of molecules in2.0mole ofH2gas
=2.0x6.02x1023
As eachH2molecule contains 2 electrons/protons, so
n= 2x2.0x6.02x1023 =24.08 x1023
q=ne= 24.08 x1023x1.6x10-19
= 0.3853 x106C = 0.3853 Me.
[1 Me=106q
7.Proceed as in Example4.
1.11 CONSERVATION OFCHARGE
18.Statethe law ofconservation of charge. Give some
examples toillustrate thislaw.
Law of conservation of charge.Ifsome amount of
matter is isolated in a certain region of space and no
matter either enters or leaves this region by moving
across its boundary, then whatever other changes may
occur in the matter inside, itstotalcharge will not
changewith time. This is thelaw of conservation of
charge which states:
1.Thetotal charge of an isolatedsystem remainsconstant.
2.The electricchargescan neither be creatednor destroyed,
they can only be transferred from one body to another.
The law of conservation of charge is obeyed both in
large scale and microscopic processes. In fact, charge
conservation is a global phenomenoni.e.,totalcharge of
the entire universe remains constant.
Examples:
1. When a glass rod is rubbed with a silkcloth, it
develops a positive charge. But at thesame
time, the silkcloth develops anequal negative
charge. Thus the net charge of the glass rod and
the silk cloth is zero, asit was before rubbing.
2. The rocksalt ionises in aqueous solution as
follows:
NaCl ~ Na+ +Cl"
As the total charge is zero before and after the
ionisation, so charge is conserved.
3. Charge is conserved during the fission of a2t~U
nucleus by a neutron.
In+235U -tI4IBa+92Kr+3 In+Energy
o 92 56 36 0
Total charge before fission(0 + 92)
=Total charge after fission(56+36+3x0)
PHYSICS-XII
4.Electric charge is conserved during the
phenomenon ofpair productioninwhich a"(-ray
photon materialises into an electron-positron pair.
y- ray ~ electron +positron
zero cha!ge ( -e) (+e)
5. Inannihilation of matter, an electron and a posi-
tron on coming in contact destroy each other,
producing two y-ray photons, each of energy
0.51MeV.
electron+positron
(-e) (+e)
~ 2 y- rays
zero charge
For Your Knowledge
~Conservation of charge implies that electric charges
canbe created or destroyed always in the form of equal
and opposite pairs but never in isolation. For example,
in the beta decay of a neutron (zero charge), a proton
(charge+e)and an electron (charge -e)are produced.
Total charge remains zero before and after the decay.
~The law of conservation of charge is an exactlaw of
nature. It is valid in all domains of nature. Evenin the
domains ofhigh energy physics, where mass changes
into energy and vice-versa, the law of conservation of
charge strictly holds good.
1.12 ELECTRIC CHARGE VS MASS
19.Compare the properties of electric charge with
thoseof mass of a body.
Table 1.2 Comparison of the properties of
electriccharge and mass
Electric charge Mass
1.
Electric charge may beMassof a body is
positive, negative or zero. always positive.
2.Electric charge is always Quantization of mass is
quantized :q=ne notyet established.
3.Charge on a body does Mass of a body
not depend on itsspeed. increases with its speed.
4.Charge isstrictly Mass is not conserved by
conserved. itself as some of the mass
may get changed into
energy or vice versa.
5.Electrostatic forces Gravitational forces
between two charges between two masses are
maybeattractive or always attractive.
repulsive.
Electrostatic forces
,
6. Gravitational forces
between different between different bodies
charges may cancel out. never cancel out.
7.A charged body always A body possessing
possesses some mass. mass may not have any
netcharge.{{{0rsxiwhvmzi0gsq

10. ELECTRIC CHARGES
AND FIELD
20.How does the speed of an electrically charged
particle afect its (i) mass and (ii) charge?
Effect of speed on mass and electric charge.According
tothespecial theory of relativity, the mass of abody
increases with its speed in accordance with the relation:
m= 1110
g2
1-
c2
where,1110=rest mass of the body,c=speed of light,
and m=mass of the body when moving with speedv.
Asv<c, therefore, m>1110.
In contrast to mass, the charge on a body remains
constant and does not change as the speed of thebody
changes.
1.13 COULOMB'S LAW OF ELECTRIC FORCE
21.State Coulomb's law in electrostatics. Express the
sameinSf units. Name and define the units of electric
charge.
Coulomb's law. In 1785, the French physicist
Charles Augustin Coulomb(1736-1806) experimentally
measured the electric forces between small charged
spheres by using a torsion balance. He formulated his
observations in the form of Coulomb's law which is
electrical analogue of Newton's law of Universal
Gravitation in mechanics.
Coulomb's law states that the force of attraction or
repulsion between two stationary point chargesis(i)directly
proportional to the product of the magnitudes of the two
charges and(ii)inversely proportionaltothe square of the'
distance between them. This force acts along theline joining
the two charges.
ql q2
• •
Fig.1.7Coulomb's law.
Iftwo point charges qlandq2are separated by
distance r,thenthe forceFof attraction or repulsion
between them is such that
1
FIXql q2andFIX-
r2
or
where kis a constant ofproportionality, called electro-
static force constant. The value ofkdepends on the
nature of the medium between the twocharges and the
system of units chosen to measureF, ql' q2andr.
.For the twocharges located in free space and in 51
units, we have
k=_1_ =9x109Nm2 C-2
411: EO
1.9
whereEOis calledpermittivityof free space. So we can
express Coulomb's law in 51 units as
F=_1_ qlq2
411: EO'r2
Units of charge.(i)TheSfunit of charge iscoulomb. In
the aboveequaticn.if ql=q~=1C and r=1m,then
1.
F=--=9x109N
.4rcEO
SOone coulomb isthat amount of charge that repels an
equal and similar charge with a force of9x109Nwhen
placedinvacuum at a distance of one metre fromit.
(ii)In electrostatic cgssystem, the unitof charge is
known aselectrostatic unit of charge(e.s.u. of charge) or
statcoulomb(statC).
One e.s.u.of charge or one statcoulomb isthat charge
which repels an identical charge in vacuum at a distance of
one centimetre fromitwith a force of1dyne.
1 coulomb=3x109stat coulomb
=3x109e.s. u. of charge
(iii)In electromagnetic cgssystem, the unit of
charge is abcoulomb or electromagnetic unit of
charge(e.m.u. of charge).
1coulomb=1~abcoulomb=1~e.m.u. of charge
For Your Knowledge
>Atorsion balance isasensitive device to measure force.
>When the linear sizesofcharged bodies are much smaller
than the distancebetween them, their sizesmay be ignored
and the charged bodies are calledpoint charges.
>Coulomb's law is valid only for point charges.
>In51units, theexactvalue of the combination411:EOis
4 107 C2N-1 -2
11:EO=? m
wherecis the speed of light in vacuum having theexact
value299792458x108ms-I.
•
>Electrostatic force constant,
k=8.98755xl 09Nm2C2 c:9x109Nm2 C2.
>Permittivity of freespace,
EO=8.8551485x10-2 C2N-1m-2
.:::9xlO-2C2N-I m-2.
>51unit of permittivity
=coulombxcoulomb=C2N-Im-2
newtonxmetre2
The unitC2N-1m-2 is usually expressed asfarad per
metre (Fm-I) .
>More strictly, the51unit of charge1coulomb is equal to
1 ampere-second, where 1 ampere is defined in terms of
the magnetic forcebetween two current carrying wires.°°°2tu}kyjxo£k2ius

11. 1.10
1.14COULOMB'
S LAW IN VECTOR FORM
22. Write Coulomb's lawinvector form. Whatisthe
importance of expressing itinvector form?
Coulomb's law in vector form.As shown in Fig. 1.8,
consider two positive point charges q1andq2placed in
vacuum at distancerfrom each other. They repel each
other.
~ ~
F12•••••I----· ---------------"·---,l·~F21
+ql +q2
Fig. 1.8 Repulsive coulombian forces forq1q2>o.
In vector form, Coulomb's law may be expressed as
-4
F21=Force on chargeq2due toq1
1 qlq2 "
=-- --r.
4nI: .?12
o
-4
" r.
wherer12=R ,is aunitvector in the direction fromql
r
toq2.
-4
Similarly, F12=Force on chargeq1due toq2
1 qlq2 "
=--·-2- r21
4n1:0r
-4
" r,
wherer21=-.11,isa unit vector in the direction fromq2
r
toq1.
The coulombian forces between unlike charges
(qlq2 <0) are attractive, as shown in Fig. 1.9.
Fig. 1.9 Attractive coulombian forces forq1q2<o.
Importance of vector form.The vector form of cou-
lomb's law gives the following additional information:
1\ 1\ -+-+
1. Asr21= -r12,thereforeF21= -F12·
Thismeans that the two charges exert equal and
opposite forces on each other. SoCoulombian
forces obey Newton's third law of motion.
-4 -4
2. AstheCoulombian forces act along F12orF21,
i.e.,along the line joining the centres of two
charges, sothey arecentral forces.
23. Whatisthe range over which Coulombian forces
can act?State the limitations of Coulomb's law in
electrostatics.
PHYSICS-XII
Range of coulombian forces.Coulombian forces act
over an enormous range of separations(r),from
nuclear dimensions(r=10-15m) to macroscopic dis-
tances as large as 1018 m.Inverse square is valid over
this range of separation to a high degree of accuracy.
Limitations of Coulomb'slaw.Coulomb's law is
not applicable inallsituations. Itis valid only under
the following conditions:
1.The electric charges must be at rest.
2.The electric charges must be point charges i.e.,
the extension of charges must be much smaller
than the separation between the charges.
3.The separation between the charges must be
greater than the nuclear size (la-15m), because
for distances<la-15m, the strong nuclear force
dominates over the electrostatic force.
1.15DIELECTRIC CONSTANT:
RELATIVE PERMITIIVITY
24.What do you mean by permittivity of amedium?
Define dielectric constant interms offorcesbetween two
charges.
Permittivity : An introduction. When two charges
are placed in any medium other than air, the force
between them is greatly affected. Permittivity isa
property of the medium which determines the electric force
between two charges situated inthat medium. For example,
the force between two charges located some distance
apart in water is about I/80th of the force between
them when they are separated by same distance in air.
This is because the absolute permittivity of water is
about 80 times greater than the absolute permittivity of
air or free space. .
Dielectric constant or relative permittivity.Accor-
ding to Coulomb's law; the force between two point
chargesqlandq2'placed in vacuum at distance rfrom
each other, is given by
F=_1_. ql~2 ...(1)
vac4nI:r:
o
When the same two charges are placed same
distance apart in any medium other thanvacuum, the
force between them becomes
Fd=_1_. ql~2 ...(2)
me 4nl:r:
The quantity I:is called absolute permittivityor just
permittivityof the intervening medium. Dividing
equation (1) by equation (2), we get
1 qlq2
Fvac =~·7 I:
Fmed _1_ ql q2 1:0
4nl:· r2°°°3xy£o~n}s«o3myw

12. ELECTRIC CH
ARGES AND FIELD
Theratio(E / EO)ofthepermittivity (E)ofthe medium tothe
permittivity (EO)of free space iscalled relative permittivity
(Er)or dielectric constant (K)of the given medium. Thus
EF
e,or K=-=~
EOFrned
So one can define dielectric constantin terms of
forcesbetween charges as follows :
Thedielectric constantorrelative permittivity ofa
mediummay be defined as theratio of the forcebetween two
charges placed somedistance apartinfree space to the force
between the same twocharges when they are placed thesame
distance apartinthe given medium.
Clearly, when a material medium of dielectric
constantKis placed between the charges, the force
between them becomes 1/Ktimes the original force in
vacuum. Thatis,
F=Fvac
rned K
Hence the Coulomb's law for any material medium
maybe written as
K(vacuum)=1
K(air) = 1.00054
K(water) = 80.
Formulae Used
1.t:=_1_. ql2q2
vac 41t Er
o
2t.__ 1_qlq2
• rned -41t E Kr2
o
Units Used
%,q2are in coulomb,Fin newton andrin metre.
Constant Used
k=_1_ =9x109Nm2c:-2
41t EO
Example5.The electrostatic forceof repulsion between two
positively charged ions carrying equalchargesis~.7x10-9N,
when they are separated by a distance of5 AHow many
electrons are missing fromeach ion?
Solution. HereF=3.7x10-9N,
r= 5A=5x10-10 m,ql=q2=q(say)
As
F=_1_. q1q2
41tEo ?
1.11
or
_99x109xqxq
3.7x10= 10 2
(5x10-)
2= 3.7x10-9x25x10-20= 10.28 x10-38
q 9x109
q=3.2x10-19 C
Number of electrons missing from each ion is
q3.2x10-19
n=-= =2.
e1.6x10-19
or
Example6.Afreepith-ball A of8g carries a positive
charge of5x10-8C.What must be thenature and
magnitude of charge that should begiven to asecond
pith-ballBfixed5cm below theformer ball sothat the upper
ballisstationary? [Haryana 01]
Solution. The pith-ball Bmust be of positive charge
i.e.,of same nature as that ofA,so that the upward
force of repulsion balances the weight of pith-ball A
When the pith-ballAremains F
stationary, ?
F=~g TA q)
or_1_q1q2=mg
41tEo? --1 5em m)g
But ~=8g=8xlO-3kg 10
B q2
q1=5x10-8C
r= 5 em =0.05 m Fig.1.10
9x109x5x10-8xq2 -3
------;;-2--= =8x10x9.8
(0.05)
8x9.8x{0.05lx10-4
q2= 9x5
or
=4.36x10-7C(positive).
Example 7. Aparticle of mass m and carrying charge -q1
ismoving around a charge+q2along acircular path of
radius r. Prove that the period of revolution of the charge - q1
about+q2isgiven by
r--=----:-
161t3 Emr3
T= 0
q1q2
Solution. Suppose charge-q1moves around the
charge+q2with speedvalong the circular path of
radiusr.Then
Force of attraction between the two charges
= Centripetal force
1q1q2 _ mv2
or41tEo7--r-
or
_1_q1q2
41tEo mr
v=°°°2tu}kyjxo£k2ius

13. 1.12
The period of revolution of charge -qla
round+q2
will be
Example 8.Two particles, each having a mass of5g and
charge1.0x10-7C ,stayin limiting equilibrium on a
horizontal table with a separation of10em between them.
The coefficient offriction between each particle and the table
isthe same. Find!.L
Solution.Hereql=q2= 1.0 x10-7C,
r=10em=0.10m,m=5g=5x10-3 kg
Themutual electrostatic force between the two
particles is
q q9x109x(1.0x10-7)2
F=k ~ 2 = 0.009 N
r (0.10)
The limiting force of friction between a particle and
the table is
f=!lxmg=!lx5x10-3 x9.8 =0.049 !l N
Asthetwo forces balance each other, therefore
0.049 !l = 0.009
= 0.009 =O.lB.
!l 0.049
or
Example9.(a) Two insulated charged copper spheres A
andBhavetheir centres separated by a distance of50cm.
Whatisthe mutual force1,electrostatic repulsion ifthe
charge on eachis6.5x10-C?The radii of A and Bare
negligible compared to the distance of separation. Also
compare this force with their mutual gravitational attraction
ifeach weighs 0.5kg.
(b) Whatisthe force of repulsion if(i) eachsphere is
charged double the above amount, and the distance between
themishalved; (ii) the two spheres are placed in water?
(Dielectric constant of water=80). [NCERT]
Solution.(a)Hereql=q2=6.5 x10-7C,
r=50 em. =0.50 m
Using Coulomb's law,
F.=k. qlq2
air r2
=9x109.6.5x10-7 x6.5x10-7N
(0.50)2
=1.5x10-2 N.
Themutual gravitational attraction,
F=G~""2
e R2
6.67x10-11 x0.5x0.5 = 6.67x10-11 N
(0.5)2
Clearly, Fe «~ir.
PHYSICS-XII
(b)(i)When charge on each sphere is doubled, and
thedistance between them ishalved, the force of
repulsion becomes
F'.=k.2ql·2q2 =16k.q1q2
arr (r /2)2 r2
=16x1.5x10-2=0.24N.
(ii)The forcebetween two charges placed in a
medium of dielectric constant Kis givenby
F=_1_!qlq2
4m,o· K·r2
For water, K= 80
F=~ir=1.5x10-2
water K 80
.=1.875 x10-4N=-1.9 x10-4 N.
Example 10. Supposethe spheres A andBin Example 9have
identical sizes. A third sphere of the same sizebut uncharged
is brought in contact with the first, then brought in contact
with thesecond, and finally removed from both. Whatisthe
new force of repulsion between A and B? [NCERT]
Solution. Charge on eachof the spheres AandBis
q=6.5xlO~7C
When a similar but uncharged sphere C is placed in
contact withsphere A,each sphere shares a charge
q /2,equally.
q Charge=0 q/2q/2
[email protected]
q q12 3q/43q/4
0+@---'~
«n 3q/4
AO--'---OB
Fig. 1.11
Now when the sphereC(withcharge q /2)isplaced
in contact with sphere B(with charge q),the charge is
redistributed equally, so that
Charge on sphere Bor C =!(q+1)=3q
2 2 4
•.New force of repulsion between AandBis
3q q
F=_l_ 4·2"
41tEo· ,2
= ~x1.5x10-2N=0.5625x10-2N
8
=-5.7x10-3 N.
Example 11. Two similarly equally charged identical metal
spheres A andBrepeleach other with aforce of2.0x10-5N.
A third identical uncharged sphereCistouchedto A,then
placed atthe midpoint between A and B.Calculate the net
electrostatic force on C. [CBSE 0003]{{{0rsxiwhvmzi0gsq

14. ELECTRIC CHARGE
S AND FIELD
Solution. Let the chargeon each of the spheresA
and B beq.If the separation betweenAand B isr,then
electrostatic force between spheresAandBwill be
2
F=k .q2=2.0x10-5N
r
When sphere C is touched toA,the spheres share
chargeq12each, because both are identical.
Force on C due toA
(qI2)2 q2
=k--=k- alongAC
.(r12)2 ,2 ,
Force on C due toB
-k q.ql2 -k 2q2 I BC
- .(rI2)2 -'7'aong
Since these forces act in opposite directions,
therefore net force on C is
, 2q2 q2 q2 -5
F=k.-2-k.2"=k2"=2.0x10N,alongBe.
r r r
Example12.Two identical charges,Qeach, are kept at a
distance rfrom eachother.Athirdchargeqisplaced on the
line joining the above twocharges such that all the three
chargesare inequilibrium. Whatisthemagnitude, sign and
position of the charge q? [CBSE OD 94, 98]
Solution. Suppose the three charges be placed in
the manner, as shown in Fig. 1.12.
14 r ~I
14 x
~IB
AI I Ie
Q q Q
Fig.1.12
Thechargeqwill be in equilibrium if the forces
exerted on it by the charges atAand C are equal and
opposite.
k Qq=k~
. x2 . (r-x)2
or
or x=r-x
r
x=-
2
or
Since thecharge at Aisrepelled by the similar
charge at C, so itwill be in equilibrium if it is attracted
bythecharge qatB,i.e.,the signof chargeqshould be
opposite tothat ofchargeQ.
Force of repulsion between charges at Aand C
=Force of attraction between charges
atAandB
k~=k Q.Q orq=Q.
(r12)2,2 4
or
Example13.Two point charges+4e and +eare 'fixed' a
distance 'a' apart. Where should a thirdpoint charge q be
placed ontheline joining thetwo chargessothatitmay be in
1.13
equilibrium?Inwhich casethe equilibrium will be stable
andin which unstable ?
Solution. Suppose the three charges are placed as
shown in Fig. 1.13. Let the chargeqbe positive.
+4e +q +e
I
•
I ~ I
F2 F1
I, x ,I,a-x--l
Fig. 1.13
For the equilibrium of charge +q,wemust have
Force of repulsionFlbetween+4eand+q
=Force of repulsionF2between+eand+q
14exq 1exq
4Tc!:O ~ =4m:o(a-X)2
4(a - x)2 =x2
2(a-x)=±x
2a
x=- or2a
3
Asthe chargeqis placed between+4eand+e,so
only x=2a13is possible. Hence for equilibrium, the
chargeqmust be placed at a distance2al3from the
charge+4e.
We have considered the chargeqto be positive. If
we displace it slightly towards chargee,from the
equilibrium position, thenFlwill decrease andF2will
increase and a net force(F2 - F1) will act onqtowards
lefti.e.,towards the equilibrium position. Hence the
equilibrium of positive qisstable.
Now if we take chargeqto be negative, the forcesFl
andF2will be attractive, as shown in Fig. 1.14.
or
or
or
+4e -q +e
I
•
I ~ I
F1 F2
I, x ,Ia-x--l
Fig. 1.14
The charge-qwill still be in equilibrium at
x=2a13.However, if we displace charge-qslightly
towards right, thenFlwill decrease and F2will
increase. A net force(F2-F1)willact on-qtowards
righti.e.,away from the equilibrium position. So the
equilibrium of the negative q will be unstable.
Example 14.Two'free'pointcharges+4eand+eare
placed a distance 'a' apart.Where should a third point charge
q beplaced between themsuch that the entire system may be
inequilibrium? What shouldbethe magnitude and sign of q?
What type of a equilibrium will itbe?°°°2tu}kyjxo£k2ius

15. 1.14
Solution.Suppose the
charges are placed as shown
in Fig. 1.15.
+4e -q +e.~.III•~
III
F F' F1
I, x
I,
F2
---ojo·I·>--a -x------l
a .1
Fig.1.15
Asthe charge+eexerts repulsionFon charge+4e,
so for theequilibrium ofcharge+4e,thecharge -q
must exert attractionF'on+4e.This requires the
chargeqto benegative.
For equilibrium of charge+4e,
F=F'
14e x e 14exq
411:Eo----;;r=411:Eo ~
ex2
q=-
a2
For equilibrium of charge -q,
AttractionF1between+4eand -q
=Attraction F2between+eand-q
14e xq 1exq
..411:Eo~ =411:Eo(a -x)2
x2 =4(a _x)2
x=2a/3
ex2 e4a2 4e
q=-=- -=-
a2 a2'9 9'
The equilibrium of the negative charge qwill be
unstable.
or
or
Hence
Example15.Two point charges of charge values Qand q
areplaced at distances x andx/2respectively from a third
chargeof charge value 4q, allchargesbeing in the same
straight line. Calculate the magnitude and nature of charge
Q,such that the net force experienced by the charge qiszero.
[CBSE D 98].
Solution. Suppose the threecharges are placed as
shown in Fig. 1.16.
~ q Q
• III· ~ •
A F8 C FA B
Fig. 1.16
For the equilibrium of chargeq,the chargeQmust
have the same sign as that ofqor4q ,so that the forces
FAandFBare equal and opposite.
As FA=FB
14qxq 1qxQ
411:Eo'(x/2)2 =411:Eo'(x/2)2
Q=4q.or
PHYSICS-XII
Example 16.AchargeQistobedivided on two objects.
What should be the values ofthe charges on the two objects
sothat the forcebetween the objects canbemaximum?
Solution. LetqandQ-qbethe charges on the two
objects. Then force between the two objectsis
F=_I_ q(Q-q)
411:EO . ,2
where risthe distance between the two objects.
ForFto be maximum,
dF =0
dq
1.~.~(qQ_q2)=0
411:EO,2dq
~ (qQ _ q2)=0
dq
Q-2q =0
q=Q
2
or
or
or
or
i.e.,the charge should be divided equally on the two
objects.
Example 17.Twoidenticalspheres,having chargesof
opposite sign attract each other with aforce of0.108Nwhen
separated by0.5m.The spheres areconnected by a conduc-
tingwire,which then removed, and thereafter they repel each
other with aforce of0.036N.What were the initialcharges
on the spheres ?
Solution. Let+q1and -q2be the initialcharges on
the two spheres.
(a)When the two spheres attract each other,
F=kq1~2 i.e.,0.108=9x109. q1q22
r: (0.5)
= 0.108x(0.5)2 =3x10-12
q1q2 9x109
(b)When the two spheres are connected by the
wire, theyshare the charges equally.
q+ (-q) q - q
:.Charge on each sphere=1 2 =_1__ 2
2 2
i.e.,
Force of repulsion between them is
F=k(¥)(¥)
,2
0.036 = 9x109.(q1-q2)2
(0.5)2 2
( _ )2 = 0.036x(0.5)2 x4=4x10-12
.. q1 q2 9x109
q1-q2=2x10-6 ...(i)or{{{0rsxiwhvmzi0gsq

16. ELECTRIC CHARGE
S AND FIELD
Now (q1+q2)2=(q1 - q2)2 +4q1q2
=(2x10-6)2 +4x3x10-12
=16x10-12
q1+q2=4x10-6 ...(ii)
On solving equations (i)and(ii),we get
q1=3x10-6Candq2=10-6C
which are the initial charges on the two spheres.
Example18.Two small spheres each having massm kg
and charge qcoulomb are suspended from a point by
insulating threads each Imetre longbutofnegligible mass. If
aisthe angle, each thread makes with the vertical when
equilibrium has been attained, show that
q2=(4mgl2sin2atana)4nEO [Punjab 95]
Solution. The given situation is shown in Fig. 1.17.
Each of the spheresAandBis acted upon by the
following forces:
(i)its weight mg, (ii)tensionTin the string
(iii)theforceofrepulsion Fgiven by
1 q2
F=--.-----:::i ...(i)
4nEO Alj-
o
~~'
,
,
F~i--------c ---------"
mg
Fig.1.17
As the forces are in equilibrium, the three forces on
sphereAcan be represented by the three sides of
t!.AOCtaken in the same order. Hence
~= mg=~
AC OCAO
AC
or F=mgx - ...(ii)
OC
From(i)and(ii),we have
1 q2 AC
--.--=mgx-
4nEoAW OC
ButAC= I sina,OC=I casa,AB=2AC=21sina
1 q2 Isina
--. =mgx--
4nEO 412sin2a I cosa
or q2=(4mg12sin2atana)4nEO'
1.15
~rOems For Practice
1.Obtain the dimensional formula ofEO'
(Ans. M-1L-3T4A2)
2.Calculate coulomb force between two a-particles
separated by a distance of 3.2 x10-15m in air.
[CBSE0092]
(Ans.90 N)
3.Calculate the distance between two protons such
that the electrical repulsive force between them is
equal to the weight of either. [CBSE094]
(Ans. 1.18 cm)
4.How far apartshould the two electrons be,if the
force each exerts on the other is equal to the weight
of the electron?Given thate=1.6x10-19Cand
me=9.1x10-31 kg. [Haryana 02]
(Ans. 5.08 m)
5.A pith-ballAof mass 9x10-5kg carries acharge of
5fie.What must be the magnitude and sign of the
charge on a pith-ballBheld2em directly above the
pith-ballA,such that the pith-ballAremains
stationary?
(Ans. 7.84 pC, sign opposite to that ofA)
6.Two identical metal spheres having equal and
similar charges repel each other with a force of
103 N when they are placed 10 em apart in a medium
of dielectric constant5.Determine the charge on
each sphere. (Ans. 23.9x10-6C)
7.The distance between the electron and proton in
hydrogen atom is 5.3 x10-11m.Determine the magni-
tude of the ratio of electrostatic and gravitational
force between them.
Givenme=9.1x10-31 kg,mp=1.67x10-27kg,
e=1.6x10-19C andG=6.67x10-11Nm 2 kg-2.
(Ans. Fe /Fc=2.27x1039)
8.Two identical metallic spheres, having unequal,
opposite charges are placed at a distance 0.90 m
apart in air. After bringing them in contact with
each other, they are again placed at the same
distance apart. Now the force of repulsion between
them is 0.025 N. Calculate the final charge on each
of them. [CBSE D 02C]
(Ans. 1.5x1O-6q
9.A small brass sphere having a positive charge of
1.7x10-8C is made to touch another sphere of the
same radius having a t;legative charge of 3.0 x10-9e.
Find the force between them when they are
separated by a distance of 20 cm. What will be the
force between them when they are immersed in an
oil of dielectric constant 3?
(Ans. 1.1x10-5N;0.367x10-5N)°°°2tu}kyjxo£k2ius

17. 1.16
10.The sum
of two point charges is71lC. They repel
each other with a force of 1 N when kept 30 ern
apart in free space. Calculate the value of each
charge. [CBSE F 091
(Ans.51lC, 21lC)
11.Twopoint chargesq1= 5x1O-6C and q2= 3 x 1O-6C
arelocated at positions (1 m,3rn,2 m) and (3 rn,
~ ~
5m,1m) respectively. Find the forces li2 andF21
using vector form of Coulomb's law.
---t 3 1\ /\ 1\
[AnS.li2 = -5xl0- (2i+2j -k)N,
---t 3 1\ /\ 1\
F21=svro: (2i+2j -k)N]
12.Three equally charged small objects are placed as
shown in Fig. 1.18. The object Aexerts an electric
force on object Bequal to 3.0 x 1O-6N.
A B C
• • • 3.
14 2cm ~14 1em
----+I
Fig. 1.18
(i) What electric force does C exert on B?
(ii)What isthe net electric force on B?
[Ans.(i)12.0 x 10-6 N,alongBA
(ii)9.0x10-6N,alongBAl
13.Two identical metallic spheres AandB,each carry-
ing a chargeq,repel each other with a force F. A
third metallic sphere C of the same size, but un-
charged, is successively made to touch the spheres
AandB,and then removed away. What is the force
of repulsion between Aand B? (Ans. 3F/8)
14.Twopointcharges+geand+eare kept at a distance
afromeach-other. Where should we place a third
charge qonthe linejoining thetwo charges so that
it may beinequilibrium ?
(Ans. 3: from+geCharge)
15.Twopoint electric charges of valuesqand2qare
keptat adistancedapart from each other in air. A
third charge Q is tobe kept along the same line in
such awaythat the net force acting on qand2qis
zero. Calculate the position of charge Qin terms of q
andd. [CBSED98]
(Ans.At a distance of(..fi-1)dfrom charge q)
16.A charge qisplaced at the centre of theline joining
two equal charges Q. Showthatthe system of three
charges will be in equilibrium ifq=-Q/4.
[CBSE OD 05]
17.Two pith-balls each weighing 10-3 kg are
suspended from the same point bymeans of silk
PHYSICS-XII
threads 0.5 m long. On charging the balls equally,
they are found to repel each other to a distance of
0.2 m. Calculate the charge on each ball.
[Haryana 2002]
(Ans. 2.357 x 10- 6C)
HINTS
1.F=_1_ .q1:}2or EO =litq22
41tEo r 41tFr
[E1= AT. AT =[~lL-3T4 A2l.
oMLT2. L2
2.Hereq1=q2=2e= 3.2x10-19C,r= 3.2x10-15 m
F=_1_ lItq2
41tEo' r2
9 x 109 x3.2x10-19x3.2x10-19
----------~,_------90N
(3.2x10 15)2 -.
For a proton, m=1.67x10-27 kg,
q=+e= 1.6x10-19C.
Weight of proton = Electrical repulsive force
qxq
mg=k.--2-
r
9 x 109 x(1.6 x10-19)2
1.67x10 27x9.8
or
r2=kq2
mg
= 23.04x10-2= 0.014
16.36
or r=0.0118 m=1.18 em.
exe
4. meg=k .--2-
r
or
9x109x(1.6 x10-19)2
____ '---,,.- __ --C-= 25.84
9.1x10 31x9.8
r=5.08 m.
5.Thepith-ball Bmust have charge opposite tothat of
Asothat the upward force ofattraction balances
the weight of pith-ball A.
When the pith-ballAremains
stationary, TQ20B
2r~A
Q)
m)g
or
F=rr;g
_1_ q1 q2=rr; g
41tEOr2
rr;=9x10-5kg,
q1=51lC= 5 x10-6 C,
r=2cm =0.02 m
9x109x5xlO-6xq2 -5
--------"2,-----'= =9xlOx 9.8
(0.02)
q2= 7.84x10-12C = 7.84 pc.
But
Fig. 1.19
or{{{0rsxiwhvmzi0gsq

18. ELECTRIC CHARGES AND FIELD
9x109xq2
103=---;
,-
5x(0.10)2
6.F=_l_ q1i2
41tEOK' ..
orq=23.9x10-6C.
7.Proceed as in illustrative problem on page 1.18.
8.The two spheres will share the final charge equally.
Letqbe the charge on each sphere.
F=_1_. qlq2= 0.025 N
.. 41t Eor2
9x109xqxq
-----<-2 ----'-= 0.025
(0.90)
q2=omsx(0~90)2= 225x10-14
9x10
q= 1.5 x10-6 C.
or
or
or
9.Charge shared by each sphere
= (17 - 3)x10-9= 7x10-9 C
2
9x109x(7x10-9)2 -5
F.= 2 =1.lx10 N
au (0.20)
9x109x(7x10-9)2 -5
E.]= 2 = 0.367x10 N.
01 3x(0.20)
10.HereF= 1 N, r= 30 m
F=k~2
9x109x1M2
1= ------,,'-'-'-"-
(0.30)2
10-11
orq1q2 =
Butq1+q2= 7!-1C= 7x10-6 C
Now(q1 - q2)2=(q1+q2)2 - 4q1q2
= 49x10-12- 4x10-11
=9x10-12
or q1 -q2= 3 x 10-6 = 3!-1C
On solving (i) and (ii), we get
q1= 5!-1C and q2= 2!-1C.
As
-+ 1\ 1\ 1\ ~ 1\ 1\ "
11.Here 1 =(i +3j+2k)m, '2 =(3i+5j+k)m
--» -+-+ 1\ 1\ 1\ 1\ 1\ 1\
12 = '2-1 =(3i+5j+k)-(i+3j +2k)
-+ ~2 2 2
112 I=2+2+ (-1)=3m
1.17
-+ 1~q2"
F21=-- -2-12
41tEo 12
9x109x5x10-6x3x10-6 (2i+2J-k)
32 3
3 " 1\ 1\
= 5x10- (2i+2j -k)N
-+ -+ -3 ~<'1\
Also, li2=-Fz1=-5xlO (21+2J-k)N.
12.HereAB= 2 em = 0.02 m, Be = 1em = 0.01 m
q q q....~.
A FBe BFBA C
I· 2ern ·1·1em----l
Fig.1.20
Letqbe the charge on each object.
F __ 1_~
BA-41tEo(AB)2
-69x109xq2
or 3.0x10 = 2
(0.02)
2410-19Cor q=-x .
3
(i)
1qxq 9 4x10-19
F=-----=9xlO X---"
BC41t1,o.( BC)2 3x(0.01)2
= 12.0x10-6N, along BA.
...(i)
(ii) Net force on charge atB,
F =FBC- FBA = (12.0 - 3.0)x10-6
= 9.0x10-6 N, alongBA.
13.Proceed as in Example 10 on page 1.12.
14.Force between+geandq= Force between+eandq
k gexq=k ~
.. .x2 . (a -x)2
3 1
or orx=3a/4.
x a-x
... (ii)
15.For equilibrium of chargesqand2q,the chargeQ
must have sign opposite to that ofqor2q.Suppose
it is placed at distancexfrom chargeq.
q Q 2q
• • •
I---x---+1·1__· -d -x-----1
Fig. 1.21
For equilibrium of chargeq,
k qQ-k qx2q
x2 - d2
For equilibrium of charge2q,
kqx2q=k Qx2q
d2 (d-x)2
...(i)
...(ii)°°°2tu}kyjxo£k2ius

19. 1.18
From(i)and(ii),we get,
k qQ=kQx2q
x2 (d -x)2
or 2x2=(d -x)2
or ..fix=d-x
or x= ~1d=(
..fi -1)d
,,2+1
i.e.,the chargeQmust be placed at a distance of
(..fi -1)dfrom the chargeq.
16.Suppose the three charges are placed as shown in
Fig.1.22.
Q q Q
• • •
A C B
14 x .14 x .1
Fig. 1.22
Clearly, the net force on chargeqis zero.Soit is in
equilibrium, the net force on other two charges
should also be zero.
Total force on chargeQat pointBis
_1_ QQ+_1_qQ=0
41t Eo .(2x)2 41t EOx2
1qQ 1 QQ
41t EO?"= -41t EO .(2x)2
q=-Q/4.
17.In/!,.DCAof forces, we have
Fmg T
-=-=-
AC DC DA
or
or
F.•..-O
A
q
mg
Fig. 1.23
AC
F=mgx-
DC
1 q2 AC
--.--=mgx-
41tEoAB2 DC
9x109xq210-3x9.8x0.1
(0.2)2 ~(0.5)2 _ (0.1)2
q=2.357 xlO-6 C
PHYSICS-XII
1.16COMPARING ELECTROSTATIC AND
GRAVITATIONAL FORCES
25.Give a comparison of the electrostatic and gravi-
tational forces.
Electrostatic forcevsgravitational force.Electro-
static forceistheforce of attraction or repulsion between two
charges at rest while the gravitational forceisthe force of
attraction between two bodies by virtue of their masses.
Similarities:
1.Both forces obey inverse square lawi.e.,
1
FCX:,z'
2.Both forces are proportional to product of
masses or charges.
3.Both arecentral forces i.e.,they act along the line
joining the centres of the two bodies.
4.Both areconservative forces i.e., the work done
against these forces does not depend upon the
path followed.
5.Both forces can operate in vacuum.
Dissimilarities:
1.Gravitational force is attractive while electro-
static force may be attractive or repulsive.
2.Gravitational force does not depend on the nature
of the medium while electrostatic force depends on
the nature of the medium between the two charges.
3. Electrostatic forces are much stronger than
gravitational forces.
Illustrative Problem. Coulomb's law for electricalforce
between two charges and Newton's lawfor gravitational force
between two masses, both have inverse-square dependence on
the distance between charges/masses.
(a) Compare the strength of theseforces by determining
the ratio of their magnitude (i) for an electron and a
proton and (ii) for two protons.
(b) Estimate the accelerations forelectron and proton
due to the electric~l forceattheir mutual attraction
when they are1A(=10-1 m)apart.
How much is the electrostatic force stronger than
the gravitational force?
(a)(i)From Coulomb's law, the electrostatic force
between an electron and a proton separated by
distanceris
Negative sign indicates that the force is attractive.
From Newton's law of gravitation, the corresponding
gravitational attraction is{{{0rsxiwhvmzi0gsq

20. ELECTRIC CHARGES AND FIELD
m m
f=-G~
G ?-
wherem
pandmeare the masses of the proton and
electron.
Hence
1 ~ 1=G~:2me
Butk=9x109Nm2 C-2, e=1.6 x10-19 C,
mp= 1.67x10-27kg,me=9.1x10-31kg,
G=6.67x10-11Nm2 kg-2
IF 1 9x109x(1.6x10-19l
F~= 6.67 x10-11x1.67x10-27x9.1x10-31
=2.27x1039
(a)(ii)Similar to that in part(i),the ratio of the
magnitudes of electric force to the gravitational force
between two protons at a distanceris given by
1Fe1-ke2 -9x109x (1.6 x10-19)2
FG -Gmpmp-6.67x10-11x(1.67x10-27)2
=1.24x1036
Thus the large value of the (dimensionless) ratio of
the two forces indicates that the electrostatic forces are
enormously stronger than the gravitational forces.
(b)The magnitude of the electric force exerted by a
proton on an electron is equal to the magnitude of the
force exerted by an electron on a proton. The magni-
tude of this force is
ke2 9x109x(1.6x10-19)2
F=-=---.....:...-~~----'----
?- (10-1°)2
=2.3x10-8N
Acceleration of the electron due to the mutual
attraction with the proton,
F2.3x10-8N 22 2
ae=-= 31=2.5x10 ms"
me9.1x10- kg
Acceleration of the proton due to the mutual
attraction with the electron,
a=£= 2.3x10-8N= 1.3x1019ms-2
p mp 1.67 x 10-27 kg
Clearly, the acceleration of an electron or a proton
due to the electric force is much larger than the accele-
ration due to gravity. So, we can neglect the effect of
gravitational field on the motion of the electron or the
proton.
1.19
26.Give two examples which illustrate that the
electrical forces are enormously stronger than the gravi-
tational forces.
Examples :(i)A plastic comb passed through hair
can easily lift a piece of paper upwards. The electro-
static attraction between the comb and the piece of
paper overcomes the force of gravity exerted by the
entire earth on the paper.
(ii)When we hold a book in our hand, the electric
(frictional) forces between the palm of our hand and
the book easily overcome the gravitational force on the
book due to the entire earth.
In the words ofFeynman, if you stand at arm's
length from your friend and instead of being electri-
cally neutral each of you had an excess of electrons
over protons by justone per cent,then the force of
repulsion between you would be enough to lift the
entire earth.
1.17FORCES BElWEEN MULTIPLE CHARGES:
THE SUPERPOSITION PRINCIPLE
27.State the principle of superposition of electrostatic
forces. Hence write an expression for the force on a point
charge due to a distribution ofN-1point charges in
terms of their position vectors.
Principle of superposition of electrostatic forces.
Coulomb's law gives force between two point charges.
The principle of superposition enables us to find the
force on a point charge due to a group of point charges.
This principle is based on the property that the forces
with which two charges attract or repel each other are
not affected by the presence of other charges.
The principle of superposition states that when a
number of charges are interacting, the total force on a given
chargeisthe vector sum of the forces exerted onitdue to all
other charges. The force between two charges isnot afected
by the presence of other charges.
As shown in Fig. 1.24, consider Npoint charges
q1' q2' Q3'..., QN placed in vacuum at points whose
--t --t --t --t
position vectors w.r.t. origin0arer1, r2, r3, ... , rN
respectively.
According to the principle of superposition, the
total force on chargeQ1is given by
,
--t --t --t --t
Fl=F12+F13+.....+F1N
where~2' ~3'.... ,~N are the forces exerted on
chargeqlby the individual chargesQ2''13' ..... ,QN
respectively.°°°2tu}kyjxo£k2ius

21. 1.20
y
o·~~=-----------------------~
X
Fig. 1.24 Superposition principle: Force on
chargeqlexerted byqzandq3'
According to Coulomb's law, the force exerted on
chargeqldue toq2is
~ 1 qlq2"
F----- t:
12-4m;T!:12
o12
~ ~
1 qlq2 r1-r2
4m;.~ ~ 2' ~ ~
o Ir1-r21 Ir1-r21
or
or
~ ql
F=--
141tE
o
~ ~
In general, forceFaon ath chargeqalocated atradue
to all other(N-1)charges may be written as
PHYSICS-XII
~
Fa=Total force on ath charge
~ q Nq" q N ti
F=_a_Lirab=_a_Lqb a - b
41tEob=1~b 41tEOb=lit _i13
b"a b"a a b
wherea=1,2, 3, ..., N.
Itmaybenoticed that for each choice of a,the
summation on b omitsthe valuea.This is because
summation must be taken only overothercharges. The
above expression canbe written in a simpler way as
follows:
~
F=Total force on chargeqdue to many point
chargesc(
F=-q- Lc(
41tEo all point
charges
~ ~
r -r'
~ ~3
1r -r'1
Examples based on
Principle ofl,Superposition~7. " :
of Electric'Forces .' . '\ ',.;;.
Formulae Used
~ ~ ~ ~ ~
Fl=F12+F13+F14+ ...+FIN
F=~Fl+F22+2FlF2coso
Units Used
Forces are in newton, chargesincoulomb and
distances in metre.
Example 19.An infinite number of charges each equal to
4IlCare placed along x-axis at x=1m, x=2m,x=4m,
x=8m and so on. Find the total force on a charge of1 C
placedatthe origin. [lIT 95]
Solution. Hereq=4IlC=4x10-6C,qo=1C
By the principle of superposition, the total force
acting on a charge of 1C placed at the origin is
-qqo[1 1 11
F -41tEo,;+tf+-1+ ...
=9x109x4x10-6x1 [ ~+ ~ + ~ + ...]
1 2 4
Sumof the infinite geometric progressio~
a 1 4
1-r=1-!='3
4
F=9x109x4x10-6 xi=4.8x104N.
3°°°3xy£o~n}s«o3myw

22. ELECTRIC CHARGES A
ND FIELD
Example20.Consider three charges ql'q2'q3each equal to
qat the vertices of anequilateral triangle of side I.Whatis the
forceon acharge Q(with the samesign asq)placed at the
centroid of the triangle ? [NCERT]
Solution. Suppose the given charges are placed as
shown in Fig. 1.2S(a).
A
ql
"[;?"OAO
"BO
(a) (b)
Fig. 1.25
LetAO= BO=CO=r
Force on chargeQdue toql'
F=_1_ QqlAD
14m, A02
o
Force on chargeQdue toq2'
F=_1_ Qq2B"o
.c.:> 24m,oB02
Force on chargeQdue toq3'
F=_1_ Qq3CD
34m,oC02
By the principle of superposition, the total force on
chargeQis
----+ ----t ----t ----t
F=Ji+Fz+F3
=~[AD+B"o+CD] [.,' lh=q2=q3=q]
4m,or
As shown in Fig. 1.2S(b), the angle between each
"" "
pair of the unitvectors AO,BO and CO is 120°, so they
form a triangle of cyclic vectors. Consequently,
" " "
AO+ BO+ CO=O
~
HenceF= 0i.e.,the total force on chargeQis zero.
Example 21.Three point charges +q each are kept at the
vertices of anequilateral triangle ofside'1'.Determine the
magnitude andsign of the charge to be kept at itscentroid so
that the charges at the vertices remain in equilibrium.
[CBSE F 2015]
Solution. At any vertex, the charge will be in
equilibriumifthe net electric force due to the
remaining three charges is zero.
1.21
I \
I \
Fig. 1.26
LetQbe the charge required to be kept at the
centroid G. Then,
-4
Ji= Force atAdue to the charge atB
1 2 ~
=--!L,along BA
4m,o12
-4 1 2 -4
Fz= Force atAdue to charge atC= --.q2'alongCA
4m,oI
~ -4 -4 1q2 -4
Ji+F2=2Jicos30°, alongGA=.Ji--·2, alongGA
41teo I
Force atAdue to charge atC
1Qq 1 Qq
4nEo'AC2=4nEo'(l /.J3)2
13Qq
4nEo'[2
~ ~
This must be equal and opposite to(Fl+F2).
•.3Qq=-.J3q2 orQ=- ~.
Example22.Consider the chargesq,qand - q placed at
the vertices of an equilateral triangle, as shown in Fig. 1.27.
Whatisthe force on each charge ? [NCERT)
Solution. The forces of attraction or repulsion
between different pairs of charges are shown in
Fig. 1.27. Each such force has magnitude,
1q2
F=-- -
4nEo' 12
q3=-q
F F
r
r
r
r
\
\
\
\
Fig. 1.27°°°2tu}kyjxo£k2ius

23. 1.22
By the parallelogram law, the net force on charge
qlis
~ = ~F2+F2+2FxFcos 120
0Be
=~2F2+2F2(-1/2)BC=F B"C
"
whereBCis a unit vector alongBC
Similarly, total force on chargeq2is
"
whereACis a unit vector along AC
Total force on chargeq3is
F;= ~F2+F2+2FxFcos 600 ~=.J3F ~
where ~ is a unit vector along the direction bisecting
LACR
Example23.Charges of+5!lC,+10!lCand-10!lCare
placed in air at the corners A, BandCof an equilateral
triangle ABC, having each side equal to5em.Determine the
resultant force on thecharge at A.
Solution. The charge at B repels the charge atA
with a force,
F=kqlq2=9x109x(5x10-6) x(10x10-6) N
I? (0.05)2
=180 N, alongBA
B..------------C
+10 IlC 5 em - 10 IlC
Fig. 1.28
The charge atCattracts the charge atAwith a force
9x109x(5x10-6) x(10x10-6)
F= N
2 (0.05)2
~ N, along AC
By the parallelogram law of vector addition, the
~
magnitude of resultant forceFon charge atAis
F=~F/+F/ +2FIF2cose
PHYSICS-XII
=~(180)2 + (180)2 + 2x180x180xcas1200 N
= 180~1+ 1+ 2 x(-1/2) N=180N
Let the resultant forceFmake an angle ~ with the
forceF2.Then
F2sin 1200 180xsin 1200
tanp= --"------
Ii+F2cas1200 180 + 180 cos 1200
=180x.J3 /2=.J3
180 + 180(-~)
..P=60°
~
i.e.,the resultant force F is parallel toBC
Example24.Four equal point chargeseach16!lCare
placed on the four corners of a square of side0.2m. Calculate
the force on anyone of the charges.
Solution. As shown in Fig. 1.29, suppose the four
charges are placed at the comers of the squareABCD.
Let us calculate the total force onq4'
0.2m
s
N
ci
o
iv
9
A
/
/
/
/
/
/
/
/
0.2m
Fig. 1.29
Here AB= BC=CD=AD=O.2 m
ql=q2=q3=q4= 16 j.lC= 16 X10-6 C
Force exerted onq4byqlis
9x109x16x10-6 x16x10-6
F. - --------,.----
1 - (0.2)2
=57.6 N, alongADproduced
Force exerted onq4byq2is
9x109x16x10-6 x16x10-6
E - ----.,,----;;;----
2 - (0.2)2 + (0.2)2
=28.8 N, along BD produced,
Force exerted onq4byq3is
9x109x16X10-6 x16X10-6
E ---------,.----
3 - (0.2)2
=57.6 N, along CD produced{{{0rsxiwhvmzi0gsq

24. Solution. As show
ninFig.1.30(b),the force exerted
on charge + 2 IlC by charge atB,
F=_1_q1q2
14rcEO1-
9x109x2x10-6x 3 x10-6
(0.20)2
= 1.35 N, along AB
Force exerted on charge + 2IlCby charge atC,
9x109x2x10-6x 3 x10-6
F---------;;,,.-----
2 - (0.20)
=1.35 N, along AC
Resultantforce of1iandFz
F=~rF-=lo...+-F2::-2 -"+'--2-F-1-F-1 -co-s-6-0-0
=~1.352 + 1.352 + 2x1.35 x1.35x0.5
= 1.35 x.J3=2.34 N,alongAM
For thechargeatAto be equilibrium, the chargeqto
be placed at pointMmust be a positive charge so that it
exerts a force on+21lCcharge along MA.
Now, AM=~202 _102
=.J300 =10.J3em
=0.1x.J3m Fig. 1.32
ELECTRIC CHARGES AND FIELD
AsF1andF3areperpendicular to each other, so
their resultant force is
F'=~r-li-=-2-+-I;-=-2=~57.62 +57.62
=57.if2=81.5N,inthedirection ofFz.
Hence total force on q4is
F=F2+F'=28.8 + 81.5
=110.3N,alongBDproduced.
Example25.Threepointchargesof+2 1lC, -3 IlC and
-3IlCare kept at the vertices A, Band Crespectively of an
equilateral triangle of side 20cm as showninFig.1.30(a).
Whatshould be the sign and magnitude ofthe charge to be
placed at the midpoint (M) of sideBCso thatthe charge at A
remains in equilibrium ? [CBSE 005]
+2J.lC
-3J.lC~~ ~~~
B--- 20cm- C B
-3J.lC -3J.lC
(a)
M
(b)
Fig. 1.30
1.23
Net force on charge atAwill be zero if
9x109xqx2x10-6
----'-:=-:::--- = 2.34
(0.1x.J3)2
_ 2.34 x0.01x3 -39 10-6 C - C
or q - 3 -.x -3.91l •
18x10
rproblems for Practice
1.Ten positively charged particles are kept fixed on
thex-axis at points x= 10 em, 20 em, 40 em, ... r
100em. The first particle has a charge 1.0 x10-8C,
thesecond 8x10-8C,third 27x10-8C,and so on.
The tenth particle has a charge 1000x10-8C.Find
the magnitude of the electric force acting on a 1 C
charge placed at the origin. (Ans.4.95x105N)
2.Charges I1J.=1.5 mC,q2=0.2 mC andq3= -0.5 mC
are placed at the pointsA,BandCrespectively, as
shownin Fig. 1.31. If1=12 m andr2=0.6m,
calculate the magnitude of resultant force onq2.
(Ans. 3.125x103N)
F2
B
A F}
q} q2
Fig.1.31
3.Two equal positive charges, each of21lCinteract
with a third positive cHarge of 3 IlC situated as
shown in Fig. 1.32. Find the magnitude and
direction of the force experienced by the charge of
3 1lC. (Ans. 3.456x10-3N,alongDCproduced)
A
.,2 J.lC
1 ,
1
1
I
I
3m; "
; "3 J.lC
1 '
01- - - - - - - - - - - - - -_':-·C
; 4m -:
I
1
3m;
1
I
I
I ,
·'2J.lC
B{{{0rsxiwhvmzi0gsq

25. 1.24
4.Four charges+q ,+q,-qand-qare
placed respec-
tively at the four cornersA,B,Cand Dof a square of
sidea.Calculate the force on a charge Q placed at
the centre of the square.
[Ans. _1_ 4fipq ,parallel toADor BC]
41t1,o a
HINTS
1.By the principle of superposition, the total force on
the1C charge placed at the origin is
Fo=FOl+F02+F03+....+lio
=3.L[ql+q2+q~+....+ql0]
41t Eo rfri'3 1{0
9 [1.0x10-8 8x10-8
=lx9xlO +--..-
(0.10)2 (0.20)2
27x10-8 1000x10-8]
+(0.30)2+ ... +(1.00)2
=9x109x10-6[1+2+3+... +10]
= 9 x55 x103= 4.95 x10s N.
1~q29x109x0.2x10-3 x9x109
2.li=41tEo -;;= (1.2)2
=1.875x 103 N, alongABproduced
E __ 1_q2q3 9x109x0.2x10-3x0.5x10-3
2 - 41tEo' ,i (0.6)2
=2.5x103N, alongBC..l.AB
As li ~ the resultant force onq2is
~2 +F22= 3.125x103N.
3.HereqA=qB=2 J.lC=2x10-6C,
qc=3 J.lC=3x10-6C
AC= BC=~32+42=5m
I
I
I
I
I
I
3m:
: 4m e
O~------------
: e
I
I
3m:
I
I
I
qBI
B
Fig. 1.33
Force exerted by chargeqAon'ic-
F -_1_qA qc
A -41tEo(AC)2
PHYSICS-XII
9x109x2x10-6x3x10-6
52
=2.16x10-3N,alongACproduced
Similarly, force exerted by chargeqBonqc'
FB=2.16x10-3N,along BC produced
Clearly,FA=FB(in magnitude)
The components ofFAandFBalong Y-axis will
cancel out and get added along X-axis.
Total force on3J.lCcharge,
F=2li case=2x2.16x10-3x~
5
=3.456x10-3N, alongCX.
4.HereAB=BC=CD=DA=a
AO=BO=CO=DO= ~~a2+a2=:h
A
a
B
+q
, ,
+q
,
, ,
,
,
, ,
,
, ,
,
, ,
,
,
, ,
,
,
" o"~
,
a
', a
FA
F
Fe
"I'M''''...
","L'"
, ,, ,
, ", ,
,
,
,
N
,
-q -q
D
a
C
Fig.1.34
LetFA' Fa> Fc andFobe the forces exerted by charges
at pointsA,B,Cand Don charge Qat point 0. Then
FA=FB=Fc=Fo
1 qxQ 12qQ
=41tEo.(a;.,fil=41tEo.7
The resultant of the forcesFAandFc'
F.=F+E=_1_ 2qQ+_1_2qQ
1 A C 41tEo' a2 41tEo' a2
14qQ
or li = -- .-2 'along OL
41tEoa
Similarly, resultant of the forcesFBandFo'
14qQ
Ii=FB+FD=--'-2 'along OM
41tEoa
Hence the resultant force on charge Q is
I2 2 14fi qQ
F=Vli+Ii=-- --2- , alongON
41tEoa
As the forces li andF2are equal in magnitude, so
their resultantFwillact along the bisector ofLCOD
i.e.,parallel toADor BC{{{0rsxiwhvmzi0gsq

26. ELECTRIC CHARGES
AND FIELD
1.18ELECTRIC FIELD
28.Briefly develop the concept ofelectric field.
Concept of electric field. Theelectrostatic force acts
between two charged bodies even without any direct
contact between them. Thenature of thisaction-
at-distance force can be understood by introducing the
concept of electric field.
Source chargeB++
+O·q +
+ +
+
Testcharge
Fig.1.35A charged body produces an electric field around it.
Consider a charged body carrying a positivechargeq
placed at point O. It is assumed that the chargeq
produces an electrical environment in the surroun-
ding space, calledelectric field.
To test the existence of electric field at any point P,
wesimply place a small positivechargeqo'called the
~
testchargeat the point P. If a force F is exerted on the
~
testcharge, thenwe say that an electric field E exists at
the point P. The chargeqis called thesource charge as it
~
produces the field E.
29.Define electric field at a point. Giveitsunitsand
dimensions.
Electric field. An electric field is said to exist at a point
ifaforceofelectrical origin isexerted on astationarycharged
body placed at that point. Quantitatively, the electric field or
~
the electric intensity or the electric field strength Eat a
pointis defined as theforce experienced by a unit positive
test charge placed at that point, without disturbing the
position ofsourcecharge.
As shown in Fig. 1.35, suppose a test chargeqo
~
experiences a force F at the point P. Then the electric
field at that point will be
-t~
E=s-
qo
There is a difficulty in defining the electric field by
the above equation. The testchargeqomaydisturb the
charge distribution of the source charge and hence
~
change the electric field E which we want to measure.
The test chargeqomust besmallenough so that it does
~
not change the value of E. It is better to define electric
field as follows:
1.25
The electric field at a point is defined as the electrostatic
forceper unit test chargeacting on a vanishingly small
positive testchargeplaced at that point. Hence
~
The electric field E is avector quantity whose
~
directionissame as that of the force F exerted on a
positive test charge.
Units and dimensions of electric field.As the
electric field is force per unit charge, so its SI unit is
newton per coulomb (NC-1). It is equivalent tovolt per
metre(Vm-1).
~
The dimensions for E can be determined as
follows:
[E]=Force=MLr2
Charge C
=~~;2=[MLr3A-l] [-:lA=~~]
30.Givethe physical significance of electric field.
Physical significance of electric field.The force
experienced by the test chargeqoisdifferent at
~
different points. So E alsovaries from point to point.
~
In general, E is not a singlevector but a set of infinite
~
vectors. Each point r is associated with a unique
~
vector E(r).So electric field is an example of vectorfield.
Byknowing electric field at any point, we can
determine the force on a charge placed at that point.
The Coulomb force on a chargeqodue to a source
chargeqmay be treated as two stage process:
(i)The source chargeqproduces a definite field
~ ~
E(r)at every pointr.
~ ~
(ii)The value ofE(r)at any pointrdetermines the
force onchargeqoat that point. Thisforceis
F=qoE(r)
Electrostatic force=ChargexElectric field.
Thus an electric field plays an intermediary role in
the forces between two charges:
Charge ~ Electric field ~ Charge.
It isin this sense that the concept of electric field is
useful. Electric field is a characteristic of thesystem of
charges and is independent of the test charge that we
place at a point to determine the field.°°°2tu}kyjxo£k2ius

27. 1.26
Exam/esbased on
.. ..
.....
..
Formulae Used
-->
--> F -->-->
E=- or F=qoE
qo
Units Used
When force is in newton, charge in coulomb and
distance in metre, electric field strength is in
newton per coulomb(NC-1) or equivalently in
voltper metre (Vm -1) .
Example 26. Calculate the electric field strength required
to just support a water drop of mass10-3kg and having a
charge1.6 x 10-19 C. [CBSEOD99]
Solution. Herem=10-3kg,q=1.6x10-19C
LetEbe the strength ofthe electric field required to
just support the water drop. Then
Force on water drop due to electric field
=Weight of water drop
orqE=mg
E=mg=10-3x9.8=6.125x1016NC-1.
q1.6x10-19
Example 27. Calculate the voltage needed to balance an oil
drop carrying10electrons when located between the plates
of a capacitor which are5 mmapart. The mass of oil dropis
3x10-16kg.Take g=10ms-2. [CBSEOD95C]
Solution. Hereq=10e= 10x1.6x10-19C
d=5mm=5 x 10-3 m,m=3x 10-16 kg,g=10ms-2
+ + +
Fig. 1.36
or
When thedrop isheldstationary,
Upward force on oil drop due to electric field
=Weight of oil drop
qE=mg
V
q.-=mg
d
V= mgd=3xl0-16xl0x5xl0-3
q 10 x1.6 x10-19
=9.375 V.
PHYSICS-XII
Example 28. How many electrons should be removed from
a coin of mass1.6g, so thatitmay just oat in an electric
field of intensity109Net, directed upward?[Pb.98C]
Solution.Herem = 1.6g= 1.6x10-3kg,
E =109 Ne1 qE
Letnbe the number of electrons?1
removed from the coin. E
Then charge on the coin,
q=+ne
mg
When the coin just floats, Fig. 1.37
Upward force of electric field=Weight of coin
qE orneE=mg
mg1.6x10-3x9.8 7
n=-= =9.8x10.
eE1.6 x 10-19 x 109
Example 29. A pendulum of mass80milligram carrying a
charge of2 x 10-8 C isat rest in a horizontal uniform
electric field of2x 104 Vm-1. Find the tension in the thread
of the pendulum and the angle it makes with the vertical.
Solution. Herem = 80mg= 80x10-6kg,
q=2 x 10-8 C, E =2x104Vm-1.
+-I~ ...•.qE
------
mg
Fig. 1.38
LetTbe the tension in the thread andebe the angle
it makes with vertical, as shown in Fig.1.38.When the
bobis inequilibrium,
Tsine=qE;Tcose=mg
tane=Tsine=3E
Tcos 8mg
2 x 10-8 x2x104
---..,,---- =0.51
80x10-6x9.8
8 =270
qE2x10-8x 2x104
T=--=------
sin8 sin270
Also,
=8.81 X10-4N.{{{0rsxiwhvmzi0gsq

28. ELECTRI
C CHARGES AND FIELD
Example30.An electron moves a distance of6em when
accelerated from rest by an electric field of strength
2x104NC-1. Calculate the timeof travel. The mass and
charge of electron are9x10-31kg and1.6x10-19 C
respectively. [CBSE D 91)
Solution. Force exerted on the electron by the
electric field,
F=eE
:.Acceleration, .
a=£ =eE= 1.6 x10-19 x2x104=0.35 x1016ms-2
m m 9x10-31
Nowu=0,s=6.0 em =0.06 m, a=0.35x1016ms-2
As s =ut+ ~at2
.. 0.06 = 0+ ~x0.35x1016xt2
t= 0.06 x2 = 0.585x10-8 s.
0.35x1016
or
Example 31 .An electron falls through a distance of1.5em
in a uniform electric field of magnitude 2.0x104Ne1
[Fig.1.39(a)}. The direction of the fieldisreversed keeping
its magnitude unchanged and a proton falls through the
same distance [Fig. 1.39(b)}. Compute the time offall in each
case.Contrast the situation (a) with that of'freefall under
gravity'. [NCERT)
+ + ++
+ + + +
(a) (b)
Fig. 1.39
Solution. (a)The upward field exerts a downward
forceeEon the electron.
eE
:.Acceleration of the electron, a=-
eme
1212
Asu=O s=ut+-at =-at
, 2 2
:.Time of fall of the electron is
.-------~------~~
t= ~=~2sme = 2x1.5x10-2x9.1x10-31
eVae eE 1.6x10-19 x2.0x104
=2.9x10-9s.
(b)The downward field exerts a downward forceeE
on the proton.
.. ap=fi
1.27
Time of fall of the proton is
t =~s =~2smp
pa eE
p
2x1.5x10-2x1.67x10-27 = 1.25x10-7s.
1.6x10-19 x2.0x104
Thus the heavier particle takes a greater time to fall
through the same distance. This is in contrast to the
situation of 'free fall under gravity' where the time of
fall is independent of the mass of the body. Here the
acceleration due to gravity'g',being negligibly small,
has been ignored.
Example 32.An electron isliberated from the lower of the
two large parallel metal plates separated by a distance of
20 mm. The upper plate hasa potential of+2400V relative
to thelower plate. How long does theelectron take to reach
the upper plate? Take ~ofelectrons 1.8x1011Ckg-1.
m
Solution. HereV= 2400V,d= 20 mm = 0.02 m,
~=1.8 x1011Ckg-1
m
Upward force on the electron exerted by electric
field is
eV
F=eE=--
d
+
..Acceleration,
FeV1.8x1011x2400-2
a=-=-= ms
m md 0.02
=2.16 x1016ms-2
Using, s= ~at2,we get
t={2S={2d= 2x0.02 s = 1.4x10-9s.
V-; V-;;2.16x1016
Example33.A stream of electrons moving with a velocity
of3x107ms-1isdeected by2mm in traversing a distance
of 0.1 m in a uniform electric field of strength 18Vem-1.
Determine elm of electrons.
Solution. HereVo=3x107ms ",
y=2 mm =2x10-3m, x=0.1m,
E= 18Vcm-1 =1800Vm-1
eE x
ma=eEora= - andt=-
m Vo
1 eEx2
y=-at2
2 2 -;;;.v2
o
e_2yv5 _2 2x10-3 x9x1014
-;;; - Ex2 - 1800x(0.1)2
=2x1011Ckg-l.
or°°°2tu}kyjxo£k2ius

29. 1.28
Example34.Anelectri
cfieldEisset up between the two
parallel plates of a capacitor, as shown inFig.1.40.An
electronenters the field symmetrically between the plates
with a speed vo'TIlelength of each plate isI.Find the angle of
deviation of the path of the electron as itcomesout of thefield.
~
•i~.~v-o--+--+----~~-~------------
'1
Fig. 1.40
Solution.Acceleration of the electron in the
upward direction,
eE
a=-
m
Time taken to cross the field,t=J.-
Vo
Upward component of electron velocity on
emerging from field region,
eEl
v=at=--
y mvo
Horizontal component remains same,Vx=Vo
Ifeis the angle of deviation of the path of the
electron, then
tane=Vy=eE;or
Vxmvo
e -1eEl
=tan --2'
mvo
Example35.A charged particle, of charge21lC and mass
10 milligram, moving with a velocity of 1000 mlsentres a
uniformelectric field of strength103Ne1 directed
perpendicular toitsdirection of motion. Find the velocity
and displacement, of the particle after10 s.
[CBSE Sample Paper 11]
Solution.The velocity of the particle, normal to the
direction of field.
~O ms-I,is constant
The velocity of the particle, along the direction of
field, after 10 s, is given by
"v="v+ayt
-0 qEy _2x10-6x103x10 -2000 -1
-+-t- 6 - ms
m lOx10-
The net velocity after 10 s,
v=~v;+v: =~(1000l+(2000l=1000.J5ms-1
Displacement, alongthe x-axis, after 10 s,
x= 1000x10m=10000 m
PHYSICS-XII
Displacement alongy-axis(in the direction of field)
after 10 s,
1 1qE21 2x10-6x103 2
y=ut+-at2=(O)t+-_Yt =-x -6x(lO)
!f2!f 2m 2 10x10
=10000 m
Net displacement,
r=~x2+y2= ~(10000)2+(10000)2 = lOOOO.Ji m.
~rOems ForPractice
1.If an oil drop of weight 3.2 x10-13N isbalanced in
an electric field of 5x105Vm-I,find the charge on
the oil drop. [eBSE D93](Ans.0.64x10-18C)
2.Calculate the magnitude of the electric field, which
can just balance a deutron of mass 3.2x10-27 kg.
Takeg=10 ms-2. [Punjab99]
(Ans. 2.0x10-7Ne1)
3.Acharged oil drop remainsstationary when
situated between two parallel plates 20 mm apart
and a p.d. of 500Vis applied to the plates. Find the
charge on the drop if it has a mass of 2x10-4kg.
Takeg= 10 ms-2. (Ans.8x10-13 C)
4.In Millikan's experiment, an oil drop of radius
10-4em remains suspended between the plates
which are 1 em apart. If the drop hascharge of 5e
over it, calculate the potential difference between
the plates. The density of oil may be taken as
1.5gem-3. (Ans.770V)
5.Aproton falls down through a distance of 2cm in a
uniform electric field of magnitude 3.34 x103NC-1.
Determine(i)the acceleration of the electron (ii)the
time taken by the proton to fall through the
distance of 2 cm, and(iii)the direction of the electric
field. Mass of a proton is 1.67 x10-27kg.
(Ans.3.2x1011ms-2, 3.54 x1O-7s,
vertically downwards)
6.Aparticle of mass 10-3 kg andcharge 5IlCisthrown
at a speed of 20 ms-1a§ainst a uniform electric field
of strength 2xlOSNC- . How much distance will it
travel before coming to rest momentarily?
(Ans. 0.2 m)
HINTS
1.UseW=qE.
2.E=mg 3.2x10-27x10=2.0xlO-7 NCl.
e 1.6x10-19
3.
V
mg=qEormg=q-
d
:.q=mgd= 2x10-4 x10x20x10-3= 8xlO-8 C.
V 500{{{0rsxiwhvmzi0gsq

30. ELECTRI
C CHARGES AND FIELD
4 3 V
4.Use"31tr pg=ned.
5.(i)a=£=eE=1.6x10-19 x3.34x103
m m 1.67x1027
=3.2x1011ms-2.
(ii)s=0+~at2
..t={¥=
2x0.02 -7
~11=3.54x10s.
3.2xlIT
(iii)The field must act vertically downwards so
that the positively charged proton falls
downward.
6.F=qE=5x10-6x2x105=1N
As the particle is thrown against the field, so
F 1 3-2
a=--;;;= -10-3= -10 ms
Asv2-if=2as .. 02-202=2x(-103)xs
or s=0.2 m.
1.19ELECTRIC FIELD DUE TO A POINT CHARGE
31.Obtain an expression for the electric field
intensity at a point at a distance r from acharge q. What
isthenatureof this field?
Electric field due to a point charge.A single point
charge has the simplest electric field. As shown in
Fig.1.41, consider a point chargeqplaced at the origin
O. We wish to determine itselectric field at a point Pat
o 7 p -->
q••--------~---------..~.~------~.~F
Source qo
charge Test
charge
Fig. 1.41 Electric field of a point charge.
a distance r from it. For this, imagine a test chargeqo
placed at point P. According to Coulomb's law, the
force onchargeqois
F=_1_. qqo;
41tEo ,1
where; is aunit vector in the direction fromqtoqo'
Electric field at point Pis
~
~F 1 «:
E=-=---r
qo 41tEo,1
~
The magnitude of the fieldEis
t:__1_!L
-41tEo . r2
Clearly, Ea:1/,1.This means that at all points on
the spherical surface drawn around the point charge,
1.29
~
the magnitude ofEis same and does not depend on
~
the direction ofr.Such a field is calledspherically
symmetricorradial field, i.e., a field whichlooksthe
sameinalldirections when seen from the point charge.
1.20ELECTRIC FIELD DUE TO A SYSTEM
OF POINT CHARGES
32.Deduce anexpression for the electric field at a
point due toasystem of Npoint charges.
Electric field due to a system of point charges.
Considerasystem of Npoint charges ql' q2' .....rqN
having position vectors r;,~,..... r~with respect to the
y
-->
qj riP qo
1--------.- ------~-rF-----I~
I ,,"/P
I ,
,'y+ 4('-+
" 1 1/r2P
I
I
~~------------------------------.x
o
Fig.1.42Notations used in the determination of electric field
at a point due to two point charges.
origin O. We wish to determine the electric field at
point P whose position vectorisf.According to
Coulomb's law, the forceon charge testqodue to
chargeq1is
~_ 1 q1qO"
F1--- .---y-r1P
41tEo 'IP
where;1Pis a unit vector in the direction fromq1to P
andr1Pis the distance betweenq1and P. Hence the
electric field at point P due to chargeq1is
~
~_Fl _1q1"
El-----Tr1P
qo41tEo r1P
Similarly, electric field at P due to charge q2is
~_ 1 q2"
£2---'Tr2P
41tEo Izp
ACC~ principle of superposition of electric
fields, theelectric field at any point due to a group of charges
isequal to the vectorsum of the electric fields produced by
each charge individually at that point, when allothercharges
areassumed to beabsent.°°°2tu}kyjxo£k2ius

31. 1.30
Hence, the
electric field at point P due to the system
ofNcharges is
or
--+
£
0-- - - - - - -~- - - - - - - - - -'),~:~---.,
q --+ 'I - £--+
I rIP ,P: I
,_--+
: ~!_4P
I
I
~r'3P
I
I
I
I
oq3
cY
q,
Fig.1.43Electric field at a point dueto asystem of
charges is the vector sum of the electric
fields at the point due to individual charges.
In terms of position vectors, we can write
N
~~
~ 1 qj r-r.
E= --L
---'
41t1;0 ~~2 ~~
j=1
Ir -'iI Ir-'iI
~ 1
N
qj~ ~
or E= --L
(r- 'i)'
41t1;0 ~
_ ~13
j=1
Ir
Examples based on
: Electric Fields of Point Charges
Formulae Used
1.E=_l_. ~
41tEOr
2.Bythe principle of superposition, electric field
due to a number of point charges,
~ ~ ~ -+
E=f1+f2+f3+ ...
Units Used
When qis incoulombandrin metre; EisinNC-1
orVm-1.
PHYSICS-XII
Example 36. Assuming that the charge on an atomis
distributed uniformly in a sphere of radius10-10m,what
will be the electric field at the surface of the gold atom?For
gold,Z =79.
Solution. The charge may be assumed to be con-
centrated at the centre of the sphere of radius 10-10 m.
r= 10-10 m,q=Ze=79x1.6x1O-19C
1q9x109x79x1.6x10-19
E=---= ----.,..".-;~--
41t EO.,z (10-10)2
=1.138x1013NC-1•
Example 37.Two point charges of2.0x10-7Cand
1.0x10-7Care1.0em apart. Whatisthe magnitude of the
field produced by either charge at the site of the other ?Use
standard value of1 /41t EO' [Punjab 98]
Solution. Here q1=2.0 x10-7C
q2= 1.0x10-7Cr= 1.0 em = 0.01 m
Electric field due toq1at the site ofq2'
E= _1_q1= 9x109x2.0x10-7
1 41tEO',z (0.10)2
=1.8x107NC-1.
Electric field due toq2atthe site ofq1'
E= _1_q2= 9x109x1.0 x 10-7
241tEO',z (0.10)2
=9x106NC-1.
Example 38. Two point charges of+5xlO-19C and
+ 20x10-19 Care separated by a distance of2m.Find the
point on the line joining them at which electric field
intensityiszero. [CBSEODOlC)
Solution.
-N -N
ql=+5x10 C q2=+20x10 C
••••• •
A £2 P £1 B
14----x----+l~11+-4 ---- 2-x ~I
Fig.1.44
The electric field at point P will be zero if
El=Ez
1 5x10-19 1 20x10-19
41t1;0 x2 4m:o' (2 -x)2
or 4x2=(2_x)2or2x=±(2-x)
or x=2/3 m or-2m
Atx= -2mi.e.,at2 m left ofql'electric fields due
toboth charges will be in same direction. So x= -2 m
is not a possible solution.
Hence electric field will be zero at 2 /3m to the
right ofql'°°°3xy£o~n}s«o3myw

32. ELECTRIC CHARGES
AND FIELD
Example 39. Two point charges of +16IlC and-9IlC are
placed8em apart in air. Determine the position of the point
at which the resultant fieldiszero. [Punjab 94]
Solution. Let Pbe the point at distancexern from
A,where the net field is zero.
q1=+16~C q2= -9 ~C
• • •
A P B
"14--- X---~~14t_-- 8-x ~I
Fig. 1.45
or
At pointP, EI +E2= 0
kx16x10-6 +kx(-9)x10-6 =0
(xx10-2)2 [(8-x)x10-2]2
16 9
x2(8 -x)2
4 3
-=+--
X -8-x
32
x=-cm,32 ern
7
or
or
Atx= 32 em, both EI and E2 will be in the same
7
direction, therefore, net electric field cannot be zero.
Hencex= 32 em
i.e.,electric fieldis zero at a point 24emto the right of
- 9IlCcharge.
Example 40. Twopoint chargesqI= + 0.2Cand
q2=+0.4Careplaced0.1m apart. Calculate the electric
field at
(a) the midpoint between the charges.
(b)apoint on the linejoining qI andq2 such that itis
0.05 m away from q2 and 0.15m away fromqI'
[CBSE D 93C]
Solution.(a)Let0be the midpoint between the
two charges.
ql=+0.2C
• •
q2=+0.4 C
•
o
£1-----.~ £2
••14------0.1 m ------ ..••~I
A B
Fig. 1.46
Electric field at0due to qI'
.E =kqI= 9x109x0.2 =7.2x1011NCI,
1rI2 (0.05)2
acting alongAO
1.31
Electric field at0due toq2'
E =kq2= 9x109x0.4
2ri (0.05)2
= 14.4x1011NCI, acting alongBO
Net field at0= ~ - EI
=7.2x1011NC1, acting along BO.
(b)Electric field atPdue to qI'
kqI9x109x0.2 .
EI = ~= 2' acting alongAP
'1 (0.15)
Electric field atPdue toq2'
kq29x109x0.4 .
E2=~= 2' actmg alongBP
'i (0.05)
q1=+0.2 C q2= +0.4 C £1-----.
• •
.p
A B £2-----.
14 0.1 m ~14 0.05m---+t
Fig.1.47
Netelectric field at pointPis
E= E + E =9X109[~+~]
12 (0.15)2 (0.05)2
=1.52x1012NC1, acting alongAP.
Examfle 41. Two point charges qI andq2 of10-8Cand
-10-Crespectively are placed0.1m apart. Calculate the
electric fields at points A, BandCshown in Fig. 1.48.
-+ [NCERT]
£1
B +1O-8C A
....--______ _1O-8C
1+--0.05m -H4- 0.05m ..•••.. 0.05m -~
Fig. 1.48
~
Solution. The electric field vector EI atAdue to the
positive charge qI points towards the right and it has a
magnitude,
E =kqI= 9x109x10-8 NC-I
1r? (0.05)2
=3.6x104NCI°°°2tu}kyjxo£k2ius

33. 1.32
~
The electri
cfieldvectorE2at A due to the negative
chargeq2points towards the right and it has a
magnitude,
E = 9x109x10-8 NC-1 =3.6x104NC-1
2 (0.05)2
Magnitude of the total electric field atA
Ea= E1+ E2
=3.6x104+ 3.6x104= 7.2x104NC-1
~
Eaisdirected towards the right.
~
The electric fieldvector E1 at B due to the positive
chargeq1points towards the left andit has a
magnitude,
E = 9x 109x10-8NC-1 =3.6x104NC1
1 (0.05l
~
The electricfield vectorE2at B due to the negative
chargeq2points towards the right andit has a
magnitude,
E= 9x109x10-8NC-1 = 4x103NC1
2 (0.15l
Magnitude of the total electric field at B
4 1
Eb=E1-E2= 3.2x10NC
~
Ebis directed towards the left.
Magnitude of each electric field vector, at point C,
of chargesq1andq2is
E=E =9x109x10-8=9x103NC-1
1 2 (o.ll
The directions in which these two vectors point are
shown in Fig. 1.48. The resultant of thesevectors is
given by
Ec=~rE-1-=-2-+-E-2-=-2-+-2-E1-E-2-c-o-s-f)
=~(9x103)2 +(9x103)2 +2x9x103x9x103cos120°
=9x103~1+1+2(-1/2) NC-1 =9x 103NC1
~ ~
Since E1 and E2 are equal in magnitude, so their
resultant~acts along the bisector of the angle
~ ~
between E1 andE2,i.e.,towards right.
Example42.ABCDisa square ofside5m.Charges of
+ 50 C, - 50 Cand+ 50 Care placed at A, CandD
respectively. Find the resultant electric field at B.
Solution. Electric field at B due to + 50 C charge at
Ais
q 50
E1=k.? =k·52 =2k,alongAB
PHYSICS-XII
A 5m ~
+50C.-------,-*::'-r"";';""-+- --.•x
,
-->
E
5m
5m
+50C-------~- 50 C
o 5m Co
o
o
Fig. 1.49 ,
y
Electric field at B due to -50 Ccharge at C is
50
E2=k'2 =2k,along BC
5
Electric field at B due to + 50 C charge at D is
E3=k.~50 =k,alorig DB
( 52 +52)2
Component of E1 alongx-axis = 2k
(as it acts alongx-axis)
Component of E2 alongx-axis =0
(asit acts alongy-axis)
Component of E3 alongx-axis
=E3cos 45° = k·1= ~.
:.Total electric field at B along x-axis
Ex= 2k+ 0 +~=k (2+ 1 J
Now,
Component of E1 along y-axis = 0
Component of E2 along y-axis = 2k
Component of E3 along y-axis
Ey= E3sin 450 =k.1= ~
But the components of E2 and E3 act in opposite direc-
tions,therefore, total electric field at B along y-axis
=2k -~=k(2-1J
:.Resultant electric field at B will be
E=JE2+E2
Vx Y
=[k(2+ 1Jr +[k(2-1Jr =J9k2
=3k=3 x9x109NC-1= 2.7x1010NC-1
1£the resultant field E makes angle ~ withx-axis, then
tan~= Ey = (2 -ll.fi)k =0.4776 or ~=25.50.
Ex(2+1I.fi)k{{{0rsxiwhvmzi0gsq

34. ELECTRIC CHARGES
AND FIELD
Example 43.Four charges +q,+q, - q, - q are placed
respectively at the four corners A, B,CandDofasquare of
side 'a'.Calculate the electricfield at the centre of thesquare.
[Punjab 96C]
Solution.LetEA,E8'11:and ED be the electric fields
at the centre°of the square due to thecharges atA,B,
C and D respectively. Their directions are as shown in
Fig. I.50(a).
+q a +q
A~-----~B
+q a +q
A------ ..•B
D~-----~C D¥--- ..••....--..,.C
-q -q -q -q
(a) (b)
Fig. 1.50
Since all thecharges are of equal magnitude and at
the same distancerfrom the centre0,so
EA= EB=11:= ED=k· ~=(aq)2=2~~
..fi[.:?+?=a2]
Because EAand11:actin the same direction, so
their resultant is
E_E P_2kq2kq _ 4kq
1 -A+L-a2+a2 - a2
Similarly, resultant of EB and ED is
4kq
Ez=EB+ ED =-2
a
Now, the resultant of EI and E2 will be
E =~EI2+E22=(:~qr+(4a~qr
=4..fik!L
a2'
directed parallel toADorBC,as shown in Fig.1.50(b).
E 1
cos~=--1=- ..~=45°
E..fi
i.e.,the resultant field is inclined at an angle of 45 °withAC.
Example 44. Two point charges +6q and -8q are placed at
the vertices'B'and'C'of anequilateral triangle ABC of side
'a' as shown in Fig. 1.51(a). Obtain the expression for (i)the
magnitude and(ii)the directionof theresultantelectric field
at the vertex A due to these two charges. [CBSE OD 14C]
Solution. (i)As shown in Fig. 1.5I(b), the fields at
~ -+
pointAdue to the charges at Band CareEBAandEAC
respectively.
1.33
A
+6q'--------->- 8q+6q__----- ...•- 8q
B C B a C
Fig.1.51 (a) ( b)
Their magnitudes are
16q 1q
EBA=--'2=6E,where E=--'2
4n~a 4n~a
18q
EAC=--'2=8E
4m,oa
The magnitude of the resultantfield is
Enet=~E~A+E~c+2EBAEACcosI20°
=(6E)2+(8E)2+2x6 Ex8 Ex (-~)
=E.J52=_1_q.J52
4m,oa2
(il)If the resultant field makes an angle ~withAC,then
EBAsinI20° 6Ex(.J3/2) 3.J3
tanf=------"'''--------
EAC+EBAcosI20° 8E+6E( _~)S-
..~ = tan-l( 3:J
<prOblems For Practice
1.An electron is separated from the proton through a
distance of 0.53 A.Calculate the electric field at the
location of the electron. (Ans. 5.1x1011NC-l)
2.Determine the electric field produced by a helium
nucleus at a distance of 1Afrom it.
(Ans.2.88 x1011NCI)
3.Two point charges +qand +4qare separated by a
distance of6a.Find the point on the line joining the
two charges where the electric field is zero.
(Ans. At a distance 2afrom charge+q)
4.Two pointcharges qlandq2of 2x1O-8C and
- 2x1O-8C respectively are placed 0.4 m apart.
Calculate the electric field at the centre of the line
joining the two charges. [.'l5E F 94C)
(Ans.900 NC-l, towards the-\"charge)°°°2tu}kyjxo£k2ius

35. 1.34
5.Tw
o pointcharges+qand -2qare placed at the
vertices'B'and'C' of an equilateral triangle ABC
of side'Iias givenin the figure. Obtain the
expression for(i)the magnitude and(ii)the
direction of the resultant electric field at thevertex
Adue to these two charges. [CBSE 00 14C]
[Ans.(i)_1_q.J3 (ii)30°withAq
41tEo a2
A
Fig.1.52
6.Find the magnitude and direction of electric field at
point P in Fig. 1.53.
A
+q
E.
B
pa
*---------------~+q
Ca
Fig.1.53
(Ans. E=_1_ 2~,alongBPproducedJ
41tEo a
7.Threecharges, each equal to qare placed at the three
corners of a square of side a.Find the electric field
at the fourth corner. (Ans. (2..fi+1)q2J
81teoa
8.Figure1.54shows four point charges at thecorners
of a squareofside2cm.Find the magnitude and
direction of the electric field at the centre0of the
square, if Q=0.02Jlc.
-2Q 2cm +2Q
A ,B
,
, ,
,
,
,
, ,
, ,
, ,
,,
A
/,'0'"
, ,
, ,
, ,
, ,
, ,
, ,
D' ,c
+Q 2em -Q
_1_ =9x109Nm2 C-2.
41tEo
2em2cm
Fig.1.54
Use
[ISCE 98]
(Ans.s-Iix105Nc1, parallel toBA)
PHYSICS-XII
HINTS
1.Electric field at the location of the electron,
_ 1 q _9x109x1.6x10-19 _ 11 C-1
E---. -2 - 102 -5.1xlO N .
41tEo r (0.53 x10- )
2.Hereq=+2eandr=1A=10-10m.
3.Suppose theelectric field is zero at distance xfrom
the charge +q.Then
1q 1 4q
41tEO .x2=41tEO.(6a-x)2
or (6a-x)2=4x2 or6a-x=2x
or x=2a
. .Electric fieldiszeroatdistance 2afrom the
charge+q.
4.Proceed as in Q. 1.8on page1.81.
5.Proceed as in the solution of Example 44on
page1.33.
6.Here EAandEcareequal and opposite and hence
cancel out.
BP=asin 45° =a /..fi
HenceE=E=_1_ q
B 41tEo· (a /..fi)2
__ 1_2qalongBPproduced.
- 41tEo . a2'
7.Refer to Fig. 1.55.
A
EVe EB
o E
a
q , A
,
,
,
,
~~,/
,
,
,
,
aa
qB•-------a------ ...•Cq
Fig.1.55
ED= ~E~+E2+EB
(qJ2+ (qJ2+ q
41t EO a2 41tEOa2 41tEo(..fi a)2
=.s:[v';+-4]=(2..fi +1)-q-2 .
41tEoa 2a 81tEoa
8.Here, AB=BC=CD=AD=2em
~22+22
AO=BO=CO=DO= =..fiem
2
=..fix10-2 m{{{0rsxiwhvmzi0gsq

36. ELECTRIC CHARGES AND FIELD
Fig.1.56
1-2Q
1 4
..EA= --.--2=--.Qx10,alongOA
47tEo(OA) 47tEo
1 2Q 1 4
EB=-- .--2 =--.Qx10,alongaD
47tEo (aB) 47tEo
1 Q 1Q 4
Ec=--'--2 =--.- x10 ,alongOC
47tEo (OC) 47tEo2
1 Q 1Q 4
andED=--. --2 =--. -x10ralongOB
47tEo (OB) 47tEo 2
Net electric field alongOA,
1Q 4
E1=E - Ec=- . - x10
A 47tEo 2
Net electric field alongOl),
1Q 4
E,= EB- ED = -- .-x10
47tEo 2
Hence, the resultant electric field at point 0,
E=~E12+ Fi
1Q 4
=47tEo ..fix10,parallel to sideBA
But,Q=0.02jlC=0.02x10-6C
.. E= 9x109.0.02x~-6 xl04
=9.fix105Ne1, paralleltosideBA.
1.21CONTINUOUS CHARGE DISTRIBUTION
33.What is a continuous charge distribution?How
can we calculate the force on a point chargeqdue to a
continuous charge distribution?
Continuous charge distribution.In practice, we
deal with charges much greater in magnitude than the
charge on an electron, so we can ignore the quantum
nature of charges and imagine that the charge is spread
in a region in a continuous manner. Such a charge
distribution is known as acontinuous charge distribution.
Calculation of the force on a charge due to a conti-
nuous charge distribution.As shown in Fig.1.57,
consider a point chargeqolying near a region of contin-
uous charge distribution. This continuous charge distri-
bution can be imagined to consist of a large number of
small chargesdq.According to Coulomb's law, the
force on point chargeqodue to small chargedqis
~
h "r. .
were r=-,ISa unit
r
vector pointing from the
small chargedqtowards
the point chargeqo'By
the principle of super-
position, the total force
on chargeqowill be the
vector sum of the forces
exerted by all such
small charges and is
given by
1.35
Fig. 1.57 Force on a point charge
q0due to a continuous charge
distribution.
~- f ~ - f1qo dq"
F -dr -47tl;o'T.r
F=~fdq.;
4m;o ~
34.Name the diferent types of continuous charge
distributions. Define their respective charge densities.
Write expression for the electric field produced by each
type of charge distribution. Hence write expression for
the electric field of a general source charge distribution.
Different types of continuous charge distributions.
There arethreetypes of continuous charge distributions :
(a)Volumecharge distribution.Itisa chargedistri-
bution spread over a three dimensional volume or region Vof
space,as shown in Fig.1.57.We define thevolume charge
density at any point in this volume as the charge contained
per unit volume at that point,i.e.,
dq
p=dV
or
The SI unit forpis coulomb per cubic metre (Cm-3).
For example, if a
chargeqis distributed
over the entire volume
of a sphere of radiusR,
then its volume charge
density is
p=-q-Cm-3
i7tR3
3
The charge con-
tained in small volume
dVis
dq=pdV
dq=pdV
Fig. 1.58 Volume charge
distribution°°°2tu}kyjxo£k2ius

37. 1.36
Total electrostatic
force exerted on chargeqodue to
the entire volumeVis given by
F;=.s:fdq;=lf ~dV ;
47tEov? 47tEoV?
Electric field due to the volume charge distribution
at the location of chargeqois
--t
E;,=Fv=_1_f~dV;.
qo47tEov?
(b)Surface charge distribution.Itisa charge
distribution spread over a two-dimensional surfaceSin
space,as shown in Fig.1.59.We define thesurface
charge density at any point on this surface as the charge per
unit area at that point, i.e.,
cr=dq
dS
The 51 unit forcris Cm-2.
dq=adS
Fig. 1.59 Surface charge distribution.
For example,if a chargeqis uniformly distributed
over the surface of a spherical conductor of radiusR,
then its surface charge density is
cr=-q-Cm-2
47tR2
The charge contained in small areadSis
dq=cdS
Total electrostatic force exerted on chargeqodue to
the entire surfaceSis given by
F;..»:f ~dS;
41IEo Sr:
Electric field due to the surface charge distribution
at the location of chargeqois
--t
--t _Fs _1fcr"
Es - - - -- 2dSr .
qo.47tEo S r:
(c)Line charge distribution.Itisa charge
distribution along a one-dimensional curve or lineLin
space,as shown in Fig. 1.60. We define the line-charge
PHYSICS-XII
density at any point on this line as the charge per unit
length of the line at that point, i.e.,
A=dq
dL
The 51 unit for Ais Cm-1.
+ +
+
+
+ ->
r
+
d~:'- dq=AdL
+
+
+
+
+
Fig. 1.60 Line charge distribution.
For example,if a chargeqis uniformly distributed
over a ring of radiusR,then its linear charge density is
A=-q- Cm-1
27tR
The charge contained in small lengthdLis
dq= AdL
Total electrostatic force exerted on chargeqodue to
the entire lengthLis given by
~..».f!:.dL;
47tEo L?
Electric field due to the line charge distribution at
the location of chargeqois
--t
E=!L =_1_f!:.dL;
Lqo47tEo L?
The total electric field due to a continuous charge
distribution is given by
;::f --t --t --t
1:.eont=e;+Es+EL
orfeont=_1_[f ~dV;+f ~dS;+f-~dL;1
47tEovr Sr: Lr:
General charge distribution.Ageneral charge distri-
bution consists of continuous as well as discrete charges.
Hence total electric field due to a general charge
distribution at the location of chargeqois given by
--t --t ;::f
Etotal=Ediscrete+1:.eont
or E =_1_ [~qi;+f ~dV ;
total 47tE.~? i r2
o1-1 I V{{{0rsxiwhvmzi0gsq

38. ELECTRIC CHARGE
SAND FIELD
"~
In all the above cases, r=r /ris a variable unit
vector directedfrom each point of the volume, surface
or line charge distribution towards the location of the
pointchargeqo'
Formulae Used
1.Volume charge density,p= ~
dV
2. Surface charge density,c= ~~
3.Linear charge density,A.= ~~
4.Force exerted on a chargeqodue to a continuous
charge distribution,
F=-'l!Lfdqr
4m:r2
o
5.Electric field due to a continuous charge distribution,
E=_l_fdq r
4m:ar2
Units Used
pis in Cm -3,oin Cm-2,A.in Cm-1 andEin NC-1.
Example 45.Acharged spherical conductor has a surface
density of0.7Cm-2 .Whenits charge isincreased by 0.44 C,
the charge density changes by 0.14 Cm-2. Find the radius of
thesphere and initialcharge on it.
Solution.c=q2
41tr
Infirstcase: O.7=-q-
41tl
or
In second case :
0.7+0.14=q+0;
41t
0.84=q+0.44
41tr2
Dividing(ii)by(i),we get,
0.84 q +0.44
or
0.7 q
~ =1+0.44
5 q
..Initial charge,q=2.2 C.
From(i),r=~crxq41t =~0.72~41t
2.2x7
1---- =O.Sm.
0.7x4x22
1.37
Example 46. Sixty four drops of radius 0.02m and each
carrying a charge of5!lCare combined to form a bigger
drop.Find how the surface density ofelectrification will
change if no charge islost.
Solution.Volume of each small drop
=i1t(0.02)3 m 3
3
Volume of 64 small drops=i1t(0.02)3 x64 m3
3
LetRbe the radius of the bigger drop formed. Then
i1tR3=i1t(0.02)3 x64
3 3
or R3=(0.02)3x43
.. R=0.02x4=0.08 m
Charge on small drop=5!lC=5x10-6C
Surface charge density of small drop,
c=_q_=5x10-6 Cm-2
141tl41t(0.02)2
Surface charge density of bigger drop,
5x10-6 x64 C-2
cr= m
2 41t(0.08)2
cr1= 5x10-6 X41t(0.08)2 =!= 1 : 4.
cr241t(0.02)2 5x106x64 4
... (i)
Example 47. Obtain the formula for the electricfield due to
a long thin wire of uniform linear charge densityA.without
using Gauss's law. [NCERT 1.30]
Solution. Electric field of a line charge from
Coulomb's law. Consider an infinite line of charge with
uniform line charge density;>..,.as shown in Fig. 1.61.
We wish to calculate its electric field at any pointPat a
distanceyfrom it. The charge on small elementdxof
the line charge will be
dq=A.dx
...(ii)
Fig. 1.61 A section of an infinite line of charge.°°°2tu}kyjxo£k2ius

39. 1.38
The electric
field at the point P due to the charge
elementdqwill be
dE=_l_ dq=_1_ ~
4m,0.?- 4n60' y2+x2
The fielddEhastwo components:
dEx=-dEsin e and dEy=dEcose
The negative sign inx-component indicates that
-+
d Exacts in the negative x-direction. Every charge ele-
ment on the right has a corresponding charge element
on the left. Thex-components of two suchcharge
elementswill beequal and opposite and hence cancel
-+
out.The resultant fieldEgetscontributions onlyfrom
y-components and is given by
X=+OO
E=Ey=fdEy=fcos edE
x=-oo
xf=OO 1 J...dx
=2 cose.--. 2 2
x=0 4n60 y+x
x=oo
J...f dx
=-- cose---
2n60 x=0 y2+x2
Now x=ytan e
dx=ysec2e de
J...6=1t/2 sec2ede
E=--fcoso~y-----;:;---
2n60 6=0 y2(1+tan2e)
J...6=1t/2 J...
=--fcosede=-- [sine]~/2
2n60 y6=0 2n60 y
or
J...(.n .0)
=2n60ysm2"-sm
E=_J..._.
2n60 y
Example48.Acharge isdistributed uniformly over aring or
of radius'a'.Obtain anexpression for the electric intensity E
at a point on the axis of the ring. Hence show that for points
at large distances from the ring, it behaves like a point
charge. [CBSE Sample Paper 90]
Solution. Suppose that the ring is placed with its
plane perpendicular to thex-axis, as shown in Fig1.62.
Consider a small elementdlof the ring.
As the total chargeqis uniformly distributed, the
chargedqon the elementdlis
dq=-q-.dl
2na
PHYSICS-XII
dl
dl
Fig. 1.62
-+
:.The magnitude of the fielddEproduced by the
elementdlat the field pointPis
aE=k.dq=kq .dl
r22nar2
-+
As shown in Fig. 1.62,the fielddEhas two
components:
1.theaxialcomponent dE cos e, and
2. the perpendicular component dE sin e.
Since the perpendicular components of any two
diametrically opposite elements are equal and
opposite, theyall cancel out in pairs. Only the axial
components will add up to produce the resultant field
Eat pointP,which is given by
2M
E=fdEcose
o [.:Only the axial components
contribute towards E]
2Mk k 2M
=f-q .dl..:.=qx.~fdl
o2na?-r2na r 0
[.: cosB=;]
E= kqx =_1_ qx
(x2+a2)3/2 41tEo'(x2+a2)3/2 •
Special case
For points at large distances from the ring,x»a
E_kq _1q
- x2 - 4n6 . x2
o
This is the same as the field due to a point charge,
indicating that for far off axial points, the charged ring
behaves as a point charge.
Example 49. A thin semicircular ring of radius a is
charged uniformly and the charge per unit length is J....Find
the electric field at its centre. [CBSE PMT 2000, AIEEE 2010]{{{0rsxiwhvmzi0gsq

40. ELECTRI
C CHARGES AND FIELD
Solution. Consider two symmetric elements each
oflengthdlatAand B. The electric fields of the two
elements perpendicular to PO get cancelled while
those along PO get added.
Electric fieldat0due to an element of lengthdlis
1dq
dE=---cosS [AlongPO]
41tEoa2
1Adl
=----cosS
4m,0a2
[dl=adS]
Fig. 1.63
Totalelectric field at the centre 0is
1(/2 1(/2 SdS
E=fdE=2f_1_Acos
-1(/2 041tl:o a
=_1_~[SinS]~/2 =_1_~.1 =_"_.
21tl:oa 21tl:oa21tEoa
rproblemsfor Practice
1.Auniformly charged sphere carries a totalcharge
of21tx1O-12C. Itsradius is 5em and is placed in
vacuum. Determine its surface charge density.
(Ans. 2x10-10Cm -2)
2.What charge would be required to electrify a
sphere of radius 15 em so as to get a surface charge
density of2."Cm-2 ?
11r- (Ans, 1.8x10-7C)
3.A metal cube of length 0.1 m is charged by 12 ~C.
Calculate its surface charge density.
(Ans,2x10-4Cm-2)
4.Two equal spheres of water having equal and
similar charges coalesce to form a large sphere. If
no charge is lost, how will the surface densities of
electrification change? (Ans. 0"1: 0"2 =22/3 :2)
1.39
HINTS
1.UseO"=~.
41tr
2.Useq=41tr20".
3.Surface area of cube=6x/2=6x0.01=0.06 m2.
4.i1tR3=2x.i1tr3orR=21/3r
.3 3
2
2 -2
0"1_q 41tR_R2_23r _2/3.
---- --------2 2
0"241t?' 2q 2r2 2r2 .
1.22ELECTRIC DIPOLE
35. Whatisan electric dipole ?Define dipole moment
and give its SI unit. Give some examples of electric
dipoles. What are idealorpoint dipoles?
Electric dipole.A pair ofequal and oppositecharges
separated by a small distanceiscalled anelectric dipole.
Dipole moment. It measures the strength of an
electric dipole.Thedipole moment of an electric dipole isa
vectorwhose magnitude iseithercharge times the separation
between the two opposite charges and the directionisalong
thedipole axis from the negative to the positive charge.
As shown in Fig. 1.64, consider an electric dipole
consisting of charges+qand -qand separated by dis-
tance 2a.The line joining the charges is called dipole axis.
-q
•
+q
•••..•
p
Fig. 1.64
Dipole moment = Eitherchargexa vector drawn
from negative to positive charge
or
~ ~
p=qx2a
~
Thus the dipole momentpis avectorquantity. Its
direction is along the dipole axis from -qto+qand its
magnitude is
p= qx2a
The SI unit of dipole momentiscoulomb metre (Cm).
When both the chargeqand separation 2aare finite, the
dipole has a finite size (equal to 2a),a location
(midpoint between+qand -q),a direction and a
strength.
Examples of electric dipoles.Dipoles are common
in nature. In molecules like Hz0' HCI, C2HSOli
C~COOli etc., the centre of positive charges does not
fall exactly over the centre of negative charges. Such
molecules are electric dipoles. They have a permanent
dipole moment.°°°2tu}kyjxo£k2ius

41. 1.40
Ideal or
point dipole. We can think of a dipole in
which size 2a ~0 and chargeq ~00in such a way that
the dipole moment,p=qx2ahas a finite value. Sucha
dipole of negligibly small sizeiscalled an ideal or point
dipole.
Dipoles associated with individual atoms or molecules
may be treated as ideal dipoles. An ideal dipole is
specified only by its location and a dipole moment, as
it has no finite size.
1.23DIPOLE FIELD
36.Whatisa dipole field? Why does the dipole field at
large distance falls of faster than 1/r2?
Dipole field.The electric field produced by an electricor
dipoleiscalled a dipole field.This can be determined by
using(a)the formula for the field of a point charge and
(b)the principle of superposition.
Variation of dipole field with distance. The total
charge of an electric dipole is zero. But the electric field
of an electric dipole is not zero. This is because the
charges+qand -qare separated by some distance, so
the electric fields due to them when added do not
exactly cancel out. However, at distances much larger
than the dipole size(r»2a),the fields of+qand-q
nearly cancel out. Hence we expect a dipole field to fall
off, at larger distance, faster than1/,1,typical of the
field due to a single charge. In fact a dipole field at
larger distances falls off as1/?
1.24ELECTRIC FIELD AT AN AXIAL POINT
OF A DIPOLE
37.Derive an expression for the electric field at any
point on the axial line of an electric dipole.
Electric field at an axial point of an electric dipole.
As shown in Fig.1.65,consider an electric dipole
consisting of charges+qand -q,separated by distance
2aand placed in vacuum. Let P be a point on the axial
line at distancerfrom the centre0of the dipole on the
side of the charge+q.
p
-0 +q s., P E+q
••---+1-- • ---.....•_--+_--_•.
-q
14--- 2a---..
14 r----~.I
Fig. 1.65Electricfieldat anaxial point of dipole.
Electric field due to charge -qat point Pis
~ -q"
E= P (towards left)
-q4nEo(r+al
wherepis a unit vector along the dipole axis from-q
to+q.
PHYSICS-XII
Electric field due to charge+qat point Pis
E= qp (towards right)
+q4nEo(r - al
Hence the resultant electric field at point P is
~ -;:; ~
E axial=1:,+q+E_q
=4:EJ(r~a)2 - (r: a)2]P
q 4ar"
=4nEO . (,1 - a2)2p
~ 1 2pr"
Eaxial=4nEo . (,1 _ a2)2p
Herep=qx2a=dipole moment.
Forr» a, a2 can be neglected compared to?
~ 1 2p"
Eaxial --- -p
- 4nEo .r3
(towards right)
Clearly, electric field at any axial point of the dipole
acts along the dipole axis from negative to positive
~
chargei.e.,in the direction of dipole momentp .
1.25· ELECTRIC FIELD AT AN EQUATORIAL
POINT OF A DIPOLE
38.Derive an expression for the electric field at any
point on the equatorial line of an electric dipole.
Electric field at an equatorial point of a dipole. As
shown in Fig.1.66,consider an electric dipole consis-
ting of charges -qand+q,separated by distance2a
and placed in vacuum. Let P be a point on the equa-
torialline of the dipole at distancerfrom it.
i.e., OP=r
_q__~ __ a__ ~o~_a __ ~.+q
A ----+ B
P
Fig.1.66Electric field at an equatorial point of a dipole.°°°3xy£o~n}s«o3myw

42. ELECTRIC CHARGES
AND FIELD
Electric field at point P due to +qcharge is
--+ 1 q .
E+q= --. ~ , directed along BP
41tEo r:+a
Electric field at point P due to-qcharge is
--+ 1 q .
E_q= --. ~, directed alongPA
41tEor:+a
--+ --+
Thus the magnitudes ofE_qandE+qare equali.e.,
1 q
E=E =-.--
-q +q41tEo?-+a2
--+ --+
Clearly, the components ofE_qandE+qnormal to
the dipole axis will cancel out. The components
parallel to the dipole axis add up. The total electric
--+ --+
fieldEequais opposite top.
--+ A
Eequa=-(E_q cos9+E+qcos9)P
=-2E_qcos9p [E_q=E+ql
=-2_l q_ ap
.41tEo?-+a2.~?-+a2
[... 00,0=hl
--+ 1 P A
E =-- P
equa 41tEo .(?-+a2)3/2
wherep=2qa,is the electric dipole moment.
If the point P is located far away from the dipole,
r»a, then
or
Clearly, the direction of electric field at any point
on the equatorial line of the dipole will be antiparallel
--+
to the dipole momentp .
39. Give a comparison of the magnitudes ofelectric
fields of ashort dipole at axial and equatorial points.
Comparison of electric fields of a short dipole at
axial and equatorial points. The magnitude of the
electric field of a short dipole at an axial point at
distance r from its centre is
E __ 1_2p
axial-41tEO r3
Electric field at an equatorial point at the same
distance r is
E =_1_£
equa 41tEo?
1.41
Clearly,
Eaxial=2 Eequa
Hence theelectric field of ashortdipole atadistance r
along its axisistwicethe electric field at the same distance
alongtheequatorial line.
1.26TORQUE ON A DIPOLE IN A UNIFORM
ELECTRIC FIELD
40. Derive an expression for the torque on anelectric
dipole placed in auniformelectric field. Hence define
dipole moment.
Torque on a dipole in a uniform electric field. As
shown in Fig.1.67(a), consider an electric dipole
consisting ofcharges+qand -qand of length2a
placed in a uniform electric fieldEmaking an angle9
with it. It has a dipole moment of magnitude,
p= qx2a
• --+ --+
Force exerted on charge+qbyfieldE=qE
--+
(alongE)
--+ --+
Force exerted on charge-qbyfieldE=-qE
--+
(opposite toE)
~otal=+qE-qE=O.
------------------------~~~
+qE
~
-qE
-q
(a)
~
p
{}
~
E
~ ~ ~
r=pxE
(b)
Fig.1.67(a)Torque on a dipole in a uniform electric field.
(b)Direction of torque as given by right hand screw rule.
Hence the net translating force on a dipole in a
uniform electric field is zero. But the two equal and
opposite forces act at different points of the dipole.
They form a couple which exerts a torque.°°°2tu}kyjxo£k2ius

43. 1.42
Torque=Either
forcexPerpendicular distance
between the two forces
t=qEx2asin8=(qx2a) Esin8
or t=pEsin 8 (p=qx2a)
--+
As the direction of torqueris perpendicular to
bothpandE ,so we can write
--+
The direction of vectortisthatinwhich aright
--+
handed screw would advance when rotated from pto
--+ --+
E.As shown in Fig.1.67(b), the direction of vectortis
perpendicular to, and points into the plane of paper.
--+
When the dipole is released, the torque r tends to
--+
align the dipole with the fieldEi.e.,tends to reduce
--+
angle 8 to O.When the dipole getsaligned with E,the
--+
torquetbecomes zero.
Clearly, the torque on the dipolewill be maximum
--+
when the dipole is held perpendicular toE.Thus
tmax =pEsin 90° =pE.
Dipole moment. We know that the torque,
r=pEsin 8
IfE=1 unit, 8 =90°, thent=p
Hencedipole moment may be defined asthetorque
acting on anelectric dipole, placedperpendicular to a uniform
electric field of unitstrength.
1.27 DIPOLE IN A NON-UNIFORM
ELECTRIC FIELD
41.What happens when an electric dipole isheld in a
non-uniform electric field? What will be theforceand the
torque when the dipole isheld parallel or anti-parallel to
the electric field ?Henceexplain whydoesacomb run
through dry hair attract pieces of paper ?
Dipole in a non-uniform electric field.In a non-
uniform electric field, the+qand-qcharges of a dipole
experience different forces (not equal and opposite) at
slightly different positions in the field and hence a net
--+
force F acts on the dipole in a non-uniform field. Also,
a net torque acts on the dipole which depends on the
location of the dipole in the non-uniform field.
--+ --+ -;t--+
t=pxc(r)
--+
whereris the position vector of the centre of the dipole.
PHYSICS-XII
--+
When the dipole is parallel or antiparallel toE.In a
--+ --+ t
non-uniform field, ifpisparallel to Eor antiparallel0
--+
E ,the net torqueon the dipole is zero (because the
forces on charges±qbecome linear). However, there
is a net force on the dipole. As shown in Fig. 1.68,when
--+ --+
pisparallel toE,a net force acts on the dipole in the
--+ --+ --+
direction of increasing E.Whenpisantiparallel to E,
--+
a net force acts in the direction of decreasing E.
E
Force on-q
~ .
Force on+ q
0-------.--.0
-q P +q
Direction ofnetforce = •
Direction ofincreasing field = •
(a)
E
•
0---+---.0
+q P -q
Force -----l.~ "'~f----- Force
on+q on-q
Direction of net force =•••••f----
Direction of increasing field = •
(b)
Fig.1.68 Forces on a dipole(a)whenpis parallel
-;7 --> -->
to1:.and(b)Whenpis antiparallel toE.
A comb run through dry hair attracts small pieces
of paper.As the comb runs through hair, it acquires
charge due to friction. When thecharged comb is
brought closer to an uncharged piece of paper, it
polarises the piece of paperi.e.,induces a net dipole
momentin the direction of the field. But the electric
field due to the comb on the piece of paperisnot uni-
form. It exerts a force in the direction ofincreasing field
i.e.,the piece of paper gets attracted towards the comb.
42. Give thephysical significance of electric dipoles.
Physical significance of electric dipoles. Electric
dipoles have a common occurrence in nature. A
molecule consisting of positive and negative ions isan
electric dipole. Moreover, a complicated array of
charges can be described and analysed in terms of
electric dipoles. The concept of electric dipole is used
(i)in the study of the effect of electricfieldon an
insulator, and(ii)in thestudy of radiation of energy
from an antenna.{{{0rsxiwhvmzi0gsq

44. ELECTRIC CHARGES AND FIELD
For Your Knowledge
~In a uniform electric field, an electric dipole
expe-
riences no net force but a non zero torque.
~As the net force on a dipole in a uniform electric field
is zero, therefore, no linear acceleration is produced.
~'Torque on a dipole becomes zero when it aligns itself
parallel to the field.
~Torque on a dipole is maximum when it is held
....
perpendicular to the field E .
~In a non-uniform electric field, a dipole experiences a
non zero force and non zero torque. In the special case
when the dipole moment is parallel or antiparallel to
the field, the dipole experiences a zero torque and a
non zero force.
•
--
A non-uniform or A B C-- • •
specifically an
increasingE-field •
may be represen-
•
Directionof
ted by field lines
as shown. Fig.1.69
increasing
E-field
Clearly, EA<EB<Ec
~The direction of the electricfield at an axialpoint of an
electric dipole is same as that of its dipole moment
and at an equatorial point it is opposite to that of
dipole moment.
~The strength of electricfield at an axial point of a short
dipole is twice the strength at the same distance on
the equatorial line.
~At larger distances, the dipole field (Eex:1/r3)
decreases more rapidly than the electric field of a
point charge (Eex:1/r2).
Formulae Used
1. Dipole moment,p=qx2a;where2ais the
distance between the two charges.
2. Dipole field at an axial point at distance r from the
centre of the dipole is
E .=_1_ 2pr
axial 41t EO .(r2_ a2)2
When r »a, E __ 1_2p
axial - 41t E . r3
o
3. Dipole field at an equatorial point at distancer
from the centre of the dipole is
1 p
~a=41tEo .(?+a2)3/2
Whenr» a,F. __ 1_L
"'JUa-41t E .r3
o
4. Torque,'t=pEsin 9, where 9 is the angle between
........
pandE.
1.43
Units Used
ChargeqISm coulomb, distance2ain metre,
dipole momentpin coulomb metre(Cm),fieldE
in NC-l orVm-l.
Example 50. Two charges, one+5 J.1Cand another-5 J.1C
areplaced1mm apart. Calculate the dipole moment.
[CBSE OD 94C]
Solution. Hereq= 5 J.1C= 5 x10-6C,
2a=1 mm =10-3 m
Dipole moment,
p= qx 2a=5x10-6 x10-3=5x10-9 Cm.
Example 51. Anelectric dipole,when held at30°with
respect to a uniform electric field of104 NC-l experiences a
torque of9x10-26Nm Calculate dipole moment of the
dipole. [CBSE D 96]
Solution. Here S·=30°, E=104 NC-l,
't=9x10-26Nm
As 't=pEsin S
:. Dipole moment,
r 9 x10-26 9x10-26
p---- -
-Esin S - 104 xsin30° - 104 x0.5
=1.8x10-29 Cm.
Example 52.An electric dipole consists of two opposite
charges of magnitude1/3x10-7C,separatedby2em. The
dipoleisplacedinan external field of3x107uc:'.What
maximum torque does the electric field exert on the dipole?
Solution. Hereq=.!.x10-7C,2a=2 em =0.02 m,
3
E =3x107NCl
'tmax=pEsin 90° =qx2axEx1
=.!.x10-7x0.02x3x107x1 = 0.02 Nm.
3
Example 53.Calculate the electric field due to anelectric
dipole of length10em having charges of1J.1Cat anequatorial
point12em from the centre of the dipole.
Solution. Hereq=1 J.1C= 10-6 C, r=12 em =0.12 m,
2a=10 em,a= 5 em =0.05 m
E =_1_ 2qa
equa 41tEo'(1+a2)3/2
9x109x2x10-6 x0.05
(0.122 +0.052)3/2
9x100
(0.13)3°°°2tu}kyjxo£k2ius

45. 1.44
Example54
.Two point charges, each of5 IlC but opposite
insign, are placed4em apart. Calculate the electric field
intensity at a point distant4emfrom the midpoint on the
axial line of the dipole. [Punjab 02]
Solution. Here q= 5X1O-6C, 2a=0.04 m,
a=0.02 m, r=0.04 m
1 2(qx2a)r
411:60 (,.z_a2)2
9x109x2x5x10-6 x0.04x0.04
[(0.04)2 -(0.02lf
144 =108 NC-1.
144x10-8
Example55.Two charges±10IlC are placed 5.00mm
apart. Determine the electric field at (a) a pointPon the axis
of thedipole15emaway from itscentre0on the side of the
positive charge, (b) a pointQ,15emaway from0on a line
passing through0and normal to the axis of the dipole.
[NCERT]
Solution. Hereq= 10IlC= 10-5C
2a=5mm=5x10-3m
r= 15 em = 15x10-2m
(a)Fieldatthe axial point P of the dipole is
~ 2p 2xqx2a
E=--
P411:60~ 411: 60r3
9x109x2x10-5 x5x10-3NC-1
(15x10-2)3
=2.66 x105NC-1 ,alongAS.
This field is directed along the direction of dipole
moment vector, i.e., from-qto+q,as shown in
Fig.1.70(a).
A 0 B e,
••---IIf--- •• ---------••---I~~
-10IlC +10IlC P
(a)
->jB
EQ "Q
-> ''
,'
EA I \
":\,',
,',
,,,
,, ,
, , ,
, , '
, , '
" I \
A" : '.B
- 10IlC 0 +10IlC
(b)
Fig. 1.70
PHYSICS-XII
(b)Field at the equatorial point Q of the dipole is
~ p qx2a
%= 411:6r3= 411:6 r3
o 0
9x109x10-5 x5x10-3Net
(15x10-2)3
=1.33x105NC-1, alongBA.
This field is directed opposite to the direction of the
dipole moment vector, i.e.,from+qto-q,as shown in
Fig.1.70(b).
Example56.Theforce experienced by a unit charge when
placedat a distance of0.10m from the middle of an electric
dipole on itsaxial line is0.025Nand when itisplaced at a
distance of0.2 m, the forceisreduced to 0.002N.Calculate
thedipole length. .
. 1 2pr
Solution, Ea ial=--',.z 2 2
XI 411:60 ( _a )
In first case: r=0.10 m,Eaxia1 =0.025 N
.. 0.025 = 9x109x2px0.10
[(0.10)2 _a2]2
... (i)
In second case: r= 0.2 m, Eaxial= 0.002N
.. 0.002 = 9x109x2px0.2
[(0.2l- a2]2
Dividing (i)by(ii), we get
0.025 0.10 [(0.2)2 - a2f
0.002 = 0.2 . [(0.1)2 _ a2]2
25 _1[(0.2l-a2]2
2-2"[(0.1)2 -a2f
5 ='0.04-a2
0.01- a2
.. a=0.05 m
Dipole length=2a=0.10m.
<::prOblems For Practice
1.An electric dipole is formedby+4IlCand -4IlC
charges at 5 mm distance. Calculate the dipole
moment and give its direction. [Haryana 011
(Ans.2x10-8Cm,from-ve to +ve charge)
2.An electric dipole of dipole moment 4x10-5C m is
placed in a uniform electric field of 10-3 N C-1
making an angle of30°with the direction of the
field. Determine the torque exertedbythe electric
field on the dipole. [Haryana 02]
(Ans. 2xto-8Nm)
...(ii)
or
or{{{0rsxiwhvmzi0gsq

46. ELECTRIC CHARGES AND FIELD
3.A dipole consisting
of an electron and a proton
separated by a distance of 4x1O-10m is situated in
an electric field of intensity 3x105N C-1 at an
angle of 30° with the field. Calculate the dipole
moment and the torque acting on it. Charge on an
electron=1.602x10-19 C. lKerala 94]
(Ans.6.41x10-29C m,9.615x10-24 Nm)
4.An electric dipole is placed at an angle of 60° with
an electric field of magnitude 4x105NC-1. It
experiences a torque ofsJ3Nm.If the length of the
dipole is 4 em, determine the magnitude of either
charge of the dipole. (Ans.1O-3q
5.An electric dipole consists of two opposite charges
of magnitude 2x10-6 C each and separated by a
distance of 3 ern. It is placed in an electric field of
2x105NC-1. Determine the maximum torque on
the dipole. (Ans.1.2x10-2N m)
6.Two point charges of+0.21.1I.ICand - 0.21.1I.ICare
separated by 10-8m. Determine the electric field at
an axial point at a distance of 0.1 m from their
midpoint. Use the standard value of &0'
[Punjab 97]
(Ans. 3.6x10-9 NC1)
7.Calculate the field due to an electric dipole of
length 10 cm and consisting of charges of±lOOI.lC
at a point 20 ern from each charge.
(Ans. 1125x107NC1)
HINTS
1.P=qx2a=4x10-6x5x10-3=2x10-a Cm.
2."t=pEsine=4x10-5x10-3xsin 30°
=2x10-8 Nm.
3.Hereq=e=1.602x10-19C,2a=4x10-10m,
E=3x105NC1,e=30°
P=qx2a=1602 x 10-19 x4x10-10
.::::6.41x10-29 Cm.
"t=pEsine=6.41 x 10-29 x3x105xsin 30°
=9.615x10-24Nm.
4."t=pEsine=qx2axEsine
"t sJ3
.. q- -
-(2a)Esine -0.04x4x105xsin 60°
=10-3 C.
5.Hereq=2x10-6 C,2a=3 ern=3x10-2m,
E=2x105 NC1
"tmax=P Esin 90° =qx2axEx1
=2x10-6 x3xlO-2 x2x105
=1.2x10-2Nm.
1.45
6.Herer»a
E __ 1_2p=_1_ 2(qx2a)
. . axial - 41tE ,3 41tE r3
o 0
9x109 x2xO.2xlO-12 x10-8
(0.1)2
=3.6x10-9 NCt.,
7.Hereq=lOOI.lC =10-4C, 2a=10 em=0.10 m
p=qx2a=10-4xO.10=10-5Cm
p
+100IlC a a - 100IlC
A 0 B
I+---- 10ern------+I
Fig. 1.71
Clearly,
(r2+a2)1/2 =20 em=0.20 m
1 p
Et,qua=47teo.(?+a2)3/2
9x109x10-59 7
---...,,-- = -x10
(0.2)3 8
=1.125x107NCt.
1.28ELECTRIC FIELD LINES
43. What are electric lines of force?Give their
important properties.
Electric lines of force. Michael Faraday (1791-1867)
introduced the concept of lines of force to visualize
the nature of electric (and magnetic) fields. A small
positive charge placed in an electric field experiences
a force in a definite direction and if it is free to move,
it will start moving in that direction. The path
along which this charge would move will be a line of
force.
An electric line of force may be defined as the curve
along which a small positive charge would tend to move
when free to do so in an electric field and the tangent to
which at any point gives the direction of the electricfield at
that point.°°°2tu}kyjxo£k2ius

47. 1.46
In Fig.
1.72, the curvePQRis an electric line of
force. The tangent drawn to this curve at the point P
~
gives the direction of the fieldEpat the point P.
Similarly, the tangent at the pointQgives the direction
~
of the field ~ at the pointQ,and so on.
p
Fig.1.72 Anelectricline of force.
The lines of force do not really exist, they are
imaginary curves. Yet the concept of lines of force is
very useful. Michael Faraday gave simple explana-
tions for many of his discoveries (in electricity and
magnetism) in terms of such lines of force.
For Your Knowledge
~The lines of force are imaginary curves, but the field
which they represent is real.
~The term'lines of force' is misleading. It will be more
appropriate to call them electric (or magnetic)'field
lines'.
~Afield line is a space curvei.e.,a curve in three
dimensions.
Properties of Electric lines of Force
1.The lines of force are continuous smooth curves
without any breaks.
2.The lines of force start at positive charges and end
at negative charges-they cannot form closed
loops. If there is a single charge, then the lines of
force will start or end at infinity.
3.The tangent to a line of force at any point gives the
direction of the electric field at that point.
4.No two lines of force can cross each other.
Reason.If they intersect, then there will be two
tangents at the point of intersection (Fig. 1.73)and
hence two directions of the electric field at the
same point, which is not p'!ssible.
PHYSICS-XII
Fig.1.73
5.The lines of force are always normal to the surface
of a conductor on which the charges are in
equilibrium.
Reason. If the lines of force are not normal to the
->
conductor, the component of the field E parallel
to the surface would cause the electrons to move
and would set up a current on the surface. But no
current flows in the equilibrium condition.
6.The lines of force have a tendency to contract
lengthwise. This explains attraction between two
unlike charges.
7.The lines of force have a tendency to expand
laterally so as to exert a lateral pressure on neigh-
bouring lines of force. This explains repulsion
between two similar charges.
8.The relative closeness of the lines of force gives a
measure of the strength of the electric field in any
region. The lines of force are
(i)close together in a strong field.
(ii)far apart in a weak field.
(iii)parallel and equally spaced in a uniform field.
9.The lines of force do not pass through a conductor
because the electric field inside a charged
conductor is zero.
1.29ELECTRICaa,o LINES FOR DIFFERENT
CHARGED CONDUCTORS
44. Sketch and explain the field lines of(i)a positive
point charge, (ii)a negative point charge,(iii)two equal
and opposite charges, (iu) two equal positive charges and
(v) a positively charged plane conductor.
Electric field lines for different charge systems:
(i)Field lines of a positive point charge.Fig. 1.74
shows the lines of force of an isolated positive point
charge. They are directed radially outwards because a
small positive charge would be accelerated in the
outward direction. They extend to infinity. The field is
spherically symmetric i.e., it looks same in all
directions, as seen from the point charge.{{{0rsxiwhvmzi0gsq

48. ELECTRIC CHARGE
S AND FIELD
Fig. 1.74 Fieldlines of a
positive point charge.
Fig. 1.75 Field lines of a nega-
tive point charge.
(ii)Field lines of a negative point charge.Like that
of a positive pointcharge, the electric field of a
negative point charge is also spherically symmetric but
the lines of force point radially inwards as shown in
Fig.1.75.They start from infinity.
(iii)Field lines of two equal and opposite point
charges.Fig.1.76 shows the electric lines of forceof an
electric dipolei.e.,asystem oftwoequal andopposite
point charges (±q)separated by a small distance. They
start from the positive charge and end on the negative
charge. The lines of force seem to contract lengthwise
asif the two charges are being pulled together. This
explains attraction between two unlike charges. The
field iscylindrically symmetricabout the dipole axis
i.e.,the field pattern is same in all planes passing
through the dipole axis. Clearly, the electric field at all
points on the equatorial line is parallel to the axis of the
dipole.
~E
I
Fig. 1.76 Fieldlines of an electric dipole.
(iv) Field lines of two equal and positive point
charges. Fig. 1.77 shows the lines of force of two equal
and positive point charges. They seem to exert a lateral
pressure as if the two charges are being pushed away
from each other. This explains repulsion between two
->
like charges. The field E is zero at the middle point N
of the join of two charges. This point is called neutral
point fromwhich no line of force passes. This field also
has cylindrical symmetry.
1.47
}---...~---;
N
Fig. 1.77 Field lines oftwo equal positive charges.
(v)Field lines of a positively charged plane
conductor.Fig.1.78 shows the pattern of lines of force
of positively charged plane conductor. A small positive
charge would tend to move normally away from the
plane conductor. Thus the lines of force are parallel
and normal to the surface of the conductor. They are
->
equispaced, indicating that electric field E is uniform
atall points near the plane conductor.
r--
+
+
+
+
+
+
---
Fig. 1.78 Field pattern of a positively charged
plane conductor.
45. Whatisthe relation between the density of lines of
forceand the electric fieldstrength ?Illustrate it in a
diagram.
Relation between electric field strength and density
of lines of force.Electric field strength is proportional
to the density of lines of forcei.e.,electric field strength
at a point is proportional to the number of lines of force
cutting a unit area element placed normal to the field at
that point. Asillustrated in Fig. 1.79,the electric field at
Pisstronger than atQ.
Regionof
weak field
Region of
strong field
Fig. 1.79Density of lines of force is proportional
to the electric field strength.°°°2tu}kyjxo£k2ius

49. 1.48
46. Show
that the 1Ir2 dependence of electric field of a
point chargeisconsistent with the concept of the electric
field lines.
Consistency of the inverse square law with the
electric field lines.As shown in Fig. 1.80,the number of
radial lines of force originating from a point chargeqin
a given solid angle.Mlis constant. Consider two points
PIandP2at distancesr1andr2from the chargeq.The
same number of lines (sayn)cut an element of area
rf11oat ~ and an element of arear?11natP2.
PHYSICS-XII
--+ ,
dS=dSn
(a) (b)
.Fig. 1.81 (a)A planar area element. (b)An area
element of a curved surface.
In case of a curved surface, we can imagine it to be
divided into a large number of very small area
elements. Each small area element of the curved
surface can be treated as a planar area. By convention,
the direction of the vector associated with every area
element of a closed surface is along theoutward drawn
~
normal. As shown in Fig.1.81(b),the area elementdSat
any point on the closed surface is equal todS~,where
Fig. 1.80 dSis the magnitude of the area element and ~ is a unit
Number of lines of force cutting unit area vector in the direction of outward normal.
n
element atPI=rf lln 1.31ELECTRIC FLUX
Number of lines of force cutting unit area element
n
atP2=~
'211n
As electric field strengthex:Density of lines of force
El_ nr?lln_r?
E2rln'-n--rf
1
Eex:?i.e.,
1.30AREA VECTOR
47.Whatisan area vector?How do we specify the
direction of a planar area vector?How do we associate a.
vector to the area of a curved surface?
Area vector. We corne across many situations
where we need to know not only the magnitude of a
surface area but also its direction.The direction of a
planar area vectorisspecified by the normal to the plane.In
Fig.1.81(a),a planar area elementdShas been repre-
~ ~
sented by a normal vectordS.The length of vectordS
represents the magnitude.dSof the area element. If ~ is
a unit vector along the normal to the planar area, then
~ "
dS=dS n
48.Define the term electric ux. Howisit related to
electric field intensity?Whatis its 51unit?
Electric flux.The term flux implies some kind of
flow. Flux is the property of any vector field. The
electric flux is a property of electric field.
The electricuxthrough a given area held inside an
electric fieldisthe measure of the total number of electric
lines offorce passing normally through that area.
As shown in Fig. 1.82,if an electric fieldEpasses
normally through an area elementllS,then the electric
flux through this area is
1l<l1;=e ss
--+
-+-•... --E
Fig. 1.82 Electric flux through normal area.
As shown in Fig. 1.83,if the normal drawn to the
area elementllSmakes an angleewith the uniform{{{0rsxiwhvmzi0gsq

50. ELECTRIC CHARGES
AND FIELD
~ ~
fieldE,then the component ofEnormal tot::,.Swill be
Ecos8,sothat the electric flux is
t::,.<It=Normal component of ExSurface area
=Ecos8xt::,.S
~~
ort::,.<It=Et::,.Scos8=E .IlS
....
E
Fig. 1.83 Flux through an inclined area.
~
In case the fieldEis non-uniform, we consider a
closedsurface 5 lying inside the field, asshown in
Fig.1.84.We can divide the surface5into small area
~ ~ ~ ~
elements: IlS1, IlS2, IlS3, ... rIlSN.Let the corresponding
~ ~ ~
electric fields at these elements beE1, ~ , rEN"
Closed surface 5
,
n
Fig. 1.84 Electric flux through a closed surface S.
Then the electric flux through the surface 5 will be
N ~ ~
=L E..IlS.
; =1 I I
When the number of area elements becomes
infinitely large (N~00)andIlS ~ 0, theabove sum
approaches a surface integral taken over the closed
surface. Thus
N ~ ~f~ ~
<It=limL E..t::,.S.=E.dS
N~oo ;=1 I I
6S...•0 S
~
Thus theelectric flux through any surface 5, open
orclosed, is equal to the surface integral of the electric
~ ~
fieldEtaken over the surface S.
1.49
Electric flux is ascalarquantity.
Unit of<It=Unit ofExunit of5
:.51unit of electric flux
=NC-1.m2 =Nm2C-1.
Equivalently, 51unit of electric flux
=Vm-1.m2 =Vm.
1.32GAUSS'S THEOREM
49. State and prove Gauss's theorem .
Gauss's theorem. This theorem gives a relationship
between the total flux passing through anyclosed
surface and the net charge enclosed within the surface.
Gauss theorem states that the total ux through a
closed surface is1/Sotimes the net chargeenclosedbythe
closedsurface.
Mathematically, it can be expressed as
<It=fE.d'S =!L
s So
Proof. For the sake of simplicity, we prove Gauss's
theorem for an isolated positive point chargeq.As
shown in Fig.1.85,suppose the surface5is a sphere of
radiusrcentred onq.Then surface5is aGaussian
surface.
«
Spherical
Gaussian
surface
Fig. 1.85 Flux through a sphere enclosing a point charge.
Electric field at any point on 5 is
E=_1_ .!J..
411:So .,2
Thisfield points radially outward at all points onS.
Also, any area element points radially outwards, so it
~
is parallel toE,i.e.,8=0°.
~
Flux through areadSis
~ ~
d<lt=E.dS=E dScos 0°=EdS°°°2tu}kyjxo£k2ius

51. 1.50
Total flux through s
urfaceSis
eIE=fdelE=fEdS=EfdS
s s s
=ExTotal area of sphere
=_1_. ~.4n?
4m;0r:
or eIE=!L
£0
This proves Gauss's theorem.
For Your Knowledge
~Gauss's theorem is validfora closed surface of any
shape andfor any general charge distribution.
~If the net charge enclosed by a closed surface is zero
(q=O~then flux through itis alsozero.
~E=-.i.=0
"0
~The net flux through aclosed surface due toa charge
lying outside theclosedsurfaceis zero.
~The chargeqappearingin the Gauss's theorem
includes the sum of all thecharges located anywhere
inside the closed surface.
->
~The electricfield Eappearing in Gauss's theorem is
due to all thecharges, bothinside and outside the
closed surface. However, the charge qappearing in
the theorem is only contained within the closed
surface.
~Gauss's theorem is based on the inverse square
dependence on distancecontained in thecoulomb's
law.In fact, it is applicabletoany fieldobeying
inverse square law. It will not hold in case of any
departure from inversesquare law.
~For a medium of absolute permittivity" ordielectric
constantK,the Gauss's theorem can be expressed as
tE.dS=1=~
sK"O
1.33GAUSSIAN SURFACE
50.Whatisa Gaussian surface?Give its importance.
Gaussian surface. Anyhypothetical closed surface or
enclosing a chargeiscalled the Gaussian surface of that
charge. It is chosen to evaluate thesurface integral of
the electric field produced by the charge enclosed by it,
which, in turn, gives the total flux through the surface.
Importance. Byaclever choice of Gaussian sur-
face, we can easily find theelectric fields produced by
PHYSICS-XII
certain symmetric charge configurations which are
otherwise quite difficult to evaluate by the direct
application of Coulomb's law and the principle of
superposition.
1.34COULOMB'S LAW FROM
GAUSS'S THEOREM
51.Deduce Coulomb's law from Gauss's theorem.
Deduction of Coulomb's law from Gauss's
theorem. As shown in Fig. 1.86,consider an isolated
positive point charge q.Weselect a spherical surface5
of radiusrcentred at chargeqas the Gaussian surface.
5
Spherical
-Gaussian
surface
E ->
--+- -----~dS
dS
Fig. 1.86 Applying Gauss's theorem to a
point charge.
->
By symmetry,Ehas same magnitude at all points
onS.AlsoEandisat any point on5are directed
->
radiallyoutward,Hence flux through area dSis
--> -->
dh=E.dS=EdScosO° =EdS
Net flux through closed surface5is
eIE=fE .dS=fEdS=EfdS
s s s
=Extotalsurface area ofS=Ex4n?
Using Gauss's theorem,
or
E=_l_ !L
4n£o .?
Theforce on the point chargeqoif placed on surface
5 will be
F=qE=_1_qqo
o 4n£o?
This provesthe Coulomb's law.www4{}·q«p¥u•q4o}z

52. ELECTRIC CHARGES
AND FIELD
Examples based on
Electric Flux and Gauss's Theorem
Formulae Used
1. Electric flux throughaplane surface area 5 heldin
-->
a uniform electric field E is
-->-->
<1>£=E.5=EScose
whereeis the angle which the normal to the
-->
outward drawn normal to surface area5makes
-->
with the field E.
2. According to Gauss's theorem, the total electric flux
through a closed surface5enclosing chargeqis
4>£=fE.dS=!L
s Eo
FI d . Total flux cjI£
3.ux ensity = = -
Area 5
Units Used
Electric flux4>£is in Nm2C-1and flux density in
NC1.
Constant Used
Permittivity constant offreespaceis
EO= 1 =8.85 x10-12C2N-1m-2
4nx9x10-9
~ " 1\ "
ExampleS? IfE=6i+3j+4k.calculaie theelectric
ux through asurface of area 20unitsinY-Zplane.
[Haryana 97]
~ /\ 1\ 1\
Solution.Electricfield vector, E=6i+3j+4k
-->
As the area vector 5intheY-Zplane points along
outward drawn normal i.e.,along positive X-direction, so
--> "
S =20i
-»~ 1\ 1\ 1\ "
Flux, <It=E.S=(6i+3j+4k).20i
=120units.
Example 58. A circular plane sheetrradius10emis
placed in a uniform electric field of5x10Ne1, making an
angle of60°with the field.Calculate electric ux through the
sheet.
Solution. Here r= 10 ern=0.1 m, E=5x105NC-1
As the angle between the plane sheet and the
electric field is 60°,angle made by the normal to the
plane sheet and the electric field ise=90° -60° =300
Flux, <It=EScose=Exn?xcose
=5x105x3.14x(0.1)2 xcos30°
=1.36x104Nm2 C1•
1.51
Example 59. A cylinder is placedinauniform electric field
-->
Ewithitsaxis parallel tothefield. Show thatthe total
electric uxthrough the cylinder is zero.
Solution.The situation is shown in Fig. 1.87.
is
Fig. 1.87
Fluxthrough the entire cylinder,
<It=fE.dS +fE.iS +fE-iS
leftplane right plane curved
face face surface
=fEdScos180°+fEdScos 0°+ fEdScos 90°
=-Ef dS+Ef dS+ 0
=-Exnr2+Exn?=O.
Example 60. Calculate the number of electric lines offorce
originating from a charge of1C. .
Solution. The number of lines of force originating
from a charge of 1C
=Electric flux through a closed
surface enclosing a charge of 1C
q 1 II
- 12=1.129x10.
EO8.85x10-
Example 61. A positive charge of17.7 ~Cisplacedatthe
centre of a hollow sphere of radius 0.5 m. Calculate the flux
density throughthe surface of the sphere.
Solution. From Gauss's theorem,
Flux, <It=!L.= 17.7 x10-6 =2x106Nm2c1
EO8.85x10-12
FI d .Total flux
ux ensity = ---~
Area
2x106 5-1
--~2 =6.4x 10NC.
4n (0.5)
Example 62. Calculate the electric flux througheach of the
six faces of a closed cubeof length l, if a charge q is placed
(a)at its centre and(b)at one of itsvertices.
Solution.(a)By symmetry, the flux through each of
the six faces of the cube will be samewhen chargeqis
placed atitscentre.
<It=~.!L.
6EO°°°2tu}kyjxo£k2ius

53. 1.52
(b)When charge qispl
aced atone vertex, the flux
through each of thethree faces meeting at this vertex
-+
will be zero, as E isparallel to these faces. As only
one-eighth of the flux emerging from the chargeq
passes through the remaining three faces of the cube,
so the flux through each suchfaceis
<Pt:=!.!.!L =~.!L
3 8EO24EO
Example63.The electric field components inFig.1.88are
Ex=axl/2,Ey=Ez=0,inwhicha=800 N /C~. Calcu-
late(i) theflux <Pt:through the cube and(ii)the charge within
thecube.Assume that a=0.1m [NCERT]
y
a
~----4---~-----+---+----~x
Z
Fig. 1.88
Solution. (i)The electric field is acting only in
X-direction and its Y-and Z-components are zero. For
-+
the four non-shaded faces, the angle between Eand
-+ -+-+
toSis+1t/2.Soflux ~=E. toS is zero through each of
thesefaces.
The magnitude of theelectric field at the left face is
EL =ax1/2 =aa1/2 [x=aat theleftface]
-+ -+
Flux, <It=EL .toS=ELtoScas9
=ELa2cas180° = - ELa2
[9= 180° for the left face]
The magnitude of the electric fieldat the right face is
ER=axl/2=a(2a)1/2
[x=2aattherightface]
Flux, <IR=ERtoScas0°=ERa2
[9=0°for the rightface]
Net flux through the cube
cJt=<It +<IR =ERa2 -ELa2
=a2(ER- EL)=aa2[(2a)1/2 -a1/2]
=aa5/2[./i -1] =800 (0.1)5/2 (./i-1)
=1.05 Nm2 C1•
PHYSICS-XII
(ii)ByGauss'stheorem, thetotalcharge inside the
cube is
q=EOcJt= 1 9x1.05 =9.27x10-12 C.
41tx9x10
Example64.An electric fieldisuniform, and inthe
positive x direction for positive xand uniform with the same
magnitudeinthe negative x direction for negative x. Itis
given that
and
-+ " 1
E= 200iNC for x>0
-+ " 1
E=-200iNCfor x<o.
A right circular cylinder of length20emand radius 5em
has its centre at the origin and its axisalongthe x-axisso
that onefaceisat x=+10emand the otherisat x= -10 em
(i) What isthe net outward flux through each flatface?
(ii) Whatisthe flux through the sideofthe cylinder?
(iii)Whatisthe net outward uxthrough the cylinder?
(iv) Whatisthe net charge inside the cylinder? [NCERT]
y
San
->
-£
->
£
o,~--------~--+-->'-~x
65
14-------- 20an
Fig. 1.89
-+ "
Solution.(i)On the left face:E= -200iNC1,
-+ " 2" 2
toS=-toSi= -1t(0.05)im
Theoutward flux through the left face is
-+ -+
cJt=E . toS
=+200x1t(0.05)2i .iNm2 C1.
=+1.57Nm2 C1• [i".t=1]
On the right face;
E=200iNCJ
~ 1\ 21\
toS=toSi=1t(0.05)i
The outward flux through the rightface is
-+ -+ 2 1
cJt=E.as=+1.57NmC.
-+ -+
(ii)Foranypointon the side of the cylinder EJ..toS ,
:.Fluxthrough the side of the cylinder,
-+ -+
cJt=E.toS=EtoScas90° =o.{{{0rsxiwhvmzi0gsq

54. ELECTRIC CHARGES AND FIELD
(iii)Net outward flux through the cylinder,
<It= 157 +
157 + 0= 3.14 Nm 2c'.
(iv)By Gauss's theorem, the net charge inside the
cylinder is
q=EO<It=8.854 x10-12x3.14 = 2.78 x10-11C.
Example 65. You are given a charge +Qat the origin0
(Refer to Fig. 1.90).Consider a sphere5with centre(2,0, 0)
of radius.J2m. Consider another sphere of radius.J2m
centered at the origin. Consider the sphericalcaps(i)P5Q
(ii)PRQ(iii)PWQ, with normals outwardtothe respective
spheres, and (iu) the at circle PTQ with normal along the
x-axis.
(a) Whatisthe sign of electricfluxthrough each of the
surfaces (i)-(iv) ?
(b) Whatisthe relation between the magnitudes of
fluxes through surfaces (i)-(iv) ?
(c) Calculate the ux through the surface(ii)directly.
Assume that the area of the cap(ii)isA.[NCERT]
y
Fig. 1.90
Solution.Forthe charge + Q situated at origin 0, the
-+
fieldEpoints along +vex-direction i.e.,towards right.
(a)Theoutward drawn normal oncapP5Qpoints
towards left while it points towards right for
capsPRQ,PWQandcirclePTQ.SOthe flux is
negative for(i)andpositive for the rest.
(b)Thesame electric field lines crossing (i)also
cross (ii), (iii).Also,by Gauss's law, the fluxes
through (iii)and(iv)add upto zero. Hence, all
magnitudes of fluxes are equal.
(c)Givenarea of the cap(ii)=A
Electric field through cap(ii)is
E=_1_.Q=9 x 109 x----.fL
41[EO? (.J2)2
=4.5x109Q NC-1
Electricfluxthrough the cap (ii)is
<It=EA
=4.5x109QANC-1m 2.
1.53
Example 66. Figure1.91shows five charged lumps of
plastic and an electrically neutral coin. The cross-section of a
Gaussian surface5isindicated. Whatisthenetelectric ux
through the surfaceif
q1=q4=+3.1nC, q2=qs=-5.9nC
and q3=-3.1nC?
Fig.1.91
Solution. The neutral coin andtheoutside charges
q4andqsmakeno contribution towards the net charge
enclosed by surface 5.Applying Gauss's theorem, we
get
<It=!L=q1+q2+q3
EO EO
+ 3.1 x 10-9 -5.9x 10-9-3.1x 10-9
8.85x10-12
=-666.67 Nm2 C-1•
Example 67. 51and52are two concentric spheres enclosing
chargesQand2Qrespectively as shown in Fig. 1.92.
(i) Whatisthe ratio of the
electric ux through51
and52?
(ii)How will the electric
ux through the sphere
51 change,ifa medium
of dielectric constantKis
introduced in the space
inside51in place of air?
Fig. 1.92
(iii)How will the electric ux through sphere 51change,
ifa medium of dielectric constantKisintroduced in
the space inside52in place of air?
[CBSE 0002,14, 14C]
Solution.(i)ByGauss's Theorem,
Flux through 51is<1\=Q
EO
th__2Q+Q_3Q
Flux through 52is't2
EO EO°°°2tu}kyjxo£k2ius

55. 2.The electric field
in a certainregionof space is
(5i+41-4k)x105NC1. Calculate electric flux
due to this field over an area of(2i-1)x10-2m2. Z
(Ans.6x103Nm2 C1) Fig.1.94
1.54
Ratio of electric flux through 51 and 52 is
4>r=Q/Eo=! =1: 3
~ 3QIEO3
(ii)If a medium of dielectric constant1Cis intro-
duced in the space inside 51' then flux through 51
becomes
(iii)The flux through 51 does not change with the
introduction of dielectric mediuminside the sphere 52'
~rOems For Practice
1.If the electric field is given by
~ 1\ 1\ 1\ 1
E=8i+4j+3kNC- ,calculate the electric flux
through a surface of area 100 m2lying in theX- Y
plane. (Ans. 300Nm2C1)
--> A
3.Consider a uniform electric field E =3x103iNC-1.
Calculate the flux of this field through a square
surface of area 10em2when
(i)its planeis parallel to they-zplane, and
(ii)the normal to its plane makes a 60° angle with
thex-axis. [CBSED13C]
[Ans. (i)30Nm2c-1 (ii)15Nm2c-1]
4.Given a uniform electric fieldE=5x103iNC-1,
findthe flux of this field through a square of 10 cm
on a side whose plane is parallel to theY-Zplane.
What would be the flux through the samesquareif
the plane makes a 30° angle with theX-axis?
[CBSE D 14]
[Ans. (i)50Nm2C1 (ii)25 Nm2c-1]
5.A point charge of 17.7 IlCis located at the centre of a
cube of side 0.03 m. Find the electric flux through
each face of the cube. [Himachal 93]
(Ans.3.3x105Nm2 C1)
6.A spherical Gaussian surface encloses a charge of
8.85x10-8C.(i)Calculatethe electric flux passing
through the surface.(ii)If the radius of the
Gaussian surface is doubled, how would the flux
change? [CBSEDOl,F07]
[Ans. (i)104Nm2 C1(ii)No change]
PHYSICS-XII
7.A chargeqis situated at the centre of animaginary
hemispherical surface, as shown in Fig. 1.93.Using
Gauss's theorem and symmetry considerations, deter-
mine the electric flux duetothischarge through the
hemispherical surface. (Ans....!LJ
2Eo
eq
Fig.1.93
8.A hollow cylindrical box of length 1 m and area of
cross-section 25 em2isplaced in a three dimen-
sional coordinate system as shown in Fig. 1.94. The
--> A
electric field in the region is given by E=saxi,
where E is in NC-1 andxisin metres.
y
~--------r-+------+~r----'X
Find
(i)net flux through the cylinder,
(ii)charge enclosed by the cylinder. [CBSED13]
-+ fux 1\
9.The electricfield in a region is given by E=-i .
b
Find the charge contained in the cubical volume
bounded by the surfaces x=a,x=a,y=a,
y=a,z=aandz=a.Takefu=5x103NC-1,
a=1em andb=2 em. (Ans. 2.2x10-12C)
10.The electric field components due to a charge inside
the cube of side 0.1 m are as shown.
Ex=ax, where a=500 N/C-m
~=O, ~=O.
y
z
Fig. 1.95{{{0rsxiwhvmzi0gsq

56. ELECTRIC CHARGES A
ND FIELD
Calculate (i) theflux through the cube, and(ii)the
charge inside the cube. [CBSE OD 08]
[Ans. (i) $E= 0.656 Nm-2C1 (ii)q=5.8x10-12 CJ
-4 A
11.Auniform electricfield E =ExiN /Cforx>0 and
-4 A
E=-ExiN/C forx<0 aregiven. A right circular
cylinder of length Iemand radius, em has its centre
at the origin and its axis along thex-axis. Find out
the net outward flux. Using Gauss's law write the
expression for the net charge withinthe cylinder.
[CBSE D 08C]
HINTS
-4 /\ 1\ 1\ 1-4 /I. 2
1.E= 8i+4j+3k NC,5 = 100km
~ ~ 1\ 1\ 1\
Flux, $E= E.5 = (8i+4j+3k).100k
=300Nm2 C1•
~....... 1\ 1\ 1\ 5 A /\ 2
2.$E=E.5 =(5i+4j-4k)xlO .(2i-j)xlO-
= [5x2+4x (-1) - 0]x103Nm2C1.
=6x103 Nm2 C-1•
3.(i) Normal to the area points in the direction of
theelectric field, 9=00•
$E =E5 cos 9 = 3x103 x(0.10)2 cos00
=30 Nm2C1.
(ii) $E = 3x103 x(0.10)2 xcos60°=15Nm2C1.
4.(I)$E= EScos9
= 5x103 x(0.10)2cosOo= 50 Nm2C1•
(il)$E=5x103 x(0.1O)2cos(900-300)
=5x103 x(0.10)2 x..! = 25 Nm2C1•
2
5.Flux through each phase of the cube
1 1 q117.740-6
="6 $E ="6 Eo="6 x8.85x10 12
=3.3x105 Nm2 C-1•
. q 8.85x 10-8 C
6.(I)$E = EO = 8.85x10-12C2N-1m-2
=104 Nm2 C-1
(ii) $E = 104Nm2C-1,because the charge enclosed
is the same as in the case (i).
7.From Gauss's theorem, total flux through entire
spherical surface is
~=..i
EE
o
From symmetry considerations, flux through the
hemispherical surface is
$=l.s.
E2'Eo
1.55
8.(i) Flux through the curved surface of the cylinder
is zero.
Magnitude of the electric field at the left face,
E = 50x1= 50NC1
:.Flux through the left face,
$1.= EScos9=50x25xlO-4cosl80°
= -1250x10-4NmZc-l
Magnitude of the electric field at the right face,
E= 5Ox2= 100NC1
:. Flux through the right face,
h= 1OOx25x10-4cosOo
= 2500x10-4Nm Zc-l
Net ~ux through the cylinder,
$E= $1.+$R=(2500-1250) x 10-4Nm Zc-l
= 1250x10-4NmZc-l
= 1.250x10-1Nm2C-1.
(ii) Totalcharge enclosed by the cylinder,
q= EO$E= 8.854 x 10-12 x 1250 x 10-4C
= 11067.5 x 1O-16C =1.107pC
9.$E=<PI.+$R
= _EI.a2+ERa2= _EU .0.a2+EU·a . a2
b b
=a3EU= 5x103x(0.01)3 = 0.25 Nm2 ct.
b 0.02
q= Eo$E= 8.85x10-12x0.25 = 2.2 x10-u C.
y
- - - - - - -"'j
,
, I
____ oJ I
I I
I I
I-r-+E
I I
I I
~----~--~--~~X
,
,
/--
I
I
I
I
I
I
I
I
z
Fig. 1.96
11.Proceed as in Example 64 on page 1.52.
(i) $E =Ex.7t(_,_)2 +S.7t(_ +0
100 100
=27tT2Ex(10)-4 NmZc-l.
(ii) q=EoE=27tT2EoEx(10)-4 C°°°2tu}kyjxo£k2ius

57. 1.56
1.35FIELD DUE
TO AN INFINITELY
LONG CHARGED WIRE
52.Apply Gauss's theorem to calculate theelectric
field of a thin infinitely long straight line of charge, with a
uniform charge densityof ):em-1.
Electric field due to aninfinitely long straight
charged wire. Consider a thin infinitely long straight
wire having a uniform linear charge densityACm -1.
By symmetry, the fieldEof the line charge is directed
radially outwards and its magnitude is same at all
points equidistant from the linecharge. To determine
the field at a distancerfrom the line charge, we choose
a cylindrical Gaussian surface of radius r,lengthIand
with its axis along the line charge. Asshown in Fig.
1.97, it has curved surface51and flat circular ends52and
53'Obviously,aS1IIE,iS2.lEandiS3.lE .Soonly
the curved surface contributes towards thetotal flux.
_----:~dS,
,- .. 90°
(s + E
'.....~ + --';:
I
I
I
I
I
I
I
I
r""] ~ ~E
~:
dS,
+
+
+
+
+
I
I
I
r---o!
I
I
I
I
I ••••. - + ... _..I
::, S3 +r~~1E
------ + --- --
+ is)
I+--r--+t
+
+
+
Fig. 1.97 Cylindrical Gaussian surface for line charge.
<It=fE.dS=fE.dS1+fE.dS2+fE.dS3
5 S:i 52 53
=fEd51 cos0°+fEd52 cos90°+fEd53 cos90°
S:i ~ ~
=Efd51+0 + 0
=Exarea of the curved surface
or<It=ExTnr!
Charge enclosed by the Gaussian surface, q=Al
Using Gauss's theorem, <It=q /EO'we get
Al A
or E .21trl=- or E =--
EO 21tEor
Thus the electric field of a line charge isinversely
proportional to the distance from the line charge.
PHYSICS-XII
1.36ELECTRIC FIELD DUE TO A UNIFORMLY
CHARGED INFINITE PLANE SHEET
53.Apply Gauss's theorem to calculate the electric
field duetoaninfinite plane sheet of charge.
Electric field due to a uniformly charged infinite
plane sheet. As shown inFig. 1.98, consider athin,
infinite planesheet ofchargewith uniform surface
charge density cr.We wish to calculate its electric field
at a pointPat distance rfrom it.
+Plane sheet,
-+--
-.•..•._.•.+-+-charge density cr
+-r+++
.,.-+-+-+--+--+-
,..-,:----1 ++..,.. .j+
,j..-+-+'\
-+-'crA '
.•. I
-+-- -- -T -- - - -- -
++ '
..•.+ .'
~~--1 ...+-+-~~r---~~
-+-+-e-,....---.-~
.-+-tt-+-"'
-+-•
E
->
E
Cross-sectional
areaA
Fig. 1.98 Gaussian surface for a uniformly
charged infinite plane sheet.
By symmetry, electric fieldEpoints outwards
normal to the sheet. Also, it must have same magni-
tude and opposite direction at two points PandP'
equidistant from the sheet andon opposite sides. We
choose cylindrical Gaussian surface of cross- sectional
areaAand length 2rwithits axis perpendicular to the
sheet.
Asthe lines of force are parallel tothe curved
surface of thecylinder, the flux through the curved
surface is zero. The fluxthrough the plane-end faces of
the cylinder is
<It=EA+EA=2EA
Charge enclosed by the Gaussian surface,
q=crA
According to Gauss's theorem,
<It.:«
EO
2EA=crAor E=~
EO 2Eo
Clearly, Eisindependent of r,the distance from the
plane sheet.
(i)If thesheetispositively charged (o>0),the field
isdirectedawayfrom it.
(ii)If thesheet is negativelycharged (o<0),the field
is directedtowards it.
For a finite large planar sheet, the above formula
will be approximately valid in the middle regions of
the sheet, away from its edges.{{{0rsxiwhvmzi0gsq

58. ELECTRIC
CHARGES AND FIELD
54.Two infinite parallel planes have uniform charge
densities ofa1anda2'Determine the electric field at
points(i)to theleft ofthe sheets, (ii)between them, and
(iii)totherightof the sheets.
Electric field of two positively charged parallel
plates.Fig.1.99shows two thin plane parallel sheets of
charge having uniformcharge densities a1anda2
'witha1>a2>O.Suppose; is a unit vector pointing
from left to right.
0"] 0"2
E] + E] + E]
• •• ••
+ +
+ +
+
II
+
III
+ +
E2
+
E2
+
E2
....--
+
....--
+--.
+ +
+ +
+.
+r--.
Sheet 1 Sheet 2
Fig. 1.99
In the region I : Fields due tothetwo sheets are
~ a1" ~ a2"
E1=---r, E=---r
2So 2 2So
From the principle ofsuperposition, the total
electric field at any point of region Iis
~ ~ ~ r
E[=E1+E2=-- (a1+ (2)
2So
In the region II : Fields due to the two sheets are
~ a1" ~_a2"
E1=--r,1:._ ----r
2So 2 2So
"~ r
.'. Total field,Ell=- (a1-(2)
2so
In the region III : Fields dueto the two sheets are
E=a2 ;
22s
o
~ r
., Total field, Em=-(a1+(2)
2So
55. Two infinite parallel planes have uniform charge
densities±a.Determine the electric field inti)theregion
between the planes, and(ii)outside it.
Electric field oftwo oppositely charged plane
parallel plates. As shown in Fig. 1.100, consider two
plane parallel sheets having uniform surfacecharge
1.57
densities of±a.Suppose; be a unit vector pointing
from left to right.
+0" -0"
III
+
+
+
+
II
.
+ r
--.
Sheet 1 Sheet 2
Fig. 1.100
In the region I : Fields due to the twosheets are
"~ r
E=--a
12'
So
" "~ ~ ~ r r
Total field, E[=El+E2= --a+-a=0
2So 2So
In the region II : Fields due to the two sheets are
~ r r a"
EII=-a+--a=-r
2So 2So So
In the region III : Fields due to the two sheets are
Total field,
E=_r_a
12 '
So
E= __ r_a
2 2S
o
~
Total field, Em=0.
Thus the electric field between two oppositely
charged plates of equal charge density is uniform
which isequal to ~ and is directed from the positive to
So
the negative plate, while the field is zero on the outside
of the two sheets. This arrangement is used for
producinguniform electric field.
1.37FIELD DUE TO A UNIFORMLY
CHARGED THIN SPHERICAL SHELL
56. Apply Gauss's theorem to show that for a
spherical shell, the electric field inside the shell vanishes,
whereas outside it, the field is as if all the charge had been
concentrated at the centre.°°°2tu}kyjxo£k2ius

59. 1.58
Electric field due to a uniformly charged thin
spherical shell.Consider a thin spher
ical shellof
charge of radiusRwith uniform surface charge density
~
cr.From symmetry, we see that the electric fieldEat
any point is radial and has same magnitude at points
equidistant from the centre of the shell i.e.,the field is
spherically symmetric. To determine electric field at any
pointPat a distance rfrom0,we choose a concentric
sphere of radiusras the Gaussian surface.
E
Gaussian
,,/ --_~ surface
.:~;::::::JC:::::::::--.... "
I ,
I \
I \
I \
I \
£..--+-' -H Ir-----'-++-.;'r--~ E=_1.i.
~ IP 4rcso ,;l
\ ,
\ I
\ I
" '~r:::::::::::;..-" ~ Spherical shell,
............. "' ....;' charge density =0'
£
Fig. 1.101 Gaussian surface for outside points of
a thin spherical shell of charge.
(a)When point Pliesoutside the spherical shell. The
total chargeqinside the Gaussian surface is the charge
on the shell of radiusRand area4n:R2.
q=4n:R2 c
Flux through the Gaussian surface,
cl>E=Ex4n:?
By Gauss's theorem,
q
cl>E=-
go
£x41t?=.!i.
go
t:__1_!L
-4n: go.?
This field is the same as that produced by a chargeq
placed at the centre 0. Hencefor points outside the shell,
the field due to a uniformly charged shellisas if the entire
charge of the shell isconcentrated atitscentre.
(b) When pointPlies on the spherical shell. The Gaussian
surface just enclosesthe charged spherical shell.
Applying Gauss's theorem,
£x41tR2=.!i.
go
or [Forr»R]
PHYSICS-XII
E=q
4n:goR2
t:=5!..
go
(c)When pointPliesinside thesphericalshell. As is
clear from Fig. 1.102, the charge enclosed by the
Gaussian surface is zero, i.e.,
or [Forr=R]
or
q=O
Gaussian
surface
r---i~-++---i~£
Fig.1.102Gaussian surface forinside points
of a thin spherical shell of charge.
Fluxthrough the Gaussian surface,
<IE=Ex4n?
Applying Gauss's theorem,
<IE=.!i.
go
Ex4n? =0
or E=0 [Forr<R]
Hence electric field dueto a uniformly charged spherical
shelliszero at allpoints inside the shell.
Figure1.103shows how Evaries with distancer
from the centre of theshell of radius r.Eis zero from
r=0tor=R ;and beyondr=R,we have
1
Eoc?.
t:
r
Fig. 1.103 Variation of EwithTfor a
spherical shell of charge.{{{0rsxiwhvmzi0gsq

60. ELECTRIC
CHARGESANDFIELD
Formulae Used
1.Electric field of a long straight wire of uniform
linear charge densityA,
E=_A_
2nEOr
where ris the perpendicular distance of the
observation point from the wire.
2.Electric field of an infinite plane sheet of uniform
surface charge densitycr,
E=~
2Ea
3.Electric field of two positively charged parallel
plates with charge densities cr1andcr2such that
crl>cr2>0,
1
E=±-(~+cr2) (Outside the plates)
2Ea
1
E=-(crl-cr2) (Inside the plates)
2Ea
4.Electric field of two equally and oppositely
charged parallel plates,
E=0 (For outside points)
E= ~ (For inside points)
EO
5.Electric field of a thin spherical shell of charge
density cand radiusR,
E=_1_ !L Forr>R(Outside points)
4nE • 1'2
o
E=O
£ _1_.s.
- 4nEO•R2
Hereq=4nR2cr.
6.Electric field of a solid sphere of uniform charge
density p and radiusR:
E=_1_!L Forr>R(Outside points)
4nEa.r2
E=_l_ 3!..
4nEa .R3
E=_l_ .i:
4nEO•R2
Here q= ~1tR3P
3
Forr<R(Inside points)
Forr=R(At the surface)
Forr<R(Inside points)
Forr=R(At the surface)
Units Used
.Herecharges are in coulomb, randRinmetre,Ain
Cm-1,oin Cm-2,pin Cm-3andelectric fieldEin
NC-l or Vm-1.
1.59
Example 68. Two long straight parallel wires carry
chargesAlandA2per unit length. The separation between
theiraxes is d.Find the magnitude of the force exerted on
unit length of one due tothe charge on the other.
Solution. Electric field at the location ofwire2 due
to charge on 1is
A
E=__l_
2nEod
Force perunit length of wire 2 due to the above
field
f=Excharge on unit length of wire 2 =EA2
f=AIA2 .
2nEod
Example 69. An electric dipole consists of charges
±2x10-8C,separated by a distance of 2mm:It is placed
neara longline charge of density 4.0x10-4Cm-1.as shown
inFig. 1.104, suchthat the negative chargeisat a distance of
2emfromthe line charge.Calculate the force acting on the
dipole.
or
+
+
+
+
+ -q +q
----------------- . .
+ 2ero ~142mm-.!
+
+
+
+
Fig. 1.104
Solution. Electric field due toa line charge at
distance rfrom it,
E=_l_2"-
4nEor
Force exerted by thisfield oncharge o,
F=qE=_1_.2qA
4nEo r
Forceexerted on negative charge(r=0.02 m),
9x109x2x2x10-8x4x10-4
~= N
0.02
=7.2 N, acting towards the line charge
Forceexerted on positive charge(r =2.2 x10- 2m),
9 x109x2x2x10-8x4x10-4
F2=----2-.2-x-10--;:;2----
=6.5 N, acting away from the line charge°°°2tu}kyjxo£k2ius

61. 1.60
Net force on the dipo
le,
F = F1 - F2 =7.2 -6.5
=0.7 N, acting towards the line charge.
Example 70. (a) An infinitely long positively charged wire
hasa linear charge density A.Cm-1. Anelectron isrevolving
around the wire asitscentre with a constant velocity in a
circular plane perpendicular tothe wire. Deduce the expre-
ssion foritskineticenergy. (b) Plot a graph of the kinetic
energy as afunction of charge density A.. [CBSE F 13]
Solution. The electrostatic force exerted by the line
charge on the electron provides the centripetal force
for the revolution of electron.
Force exerted by electricfield=Centripetal force
mv2
eE=--
r
Herevis the orbital velocity of the electron
But E=_A._
21tcor
mv2eA. 2 eA.
v=---
21tco m
or
r
Kinetic energyof the electron will be
1 2eA.
Ek=-mv =--
2 41tco
(b)AsEko:A.,the graph of
kinetic energy Ekvs.charge
density A.will be a straight line
as shown in Fig. 1.105.
Fig. 1.105
Example 71. A charge of17.7 x10--4C isdistributed
uniformlyover a large sheet of area 200~.Calculate the
electric field intensity atadistance of20 emfromitinair.
[CBSEOD 03C]
Solution. Surface charge density of the sheet,
CJ=!L= 17.7 x10-4C =8.85 x10-6Cm-2
A 200 m2
Electric field at a distance of 20 cm from it in air,
E=~ = 8.85 x10-6 =5x105NC-1.
2co 2 x8.85x10-12
Example 72.A charged particle having a charge of
- 2.0 x 10-6 Cisplacedcloseto a non-conducting plate
having a surface charge density of4.0x10-6Cm-2.Find the
force of attraction between the particle and the plate.
Solution.Hereq=-2.0x1O-6C
CJ=4.0x1O-6Cm-2
Field produced by charged plate,
E=~
2co
PHYSICS-XII
Force of attraction between the charged particle
and the plate,
F=qE= CJq=4x10-6x2.0xlO-6
2Co2x 8.85x10-12
=0.45 N.
Example 73. A particle of mass 9x10-5g iskeptovera
large horizontal sheet of charge density5x10-5Cm-2. What
chargeshould be given to theparticle, so thatif released, it
doesnot fall?
Solution. Here m=9x10-5g =9x10-8kg,
CJ=5 x10-5Cm-2
The particle must be given a positive chargeq.It will
not fall if
Upward force exerted on the =Weight ofihe particle
particle by electric field
qE=mg
CJ
q.-=mg
2co
2comg
q=--
CJ
2x8.85x10-12x9x10-8x9.8
5x10-5
=3.12x10-13C.
Example 74. Alargeplane sheet of charge having surface
chargedensity 5.0x10-16Cm2liesin theX-Yplane. Find
theelectricux through a circular area of radius 0.111'(. if the
normal to the circular area makes an angle of60°with the
Z-axis.
Given that: Co=8.85x10-12C2N-1m-2.
Solution. Here CJ=5.0x10-16Cm-2,r= 0.1 m,
or
or
or
8 =60°
Field due to a plane sheet of charge,
E=~
2co
Flux throughcirculararea,
<If:=EllS cos 8=~xn,1cos8
2co
5.0x10-16 x3.14x(0.1)2 cos 60°
2x8.85x10-12
=4.44x10-7Nm2C-1•
Example 75. Aspherical conductor of radius 12emhas a
charge of1.6x10-7 Cdistributed uniformly overits
surface. What istheelectric field (i) inside the sphere,
(ii)just outside thesphere, (iii)at a point18cm from the
centreof the sphere? [NCERT]°°°3xy£o~n}s«o3myw

62. ELECTRIC CHARGESAND FI
ELD
Solution. Here q= 1.6x10-7C,
R=12 cm=0.12 m
(i)Inside thesphere, E=O.This is becausethe charge
resides on the outer surface of the spherical conductor.
(ii)Justoutside the sphere, r=R=0.12 m. Herethe
charge may be assumed to be concentrated at the
centre of the sphere.
E=_l_ 3..-
41tEO'R2
9x109x1.6x10-7=105NC1.
(0.12)2
(iii)At a point18em from the centre,
r=18cm=0.18m.
1q9x109x1.6x10-7
E=-- -=-----;;--
41tEO.r2 (0.18)2
=4.44x104NC1•
~rOems For Practice
1.An infinite line charge produces a field of
9x104NC-1 at a distance of 4 cm. Calculate the
linear charge density. [Haryana 01]
(Ans. 2x10-7Cm-1)
2.A cylinder of large length carries a charge of
2x10-8Cm -1.Findthe electric field at a distance of
0.2 m from it. (Ans. 1800 Ym-1)
3.An infinitely long wire is stretched horizontally
4 metre above the surface of the earth. Itcarries a
charge l/-!C per cm of its length. Calculate its
electric field at a point on theearth's surface
vertically below the wire. (Ans. 4.5xlOSYm -1)
4.Twolarge metal plates each of area 1 m2 areplaced
facing each other at a distance of 10 ern andcarry
equal and opposite charges on their faces. If the
electric field between the plates is 100 NC-l, find
the charge on each plate. (Ans. 8.85x10-10 C)
5.An electron is revolving around a long line charge
having charge density 2 xlO-BCm -1.Find the kinetic
energy of the electron, assuming that it is
independent of the radius of electron's orbit.
(Ans.2.88xlO-17J)
6.A particle of mass 5x10-6g is kept over a large
horizontal sheet ofcharge density 4xl 0-6Cm -2.
What charge should be given to this particle, so that
ifreleased, it does not fall down. How many
electrons should be removed to give this charge?
(Ans.2.16x10-13C, 1.355x106)
1.61
7.A spherical shellofmetal has a radius of 0.25 m and
carries a charge of 0.2 /-!c. Calculate the electric field
intensity at a point (i)inside theshell, (ii)just
outside the shell and (iii)3.0mfromthe centre of
the shell. [Ans.(i)0(ii)2.88x104NC1
(iii)200C1l
HINTS
Er9x104x0.04
1.A=21tEoEr= 41tE0-= 9
2 9 x10x2
=2x10-7Cm-1•
2.Here A= 2x10-BCm -I,r=0.2 m
:.E=_A_ =_1_.2A =9x 109x2x2xlO-B
21tEor41tEor 0.2
=1800 Vm-'l..
3.E=_l_. 2A = 9x109 x2x10-4 = 4.5x10sVm-1.
41tEor 4
4.E=.5!..=-q-
EO Eo tJ.S
:.q=EOtJ.SE=8.85x10-12 x1x100
=8.85x10-10C
5.From Example 70,
~ =~= 9x109x1.6x10-9 x2.0x10-8
41tEo
=2.88x10-17J.
6.Upward electric force on particle
= Weight ofthe particle
(J
mg=qE=q.-
2Eo
2Eomg
or q=-.--
(J
2x8.85x10-12x 5 x10-9x9.8
4x106
=2.16x10-13 C.
Number of electrons required tobe removed,
q2.16x10-13 6
n=-= 19 =1.355x10.
e1.6x10-
7.(i)Electric field at any point insidethe shell =O.
(ii)E=_1_ .!L
41tEo.R2
9x109x0.2x10-6 =2.88x104NC-1.
(0.25)2
("')E- 1q
III -41tEo.,z
9x109xO.2x10-6 =200 NC-1.
(3.0)2°°°2tu}kyjxo£k2ius

63. 1.62 PHYSICS-XII
VERY SHORT ANSWER
CONCEPTUAL PROBLEMS
Problem 1. The electric charge of any body is actually
a surplus or deficit of electrons. Whynot protons?
Solution.Electrons are loosely bound to atoms and
can be readily exchanged during rubbing. Protons are
firmly bound inside thenucleus. They cannot be easily
detached. Hence electric charge of any body is just a
surplus ordeficit of electrons and not protons.
Problem 2.When a glass rodisrubbedwith silk,
both acquirecharges. Whatisthe source of their electri-
fication?
Solution.For the electrification of a body, only
electrons are responsible. During rubbing electrons are
transferred from glassrod to silk. The glass rod acquires a
positive charge and silk acquires an equal negative charge.
Problem 3.Is the mass of a body affected on charging?
Solution.Yes.Electrons have adefinite mass. The
mass of abody slightly increases if it gains electrons while
the massdecreases if the body loses electrons.
Problem 4.Twoidentical metallic spheres of exactly
equal masses are taken. One is given a positivechargeq
coulombs and other an equal negative charge. Are their
masses after charging equal ? [lIT]
Solution.No. The positive charge of a body is due to
deficit of electrons while the negative charge is due to
surplus of electrons. Hence the mass of the negatively
charged sphere will be slightly more than that ofthe
positively charged spheres.
Problem 5.A positively charged rod repels asus-
pended object. Can we conclude that the object is posi-
tively charged?
Solution.Yes, the object is positively charged.
Repulsion is the surest test of electrification.
Problem 6.A positively charged rod attracts a
suspended object. Can we conclude that the object is
negatively charged?
Solution.No. A positively charged rod can attract
both a neutral object and aless positively charged object.
Problem 7. How does a positively charged glass rod
attract a neutral piece of paper?
Solution.The positively charged rodinduces negative
charge on thecloser end and positive charge on the
Fig. 1.106
farther end ofthe paper.Therod exerts greater attraction
than repulsion on thepaperbecause negative charge is
closer to the rod than the positive charge. Hence therod
attracts thepiece of paper.
Problem 8.Can two like charges attract each other? If
yes,how?
Solution. Yes.If one charge is larger than the other, the
largercharge induces equal and opposite charge on the
nearer end of the body with smaller charge. The opposite
induced charge is larger than the small charge initially
present onit.
Problem 9.Why do the gramophone records get
covered with dust easily?
Solution.The gramophone records get charged due to
the rubbing action oftheneedle. So they attract the dust
particles from the air.
Problem 10. An ebonite rod held in hand can be
charged by rubbing with flannel but a copper rod
cannot be charged like this. Why ? [Himachal 97]
Solution. Ebonite rod is insulating. Whatever charge
appears on it due to rubbing, stays on it. Copper is good
conductor. Any charge developed onitflowstothe earth
through our body. So copper rod cannot be charged like this.
It can be charged by providing it aplastic or rubber handle.
Problem 11.Electrostatic experiments do not work
wellon humid days. Give reason.
Solution.Electrostatic experiments require accumu-
lation of charges. Whatever charges appear during the
experimentation, they are drained away through humid
air which is more conducting than dry air due to the
presence of a larger number of charged particles in it.
Problem 12. A comb run through one's dry hair
attracts small bits of paper. Why ? What happens if the
hair is wet or if it is a rainy day? [NCERT]
Solution.When the comb runs through dry hair,itgets
chargedby friction. The molecules in the paper get
polarizedbythe charged comb, resulting in anetforceof
attraction. Ifthe hair is wet, or if it is rainy day,friction
between hair and the comb reduces. The comb does not
get charged and thus it will not attract small bits of paper.
Problem 13. Ordinary rubber is an insulator. But the
special rubber tyres of aircrafts are made slightly
conducting. Why is this necessary? [NCERT]
Solution.During landing, thetyres ofaircraft may get
highly charged due tofriction between tyres and the air
strip. Ifthe tyres are made slightly conducting, they will
lose the charge to theearthotherwise too much of static
electricity accumulated mayproduce spark and resultin
fire.{{{0rsxiwhvmzi0gsq

64. ELECTRIC CH
ARGES AND FIELD
Problem 14. Vehicles carrying inflammable materials
usually have metallic ropes touching the ground during
motion. Why ? [Himanchal 98;Punjab 99;NCERT]
Solution. Moving vehicle gets charged dueto friction.
Theinflammable material may catch fire due to thespark
produced by charged vehicle. Whenmetallic rope isused,
thecharge developed on the vehicle is transferred to the
ground and so the fire is prevented.
Problem15.An inflated balloon is charged by
rubbing with fur. Willitstick readily to a conducting
wall or to an insulating wall? Give reason. [Roorkee]
Solution. It will stick readily to the conducting wall. It
induces an equal amount of charge on the conducting
walland much smallercharge on insulating wall. So a
largeforce of attraction acts between the balloon and the
conducting wall.
Problem16.A metal sphereisfixed on a smooth
horizontal insulating plate. Another metalsphere is
placed a small distance away. Ifthe fixed sphere is given
a charge, how will the othersphere react?
Solution. The charge on the fixed sphere induces
unlike charge at the closer end and like charge on the far
endofthe free sphere. etattraction acton the free
sphere andsoit gets accelerated towards the fixed sphere.
Problem17.Is theresome way of producing high
voltage on your body without getting a shock?
Solution.Ifwe stand onaninsulating surface and
touch the live wire of ahigh power supply, ahigh poten-
tial is developed on our body, without causing any shock.
Problem 18. A charged rodattracts bits of dry cork
which after touching the rod, oftenjump away from it
violently. Why?
Solution.The charged rod attracts the bitsofdrycork
byinducing unlike chargeat their nearends and like
charge attheirfar ends. When the cork bits touch the rod,
theyshare thechargeof the rod ofthe same sign andso
getstrongly repelled away.
Problem19.What does q1+q2=0signify in
electrostatics? [CBSE0001C]
Or
Two chargesq1andq2'separated by a small distance
satisfy the equationq1 +q2 =0.What does it tell about
the charges? [CBSEF03]
Solution. The equation signifies that the electric
charges are algebraically additive and here q1andq2are
equal andopposite.
Problem20.amethe experiment which established
the quantum nature ofelectric charge. [CBSE0098]
Solution. Millikan's oil drop experiment for deter-
mining electronic charge.
Problem 21. Can a body have acharge of 0.8x10-19 C?
Justify your answer by comment? [Himachal 99C]
1.63
Solution. Thecharge on any body isalways an integral
multiple ofe.Here
0.8x10-19C
n=:J.= =0.5
e1.6x10-19 C
This isnotaninteger.Soa body cannot have a charge
of0.8x10-19C. '
Problem22.If the distance between two equal point
charges is doubled and their individual charges are also
doubled, what would happen to the force between
them ? [ISCE 95]
Solution. Theoriginal force between the two charges is
F __1_qxq
-41tEo·. ?
When the individual charges and the distance
between them aredoubled, theforcebecomes
Hencethe force will remain same.
Problem23.The electrostatic force between two
charges is a central force. Why?
Solution. The electrostatic forcebetween two charges
actsalong thelinejoining thetwocharges. So it is a central
force.
Problem24.How is the Coulomb force between two
charges affected by the presence of a third charge?
Solution. The Coulomb force between two charges
doesnotdepend on the presence of a thirdcharge.
Problem25.Two equal balls having equal positive
charge'q'coulombs are suspended by two insulating
strings of equal length. What would be the effect on the
force when a plastic sheet is inserted between the two ?
[CBSEOD 14]
Solution.The force between the two balls decreases
because x{Plastic) >1andFIX:1/K.
Problem26.Force between two point charges kept at
a distantdapart in air isF.If these charges are kept at
the same distance in water, how does the electric force
between themchange? [CBSE 0011]
Solution. Dielectric constant forwater, K=80
F =Fair=£
water K 80
Thus the force in water is 1/80 times the original force
inair.
Problem 27. The dielectric constant of water is 80.
What is its permittivity? [Haryana 97C]
Solution. Dielectric constant, K=~
Eo°°°2tu}kyjxo£k2ius

65. 1.64
:.Perm
ittivity, E=KEO=8.854x10-12 x80
=7.083x10-10C2N-lm-2.
Problem28.Give an example to illustrate that electro-
static forces are much stronger than gravitational forces.
Solution. A charged glass rod can lift a piece of paper
against the gravitational pull of the earth on this piece.
This shows that the electrostatic force on the piece of
paper is much greater than the gravitational force on it.
Problem29.Two electrically charged particles,
having charges of different magnitude, when placed ata
distance'd'from each other, experienceaforce of
attraction'F'.These two particles are put in contactand
again placed at the same distancefrom each other.
What is the nature of new force between them ?
Is the magnitude of the force ofinteraction between
them now more or less than F?[CBSE Sample Paper 11)
Solution.When the two particles are put in contact,
they share the difference of charge identically. Hence the
twoparticles repel,witha forcelessthan F.
Problem 30.An electron moves along a metaltube
with variable cross-section, as shown inFig.1.107.How
will its velocity change when it approaches the neck of
the tube?
"---~----
----~/
Fig. 1.107
Solution. The positive charge induced on the neck of
the tubewill accelerate the electron towards the neck.
Problem 31. Why should a test charge be of negli-
gibly small magnitude?
Solution. Themagnitude of the test charge must be
small enough so that it doesnot disturb the distribution of
thecharges whose electric fieldwewish to measure
otherwise themeasured fieldwillbedifferent from the
actual field.
Problem32.In defining electric field due to a point
charge, the test charge has to be vanishinglysmall. How
this condition can be justified, when we know that
charge less than that on an electron or a proton is not
possible?
Solution.Because of charge quantisation, thetest
charge qocannotgobelowe.However, inmacroscopic
situations, thesource charge ismuchlargerthan the
charge on an electronor proton, so the limit qo ~0forthe
testcharge is justified.
Problem 33.What is the advantage of introducing the
concept of electric field ?
Solution.Byknowing the electrical fieldatapoint, the
forceona charge placed at that point can be determined.
PHYSICS-XII
Problem 34.How do charges interact ?
Solution.The electric field of one charge exerts a force
on the other charge and vice versa.
Charge:;::::':Electric field :;::::':Charge.
Problem 35.An electron and a proton are kept in the
same electric field. Will they experience same force and
havesame acceleration ?
Solution. Both electron and proton will experience
force of same magnitude, F=e£Since aproton has 1836
times more mass than an electron, so its acceleration will
be 1/1836 times that of the electron.
Problem 36.Why direction of an electric fieldis
taken outward (away) for a positive charge and inward
(towards) for a negative charge?
Solution.By convention, the direction ofelectric field
isthe same as that of force on a unitpositive charge. As
this force is outward in the field of a positive charge, and
inward in the field of a negative charge, so the directions
are taken accordingly.
Problem 37. A charged particle is free to move in an
electric field. Will it always move along an electric
field? [lIT)
Solution.The tangent at any point to the line of force
gives the direction of electric fieldand hence of force on a
charge at that point. Ifthe charged particle starts from
rest,it will move along the line of force. If it is in motion
and moves initially at an angle with the line of force, then
resultant path is not along thelineof force.
Problem 38.A small test charge is released at rest at a
point in an electrostatic field configuration. Will it
travel along the line of force? [.'CERT)
Solution.Not necessarily. The test charge will move
along the line of force only if it is a straight line. This is
because a line of force gives the direction of acceleration
and notthat of velocity.
Problem 39.Whydo charges reside on the surface of
the conductor?
Solution.Charges lie at the ends of lines of force.
These lines of force have a tendency to contract in length.
The lines of force pull charges from inside a conductor to
its outer surface.
Problem 40. Why is electric field zero inside a
charged conductor?
Solution. This isbecause charges reside on the surface
of a conductor and not inside it.
Problem 41. Why do the electrostatic field lines not
form closed loops? [CBSEOD14, 15)
Solution. Electrostatic field lines start from apositive
charge and end on anegative charge or they fade out at
infinity in case of isolated charges without forming any
closed loop.{{{0rsxiwhvmzi0gsq

66. ELECTRIC CHARGES AND FIELD
Alternative
ly, electrostatic field is a conservative field.
The work done in moving a charge along a closed path
must be zero. Hence, electrostatic field lines cannot form
closed loops.
Problem 42.Do the electric lines of force really exist?
What is about the field they represent?
Solution.Lines of force do not really exist. These are
hypothetical curves used to represent an electric field. But
the electric field which they represent is real.
Problem 43.Draw
lines of force to represent
a uniform electric field.
->
------------~--~~ E[CBSE00 95]
Solution. The lines of
force of a uniform electric
field are equidistant
parallel lines as shown in Fig.1.108Uniform electric field.
Fig.1.108.
Problem 44. Fig.1.109shows electric lines of force
due to point charges qlandq2placed at pointsAand B
respectively. Write the nature of charge on them.
[CBSEF03]
Fig.1.109
Solution. As the linesofforcearepointing towards ql
aswell asQ2'so both qlandQ2mustbenegative charges.
Problem 45. A positive point charge (+q)iskeptin
the vicinity ofan uncharged conducting plate. Sketch
electric field lines originating from the point charge on
to the surface of the plate. [CBSE0009]
Solution. Starting from the charge +q,the lines of
force willterminate at the metal plate, inducing negative
charge on it. At all positions, the lines of force will be
perpendicular tothemetal surface, as shown in Fig.1.110.
+q
Fig.1.110
1.65
Problem 46. Why isit necessary that the field lines
from a point charge placed in the vicinity of a conductor
must be normal to the conductor at every point.
[CBSEF09]
Solution.If the field lines are notnormal, then the field.
-4
Ewould have a tangential component which will make
electrons move along the surface creating surface currents
and the conductor will not be in equilibrium.
Problem 47. Fig.1.111 shows two
large metal plates, Pl andP2,tightly held
against each other and placed between
two equal andunlike point charges
perpendicular to the line joining them.
(i)What willhappen to the plates
whenthey are released ?
(ii)Drawthe pattern of the electric
field lines forthe system.
[CBSEF09]Fig.1.111
P,
+Q -Q
Solution.
(i)When released, the two platestend to move
apartslightly due to the charges induced in them.
(ii)The pattern of the electric field lines for the
system is shown in Fig. 1.112.
-+
+Q(1--------+-----+-11++----- .•..-------:0 -Q
-+
Fig.1.112
Problem 48.In the electric field shown in Fig. 1.113,
the electric field lines on the left have twice the
separation as that between those onthe right. Ifthe
magnitude ofthefieldat pointAis40NC-l, calculate
the force experienced by a proton placed at pointAAlso
find the magnitude of electric fieldatpointB.
:
~:
-A
·B;;-:
:
Fig.1.113°°°2tu}kyjxo£k2ius

67. 1.66
Soluti
on.Force on protonat point A,
F =eEA= 1.6 x10-19x40=6.4x10-18N
As the separation between the lines of force at point B
is twice of that at point A,so
EB=.!EA=.! x40=20NC1.
2 2
Problem 49. The electric lines offorce tend to
contract lengthwise and expand laterally. What do they
indicate?
Solution. The lengthwise contraction indicates
attraction between unlike charges while lateral expansion
indicates repulsion between likecharges.
Problem 50.A point charge placed at any point on the
axis of an electric dipole at somelarge distance
experiences a forceF.What will be the force acting on
the point charge when its distance fromthe dipole is
doubled? [CPMT91]
Solution. At any axial pointof adipole, electric field
varies as
1 F 1 1
Eoc3"or-=3"orFoc3"
r qr r
..When the distance of the point charge is doubled,
the force reduces toF/8.
Problem 51. As shown in Fig. 1.114, a thinspherical
shell carries a charge Q on its surface. A pointcharge
Q12is placed at its centre0and anothercharge 2Q
placed outside. If all the charges are positive, whatwill
be the force on the charge at the centre?
Q
e2QeQ/2
o
Fig. 1.114
Solution. Zero, because the electric field inside the
spherical shell is zero.
Problem 52. What is the number of electric lines of
force that radiate outwards from one coulomb of charge
in vacuum?
Solution. Hereq= 1C, Eo= 8.85 x10-12C2N-1m-2
Number of lines of force=Electric flux
=!L= 1
Eo8.85x10-12
= 1.13x1011.
PHYSICS-XII
Problem 53. Consider the situation shown in Fig. 1.115.
What are the signs of qlandq2?Ifthe lines are drawn in
proportion to the charge,
what isthe ratio q1 /q2?
Solution. Here ~ is a
negative charge andq2is
a positivecharge.
!!l=.i.
q218
=1:3.
Fig. 1.115
Problem 54. An arbitrary surface encloses a dipole.
What is theelectric flux through this surface?
[~xemplarProblem]
Solution. As the total charge of a dipole is zero, so by
Gauss's theorem, the electric flux through the closed
surface is zero.
Problem 55.Theforce on an electron kept in an
electric field ina particular direction isF.What will be
the magnitude and direction of the force experienced by a
protonatthesame point in the field? Mass of the proton
is 1836 times the mass of the electron. [CBSE F07]
Solution. A proton has charge equal and opposite to
that of an electron. Hence the proton will experience a
force equal and opposite to that ofF.
Problem 56. Figure 1.116shows three charges +2q,-q
and+3q.Twocharges +2qand-qare enclosed within a
surface '5'.What is the electric fluxdue to this con-
figuration through the surface '5'? [CBSE010]
+3q
•
Fig. 1.116
Sol. Net chargeenclosed by the surface S
ution.<PE=------"'-------~-----
EO
=+2q-q=!L
EO EO
Problem 57. Twocharges of magnitudes-2Q and+Q
arelocatedat points (a,0) and(4a,0)respectively. What
isthe electric fluxdueto thesecharges through a sphere of
radius'3a'withitscentre at the origin? [eBSEOD 13]
Solution. Only the charge-2Qis enclosed by the
sphere of radius3a.By Gauss's theorem.
'"__2Q
~'E- .
EO{{{0rsxiwhvmzi0gsq

68. 1.80 PHYSICS-XII
GI DELINESTONCERT
EXERCISES
1.1.What istheforcebetween two smallcharged spheres
having charges of2x10-7Cand3x10-7Cplaced30em apart
in air?
Ans.Hereq1= 2x10-7C,q2= 3x10-7C,
r= 30 cm = 0.30 m
According to Coulomb's law, .
F=_1_.q1q2=9x109 x2x10-7 x3xlO-7
4m;0 r2 (0.30)2
=6x10-3 N(repulsive).
1.2.The electrostatic force on asmall sphere of charge
0.4 flCdue to another small sphere of charge -0.8flCin airis
0.2 N. ti)Whatisthe distance between two spheres? (ii)What
istheforceon the second sphere due to the first?
Ans.(i)Hereql= 0.4 flC = 0.4x10-6C
q2= - 0.8 flC = - 0.8x10-6C, F = 0.2 Nr=?
AsF __ 1_qlq2
- 41tEo'1'2
.. 1'2 =_1_.qlq2
41tEoF·
9x109x0.4x10-6x0.8x10-6= 144x10-4
0.2
r= 12x10-2= 0.12 m =12 em.
(ii)The two charges mutually exert equal and opposite
forces.
.'.Force on the second sphere due to the first
=0.2N(attractive).
1.3.Check that the ratio ke2/Gmemp isdimensionless. Look
up a table of physical constants and determine the value of this
ratio. What does this ratio signify?
.[e21[NmZc-2] x[C]2 .
Ans. k--- = = nounit
Gmemp [Nm2kg-2] x[kg][kg]
As'the ratiok e2 /Gmemp has no unit, soit is
dimensionless.
Now k= 9 x109NmZc-2
G=6.67x10-11Nm2 kg-2
e=1.6x10-19kg
me=9.1x10-31 kg
mp=1.66x10-27kg
; 9x109x(1.6x10-19)2
..k Gmcmp =6.67x10-11x9.1x1031x1.66x1027
or
and
=2.287x1039 •
The factor k e2!Gmemp represents the ratio of
electrostatic force to the gravitational force between an
electron and a proton. Also, the large value ofthe ratio
signifies that the electrostatic force is much stronger than
thegravitational force.
1.4.(i)Explain the meaning of the statement 'electric
charge of a body isquantised.'
(ii)Why can one ignore quantisation ofelectric charge
when dealing with macroscopic i.e., largescale
charges?
Ans. (i)Quantisation of electric charge means that the
total charge(q)of a body is always an integral multiple of
a basic charge(e)which is the charge on an electron. Thus
q=ne,wheren= 0,±I,±2,±3,.
(ii)While dealing with macroscopic charges(q=ne),
wecan ignore quantisation of electric charge. This is
because eis very small and nis very large and soqbehaves
as if it were continuousi.e.,as if a large amount of charge
is flowing continuously.
1.5.When a glass rodisrubbed with a silk cloth, charges appear
on both. A similar phenomenonisobserved with many other
pairs ofbodies. Explain how this observation isconsistent with
the law of conservation of charge.
Ans.It is observed thatthepositive charge developed
on the glass rod has the same magnitude as the negative
charge developed on silk cloth. So total charge after
rubbing is zero as before rubbin. Hence the law of
conservation of charge is being obeyed here.
1.6.Four pointcharges q A=2 flCqB=-5tiC'tc= 2 flC
qD=-5flCare located at the corners of a square ABCD of side
10c. Whatistheforce on a charge of1 flCplaced at the centre
of the square?
~102+102
Ans.HereOA= OB=OC = OD =-'----
2
=5.ficm=5.fix10- 2m
qo=-5flC 10 em qc~2flC
D---------------~C
...•
o
9
Fig. 1.149www.notesdrive.com

69. ELECTRIC CH
ARGES AND FIELD
Forces exertedon thecharge of 1 IlC located at the
centre are
~
=3.6 N, along OC
~ 9x109 x5xlO-6 x1xlO-6
F------;=------,~--
B- (5.fix10-2)2
~
=9 N,along OB
~9x109 x2xlO-6 x1xlO-6
F,------;=-----;;,....,.---
e- (5.fi x10-2)2
~
=3.6 N, alongOA
~ 9x109 x5x10-6 x1x10-6
F------;=------,~--
0- (5.fi x10-2)2
~
=9N,along OD
-+ -+- -+ -+
Clearly, Fe=-FAandFo= -FB
Hence total force on 11lC charge is
-+-+ -+-+---t
F=FA+FB+Fe+Fo
~~~~
=FA+FB- FA- FB =zeroN.
1.7.(a) An electrostatic fieldlineisacontinuous curve.
Thatis,afield line cannot have sudden breaks. Why not ?
(b) Explain whytwo field lines never cross each other at any
point? [Punjab 01,02;CBSE 0OS,03; 0014]
Ans.(a)Electric lines of force exist throughout the
region of an electric field. The electric field of a charge
decreases gradually with increasing distance from it and
becomes zero at infinity i.e.,electric field cannotvanish
abruptly. So a line of force cannot havesudden breaks, it
must be a continuous curve.
(b)If two lines of force intersect, then there would be
two tangents and hence two directions of electric field at
the point of intersection, which is not possible.
1.8.Two point charges q A=+3 IlCand qB= -3 IlCare
located20emapartinvacuum. ti)Find the electric field at the
midpoint0of the line AB joining the two charges. (ii)If a
negative test charge of magnitude 1.5x10-9Cisplaced at the
centre, find the force experienced by the test charge.
[CBSEOO03]
Ans. The directions of the fieldsEAandEBdue to the
chargesqAandqBat the midpointPare as shown in
Fi. 1.150.
Electric field at the midpoint0due toqA'
EA
qA=+3IlC ----. qB=-3fiC
• • •
A 10em 0----. 10em B
EB
Fig.1.150
1.81
1qA9x109x3x10-6
E=-- - =-------.---
A 41tEo' r2 (0.10)2
=2.7x106Ne1, along OB
Electric field at the midpoint0due toqB'
1 qB 9x109x3'xlO-6
E=-- -=-------,._-
B 41tEo' r2 (0.10)2
.=2.7 x106NC-1, along OB
Resultant field at the midpoint0is
E=EA+EB=(2.7+2.7)x106
=5.4x106NC-1, alongOB.
(ii)Force on a negative charge of 1.5 x1O-9Cplaced at
the midpoint0,
F=qE=1.5.x10-9 x 5.4x106
=8.1x10-3N, alongOA
The force on a negative charge acts in a direction
opposite to that of the electric field.
1.9.Asystem has two charges qA=25x10-7 Cand
qB=-25x10-7C,located at points A(0,0, -15em)and
B(0,0, +15em)respectively. What isthetotalcharge and
electric dipole moment ofthe system ?
Ans. Clearly, the two charges lie on Z-axis oneither,','
side of the origin and at 15 em from it, asshown in'v
Fig.1.151. .
2a=30cm=0.30 m,q=2.5x10-7C
z
-7
qB= -2.5x10 CB (0, 0,+15 em)
o y
25 10-7C A(0,0,-15 em)
qA=.X
x
Fi. 1.151
Total charge=qA+qB=2.5x10-7 -2.5x10-7=0
Dipole moment,
p=qx2a=2.5x10-7 x0.30
=0.75x10-7 Cm
The dipole moment acts in the direction from B to A
i.e.,along negative Z-axis.www6notesdrive6com

70. 1.82
1.10.An electric
dipole with dipole moment4x10-9Cm is
aligned at30°with the direction of auniformelectric field of
magnitude 5x104NC1.Calculate the magnitude of the torque
acting on the dipole.
Ans.Herep= 4x10-9Cm,e= 30°,E = 5x104NC-1
:.Torque, 't=pEsine
=4x10-9 x5x104 xsin 30°
=10-4 Nm.
1.11.A polythene piecerubbed with wool is found to havea
negative charge of3.2x10-7C.(i) Estimate thenumber of
electrons transferred. (ii)Isthere a transfer of mass from wool to
polythene?
Ans.(i)Hereq= 3.2x10-7C,e=1.6x10-19C
Asq=ne,therefore
Number of electrons transferred,
n=1= 3.2x10-7 = 2x1012
e1.6x10-19
Since polythene has negative charge, so electrons are
transferred from wool to polythene during rubbing.
(ii)Yes,there is a transfer of mass from wool to
polythene because each electron has a finite mass of
9.1x10-31 kg.
Mass transferred
=mexn= 9.1x10-31x2x1012
=1.82x10-18kg
Clearly, the amount of mass transferred is negligibly
small.
1.12.(a)Twoinsulated charged copper spheres A and B
have their centres separated by a distance of 50em.What is the
mutual force ofelectrostatic repulsionif the charge on each is
6.5x10-7C?The radii of A and B are negligible compared to
the distance of separation. (b) Whatistheforce ofrepulsion if
eachsphere is charged double theaboveamount, andthe
distance between them is halved ?
Ans.Refer to the solution of Example 9 on page 1.12.
1.13. Suppose the spheres A and B in Exercise 1.12have
identical sizes. A third sphere of the samesize but uncharged is
brought in contact with the first, then broughtincontact with
the second, and finally removed from both. What is thenew
force of repulsion between A and B?
Ans.Refer to the solution of Example 10 on page 1.12.
1.14.Figure1.152shows tracks of three charged particles in
a uniform electrostatic field. Give the signs of the three charges.
Which particle has the highest charge to mass ratio?
PHYSICS-XII
Fi. 1.152
Ans.Refer to thesolution of Problem 9 on page 1.73.
1.15.Consider a uniform electric field:
E=3x103iNC1 (i)What istheflux of this field through a
square of10emon a side whose plane isparallel to the
Y-Z-plane ?(ii)Whatisthe flux through the same square if the
normal toits plane makes a60°angle with the X-axis?
Ans. (i)Normal to a plane parallel to Y-Zplane points
in X-direction, so
I1S=0.10x0.10 £m2=0.01£ m2
Electric flux,
~ ---1- 3 ~ ~
4>£=E . I1S = 3 x10I.0.011
=30i,£=30 Nm2C-1.
(ii)Heree=60°
4>£=EI1Scos60° = 3 x103x0.01cos60°
=30x~=15 Nm2C-1.
1.16.Consider a uniform electric field:
E=3x103luc».Whatisthenet flux of this field through a
cube of side20cmoriented so that its faces are parallel tothe
coordinate planes?
Ans.The flux entering one face parallel to Y-Zplane is
equal to the flux leaving other face parallel to Y-Zplane.
Flux through other faces is zero. Hence net flux through
the cube is zero.
1.17.Careful measurement of the electric field at the surface
of a black box indicates that the net outward flux through the
surface of the boxis8.0x103Nm2C-1. (i)What is the net charge
inside the box? (ii)Ifthenet outwardflux through the surface of
thebox were zero, could you conclude that there were no charges
inside thebox?Whyor why not ?
Ans.(i)4>£= 8.0x103Nm2c-2
Using Gauss theorem,
<IE=!L
eo
3 1
Charge,q=eo.4>£= 8.0x10x 9C
41tX9x10
= 0.07x10-6C = 0.07llC
(ii)No,we cannot say that there areno charges at all
inside the box. We can only say that the netcharge inside
the box is zero.www4yztp•o°tvp4nzx

71. ELECTRIC CHARGES
AND FIELD
1.18.A point charge +10 u.C is adistance 5em directly
abovethe centre ofa square of side 10em as shown in
Fig.1.lS3(a).Whatisthemagnitude of the electric fluxthrough
the square? (Hint: Think of the square as one face of a cube with
edge10em)
Ans. We can imagine the square as face of a cube with
edge 10 cm and with thecharge of+10 j!C placed at its
centre, asshown inFig.1.153(b).
.....,-- - - ---- -::".
"," I '" I
I I
,I I
"'------- .•.--- I
: : +q: :
I I I I
I '8 I I
I I I I
: /=~~:-----7
I I
I I
I I
10em
(a)
IDem
(b)
Fi. 1.153
Symmetry of six faces of acube about its centre
ensures that the flux 45 through each square face is same
when the chargeqis placed at the centre.
:.Total flux,
<I>£=6x45=.i.
EO
45 = ~ =1.x10x10-6 x41tx9x109
6EO 6
= 1.88x10s Nm2C-1.
or
1.19.Apoint charge of2.0 j!Cisat the centre of a cubic
Gaussian surface9.0emonedge.What isthenet electric flux
through the surface ?
Ans. Hereq= 2.0 j! C=2.0xlO--6c,
EO=8.85x1O-12C2N-1m-2
ByGauss's theorem, electric flux is
q 2.0x10--6 5 2-1
<1>£= - = 12 = 2.26x10 Nm C
EO 8.85x10-
1.20.Apointcharge causes anelectric flux of-1.0x103
Nm2C'1to pass through a spherical Gaussian surface of
10.0 em radiuscentred on the charge. (i)If the radius of the
Gaussiansurface were doubled, howmuch flux would pass
through the surface? (ii)What is the value of the point charge?
Ans.(i)<1>£= _103 NmZc-1, because the charge
enclosed is the same in both the cases.
(ii)Charge,
q= Eo<1>£
19x(-1.0x103)
41tx9x10
= - 8.84 x10-9C=- 8.84 nC.
1.83
1.21.Aconducting sphere of radius 10cm has an unknown
charge. If the electric field 20emfrom the centre of thesphere is
15x103NC'l and points radially inward, whatisthenet
chargeon the sphere ?
Ans. Electric field at the outside points of a conducting
sphere is
E- _1_ s.
-41tE .r2
o
q=41tEoEr2 = __ 1-9 x1.5x103x(0.20)2 C
9x10
= 6.67 x10-9C = 6.67 nC
As the field acts inwards, the charge qmust be
negative.
. .q=-6.67 nC. .
1.22.Auniformly charged conducting sphere of 2.4m
diameter has a surface charge density of80.0j!Clm2. (i)Find
the charge on the sphere. (ii)What isthetotalelectric flux
leaving the surface of the sphere ? ICBSE D 09C)
2.4
Ans.HereR=-- =1.2 m
2
(J=80.0 j!Cm-2 =80x10--6Cm-2
(i)Charge on the sphere is
q= 41tR2 (J= 4x3.14x(1.2)2 x 80x1O-6C
=1.45x10-3 C.
(ii)Flux,
<1>£=.i.= 1.45 x10-3 x41tx9x109
EO
= 1.6x108 Nm2 C-1•
1.23.An infinite line charge produces a field of
9x104NC'l at a distance of2em.Calculate the linear charge
density.
Ans.E=9x104NC-1, r=2cm=0.02m
Electric field of a line charge, E =_A._
21tEor
..Linear charge density,
1 4
A.=21tEoEr=21tx 9 x9x10 xO.02
41tx9x10
= 0.01x10-5Cm-1= O.l1lCm-1.
1.24.Two large, thinmetal plates areparallel and closeto
each other. On their inner faces, the plates have surface charge
densities of opposite signsand of magnitude 17.0x10-22Cm-2.
What is E(a)totheleft of the plates, (b) totheright of the plates,
and(c)between the plates?
Ans. Here(J= 17.0x10-22Cm-2
(a)On the left, the fields of the two plates are
equal and opposite, so E =Zero.www6notesdrive6com

72. 1.84
(b)On the right,the fields of
the two plates are
equal and opposite, soE=Zero.
(c)Between the plates, the fields due to both
plates are in same direction. So the resultant
field is
E=~+~= ~ =17x10-22 x41tx9x109
2f:O 2EO EO
=19.2x10-10 NC-1•
1.25.An oil drop of12excess electrons isheldstationary
under a constant electric field of255 x104Vm-1 inMillikan's
oil dropexperiment. The density of the oil is
1.26g cm-3. Estimate the radius of the drop. ( g =9.81-2 ;
e=160x10-19C)
Ans.Force on the oil drop due to electric field
=qE= neE
Weight of oil drop
=mg=volumexdensityxg=~1tr3pg
3
The fieldEmust actvertically
downward so that the negatively
charged oil drop experiences an
upward force and balances the weight
of the drop.
When the dropisheld stationary,
mg
Fig.1.154
or
Weight of oil drop
=Force on the oil drop due to electric field
4 r=[3neE]1/3
- 1tr3pg=neE ..
3 41tpg
Now n=12,e=1.6x10-19C,
E=2.55x104Vm-1,g=9.81ms-2
p=1.26gem-3=1.26x103kgm-3
r=[3x12x1.6x10-19 x2.55x104]1/3
.. 4x3.14x1.26x103x9.81
= [9x16x255x10_15]1/3
314x126x981
=(9.46x10-4)1/3 x10-5
=0.0981x10-5m=9.81x10-4 mm.
1.26.Which among the curves shown in Fig. 1.155, cannot
possibly represent electrostatic field lines?
Ans.Only Fi.1.155(c)is right and the remaining
figures cannot represent the electrostatic field lines.
Figure1.155(a)is wrong because field lines must be
normal to a conductor.
Figure1.155(b)is wrong because lines of force cannot
start from a negative charge.
PHYSICS-XII
(b)
(c)
(d) (e)
Fi. 1.155
Figure1.155(c) isright because it satisfies all the
properties of lines of force.
Figure1.155(d) is wrong because lines of force cannot
intersect eachother.
Figure1.155(e) is wrong because electrostatic field
linescannot formclosed loops.
1.27.Ina certain region of space, electric field is along the
Z-direction throughout. The magnitude of electric field is,
however, not constant butincreases uniformly along the
positiveZ-direction at the rate of 105NC1m-1. What are the
force and torque experienced by a system having a total dipole
moment equal to10-7 Cm inthe negative Z-direction?
Ans.The situation is shown inFi.1.156.
Asthe electric fieldchanges uniformly in the positive
Z-direction, so
BE,=+105NC1m-1 BEx=0a~=0
Bz 'ax 'By
Asthe system has a total dipole moment in the
negative Z-direction, so
pz= -10-7Cm,Px=0,Py=0www4yztp•o°tvp4nzx

73. ELECTRIC CHARGES AND
FIELD
z
x
+q
-q
~P
.J--------y
Fig.1.156
Inanon-uniform electric field, the force on the dipole
will be
8E 8E 8rc
F=p_x+p _Y+p _'""'z_'
x8x Yay z8z
=0+0-10-7 x105=-10-2 N
The negative sign shows that the force on the dipole
actsin the negative Z-direction.
Asthedipole momentpacts in the negative
Z-direction while the electric field Eactsin the positive
Z-direction, so e=180°.
Torque, 't=pEsin180° =pEx0=O.
1.28.(i)A conductor A with a cavity [Fig. 1.157(a)] is
given achargeQ.Showthat the entire charge must appear on
the outer surface of the conductor.
(ii)Another conductor Bwith charge qisinserted into
the cavity keepingBinsulated from AShowthat the total
charge on the outside surface of A isQ+q [Fig.1.157(b)].
(iii)Asensitive instrument isto beshielded from the strong
electrostatic fields in its environment. Suggest a possible way.
Q Q+q
(a) (b)
Fig.1.157
Ans,(i)Refer answer to Q.25(6) on page 2.25.
(ii) Consider a Gaussian surface inside the conductor
but quite close to the cavity.
Inside the conductor, E=O.
1.85
Fi.1.158
By Gauss's theorem,
J.._!--"Ed--"S_Total charge _
'i'E-r' - -0
EO
i.e.,the total charge enclosed by the Gaussian surface must
be zero. This requires acharge of -qunits to be induced
on inner surface of conductorA.But an equal and
opposite charge of+qunits must appear on outer surface
Aso that charge on the surface ofAisQ+q.
Hence the total charge on the surface ofAis Q+q.
(iii) The instrument should be enclosed in a metallic
case. This will provide an electrostatic shielding to the
instrument.
1.29.A hollow charged conductor has a tiny hole cut intoits
surface.Showthat theelectric fieldintheholeis ~ ~, where ~
2Eo
istheunitvector in the outward normal direction, and(Jisthe
surface charge density near the hole.
Ans. Consider the charged conductor with the hole
filled up, as shown by shaded portion in Fi. 1.159.
Applying Gauss's theorem, we find that field just outside
is ~;; and is zero inside. This field can be viewed as the
EO
superposition of the fieldf2due to the filled up hole plus
B
A
Fig.1.159
the fieldf1due to the rest of the charged conductor. Since
inside the conductor the field vanishes, the two fields
must be equal and opposite, i.e.,
...(1)www6notesdrive6com

74. 1.86
And outside theconductor
,the fields are added up:
c
11+E2=-
EO
Adding equations (1)and(2),we get
211= ~ or 11 =~
EO 2ea
Hence the field due to the rest of the conductor or the
fieldin the holeis
E=~n
2Eo
wherenis a unit vector in the outward normal direction.
1.30.Obtain the formula for the electric field due to along
thinwire of uniform linear charge density Awithout using
Gauss's law.
[Hint. Use Coulomb's lawdirectly and evaluate the
necessary integral.]
Ans.Refer to thesolution of Example 47 on page 1.37.
1.31.Itisnowbelieved that protons and neutrons are
themselves builtout of more elementary units called quarks. A
proton and a neutron consists of three quarks each. Twotypes of
quarks, the socalled'up'quark (denoted by u)ofcharge+(213) e,
and the 'down' quark(denoted by d) of charge ( -1/3)e,together
withelectrons build up ordinary matter. Suggest a possible
quarkcomposition of a proton and neutron.
Ans.Charge on'up'quark(u)=+ ~e
Charge on 'down' quark(d)=-~e
Charge on a proton=e
Charge on a neutron=0
Let a proton containx'up'quarks and(3-x)'down'
quarks. Then total charge on a proton is
ux+d(3-x)=e
or '£ex-1e(3-x)=e
3 3
2 x
or -x-1+-=1
3 3
or x= 2 and 3 -x=3 -2 =1
Thus a proton contains 2 'up' quarks and 1'down'
quark. Its quark composition should be:uud.
Let a neutron containy'up' quarks and(3 -y)'down'
quarks. Then total charge on a neutron must be
uy+d(3 -y)=0
or .£ey_1e(3 -y)=0
3 3
ill ~y-1+~=0
or y= 1 and 3 - Y= 3 - 1= 2
Thus a neutron contains 1'up' quark and 2'down'
quarks. Its composition should be:udd.
PHYSICS-XII
...(2)
1.32.(a)Consider an arbitrary electrostatic fieldconfigu-
ration. A small test charge isplaced at a null point(i.e.,where
-->
E=0)of the configuration. Show that the equilibrium of the
test charge isnecessarily unstable.
(b) Verify this result for the simple configuration of two
charges ofthe same magnitude andsign placed a certain
distance apart.
Ans.(a)We can prove it bycontradiction. Suppose the
testcharge placed at null point be in stable equilibrium.
Since the stable equilibrium requires restoring force inall
directions, therefore, the test charge displacedslightly in
any direction will experience a restoring force towards the
null point. That is, all field lines near the null point should
bedirected towards the null point. Thisindicates that
there is a net inward flux of electric field through a closed
surface around the null point. But,byGauss's law, the
flux of electric field through a surface enclosing no charge
must be zero. This contradicts our assumption. Hence the
test charge placed at the centre must be necessarily in
unstable equilibrium.
(b)The null point lies on the midpoint of the line
joining the two charges. If the test charge is displaced
slightly on either side of the null point along this line, it
will experience a restoring force. But ifit isdisplaced
normal to thisline,the net force takes it away from the
null point. That is, no restoring force acts in the normal
direction. But stable equilibrium demands restoring force
inall directions, hence test charge placed at null point will
not beinstableequilibrium.
1.33.A particle of massm andcharge (- o)enterstheregion
between the two charged plates initially moving along x-axis
with speed Vx(likeparticle 1inFig.1.152). Thelength of plate is
Land a uniform electric field Eismaintained between theplates.
Showthat the vertical deflection of theparticle at the far edge of
theplate isqEL2/(2mv;).
Compare this motion with motion of a projectile in
gravitational field.
Ans.The motion of the charge-qin the region of the
electric field Ebetween the two charged plates is shown in
Fig.1.160.
Fig.1.160www4yztp•o°tvp4nzx

75. ELECTRIC
CHARGES AND FIELD
Force on the charge -qin the upward direction is
ma=qE
a=qE
m
:. Acceleration,
Time taken to crossthe field, t= ~
Vx
Vertical deflection at the faredge of the platewill be
lIE L2 EL2
Y=ut+-at2= 0+_.!L. -=-q-
22m v22mv2
x x
Likethemotion of a projectile in gravitational field,
thepathofa charged particle inanelectric field is
parabolic.
1.34.Suppose that the particle inExercise 1.33isanelectron
projected withvelocity Vx=2.0x106ms-1. IfEbetween the or
1.87
plates separated by 0.5 cm is 9.1x102N/C where will the
electron strike the upper plate ?
(Ie1= 1.6 x10-19 C,me= 9.1 x10-31 kg).
Ans.Herey= 0.5em=0.5x10-2m,
Vx=2.0x106 ms",E=9.1x102 NC-1, L=?
From the above exercise, the vertical deflection of an
electron is given by
eEr!
y=--2
2mevx
2 2
L2=ymevx
eE
2x0.5x10-2 x9.1x10-31 x4x1012
1.6x1019x9.1x102
=2.5x10-4
L=1.58x10-2 m ~1.6em.www6notesdrive6com

76. Text Based Exercises
r/+YPE
A :VERY SHORT ANSWER QU ESTIONS (1mark each)
1.What is the cause of charging a body?
2.Anebonite rod is rubbed with wool or fur. What
typeof charges do they acquire?[Haryana 93]
3.Aglassrod is rubbed with silk. What type of
charges do they acquire? [CBSEOD90]
4.Whydoes an ebonite rod get negatively charged on
rubbing with wool ?
5.Consider three charged bodies P, Qand R If P and
Qrepel each other and P attracts R, what is the
nature of the force between Q and R?
6.Apositively charged glass rod is brought near an
uncharged pith ball pendulum. What happens to
the pith ball ?
7.When a polythene piece is rubbed with wool, it
acquires negative charge. Is there transfer of mass
fromwool to polythene ?
8.Is the force acting between two point electric
charges qlandq2kept at some distance in air,
attractive or repulsive when:
(i)ql q2>0(ii)ql q2<0 ? [CBSE03, 07]
9.Name any two basic properties of electric charges.
[CBSED 95C ; Punjab 05C]
10.What doyou understand by quantisation of
electric charges?[Punjab 07, lOC ; CBSE OD92]
11.What is the cause of quantisation of electric
charge? [Punjab lOC]
12.What do you mean by additivity of electric charge?
13.What do you mean by conservation of electric charge?
14.Is the total charge of the universe conserved?
15.A glass rod, when rubbed with silk cloth, acquires a
charge of 1.6 x10-13 C What is the charge on silk
cloth ? [CBSED 91 ; Himachal 99 ; Haryana 99]
16.Two insulated charged copper spheresAand B of
identical size have the chargesqAandqBrespec-
tively. A third sphere C of the same size but
uncharged is brought in contact with the first and
then with the second and finally removed from both.
What are the new charges onAand B ?[CBSEF 11]
17.What is the least possible value of charge?
[Haryana 02]
18.State Coulomb's law of force between charges at
rest.Express the same in SI units.
[CBSEOD94 ; ISCE93 ; Haryana 02]
19.In Coulomb's law,F=kql~2,what are the factors
r
on which the proportionality constantkdepends?
[Himachal 02;CPMT93]
20.Name and define the SI unit of charge.
[Punjab 09C, 11]
21.In the relationF=kql~2,what is the value ofkin
r
free space? [Haryana 02]wwwxnotesdrivexcom

77. "
1.88
22.Give the SI unit of electrical permittivity of free
space. [Haryana02]
23.Write
down the value of absolute permittivity of
free space. [Punjab96]
24.Deduce the dimensional formula for the propor-
tionality constantkin Coulomb's law.
25.Write the dimensional formula for the permittivity
constantEOof free space.
26.What is the force of repulsion between twocharges
of 1 C each, kept 1 m apart in vaum ?
27.Two point charges''II'and'q2'are placed at a
distance"d'apart as shown in the figure. The
electric field intensity is zero at a point' P'on the
line joining them as shown. Write two conclusions
that you can draw from this. [CBSED 14C]
•••• --- d---....,._ •........• p
ql q2
28.Define dielectric constant of a medium in terms of
force between electric charges.
[CBSED 05 llC; F 10; Punjab11]
29.In a medium the force ofattraction between two
point electric charges, distancedapart, isF.What
distance apart should these be kept in the same
medium so that the force between them becomes
3 F? [CBSEOD 98]
30.The force between two charges placed in vaum is
F.What happens to the force if the twocharges are
dipped in kerosene oil of dielectric constant,K=2?
31.State thesuperposition prinple for electrostatic
force on a charge due to a number of charges.
[NCERT;Haryana01]
32.A forceFis acting between two point charges'q1
andq2'If a third chargeq3is placed quite close toQ2'
what happens to the force betweenQ1andQ2?
33.How many electrons are present in 1 coulomb of
charge ? [Himachal92; Punjab99]
34.Define volume charge density at a point. Write its
SI unit.
35.Define surface charge density at a point. Write its SI
unit.
36.Define line charge density at a point. Write its SIunit.
37.Define electric field at a point.
[CBSEOD 95; Punjab2000]
38.Is electric field intensity a scalar or vector quantity ?
Give its SI unit. [CBSED 99C]
39.Write the dimensional formula of electric field.
40.Name the physical quantity whose SI unit is
newton coulomb-1. [CBSED 98]
41.Draw the pattern of electric field around a point
charge(i)Q>0 and(ii)Q<O. [CBSE095, 95C]
PHYSICS-XII
42.Sketch the lines of force due to two equal positive
charges placed near each other. [CBSED 96C, 03]
43.Sketch the lines of force of a+ve point charge
placed near a -ve point charge of the same
magnitude. [CBSED 96C]
44.Draw the lines of force of an electric dipole.
[CBSEOD95C]
45.Two pointchargesQ1andQ2placed a distanced
apart are such that there is no point where the field
vanishes. What can be concluded from this?
46.A proton is placed in a uniform electric field
directed along the positivex-axis. In which
direction will it tend to move?[CBSED llC]
47.Whatis an electric dipole? [CBSEOD 08,11]
48.Define electric dipole moment. Write its SI unit.
[CBSEOD 08, 11;F 13]
49.Is electric dipole moment a scalar or vector
quantity? [CBSE06C; F 13]
50.What is a point (ideal) dipole? Give example.
51.How much is the dipole moment of non-polar
molele?
52.An electric dipole is placed in a uniform electric
field. What is the net force acting on it ?
[CBSED 92C; F 94C]
53.When is the torque on a dipole in a field maximum ?
54.What is the effect of torque on a dipole in an electric
field?
55.When does an electric dipole placed in a
non-uniform electric field experience a zero torque
but non-zero force?
56.What is the nature of symmetry of dipole field?
57.Will an electric dipole have translational motion
when placed in a non-uniform electric field? Give
reason for your answer.
58.Does the torque exerted on a dipole in a
non-uniform field depend on the orientation of the
dipole with respect to the field?
59.What is the charge of a dipole?[CBSED 10C]
60.Under what condition will acharged rlar loop
behave like a point charge in respect of its electric
field?
61.Define electric flux.[Punjab2000, 01;CBSED 13C]
62.Name the prinple which is mathematical
equivalent of Coulomb's law and superposition
prinple.
63.What is the relation between electric intensity and
flux? [Punjab97, 98,99]
64.How is electric flux expressed in terms of surface
integral of the electric field?
65.State Gauss theorem in electrostatics.
[Punjab02;CBSED 08C]www7notesdrive7com

78. ELECTRICCHARGE
SAND FIELD
66.Iselectric flux a scalaror avector7
[CBSE Sample Paper 96]
67.Give the 51unit of electric flux7 [CBSE D13C]
68.Give the 51 unit of surface integral[fE.is1of an
5
electric field7
69.What is thedirection of an area vector7
70.What is a Gaussian surface7
71.What is the use of Gaussian surface 7
72.How much isthe electric flux through aclosed
surface duetoa charge lying outside the closed
surface7
73.Two plane sheets of charge densities +(Jand-(Jare
kept in air as shown in Fig. 1.161. What are the
electric field intensities at pointsAandB7
[CBSE D03C]
---------------------+cr
•B
----------------------cr
Fig. 1.161
74.Twosmall balls, having equal positive charge q
coulomb are suspended by twoinsulating strings of
equal length 1metre from a hook fixed to a stand.
The whole set up is taken in a satellite into space
where there is no gravity. What is the angle
between the two strings and the tension in each
string7 [TIT86]
An electric dipole of dipole moment20x10-6Cm is
enclosed by aclosedsurface. Whatisthe net flux
corning out of the surface7 [CBSE D05]
How does the coulombforce between two point
charges depend upon the dielectric constant of the
medium 7 [CBSE OD05]
77.Two fixed point charges +4eand+eunits are
separated by a distancea.Where should a third
chargeqbe placed for it to be in equilibrium7
[CBSEOD05]
What is the angle between the directions of electric
field at any (i)axial point and(ii)equitorial point
due to an electric dipole7[CBSE Sample Paper08]
75.
76.
78.
Answers
1.89
79.If the radius of the Gaussiansurface enclosing a
chargeqishalved, howdoesthe electricflux through
the Gaussian surface change7 [CBSE OD08]
A dipole, of dipole momentp,ispresent in a uni-80.
81.
-->
form electric fieldE.Write the value of the angle
--> -->
betweenpandEfor which the torque, expe-
rienced by the dipole, is minimum. [CBSE D09C]
Acharge' c(isplaced at the centre of a be. What is
the electric flux passing through thecube7
[CBSEOD 12]
A charge' c(isplaced at the centre of a be of side I.
What is the electric flux passing through each face
of the be 7. [CBSE 00 12]
A charge'c(is placed at the centre of a be of sideI.
What is the electric flux passing through two
opposite faces of the be 7 [CBSE OD 12]
A chargeQI-lCisplacedat the centre of a be.
What is the flux corning out from anyone surface 7
[CBSE F10]
Charges of magnitudes 2Qand-Qare located at
points(a,a,a) and(4a,a,a). Find the ratio of the
flux of electric field, due to these charges, through
concentric spheres of radii2aand8acentered at the
origin. [CBSE Sample Paper 11]
Two charges of magnitudes-3Qand+2Qare
located at points (a,O)and(4a,O)respectively. What
is the electric flux due to thesecharges through a
sphere of radiusSawith its centre at the origin 7
[CBSE OD13]
Two concentric spherical shells of radii R and 2R
are given chargesQ1andQ2respectively. The
surface charge densities on the outer surfaces are
equal. Determine the ratioQ1:Q2. [CBSEF13]
Write the expression for the torque-tacting on a
dipole of dipole momentpplacedin an electric
82.
83.
84.
85.
86.
87.
88.
-->
fieldE.
89.
[CBSE F 15]
What is the electric flux through a be of side 1 ern
which encloses an electric dipole 7 [CBSE D 15]
•
1.Charging ors due to the transfer of electrons
from one body to another.
2.The ebonite rod acquires a negative charge and fur
or wool acquires an equal positive charge.
3.The glass rod acquires positive charge and silk
acquires an equal negativecharge.
4.This is because electrons in wool are less tightly
bound than electrons in ebonite rod.
5.Attractive.
6.The pith ball is attracted towards the rod, touches it
and then thrown away.www7notesdrive7com

79. 1.90
7.The polythene
piece acquires negative charge due
to transfer of material particles like electrons from
wool to it, so there is a transfer of mass from wool
to polythene.
8.(i)Whenq1q2>0,the force is repulsive
(ii)Whenq1 q2<0, the force is attractive.
9.Electric charges are(i)quantised,(ii)additive and
(iii)conserved.
10.Quantisation of electric charge means that the total
charge(q)of a body is always an integral multiple of
a basic charge (e)which is the charge on an electron.
Thus q=ne,wheren= 0,±1,±2,±3, .
11.The basic cause of quantisation of electric charge is
that during rubbing only an integral number of
electrons can be transferred from one body to another.
12.Additivity of electric charge means that the total
charge on a system is the algebraic sum (taking into
acount proper signs) ofall individual charges in
the system.
13.Conservation of electric charge means that the total
charge of an isolated system remains unchanged
with time.
14.Yes,charge conservation is a global phenomenon.
15.To conserve charge, the silk cloth acquires negative
charge of 1.6 x10-13c.
16.New charge on sphereA,
q' _ qA
A-2
New charge on sphereB,
, qB+qA/2 2qB+qA
qB= 2 4
17.The least possible value of charge is the magnitude
of the charge on an electron or proton and it is
e=1.6x10- 19C.
18.Refer to point 14 of Glimpses on page 1.100.
19.The proportionality constantkdepends on the
nature of the medium between the two charges and
the system of units chosen.
20.The51unit of electric charge is coulomb. One
coulomb is that amount of charge which repels an
equal and similar charge with a force of 9x109N
when placed in vaum at a distance of 1 metre
from it.
21.k= 9x109Nm2c-2.
22.51unit of &0= C2N-1m -2.
23.Permittivity of free space,
&0= 8.85x10-12C2·N-1m-2.
24.[k]=Fr2= MLT-:L2= [ML3T-4A -2].
M2 (AT)
PHYSICS-XII
25.[&]=_1_.M2= (AT)2 =[M-1L-3T4A2].
o41tFr2 [MLT2L2]
26.F=9x109N.
27.(i)The point chargesq1andq2must be of opposite
nature or signs.
(ii)The magnitude of chargeq1must be greater
than that of chargeq2'
28.The dielectric constant of a medium is the ratio of
the force between twocharges placed some
distance apart invaum to the force between the
same two charges when they are placed thesame
distance apart in the given medium.
1 3F d2 d'= ~d.
29.AsFoc2 ..-=-2 or r:
d F d' ~3.
30.
F.F
F. =-.i!!!. =-
kerosene K 2
The prinple ofsuperposition states that the total
force on a given charge is the vectorsum of the
individual forces exerted on it by all other charges,
the force between two charges being exerted in
such a manner as if all other charges were absent.
-4 -4 -4 -4
F=1i2+1i3+ +fiN
Bythe superposition prinple, the force between
two charges does not depend on the presence of
third charge. Hence the force betweenq1andq2
remains equal toF.
q 1C 18
33.n=-= 19 = 6.25x10electrons.
e1.6x10C
31.
32.
34.The volume charge density(p)at a point is defined
as thecharge contained per unit volume around
that point.
p=!!!L
dV
The51unit ofpis coulomb per bic metre (C m-3).
35.The surface charge density(c)at a point is the
charge per unit area around that point
cr=dq.
dS
The51unit for cris Cm-2.
36.The line charge density at a point on a line is the
charge per unit length of the line at that point
A=dq
dL
The51unit forAis Cm -1.
37.The electric field at a point is defined as the electro-
static force per unit positive charge acting on a
vanishingly small test charge placed at that point.
F
E=lim
qo-40qo
Mathematically,www6notesdrive6com

80. ELECTRIC CHARGESAND FIELD
38.Electri
c field is a vector quantity. Itsdirection is
same as thatof the force onaunit positive test
charge.
SI unit of electric field=NC1 or Vm.1.
39.[Electric Field]
Force MLy-2
---=---
Charge C
=MLy-2 =[MLT-3A -1].
AT
40.Ne1 istheSI unit of electricfield.
41.(i)See Fig. 1.74on page1.47 (ii)See Fig. 1.75on
page1.47.
42.SeeFig.1.77on page1.47.
43.See Fig. 1.76on page1.47.
44.See Fig. 1.76on page1.47.
45.The point charges %andq2are equal and opposite.
46.Asthe proton has a positive charge, itwill tend to
move along positive x-axisi.e.,along the direction
of the electric field.
47.An electric dipoleisa pair ofequal and opposite
chargesseparated by some distance.
48.Electric dipole moment is the product of either
....
chargeqand thevector2adrawn from the -ve
charge to the+vecharge.
.... ...•
p=qx2a
Its SI unit is coulomb metre (Cm).
49.Electric dipole momentisavector quantity.
50.A point dipoleisonewhich has negligibly small
size.In such a dipole, charge q-+00and size2a-+0
in such a way that the productp=qx2ahas a finite
value. Atomic dipoles are point dipoles.
51.Zero.
52.Zero.
53.Torque is maximumwhen dipole is held perpen-
dilar to the electric field.
54.Torque tends toalign the dipole along the direction
of the electric field.
55.When the dipole is placed parallel to the non-
uniform electric field.
56.The dipole field is cylindrically symmetric.
57.Yes. In a non-uniform electric field, an electric
dipole experiences unequal forces at its ends. The
two forces add up to give a resultant force which
gives a translatory motion to the dipole.
58.Yes.In a non-uniform electric field, the field vector
E(;7)changes from point to point, either in magni-
tude or in direction or both. Therefore, thetorque
1.91
----+ ---+ ---t ---t -+
r=PxE(r )for a dipole located atrchanges
with the change in orientation of the dipole with
respect to the field.
59.Zero.
60.When the observation point on the axis of the
cirlar loop lies at a distance much greater than its
radius, the electric field of the rlar loop is
similar to that of a point charge.
61.Electricfluxover an area in an electric field repre-
sents the total number of electriclines of forcecrossing
thisarea normally. If the normal drawn to the
.... ....
surface areatJ.S makes an angle 9with the field E,
then the electric flux through this area is
.......•
<1>£=EtJ.Scos 9=E. tJ.S
62.Gauss's theorem of electrostatics.
63.
....
The relation between electricintensity E and flux
<1>£is<1>£=EtJ.Scos 9
The electric flux <1>£throughany surface, open or
closed, is equal to thesurface integral of the electric
.... ....
field E over the surface S,
64.
65.
f...•...•
<1>£=E.dS
s
Gauss's theorem states that the flux of electric field
through any closed surface Sis 1/EOtimes thetotal
chargeqenclosed by S.
Mathematically,
<1>£=fE.is=.1..
s EO
Electric flux is a scalar.
SI unit of electric flux=Nm2C-1.
SI unit offE.dS=Nm 2c-1.
66.
67.
68.
69.The direction of an area vector is along the outward
drawn normal to the surface.
70.An imaginary closed surface enclosing acharge is
called the Gaussian surface of that charge.
71.Bya clever choice of Gaussian surface, we can
easily find the electric field produced by certain
charge systems which are otherwise quite diffilt
to determine by the application of Coulomb's law
and superposition prinple.
72.Zero.
a
73.EA=0 andEB= - .
Eo
74.As the two balls are in the state of weightlessness,
the strings would become horizontal due to the
force of repulsion.www7notesdrive7com

81. 1.92
:.Angle betw
een the two strings =180°
1q2
Tension in eachstring= -- --N
41tSo . (2/)2 .
75.Zero, because the net charge on the dipole is zero.
76F Fvac· F 1
.med=--I.e., medoc-
K K
,
77.Refer to the solution of Example 13 on page 1.13.
78. (i)At any axial point, Eacts in the directionp.
(ii)At any equatorial point, Eacts in the opposite
-->
direction ofp
Hence the. angle between the directions of the
above two electric fields is 180°.
79.Electric flux(~£=q /EO)remains unchanged
because the charge enclosed by the Gaussian
surface remains same.
80.Torque experienced by the dipole is minimum
--> -->
when angle between pand E is 0°.
1:=pEsin 0° =O.
PHYSICS-XII
81A...«
.'t'£-
EO
A. -~ ~_~ s.
83.'t'£ - . - .
6EO 3EO
84.Flux through each face of the be,
~ =~. QIlNm2C1
E6E
o
85.~('=2a) =~...i=2: 1
~(, =8a) 2Q-Q
Netcharge enclosed
86.~£=-----'''-------
EO
-3Q+2Q Q
EO EO
87.q=41tR2~ =1 : 4.
Q2 41t(2R)cr'
--> -->-->
88; 1:=PxE
89.Zero, because the net charge on the dipole is zero.
"YPE B:SHORT ANSWER QUESTIONS (2or3 marks each)
1.What is frictional electrity? Briefly describe the
electronic theory of frictional electrity.
2.What is electric charge ? Is it a scalar or vector ?
Name its SI unit.
3.How will you show experimentally that there are
onlytwo kinds pf electriccharges?
4.Define electrostatic induction. Briefly explain how
an insulated metal sphere can be positively charged
by induction.
5.What is meant by quantization of electric charge?
What is its cause? [Haryana2000; Punjab01]
6.Give six properties of electric charges.[Punjab99C]
7.State the law of conservation of charge. Give two
examples to illustrate it.
[Himachal96 ;Haryana98,2000 ;Punjab06C, 10C]
8.Howdoesthe speed of an electricallycharged
particle affects its(i)mass and(ii)charge?
[CBSE D 93]
9.State Coulomb's law of force between two electric
charges and state its limitations. Also define the
SI unit of electric charge.[Haryana96 ;Punjab2003]
10.Write Coulomb's law in vector form. What is the
importance of expressing it in vector form ?
[Haryana91, 95 ;Punjab98C, 2000]
11.Write the vector form of force acting between two
--> -->
chargesqlandq2having1.and'2as their position
vectors respectively. [Himachal2000]
12.State Coulomb's law in vector form and prove that
--> -->
F21=-li2
where letters have their usual meanings.
[Haryana97]
13.Define electric field intensity. What is its SI unit?
What is relation between electric field and force?
[CBSE OD 91]
14.Define electric field at a point. Give its physical
Significance.
15.Derive an expression for electric field intensity ata
point at distance, from a point chargeq.
[CBSE OD 94;Haryana95,99]
16.Write an expression for the force exerted on a test
charge by a continuous charge distribution.
17:Define the term electric dipole moment of a dipole.
State its SI unit. [CBSE OD 08, 11]
18.Define electric field intensity and derive an
expression for it at a point on the axial line of a
dipole. Also determine its direction.
[Punjab2000, 01 ;Haryana98, 02 ; CBSED92,95]···1uv~lzkyp°l1jvt

82. ELECTRIC CHARGES AND FIELD
19.Define the term 'electric dipole moment'. Is it a
scalar or vector ?
Deduce an expression for the electric field at a
point on the equatorial plane of an electric dipole
of length2a. [Haryana 02 ; CBSEF09 ; OD 13]
20.Define electric field intensity. Write its SI unit.
Write the magnitude
and direction of electric field
intensity due to an electric dipole of length2aat the
midpoint of the line joining the two charges.
[CBSE OD 05]
21.What is an electric dipole? Derive an expression
for the torque acting on an electric dipole, when held
in a uniform electric field. Hence define the dipole
moment. [Haryana 01, 02 ; CBSE D 08 ; OD 03C]
22.Define the term electric dipole moment. Give its
unit.Derive an expression for the maximum torque
acting on an electric dipole, when held ill a uniform
electric field. [CBSE D 02]
23.An electric dipole is placed in uniform external /
-+
electric field E . Show that the torque on the dipole
-+ -+ -+
is given by 't=PxE
wherepis the dipole moment of the dipole. What
is the net force experienced by the dipole? Identify
two pairs of perpendilar vectors in the
expression. [CBSE DISC]
24.Draw a labelled diagram showing an electric dipole
-+
making an angleewith a uniform electric field E.
Derive an expression for the torque experienced by
the dipole. [rSCE 95; CBSE OD 14]
25.An elecfric dipole is held in a uniform electric field.
(i)Using suitable diagram, show that it does not
undergo any translatory motion, and
(ii)Derive an expression for the torque acting on it
and spefy its direction. When is this torque
maximum ? [CBSE DOS, 08]
26.In a non-uniform electric field, is there any torque or
force acting on a dipole held parallel or antiparallel
to the field. If yes, show them by suitable diagrams.
27.Briefly explain how does a comb run through dry
hair attract small pieces of paper.
28.Define an electric field line. Draw the pattern of the
field lines around a system of two equal positive
charges separated by a small distance.
[CBSE D 03 ; Sample Paper 11]
29.Define electric line of force and give its two
important properties. [CBSE DOS]
30.What do electric lines of force represent? Explain
repulsion between two like charges on their basis.
[Punjab 97C]
1.93
31.Define electric flux. Write its SI unit.
A chargeqis enclosed. by a spherical surface of
radius R If the radius is reduced to half, how
would the electric flux through the surface change?
[CBSEOD 09]
32.Prove that 1/r2dependence of electric field of a
point charge is consistent with the concept of the
electric field lines.
33.State and prove Gauss's theorem in electrostatics.
[Punjab 03 ; CBSE OD 92C, 95]
34.Using Gauss's theorem, obtain an expression for
the force between two point charges.[CBSE OD 91]
35.State Gauss's theorem and express it mathe-
matically. Using it, derive an expression for the
electric field intensity at a point near a thin infinite
plane sheet of charge density O'Cm-2. [Punjab 03;
CBSE D 07, 09, 12 ; CBSE OD 01, 04, OS,06C]
36.Using Gauss's law establish that the magnitude of
electric field intenisty, at a point, due to an infinite
plane sheet with uniform charge density a; is
independent of the distance of the field point.
[CBSE Sample Paper 11]
37.Use Gauss's law to derive the expression for the
electric field between two uniformly charged large
parallel sheets with surface charge densities0'and
-0'respectively. [CBSE OD 09]
38.State Gauss's theorem in electrostatics. Uskg this
theorem,-prove that no electric field exists inside a
hollow charged conducting sphere.
[Punjab 03 ; CBSE D 02! 03 C ; CBSE OD 97]
39.A thin conducting spherical shell.of radius R has
chargeQspread uniformly over its surface. Using
Gauss's law, derive an expression for an electric
field at a point outside the shell. Draw a graph of
electricfield E(r)with distancerfrom the centre of the
shell for 0::;r::;00. [CBSE D 04, 08, 09; OD 06C, 07]
40.Using Gauss's law obtain the expression for the
electric field due to a uniformaly charged thin
spherical shell of radius R at a point outside the
shell. Draw a graph showing the variation of electric
field withr,forr>Randr<R [CBSE D 11]
41.A thin straight infinitely long conducting wire
having charge densityA.is enclosed by a lin-
drical surface of radiusrand length I, its axis
coinding with the length of wire. Find the
expression for the electric flux through the surface
of the linder. [CBSE OD 11]
42.State Gauss's theorem in electrostatics. Using this
theorem, derive an expression for the electric field
intensity due to an infinitely long, straight wire of
linear charge densityA.Cm-1.
[CBSE D 04, 08, 09 ; OD OS,06C, 07]www7notesdrive7com

83. 1.94
Answers
PHYSICS-XII
••
1
.Refer to points 2 and 6 of. Glimpses on page 1.99.
2.Refer to point 3 of Glimpses on page 1.99.
3.Refer answer to Q. 5 on page 1.2.
4.Refer answer toQ.11 on page 1.4 and Q.13 on
page 1.5.
5.Refer answer toQ.16 on page 1.6.
6.The properties of electric charges are· as follows :
(i)Like charges repel and unlike charges attract
each other.
(ii)Electric charges are quantized.
(iii)Electric charges are additive.
(iv)Electric charges are conserved.
(v)The magnitude of elementary negative charge
is same as that of elementary positive charge
and is equal to 1.6 x10-19C.
(vi)Unlike mass, the electric charge on a body is
not affected by its motion.
7.Refer answer to Q. 18 on page 1.8.
8.Refer answer to Q. 19 on page 1.8.
9.Refer to point 14 of Glimpses and the solution of
Problem 3 on page 1.67.
10.Refer answer to Q. 22 on page 1.10.
11.Refer answer toQ.22 on page 1.10.
12.Refer answer toQ.22 on page 1.10.
13.Refer answer to Q. 29 and Q. 30 on page 1.25.
14.Refer answer to Q. 29 and Q. 30 on page 1.25.
15.Refer answer toQ.31 on page 1.29.
16.Refer answer toQ.33 on page 1.35.
->-> ->
Add the forcesFv 'FsandFL.
17.Refer answer to Q. 48 on page 1.91.
18.Refer answer to Q. 37 on page 1.40.
19.Refer answertoQ.38 on page 1.40.
20.Refer answer toQ.29 on page 1.25. At any
equatorial point of a dipole,
-> 1 p"
~a =-47t EO•(,2+a2)3/2 p
At the midpoint of the dipole (,=0), the magnitude
of the field is
1p
E;.qua =47te a3
o
The direction of the field is from +ve to -ve charge.
21.Refer answer toQ.40 on page 1.41.
22.Refer answer to Q. 40 on page 1.41.
23.Refer answer to Q. 40 on page 1.41.
24.Refer answer to Q. 40 on page 1.41.
25.Refer answer toQ.40 on page 1.41.
26.Refer answer toQ.41 on page 1.42.
27.Refer answer to Q. 41 on page 1.42.
28.See Fig. 1.77 on page 1.47.
29.Refer answer to Q. 43 on page 1.45.
30.Refer answer to Q.44(iv) on page 1.47.
31.Refer to point 33 of Glimpses. If the radius of the
spherical surface is reduced to half, the electric flux
would not change as the charge enclosed remains
the same.
32.Refer answer toQ.46 on page 1.48.
33.Refer answer toQ.49 on page 1.49.
34.Refer answer to Q. 51 on page 1.50.
35.Refer answer toQ.53 on page 1.56.
36.Refer answer toQ.53 on page 1.56.
37.Refer answer to Q. 55 on page 1.57.
38.Refer answer to Q. 56(c) on page 1.58.
39.Refer answer toQ.56(a)on page 1.58 and see
Fig. 1.103.
40.Refer answer to Q.56(a)on page 1.58 and see
Fig. 1.103.
41.Refer for answer to Q. 52 on page 1.56.
42.Refer answer toQ.52 on page 1.56.
rJlTYPE C:LONG ANSWER QU ESTIONS (5marks each)
1.State the prinple of superposition and use it to
obtain the expression for the total force exerted on a
point charge due to an assembly of(N -1)discrete
point charges. [Haryana 02]
2.Obtain an expression for the electric field at any
point due to a continuous charge distribution.
Hence extend it for the electric field of a general
source charge distribution.
3.(a)Consider a system of nchargesQl,Q2, ...,qnwith
.. ->->-> -> I
position vectors1."2''3' ..·,'nre ative to
some origin'0'.Deduce the expression for the
->
net electric field E at a point P with position
vector ~,due to this system of charges.···1uv~lzkyp°l1jvt

84. ELECTRIC CHARGES AND FIELD
(b)Find the resultant electric field due to an
electric dipole of dipole moment2aq, (2abeing
the separation between the charges±q)at a
-point distant'x'on its equator
. [CBSEF 15]
4.A dipole is made up of two charges+qand-q
separated by a distance2a.Derive an expression for
~
the electricfieldEedue to this dipole at a point distantr
from the centre of the dipole on the equatorial plane.
--->
Draw the shape of the graph, betweenIEelandr
whenr»a.If this dipole were to be put in a uniform
~
external electric field E, obtain an expression for the
torque acting on the dipole. [CBSE SP15]
5.(a)An electric dipole of dipole momentpconsists
of point charges+qand-qseparated by a
distance2aapart. Deduce the expression for
--->
the electric field E due to the dipole at a
distancexfrom the centre of the dipole on its
--->
axial line in terms of the dipole momentp.
Hence show that in the limitx»a,
E~2p/(41tEox3). y
(b)Given the electric
field in the region
---> A
E=2xi,find the net
electric flux through
the be and the z
charge enclosed by it.
[CBSE D15]
6.(a)State the theorem which relates total charge
enclosed within a closed surface and the
electric flux passing through it. Prove it for a
single point charge.
(b)An'atom' was earlier assumed to be a sphere
of radiusahaving a positively charged point
nucleus ofcharge+Zeat its centre. This nucleus
was believed to be surrounded by a uniform
density of negative charge that made the atom
neutral as a whole. Use this theorem to find the
electric field of this'atom' at a distancer(r<a)
from the centre of the atom. [CBSE SP15]
7.(a)Define electric flux. Write its SI units.
Fig.1.162
Answers
1.95
(b)Using Gauss's law, prove thatthe electricfield at
a point due to a uniformly charged infinite plane
sheet is independent of the distance from it.
(c)How is the field directed if(i)the sheet is
positively charged, (ii)negatively charged?
[CBSE D 12]
8.State Gauss's law in electrostatics. Using this
theorem, show mathematically that for any point
outside the shell, the field due to uniformlycharged
thin spherical shell is the same as if entire charge
of the shell is concentrated at the centre. Why do
you expect the electric field inside the shell to be zero
acording to this theorem ?[CBSED 92;OD 06]
9.Using Gauss' law, deduce the expression for the
electric field due to a uniformly charged spherical
conducting shell of radius R at a point(i)outside
and(ii)inside the shell.
Plot a graph showing variation of electric field as a
function ofr>R andr<R(rbeing the distance
from the centre of the shell).[CBSEOD 13, 13C]
10.(a)Using Gauss' law, derive an expression for the
electric field intensity at any point outside a
uniformly charged thin spherical shell of
radius R and the density(JC/m2.Draw the
field lines when the charge density of the
sphere is(i)positive,(ii)negative.
(b)A uniformly charged conducting sphere of
2.5 m in diameter has a surface charge density
of 100IlC/ m2.Callate the(i)charge on the
sphere(ii)total electric flux passing through
the sphere. [CBSED 08]
11. (a)Define electric flux. Write its SI unit.
(b)State and explain Gauss's law. Find out the out-
ward flux due to a point charge+qplaced at
the centre of a be of side'a'.Why is it found
to be independent of the size and shape of the
surface enclosing it ? Explain.[CBSEOD 15]
12. (a)Define electric flux. Write its SI unit.
"Gauss's law in electrostatics is true for any
closed surface, no matter what its shape or size
is".Justify this statement with the help of a
suitable example.
(b)Use Gauss's law to prove that the electric field
inside a uniformly charged spherical shell is
zero. [CBSEOD 15]
1.Refer answer toQ.27 on page 1.19.
2.Refer answer toQ.34 on page 1.35.
3.(a)Refer answer to Q. 32 on page 1.29.
(b)Refer answer toQ.38 on page 1.40.
•
4.Refer answer to Q. 38 on page 1.40 and Q. 40 on
page 1.41.
1
Forr»a,l1,quaex;r3'www7notesdrive7com

85. Fig. 1.165
(b)(i)q= 41tR
2cr=4x3.14x(1.25)2 x10-4
=1.963xlO-3 C
(ii)<l>E=!L= 1.963 x10-3 x41tX109
&0
=2.465xl07Nm2c-1
11.(a)Refer to the solution of Problem 17 on page 1.71.
(b)For Gauss's law, refer to point 35 of Glimpses
on page 1.102.
Outward flux due to a point charge+qplaced
at the centre of a be of sideais given by
Gauss's law as
<l>E= Total charge enclosed=+!L
&0 &0
<l>Edepends only the total charge enclosed by
the closed surface and not on its size and shape.
12.(a)Refer to the solution of Problem 17on page 1.71.
Acording to Gauss's law, the electric flux
through a closed surface depends on the net
charge enclosed by the surface and not upon
the size of the surface.
For any closed surface of arbitrary shape
enclosing a charge, the outward flux is same as
that due to a spherical Gaussian surface
enclosing the same charge. This is because of
the fact that:
(i)electric field is radial, and
(ii)the electric field, Eex:~.
r
(b)Refer answer toQ.56(c) on page 1.58.
'-"'YPE D:VALU E BASED QU ESTIONS (4marks each)
1.96
So the graph between./i,quaandris of the type as
shown in the figure given below.
Fig. 1.163
5.(a)Refer answer toQ.37 on page 1.40.
(b)Only the faces perpendilar to thex-axis
contribute towards the electric flux. The
contribution from the remaining faces is zero.
y
a
...•
E
x
z
a
Fig. 1.164
Flux through the left face,
<l>L=EScos1800= 2(0)a2(-1) = 0
Flux through the right face,
<l>R=EScosOo=2a xa2xl=2a3
..Net flux through the be, h=<l>L+<l>R=2a3
6.(a)Refer answer to Q. 49 on page 1.49.
(b)Refer to the solution of Problem 29 on page 1.79.
7.(a)Refer answer to Problem 18 on page 1.71.
(b),(c), Refer answer toQ.53 on page 1.56.
8.Refer answer toQ.56 on page 1.57. Any Gaussian
surface lying inside spherical shell does not enclose
any charge. So by Gauss's theorem, electric field
inside the shell is zero.
9.Refer answer toQ.56 on page 1.57.
1.Aneesha has dry hair. She runs a plastic comb
through her hair and finds that the comb attracts
small bits of paper. But her friend Manisha has oily
hair.The comb passed to Manisha hair could not
attract small bits of paper. Aneesha goes to her
Physics teacher and gets an explanation of this
phenomenon from her. She then goes to different
PHYSICS-XII
10.(a)Refer answer toQ.56(a)on page 1.58. The lines
of force for positively and negatively charged
spherical shells are shown below :
juniorclasses and demonstrates this experiment to
the students. The junior students feel very happy
and promise her to join her sence club set up for
searching such interesting phenomena of nature.
Answer the following questions based on the above
information:
(a)What are the values displayed by Aneesha ?···1uv~lzkyp°l1jvt

86. ELECTRIC CHARGES AND FIELD
(b)A comb run through one's dry
hair attracts
small bits of paper. But it does not attract when
run through wet hair. Why ?
2.Neeta's grandmother, who was illiterate, was
wrapping her satin saree. She found some sparks
coming out from it. She frightened and called
Neeta. Neeta calmed down her grandmother and
Answers
1.97
explained to her the sentific reason behind these
sparks.
Answer the following questions based on the above
information:
(a)What acording to you, are the values
displayed by Neeta ?
(b)Why do sparks appear when a satin cloth is
folded?
•
1.(a)Curiosity, leadership and compassion.
(b)When the comb runs through dry hair, it gets
charged by friction and attracts small bits of
paper. The comb does not get charged when
run through wet hair due to less friction and so
it does not attract bits of paper.
2.(a)Awareness and sensitivity.
(b)The different portions of thecloth getcharged
due to friction. Then the flow of charge gives
rise to sparks..www7notesdrive7com

87. Electric Charges and Field
GLIMPSES
1.Electrostatics.It is the study of electric charges
at rest.
2.Frictional electricity.The property
of rubbed
substances due to which they attract light objects
is called electricity. The electricity developed by
rubbing or friction is called frictional or static
electricity. The rubbed substances which show
this property of attraction are said to be
electrified or electrically charged substances.
3.Electric charge. It is an intrinsic property of
elementary particles of matter which gives rise
to electric force between various objects. It is a
scalar quantity and its51unit is coulomb (C).
4.Positive and negative charges.Benjamin Franklin
introduced the present day convention that
(i)The charge developed on a glass rod when
rubbed with silk is called positive charge.
(ii)The charge developed on a plastic/ebonite
rod when rubbed with fur is called negative
charge.
5.Fundamental law of electrostatics.Like charges
repel and unlike charges attract each other.
6.Electronic theory of frictional electricity.During
rubbing, electrons are transferred from one
object to another. The object with excess of
electrons develops a negative charge, while the
object with deficit of electrons develops a
positive charge.
7.Electrostatic induction.It is the phenomenon of
'temporary electrification of a conductor in
which opposite charges appear at its closer end
and similar charges appear at its farther end in
the presence of a nearby charged body. An
insulated conductor can be positively or
negatively charged by induction.
8.Electroscope.A device used for detecting an
electric charge and identifying its polarity is
called electroscope.
9.Three basic properties of electric charges.These
are:(i)quantization, (ii)additivity, and
(iil)conservation.
10.Additivity of electric charge.This means that
the total charge of a system is the algebraic sum
of all the individual charges located at different
points inside the system.
11.Quantization of electric charge.This means that
the total charge(q)of a body is always an integral
multiple of a basic quantum of charge(e)i.e.,
q=ne,wheren=0,±1,±2,±3, .
Faraday's laws of electrolysis and Millikan's oil
drop experiment established the quantum
nature of electric charge.
For macroscopically large charges, the
quantization of charge can be ignored.
12.Basic quantum of charge.The smallest amount
of charge or the basic quantum of charge is the
charge on an electron or proton. Its exact
magnitude is
e=1.602182x10-19 C.
13.Law of conservation of charge.It states that the
total charge of a system remains unchanged
with time. This means that when bodies are
charged through friction, there is only transfer
of charge from one body to another b It no net
creation or destruction of charge takes:'-lace.
(1.99)wwwhnotesdrivehcom

88. 1.100
14.Coulomb's law.The force of attraction or
repulsion be
tween two stationary point charges
qlandq2is directly proportional to the product
qlq2and inversely proportional to the square of
the distance, between them. Mathematically,
F=kqlq2
?
The proportionality constantkdepends on the
nature of the medium between the two charges
and the system of units chosen to measure F,ql'q2
andr.For free space and in51units,
k=_1_ =9x109Nm2C-2,
41tEo
EOis called permittivity of free space and its
value is8.854x10-12C2N-1m-2.
Hence Coulomb's law in51units may be
expressed as
F __ 1_q1q2
-41tEO' ?
15.51unit of charge is coulomb(C).It is that
amount of charge that repels an equal and
similar charge with a force of9x109N when
placed in vacuum at a distance of one metre
from it.
16.Permittivity(E).It is the property of a medium
which determines the electric force between
two charges situated in that medium.
17.Dielectric constant or relative permittivity.The
ratio(E / EO)of the permittivity of the given
medium to that of free space is known as
relative permittivity(Er)or dielectic constant(K)
ofthe given medium,
EF
ErorK=-=-.Yl!f...
EOFrned
The dielectric constant of a medium may be
defined asthe ratio of the force between two
charges placed some distance apart in free space
tothe force betweenthe same two charges when
they are placed the same distance apart in the
givenmedium.
Coulomb's law for any medium other than
vacuum can be written as
F_1qlq2 __ 1_q1q2_Fvac
rned - 41tE ----; -41tEOK? -K
18.Electrostatic forcevs.gravitational force.
Electrostatic forces are much stronger than
PHYSICS-XII
gravitational forces. The ratio of the electric
force and gravitational force between a proton
and an electron is
Fe=~-:::'227x 1039
FG Gmpme
19.Principle of superposition of electrostatic forces.
When a number of charges are interacting, the
total force on a given charge is the vector sum of
the forces exerted on it due to all other charges.
The force between two charges is not affected
by the presence of other charges. The total force
on chargeqldue to the chargesq2' q3' , qN
will be
~ ~
Fl=F12+
.x:
41tEo
~ ~
FI3+ , +FIN
N
L
i=2
qi(ii-r;)
I~~13
'1-'i
~ ~
"1:-r
where1:.=1 I
11 ~ ~
l'i -'iI
=a unit vector pointing fromqitoql'
20.Electric field. An electric field is said to exist at a
point, if a force of electrical origin is exerted
on a stationary charge placed at that point.
Quantitatively, it is defined as the electrostatic
force per unit test charge acting on 'a
vanishingly small positive test charge placed at
the given point. Mathematically,
~
E==lim~
qo~aqo
Electric field is a vector quantity whose
direction is same as that of the force exerted on a
positive test charge.
21.Units and dimensions of electric field. The51
unit of electric field is newton per coulomb
(NC-1) or volt per metre (Vm-1). The
dimensions of electric field are
[E]=Force=MLr2
Charge C
=MLr2 =[MLr3A-1]
ATwwwFnotesdriveFcom

89. ELECTRIC CHARGES AND
FIELD (Competition Section)
22.Electric field due to a point charge.The electric
field of a pointcharge qat distance rfrom it is
given by
1q
E=-- -
471: EO .r2
Ifqispositive, Epoints radially outwards and if
qisnegative, Epoints radially inwards. This
field is spherically symetric.
23.Electric field due to a system of point charges :
Superposition principle for electric fields.The
principlestates that the electric field at any
point due to a group of point charges is equal to
the vectorsum of the electric fields produced by
each charge individually at that point, when all
other chargesare assumed to be absent.
N
L
i= 1
qict-1;)
1r-1;13
24.Continuous charge distribution.When the
charge involved ismuchgreater than the charge
onan electron, we can ignore its quantum
nature andassume that thecharge is distributed
in a continuous manner. This is known as a
continuous charge distribution.
Volume charge density, p=!!iCm-3
dV
Surface charge density, c=dqCm-2
dS
Linearcharge density, A.=dqCm-1
dL
25.Electrostatic force and field due to a continuous
charge distribution.The total force on a charge
qodue to a continuous charge distribution is
given by
~
~ F
E=cont
cont q
. 0
=_1_[f£..~dV+f ~~dS+fL~ ~dL1
471:Eo v? 5r: r:
1.101
26.Electric field due to a general charge distribution.
It is given by
~ ~ ~
Etota!=Ediscreat +Econt
=_1_[fq~~+f£..;dV
471:EO i=1'i v?
+f?;dS+f ~; dL]
5 L
27.Electric dipole and dipole moment.An electric
dipole is a pair of equal and opposite charges
+qand -qseparated by some distance2a.Its
dipole momentis given by
p=Either chargexvector drawn from -qto+q
=qx2ii
Magnitude of dipole moment,p=qx2a
Dipole moment is a vector quantity having
direction along the dipole axis from-qto+q.lts
SI unit is coulomb metre(em).
28.Electric field at an axial point of a dipole.The
dipole field on the axis at distancerfromthe
centre is
1 2 pr _ 12p
Eaxia1=-- . 2 2 2 --- . - forr»a.
471:Eo (r-a 471:EOr3
At any axial point, the direction of dipole field is
along the direction of dipole momentp
29.Electric field at an equatorial point of a dipole.
The electric field at a point on the perpendicular
bisector of the dipole at distancerfrom its
centre is
_ 1 P _ 1P
Eequa ---. 2 23/2---. ::Iforr» a.
471:EO(r"+a ) 471: EO r:
At any equatorial point, the direction of dipole
field is antiparallel to the direction of dipole
~
momentp.
In contrast to 1/? dependence of the electric
field of a point charge, the dipole field has 1/r3
dependence. Moreover, the electric field due to
a short dipole at a certain distance along the axis
is twice the electric field at the same distance
along the equatorial line.
30.Torque on a dipole in a uniform electric field.
The torque on a dipole of momentpwhen placedwww9notesdrive9com

90. 1.102
in a uniform electric field at an angleewith it is
given by
1=pEsine
--t --t --t
In
vector rotation,1=PxE
When the dipole is released, the torquettends
--t
to align the dipole along the field E.
IfE=1 unit ande=90°, then 1=p.So dipole
moment may also be defined as the torque
acting on an electric dipole placed perpendi-
cular to a uniform electric field of unit strength.
31.Electric lines of force.An electric line of force
may be defined as the curve along which a
small positive charge would tend to move when
free to do so in an electric field and the tangent
to which at any point gives the direction of
electric field at that point.
32.Important properties of electric lines of force.
These are:
(i)Lines of force are continuous curves
without any breaks.
(ii)No two lines of force can cross each other.
(iii)They start at positive charges and end at
negative charges-they cannot form closed
loops.
(iv)The relative closeness of the lines of force
indicates the strength of electric field at
different points.
(v)They are always normal to the surface of a
conductor.
(vi)They have a tendency to contract length-
wise and expand laterally.
33.Electric flux.The electric flux through a given
area represents the total number of electric lines
of force passing normally through that area. If the
--t
electric field E makes an angleewith the normal
to the area elementsL\5,then the electric flux is
--t --t
L\4>r=EL\5cose=E .L\ 5
The electric flux through any surface 5, open or
closed, is equal to the surface integral ofEover
the surface 5.
<PE=fE.;is
5
Electric flux is a scalar quantity.
51 unit of electric flux=Nm2 C-1.
PHYSICS-XII
34.Gaussian surface.Any hypothetical closed
surface enclosing a charge is called the Gaussian
surface of that charge.
35.Gauss's theorem.The total flux of electric field
Ethrough a closed surface5is equal to 1/EO
--t
times the chargeqenclosed by the surface5.
--t--tq
<%=fE.d5=-
5 EO
36.Electric field of a line charge.The electric field
of a long straight wire of uniform linear charge
densityA,
E=_A_ .
t.e.,
21tEOr
1
Eoc-
r
whereris the perpendicular distance of the
wire from the observation point.
37.Electric field of an infinite plane sheet of charge.
It does not depend on the distance of the
observation point from the plane sheet.
E=~
2EO
wherec=uniform surface charge density.
38.Electric field of two positively charged parallel
plates.If the two plates have surface charge
densities°1and°2such that°1>°2>0, then
E=±_1_(°1+°2)(Outside the plates)
2EO
1
E=-(01-02) (Inside the plates)
2EO
39.Electric field of two equally and oppositely
charged parallel plates.If the two plates have
surface charge densities±0,then
E=O (For outside points)
(For inside points)
40.Electric field of a thin spherical shell.If R is the
radius and0,the surface charge density of the
shell, then
E __ 1_.i.
- 41tEo.?
Forr»R (Outside points)wwwFnotesdriveFcom

91. ELECTRIC CHARGES AND FIELD(Competition Section)
E=0 Forr<R (Inside points)
E=_1_ ---.i.. Forr=R
(At the surface)
47tEO'R2
whereq=47tR2c
41.Electric field of a uniformly charged solid
sphere.Ifpis the uniform volume charge
density and R radius of the sphere, then
E __1_!L
- 47tEo•~
E __1_ ~
- 47tEo .R3
E __ 1_---.i..
- 47tEo .R2
whereq=.!7tR3P
3
1.103
Forr»R (Outside points)
Forr<R (Inside points)
Forr=R (At the surface)www9notesdrive9com

92. ELECTROSTATIC POTENTIAL
AND CAPACITANCE
C H A PT E R
2.1ELECTROSTATIC POTENTIAL
AND
POTENTIAL DIFFERENCE
Introduction.The electric field around a charge can
be described in two w s:
-t
(i)by electric field (E),and
(ii)by electrostatic or electric potential(V).
-t
The electric field E is a vector quantity, while
electric potential is a scalar quantity. Both of these
quantities are the ~haracteristic properties of any point
ina field and are inter-related.
1. Develop the concepts of potential difference and
electric potential. State and define their51units.
Potential difference.As shown in Fig. 2.1, consider
a point charge +qlocated at a point O. LetAand Bbe
two points in its electric field. When a test charge qois
moved fromAto B, a workWABh tobe done in
moving ainst the repulsive force exerted by the or
Source
charge
+q
•
Test
charge
+qo
••••
o B A
Fig. 2.1To define potential difference.
charge+q.We then calculate the potential difference
between points A and Bby the equation:
W
V-v -V-~
-B A-
qo
...(2.1)
So thepotential difference between two points in an
electric field may be defined as the amount of work done in
moving a unit positive charge from one point to the other
against the electrostatic forces.
In the above definition, we have sumed that the
test charge is so small that it does not disturb the
distribution of the source charge. Secondly, we just
apply so much external force on the test charge that it
just balances the repulsive electric force on it and hence
does not produce any acceleration in it.
SI unit of potential difference is volt (V).It h been
named after the Italian scientistAlessandro Volta.
11 1joule
vot=--"---
1 coulomb
1V=1Nm CI =1JCI
Hence thepotential differencebetween two points
in an electric field is said to be 1voltifljoule of work has
to be done in moving a positive charge of 1coulomb from one
point to the other against the electrostatic forces.
Electric potential. The electric potential at a point
located far from a charge is ten tobezero.
(2.1)wwwCnotesdriveCcom

93. 2.2
In Fig. 2.1
,ifthe point Alies at infinity, thenVA=0, so
that
w
V= VB=-
qo
where Wis the amount of work done in moving the
test chargeqofrom infinity to the point BandVBrefers
to the potential at pointB.
Sotheelectric potential at a point in anelectricfield is
the amount of work done inmoving a unit positive charge
from infinity to that point against the electrostatic forces.
. .Work done
Electricpotential =----
Charge
SI unit of electric potential is volt(V).Theelectric
potential at a point in an electric fieldissaidtobe1voltif
one joule of work has to bedone in moving a positive charge
of1coulomb from infinity to that point against the
electrostatic forces.
2.2 ELECTRIC POTENTIAL DUE TO
APOINT CHARGE
2.Derive an expression for the electric potential at a
distance r from a point charge q.Whatisthe nature of
thispotential?
Electric potential due to a point charge. Consider a
positive pointchargeqplaced at the origin O. We wish
to calculate its electric potential at a point P at distance
rfrom it, shown in Fig. 2.2.By definition, the electric
potential at point P will be equal to the amount of work
done in bringing a unit positive charge from infinity to
the point P.
q ~ qo F
(9~--------•• ----~.~--~.~--~.__ ----oo
o P
'1+--- r-----+t
B A
14 x-----.t~1
Fig. 2.2Electric potential due to a point charge.
Suppose a test chargeqois placed at pointAat
distancexfrom O. By Coulomb's l, the electrostatic
force acting on chargeqois
F __ 1_qqo
-4rcEO•x2
~
The force F acts from the chargeq.Thesmall
work done in moving the test charge qofromAto B
~
through small displacementdxainst the electro-
static force is
~~
dW=F .dx=Fdxcos 1800 =-Fdx
PHYSICS-XII
r r
W=fdW=-fFdx=-fl. qqodx
00 co4rcEOx2
= _~fX-2dx=_~[_.!.]r
4rcEOco 4rcEO Xco
_~[~_~] __1_q%
-4rcEOr00 -4rcEO·r .
Hence thework done in moving a unit test charge
from infinity tothe point P, or the electric potential at
point Pis
V=W or V=_l_.2.
qo 4rcEOr
Clearly, Vcc1/r. Thus the electric potential dueto a
point charge isspherically symmetric it depends only
on the distance of the observation point from the
charge and not on the direction of that point with
respect to the point charge. Moreover, wenotethat the
potential at infinity is zero.
Figure 2.3 shows the variation of electrostatic
potential(Vcc1/r) and the electrostatic field (Eo;1/1)
with distance rfrom a charge q.
5~-w--r--.---~--r-~--~---r--'---'
4.5
4
3.5
3
t2.5
t<l
2:::':
1.5
I
0.5
00 0.5
V=_l_. !!..
41[1:0 r
L5 2 2.5 3 3.5 4 4.5 5
r~
Fig.2.3Variation of potentialVand fieldE
withrfrom a point charge q.
2.3 ELECTRIC POTENTIAL DUE TO A DIPOLE
3.Derivean expression for the potential at a point
alongthe axial line of a short dipole.
Electric potential at an axial point of a dipole.As
shown in Fig. 2.4,consider an electric dipole consisting
of two point charges -qand+qand separated by
distance 2a.Let P be a point on the axis of the dipole at
a distance rfrom its centre0.
-q +q
e======t:1 ===:::::::emummu-l
A 0 B P
I+--- a-~~1"4-- a-----+I
The total work done in movingthe chargeqofrom 14 ~I
infinity to the point P will be Fig.2.4Potential at an axial point of a dipole.···3wx«n£m~r±n3lxv

94. ELECTROSTATI
C POTENTIAL ANDCAPACITANCE
Electric potential at pointPdue to the dipole is
1-q 1q
V=V +V =--.-+--.-
1 2 41t EO AP 41t EOBP
1 q 1 q
=---.--+--.--
41t~r+a 41t~ r-a
__ q[_1 __1]
41t EO r - ar+a
=_q_[(r+a)-(r-a)]=_l_ qx2a
2 2 . 2 2
41t EO r-a 41t EO r:-a
or
V- 1 P
---'--2
41t EO?-a
[.: p=qx2a]
For a short dipole,a2< <?,soV=_1_E
41tEO.?.
4.Show mathematically that the potential at a point
on the equatorial line of anelectricdipole iszero. or
Electric potential at an equatorial point of a dipole.
As shown in Fig. 2.5, consider an electric dipole
consisting of charges-qand+qand separated by or
distance2a.Let P be a point on the perpendicular
bisector of the dipole at distance rfrom its centre O.
p
Fig. 2.5Potential at an equatorial point of a dipole.
Electric potential at point P due to the dipole is
1 -q 1 q
V=V +V =--.-+--.-
1 2 41t EO AP 41t EO BP
=__1_ q+_1_ q=0
41t EO.~?+a2 41t EO.~?+a2 .
5. Derive an expression for the electric potential at
any general point at distance rfrom the centre of a dipole.
Electric potential at any general point due to a
dipole.Consider an electric dipole consisting of two
point charges -qand+qand separated by distance 2a,
shown in Fig. 2.6. We wish to determine the potential
at a point P at a distancerfrom the centre0,the direc-
~
tion OP ming an angleewith dipole momentp.
2.3
LetAP=r1and BP=r2. P
Net potential at pointPdue to the dipole is ,":
" I
V=V1+V2 rl,'" /
=_1_.-q+_1_.!1. ""/r2
41t EO r1 41tEOr2 ,,' /
,rv! I
q[1 1] ~.: "v" :
=41tEO ~-~ ",Le···-./
A B
[ ] -q .0 +q
-41tQEo \~2r2 I+-- 2a--.t
Fig. 2.6
If the pointPlies far away from the dipole, then
r1-'2'"ABcose=2acoseand'lr2"'?
V=-q- 2acose
41tEO.?
V-1pcose
-41tEo.-?-
--t ~ ~ 1\
V- 1p.r_1p.r
- 41tEo--r- 41tEo ~
Herep=qx2a,isthe dipole moment and
"~
'='/',
~ ~
is a unit vector along the position vector OP=r .
Special Ces
(i)When the point P lies on theaxial lineof the
dipole,e=0°or 180°, and
1 p
V=+-- -
- 41t EO.?
i.e., the potential h greatest positive or the
greatest negative value.
(ii)When the point P lies on theequatorial lineof the
dipole,e=90°, and V=0,i.e.,the potential at
anypoint on theequatorial line of the dipole is
zero. However, the electric field at such points
is non-zero.
6. Give thecontrasting features ofelectric potential of
a dipole from that due to a single charge.
Differences between electric potentials of a dipole
and a single charge.
1. The potential due to a dipole depends not only
on distancerbut also on the angle between the position
~
vector, of the observation point and the dipole moment
~
vectorp.The potential due to a single charge depends
onlyonr.
2. The potential due to a dipoleislindrically symmetric
about the dipole axis. If we rotate the observation point···3wx«n£m~r±n3lxv

95. 2.4
Pabo
utthe dipole axis (keeping rand 8 fixed), the
potential Vdoes not change.The potential due to a
single charge isspherically symmetric.
3.Atlarge distance, the dipole potential falls off
1/?while the potentialdue to a single charge falls off
1/r.
2.4ELECTRIC POTENTIAL DUE TO
ASYSTEM OF CHARGES
7.Derive an expression for the electric potential at a
pointdueto a group ofNpoint charges.
Electric potential due to a group of point charges.
As shown in Fig. 2.7,supposeNpointcharges
ql' q2' q3' ..... , qN lie atdistancesr1,r2,r3, ·····"N from a
point P.
p~~----------~--------------~q4
Fig. 2.7Potential at a point due to a
system ofNpoint charges.
Electric potential at point P due to charge qlis
V=_1_!ll
1 .
41tEOr1
Similarly, electric potentials at point Pdue to other
charges will be
_1q2 _1q3 _1qN
V2---.-, V3---·-, ...,VN---.-
41t EOr2 41tEor3 41t EOrN
Aselectric potential is a scalar quantity, so thetotal
potential at point P will be equal tothe algebraic sum
of all the individual potentials, i.e.,
V=VI+V2+V3+ ...+VN
=_I_[ql +q2+q3+ ...+qN]
41t EO11'2r3 rN
V=_I_ ~ qi
41tEo i=1 ';
~ ~ ~ ~
Ifr1,r2,r3,...,rNare the position vectors of theN
or
pointcharges, the electric potential at a point whose
~
position vector isr ,would be
V=_I_ ~ qi
41tEOi=117_~1
or
PHYSICS-XII
2.5ELECTRIC POTENTIAL DUE TO A
CONTINUOUS CHARGE DISTRIBUTION
8. Deduce an expression for the potential at a point
dueto acontinuous chargedistribution. Hence writethe
expression for the electric potential due to a general source.
Electric potential due to a continuous charge
distribution.We can imine that a continuous charge
distribution consists of a number of small charge
~ ~
elements located at positions ~.Ifris the position
vector of pointP,then the electric potential at point P
due to the continuous charge distribution can be
written as
V- 1fdq
41tE ~ ~
o 1r - '; 1
Whenthe charge is distributed continuously in a
volumeV,dq=PdV,wherepis volumecharge density.
The potential at point P due to the volume charge
distribution will be
V_I fpdV
v -41tE ~ ~
oVlr-~1
When the charge is distributed continuously over
an area5,dq=odSwhereoissurface charge density.
V_IfcrdS
5- 41tE ~ ~
oSir-';1
When the charge is distributed uniformly along a
lineL,dq='AdL,whereAis line charge density.
V_IfAdL
L - 41tE ~ ~
. oLlr-~1
The net potential at the point P due to the conti-
nuous charge distribution will be the algebraic sum of
theabove potentials.
V::ont=Vv+Vs+VL
orV=_l_[J~+J ~+J~l
cont4 ---+ ---+- ~ ~ ~-t
11:EOvir-rjlsir-rjl LIr -rjl
Electricpotential due to a general source. The
potential due to a general source charge distribution,
which consists of continuous well discrete point
charges, can be written
V=V::ont+Vdiscrete
V=__1__[f pdV+fcrdS
41tEOv17 - ~1517 - ~1
+fAdL " qi1
~~+L.. ~~
L1r-';1Allpoint1r-~1
charges···3wx«n£m~r±n3lxv

96. ELECTROSTATIC P
OTENTIAL AND CAPACITANCE
2.6ELECTRIC POTENTIAL DUE TO A
UNIFORMLY CHARGED THIN
SPHERICAL SHELL
9. Writeexpression for the electric potential due to a
uniformly charged spherical shellat a point (i)outside the
shell, (ii)on the shell and (iii)inside theshell.
Electric potential due to uniformly charged thin
spherical shell. Consider a uniformly charged spherical
shell of radius Rand carrying charge q.We wish to
calculate its potential at point P at distance rfrom its
centre 0, as shown in Fig. 2.8.
p
v
V=_l_.~
~ __4rc.•.•EoRShellwith
charge Q
Fig. 2.8Potential due to
a spherical shell.
Fig. 2.9Variation of potential
due to charged shell with distance
Tfromits centre.
(i) Whenthe point Plies outside the shell. We know
that for a uniformly charged spherical shell, the electric
field outside the shell is if the entirecharge is
concentrated at the centre. Hence electric potential at
an outside point is equal tothat ofa point charge
located at the centre, whichisgivenby
V=_I_!i [Forr>R]
4m;0r
(ii)When pointPlies on the surface of the shell. Here
r=R.Hence the potential on thesurface of the shell is
V=_I_!L [Forr=R]
4m;0 R
(iii)When point Plies inside the shell. The electric field
at any pointinside the shell is zero. Hence electric
potential due to a uniformly charged spherical shell is
constant everywhere inside the shell anditsvalue is
equal to that on the surface. Thus,
V=_I_!L [Forr«R]
4m,0 R
Figure 2.9 shows the variation of the potential V
due toauniformly charged sphericalshell with
distancermeured from the centre of the shell. Note
thatVisconstant (=q/4m;oR) from r=0 tor= R along
a horizontal line and thereafter VIX1/rfor points
outside the shell.
2.5
r>Electric ;::::~r~:~'O:::::hil'pot~ti.~
gradient is a vector quantity.
~The electric potential near anisolated positive charge is
positive because work h to be done by an external
ent to push a positive charge in, from infinity.
~The electric potential near anisolated negative charge is
negative because the positive test charge is attracted by
the negativecharge.
~The electric potential due to a chargeqat its own
location is not defined - it is infinite.
~Because of arbitrary choice of the reference point, the
electric potential at a point is arbitrary to within an
additive constant: But it is immaterial because it is the
potential difference between two points which is
physically significant.
~For defining electric potential at any point, generally a
point far ay from the source charges is ten the
reference point. Such a point is assumed to be at infinity.
~As the electrostatic force is a conservative force, so the
work done in moving a unit positivecharge from one
point to another or the potential difference between
two points does not depend on the path along which
the test charge is moved. )
Examples basedon
..
Formulae Used
Work done W
1. Potential difference= orV= -
Charge q
2. Electric potential due to a point chargeqat
distancerfrom it,
V=_1_.1
4rcEOr
3. Electric potential at a point due to N point charges,
V=_1_ ~ !iL
4rcEOi=11j
4. Electric potential at a point due to a dipole,
~
V_I pcose_1P.r
- 4rcEO-r-2- -4rcEO ~
Units Used
Chargeqis in coulomb, distancerin metre, work
done W in joule and potential differenceVin volt.
Example1.lfl00Jof work has tobe done in moving an
electric charge of4Cfrom a place, where potentialis-10V
toanother place, where potentialisVvolt,find the value
ofV.···3wx«n£m~r±n3lxv

97. 2.6
Solution.HereWAB= 100J,qo=
4CVA=-10 V,
VB=V,
As v:-V - WAB
B A-
qo
V-(-10)= 100 =25
4
V=25-10=15V.or
Example 2. Determine the electric potential at thesurface
of a gold nucleus.Theradiusis 6.6 x10-15m and the atomic
numberZ=79.Given charge on a proton=1.6x1O-19c.
[Himachal 96)
Solution.Asnucleus is spherical, itbehaves like a
point charge for externalpoints.
Here q=ne=79 x 1.6 x10-19C,
r=6.6x10-15m
1q9 x 109 x79x1.6x10-19
..V=--.-= V
4nEor 6.6x10-15
= 1.7x 107V.
Example 3.(i)Callate the potential at a point Pdue to a
charge of4x10-7 Clocated9em away.(ii)Henceobtain
the work done in bringing a charge of 2x10-9 Cfrom
infinitytothe pointP.Does the answer depend on the path
along which the charge is brought ? [NCERT)
Solution. (i)Hereq= 4 x10-7C,r=9 em = 0.09 m
Electric potential at point P is
V=_1_ .!I= 9x109x4x10-7=4x104V.
4nEOr 0.09
-7
q=4x10 C
••------- ••- - - - - - - - - --00
o P
'4 9 m---+I.'
Fig. 2.10
(ii)Bydefinition, electric potential at point Pis
equalto the work done in bringing a unitpositive
charge from infinity to the point P.Hence the
workdone in bringing a charge of 2 x10-9Cfrom
infinity to the pointP is
W=qo V=2x10-9x4x104=8x10-5J
No,the answer does notdepend on the path along
which the charge is brought.
Example 4. Ametalwireisbentin a rcle of radius 10em
Itisgivena charge of200IlCwhich spreads on it uniformly.
Callate the electric potential at its centre.
[CBSE OD 9SC)
Solution. Here q=200IlC= 2x10-4C
r=10em =0.10 m
PHYSICS-XII
We can consider the circular wire to be made of a
large number of elementary charges dq.Potential due
to one such elementary chargedqat the centre,
dV=_l_. dq
4nEOr
Total potential at the centre of the circular wire,
V= LdV=L-1-. dq =_l_Ldq
4nEOr4nEOr
1q9x109x2x10-4 6
. -= =18x10V.
4nEOr 0.10
Example 5. Electric field intensity at point'B'due to a
pointcharge'Q'kept at point'A'is 24 NC-1and the
electric potential at point'B'due to same chargeis12fC-1.
Callate the distanceABand also the magnitude of charge
Q. [CBSE OD 03C)
Solution.Electric field of a pointcharge,
E=_l_. Q =24NC-1
4nEO?-
Electric potential of a point charge,
V=_1_. Q ~ 12 JC-1
4nEOr
Thedistance ABisgiven by
V12
r=-=-=O.5m
E24
The mnitude of the charge,
Q=4nEOVr= _1-9 x 12x0.5 =0.667 x10-9C
9x10
Example 6. To what potential wemustcharge an insulated
sphere ofradius14emsothatthesurface charge densityis
equal to IIICm'2?
Solution.Here r =14 cm=14x10-2m,
c=IIICm-2=10-6 Cm-2
1q 14n?-cr 1
..V=--.-=--.--=--.4nrcr
4nEo r4nEo r 4nEo
=9x109x4x22x14x10-2x10-6V
7
=15840 V.
Example 7.A charge of241lCisgiven to ahollow metallic
sphere of radius0.2m Find the potential [CBSE D 95)
(i)at the surface of the sphere, and
(ii)at a distance of0.1emfrom the centre of the sphere.
Solution. (i)q =241lC=24 x 10-6 C,R=0.2 m
Potential at the surface of the sphere is
V=_1_ .!L= 9x109x24x10--6 V=1.08x106V.
4nEoR 0.2~~~1styjxiwn}j1htr

98. ELECTROSTATIC POTE
NTIAL AND CAPACITANCE
(ii)As potential at any point inside the sphere
=Potential on the surface
:.Potential at adistance of 0.1 em from the centre
=1.08x106v.
Example8.Twenty seven drops of same size are charged at
220Veach. They coalesce toform a bigger drop. Callate
thepotential of the bigger drop. [Punjab 01)
Solution. Let radius of each small drop=r
Radius of large drop=R
Theni1tR3=27xi1tr3
3 3
or R=3r
Potentialof each small drop,
V=_l_.!1.
41tEOr
:.Totalcharge on 27 drops,
Q=27q=27X41tEOrV
Potential of large drop,
V'=_1_.Q=_1_ ._27_x_4_1t-,Eo,,-r_V_
41tEOR 41tEO 3r
=9V=9x220 =1980 V.
Example9.Two charges 3x10-8Cand -2x10-8Care
located 15emapart.Atwhat point on the line joining the
two charges is the electric potential zero? Take the potential
at infinity tobe zero. [CERT)
Solution. As shown in Fig. 2.11, suppose thetwo
point charges areplacedonX-axis with the positive
charge located on the origin0.
-8 -8
q}=3x10C q2=-2 x10 C
o-----------+--------~o
o P A
I--- X .,.0.15 -x---+l
Fig. 2.11 Zero of electric potential for two charges.
Let the potential be zero at the pointPandOP=x.
Forx<0(i.e.,tothe left of0),the potentials of the two
charges cannot add up to zero.Clearly, xmust be
positive.Ifxlies between 0andAthen
VI+V2=0
1[ql+q2 ]-0
41tEOx 0.15-x
or 9X109[3X10-8 _2XlO-8]=0
x 0.15-x
3 2
or -----=0
x0.15-x
which givesx=0.09 m=9em
2.7
The other possibility is thatxmayalso lie onOA
produced, shown in Fig. 2.12.
-8 -8
q}=3x10 C Q2=-2xlO C
o--------------------~o~------------~I
o A P
I.....: 0._15__- -_-_-_-x-~'Io-, --x - 0.15=1
Fig. 2.12
As VI+V2=0
.. _1_ [3X 10-8 _2x10-8] =0
41tEO X x-0.15
which givesx=0.45 m=45em
Thus the electric potential is zero at9ern and 45 em
ay from the positive charge on the side of the
negative charge.
Example 10. Calculate the electric potential at the centre
of asquare ofside.J2m, having charges 100J.lc,-50J.lc,
20J.lc,and -60J.lCat thefour corners of the square.
[CBSE OD 06C]
Solution. Dional of the square
=~(.J2)2+(.J2)2=2 m
Distance of each charge from the centre ofthe
square is
r=Half dional=1m
.,Potential at the centre of the square is
V=_l_[ql +q2+q3+q4]
41t EO r r rr
V= 9x109[100x10-6 50x10-6
1 1
20x10-6 60x10-6]
+ ----
I 1
=9x109x10-6 x10=9x104v.
Example 11. Fourcharges+q,+q,- qand-qareplaced
respectively atthe corners A, B,CandDof asquare of side
'a'arranged inthe given order.Callate the electric
potential at the centre o.If
EandFarethemidpoints
ofsidesBCandCDrespec- a/2
tively, what will be the
workdoneincarrying a
charge 'e'from0toEand a/2
from0toF?
Solution. Let OA=
OB=OC=OD=r.
Fig. 2.13···3wx«n£m~r±n3lxv

99. 2.8
Then th
epotential atthe centre 0is
V=_1_[!l.+!. _ !. _ !l.] =0
o4TCEO r r r r
Again, thepotential at pointEis
1[q q q q]
VE=4TCEO AE+BE-CE-DE=0
[.:AE=DE,BE=CEl
Now, AF=BF=)a2+(~r=~a
.. The potential at point Fis
1[q q q q]
VF=4TCEO AF+BF - CF - DF
-.ss.[_1__.2..][.:AF=BF,CF=DFl
4TCEO AF CF
2q[2 2] q(1)
=4TCEO .Jsa--;;=TCEOa.Js-1
Work done in moving the charge' e'from 0toEis
W=e[VE - Vol =ex0=0
Work done in moving the charge 'e'from 0toFis
W=e[VF -vol=e[-q (~-1)-0]
TCEoa,,5
=TC::aCk-1).
Example 12. A short electric dipole has dipole moment of
4x10-9CmDetermine the electric potential due to the
dipole at a point distant0.3mfrom the centre of thedipole
situated (a) on the axial line (b) on equatorial line and (c)on
a linemaking an angle of60°withthe dipole axis.
Solution. Here p= 4x10-9Cm, r= 0.3 m.
(a)Potential at a point on the axial line is
1p9x109x4x10-9
V=--.-= =400 V.
41tEo? (0.3)2
(b)Potential at a point on the equatorial line = o.
(c)Potential at apointona line that mes an angle
of 60° with dipole axis is
V __1_pcosS
- 4TCEO• ?
9x109x4x10-9cos 60°
2 =200 V.
(0.3)
Example 13. Twopointcharges of +3~C and-3~Care
placed2x10-3mapari fromeach other. Calculate (i) electric
field and electric potential at adistance of0.6m from the
PHYSICS-XII
dipole in broad-side-on position (ii)electric field andelectric
potential at the same point after rotating thedipolethrough
90°.
Solution. Dipole moment,
p=qx21=3x10-6x2x10-3=6x10-9Cm
(i)Electric field in broad-side-on position is
E= _1_. E=9x109x6x10-9=250 Net
4TCEOr3 (0.6)3
Electric potentialin broad-side-on position, V=O.
(ii)When the dipole is rotated through 90°, the
same point is now in end-on-position with respect to
thedipole.
E=_l_. 2p =500Net
4TCEOr3
1p9x109x6x10-9
V=--.-= =150V.
4TCEO? (0.6l
Example 14. Two charges -qand +qarelocatedat points
A(O,O,-a) and B(O,O,+a) respectively. How much workis
done in moving a test charge from point P(7,O,0) to
Q(-3,0,0) ? [CBSE 009)
Solution. Points P andQare located on the equa-
torial line of the electric dipole and potential of the
dipole at any equatorial point is zero.
:. Work done in moving a testchargeqofromPtoQ,
W=qO(VQ- Vp)=qo(O-O)=O.
-q
B(O,0,a)
Q(-3,0,0)
X
P(7,0, 0)
+q
~--~~------~-------'----.Z
A(O,0, -a)
y
Fig. 2.14
r-problems For Practice
1.The work done in moving a charge of3 Cbetween
two points is 6J.What is the potential difference
between the two points? (Ans.2V)
2.The electric potential at 0.9m from a point charge is
+50 V.Whatis the mnitude and sign of the
charge? [CBSE D 95C]
(Ans.5x10-9C,positive)···3wx«n£m~r±n3lxv

100. ELECTROSTATIC POTENT
IAL AND CAPACITANCE
3.The electric field at a point due to a point charge is
20 NC-1 and the electric potential at that point is
10 JC-1. Calculate the distance of the point from the
charge and the mnitude of thecharge.
[CBSE D06]
(Ans. 0.5 m, 0.55x 1O-9q
4.Two pointsAand B are located in diametrically
opposite directions of apointcharge of+2 J.lCat
distances 2.0 m and 1.0 m respectively from it.
Determine the potential difference VA-VB'
(Ans.- 9x103V)
5.A hollow metal sphere is charged with 0.4 J.lCof
charge and h a radius of 0.1 m.Find the potential
(i)at the surface(il)inside thesphere(iil)at a
distance of 0.6 m from the centre. The sphere is
placed in air. (Ans.36 kV, 36 kV, 6 kV)
6.Two point charges of+10 J.lC and +20 J.lC are
placed in free space 2 cm apart. Find the electric
potential atthe middle point of the line joining the
two charges. (Ans.27 MV)
7.Two point chargesqand-2qare kept 'd'distance
apart. Find the location of the point relative to
charge 'q'at which potential due to this system of
charges is zero. [CBSE OD14C]
(Ans. At distance d /3 from chargeq)
8.Twopoint charges, one of +100 J.lCand another of
- 400 J.lC, are kept 30 cm apart. Find the points of
zero potential on the line joining the twocharges
(sume the potential at infinity to be zero).
(Ans.6 cm from+100 J.lCcharge)
9.A charge q=+1J.lCisheld at0between the points
Aand B such that AO=2 mand80=Irn,as shown
in Fig. 2.15(a). Calculate thepotential difference
(VA - VB)'What will be thevalue ofthe potential
difference (VA - VB)if position of B is changed
shown in Fig.2.15(b)? (Ans. - 4500 V,-4500 V)
q
• • •
B 1m 0 2m A
(a)
B
1m
q
0 2m A
(b)
Fig. 2.15
10.Twosmall spheres ofradius 'a'each carrying
charges+qand-qare placed at pointsAandB,
distance'd' apart. Calculate the potential difference
between points Aand B. (Ans.2q/4TtEOd)
2.9
11.The sides of rectangle ABCD are 15 cm and 5 ern,
shown in Fig. 2.16. Point charges of - 5 J.lC and
+2J.1Care placed at the vertices Band D respec-
tively. Calculate electric potentials at the vertices A
and C. Alsocalculate the work done in carrying a
charge of 3 J.lCfrom Ato C. (Ans.2.52 J)
A 0
Fig. 2.16
12.Charges of 2.0 x10-6C and1.0x10-6C are placed at
the comersAand B of a square of side 5.0 cm
shown in Fig. 2.17.How much work will be done in
moving a charge of 1.0 x1O-6C from C to D ainst
the electricfield? (Ans.0.053 J)
o 5cm C
,,0:,
A B
Fig. 2.17
13.Calculate the potential atthe centre of a square
ABCDof each side.fim due to charges 2, -2,- 3
and 6 J.lCat four comers of it. [Haryana97]
(Ans.2.7x104V)
14.Charges of +1.0x10-11C, - 2.0 x10-11 C,
+1.0x10-11C are placed respectively at the comers
B,C and D of a rectangleABCD. Determine the
potentialat the comerA.GivenAB=4 em and
BC=3cm.
(Ans.1.65 V)
15.ABCDis asquare of side 0.2 m. Charges of 2 x 10-9,
4x10-9, 8 x10-9C are placed at the comers A,B
and C respectively. Calculate the work required to
transfer a charge of 2x10-9CfromD to the centre a
of the square. [Kamata 88]
(Ans. 6.27x10-7J)
16.Positivecharges of 6, 12 and 24 nC are placed atthe
three vertices of a square. Whatcharge must be
placed at the fourth vertex sothat total potential at
the centre of the square is zero? (Ans. -42nq
17.Two equal charges, 2.0x10-7C each are held fixed
at a separation of 20 cm. A third charge of equal
mnitudeis placed midway between the two
charges. It is now moved to a point 20 ern from both
the charges. How much work is done by the electric
field during the process? (Ans. -3.6x10-3J)···3wx«n£m~r±n3lxv

101. 2.10
18.ABCi
s a right-angled triangle, where ABand BC
are 25 em and60 em respectively; a metal sphere of
2emradius charged to a P9tential of 9x105Vis
placed at B.Find the amount of work done in
carrying a positive charge of 1 C from C to A.
A (Ans.0.042J)
~c
Fig. 2.18
HINTS
w6J
1.V=-=-=2V.
q3C
2.As V= _1_}. . . 50=9x109x-.!L.
41tEor 0.9
_50x0.9_5 lO-9C
orq----9 - x
9x10
Asthe potential ispositive, the chargeqmust be
positive.
3.Electric fieldofapointcharge,
E=_1_ .!L=20NCI
41tEor2
Electric potential of a point charge,
V=_1_ ..1= 10JCI
41tEor
Clearly, distancer=V= 10=0.5 m
E20
Mnitude of charge,
10x0.5 -9
q=41tEo' V.r=---9 =0.55x10 C.
9x10
4.Here' q=2f,lC=2x10-6C,rA=2m,rB=lm
V -V=-q-[~_1.]
A B 41tEo rA rB
= 2x10-6x9x109[~-~JV
=-9x103V.
5.(i)Potential at the surface,
1q4x10-7 x9x109
V=--.-=-----
41tEOr 0.1
= 36000 V=36 kV.
(ir)Potential inside a hollow conductor isthe same
ason its surface.
(iii)When r=0.6 m,
9x109x4x10-7
V= =6000 V=6 kV.
0.6
PHYSICS-XII
6.V=t'J+V2=_1_ [ql+q2 ]
41tEO '1r2
9[10x10-6 20x10-6 ]
=9x10 + --:--:- _
0.01 0.01
=27x106V=27 MV.
7.Let the point P of zero potential lie at distance x
fromthe chargeq.
q x d-x -2q
o~--~I--~~--------~o
A P B
.._1_.9 ..+_1_.(-2q)=0 or.! =_2_ orx=:!..
41tEo X41tEo(d -x) x d -x 3
8.Suppose the point of zero potential is located at
distance xmetre from the charge of + 100 uC, Then
V= _1_ [100 x10-6 _400x10-6] = 0.
41tEo x 0.30 -x
This givesx= 0.06 m = 6 emi.e.,the point of zero
potential lies at 6 emfromthe charge of+100f,lc.
VV_ 9 [1.0x10-6 1.0x10-6]
9. A-B-9x10 - ---
2.0 1.0
=-4500V
As potential is a scalar quantity, so change in position
of the charge does not affectthe value of potential.
10.VB-VA=_1_.~ __l_.=.!L-.-3L
41teod41tEo d41teod
[2 10-6 5 10-6]
11.VA= 9x109 x - x =-7.8x105V
0.15 0.05
[2x10-6 5x10-6]
Ve= 9x109 - = 0.6x105 V
0.05 0.15
W=q(Ve -VA)= 3.0 x10-6(0.6x105+7.8x105)
=2.52J.
12.V -_1_ [~ +.31...J
e- 41tEo AC BC
9 9[2.0x10- 61.0x10-6 ]
=x10 + -----
-Iix0.05 0.05
=9000[ 2+.fi]V
.fix0.05
V 1[ql q2J
D=47tEOAD+BD
. 0:9[2.0x10-6 1.0x10-6]
=9x1 +-;=--
0.05.fix0.05
=9000 [2.fi+1]V
.fix0.05«««2uv{lzkyp~l2jvt

102. ELECTROSTATIC POTENTIAL AND CAPACITANCE
W=q(Vo - Vc)
= 1
.0x10-6x9000[2..fi+1 - 2 -..fi]
-Iix0.05
=0.053J.
13.Dional of the square = ~(..fi)2+(..fi)2= 2 m
Distance of each charge from the centre,
r= Half dional=1 m
:. Potential at the centre of the square is
9[2x10-6 2x10-6 3x10-6 6x10-6]
V=9x10 -----------+---
1 1 1 1
=2.7x104V.
14.AC= ~42+32= 5crn = 0.05 m,AD=BC= 0.03 m
1 [ 1.0 x 10-11 2.0x10-11 1.0x10-11 ]
V=-- - +----
41teo 0.04 0.05 0.03
=1.65V.
9[2x10-9 4x10-9 8x10-9 ]
15.V0=9x10 + + ---
0.2 0.2 Ji 0.2
= 577.26 V
9[2x10-9 4x10-9 8x10-9 ]
V=9xlO + +----r~
o 0.1..fi 0.1..fi 0.1..fi
= 890.82 V
W=q[V0 -VoJ= 2x10-9[890.82 - 577.26]
=6.27x10-7J.
16.Suppose a charge ofqnC be placed at the fourth
vertex.
Let length of half dional bexmetre.
9 [6x10-9 12x10-9 24x10-9
Vo=9x10 + + ---
x x x
qx10-9]
+ =0
x
~ +12+24+3..=0
x x x x
3..= _42
x x
or q=-42 nC.
17.The situation is shown in Fig. 2.19.
" 1 [ 2x10-7 2 x 1O~7]"
V-V =-- +---
C 041tto0.20' 0.20
1 [2x10-7 2x10-7]
- 41tto0.10+0.10
= -1.8 x10-4V
or
or
2.11
c
Fig. 2.19
W=q(Vc - Vo)
= -2x10-7 x1.8x104= - 3.6x10-3J.
18.Potential of the charged sphere is
V=_l_.!1.
41tEOr
. .9x105 = 9x109x-q-
0.02
0.02 2 -6 C
q=104=x10 = 21-1
Potential atAdue to chargeqis
1q9x109x2x10-6
VA=--.-= V
41tEOr 0.25
or
Potential atCdue to chargeqis
9x109x2x10-6
Vc= V
0.60
Potential difference betweenAandCis
VA- Vc= 1.8 x10-3[_1 1_] V
0.25 0.60
1.8x7
=--V = 0.042V
300
Work done in moving a charge of+1C fromCtoA
W=q(VA - Vc) = 1x0.042 =0.042J.
2.7RELATION BETWEEN ELECTRIC FIELD
AND POTENTIAL
10. Show that the electric field at any pointisequal to
the negative of the potential gradient at that point.
Computing electric field from electric potential. As
shown in Fig. 2.20, consider the electric field due to
charge+qlocated at the originO.LetAandBbe two
adjacent points separated by distancedr.The two
+q
•
a B A
Fig. 2.20 Relation between potential and field.···3wx«n£m~r±n3lxv

103. 2.12
~
points are
so close that electric fieldEbetween them
remains almost constant. LetVandV+dVbe the or
potentials at thetwo points.
The external force required to move the testcharge
~
qo(without acceleration) ainst the electricfieldEis
given by
~ ~
F=-qoE
111ework done to move the testcharge fromAtoBis
W=F .dr= -qoE.dr
Also, the work in moving the test charge from Ato
Bis
W=Chargexpotential difference
= qo(VB-VA)=qo dV
Equating the two works done, we get
-qoE .dr=qo.dV
E=- dV
dr
The quantitydVis the rateof changeofpotential
dr
with distance and is called potential gradient. Thusthe
electric field at any pointisequal to the negative of the
potential gradient at that point. The negative sign shows
that the direction of the electric field is in the direction
of decreing potential: Moreover, the field is in the
direction where this decree is steepest.
From the above relation between electricfield and or
potential, we can dr the following important con-
clusions:
or
(i)Electric field is in that direction inwhich the
potential decree is steepest.
(il)The mnitude of electric field is equal tothe
change in the mnitude of potential per unit
displacement (called potential gradient) normal
to the equipotential surface at the givenpoint.
11.How can we determine electric potential ifelectric
fieldisknown at any point?
Computing electric potential from electric field.
Therelation between electric field and potentialis
~ dV ~ ~
E= --- ordV= -E.dr
~
dr
~
Integrating the above equation between points r1
~
andr2,we get
PHYSICS-XII
...• ~
where VIandV2are the potentials at'1andr2
~ -
respectively. If we te'iat infinity, thenVI=0and
~~
putr2=r,weget
...•
r
~ f ~ ~
V(r)=- E.dr
Hence by knowing electricfield at any point, we
can evaluate the electric potential at that point.
12.Showthat the units volt/metre and newton/
coulomb areequivalent. To which physical quantitlj do
thelj refer ?
SI units of electric field.Electric field at any point is
equal to the negative of the potential gradient. It
suggests that the SI unit of electric field isvolt per metre.
But electric field is also defined the force
experienced by aunitpositive charge, so SI unit of
electric field isnewton per coulomb. Both of these units
are equivalent shown below.
volt joule/coulomb
metre metre
newton - metre newton
coulomb - metre coulomb
1Vm-1 =lNC1
Exam/esbased on
.•.-..•. Electric F.ield
-..•.
Formulae Used
1.Electric field in a region can be determined from
the electric potential by using relation,
E=_dV
dr
or E= _avF= _avr:-=_av
xax '-y Dy ,'-zaz
2.Electric field between two parallel conductors,
E=V
d
3. Electric potential in a region can be determined
from the electric field by using the relation,
r ~ ~
V=-fE.dr
00
Units Used
Eis in NC-lor Vm-\Vin volt, rin metre.~~~1styjxiwn}j1htr

104. ELECTROSTAT
IC POTENTIAL AND CAPACITANCE
Example 15.Find the electric field between two metal
plates3mm apart,connected to12V battery.
Solution. Electric field,
E ==V==12 V ==4 x103Vm-1
d3x10-3 m
Example 16.Calculate thevoltage needed to balance an oil
drop carrying10electrons when located between the plates
of a capacitor which are 5mmuparitg«10ms"2).The mass
of oildropis3x10- 16kg.
Solution. q==ne==10x1.6x10-19C
m==3x10-16 kg,d==5mm ==5x10-3m
E==V==V Vm-1
d5x10-3
For thecharged oil drop to remain stationary in
electric field,
qE==mg
10x1.6x10-19x V3==3x10-16x10
5x10-
3x10-16x10x5x10-3
V== ==9.47 V.
10x1.6x10-19
or
Example 17.An infinite plane sheet of charge densitJj
10-8Cm2is held in air. In this situation how far apart are
two equipotential surfaces, whosep.d.is5V?
Solution. Electric field of aninfinite plane sheet of
charge,
E==~
21:0
If Mistheseparationbetween two equipotential
surfaces having potential difference to.V,then
E==to.V
M
(jto.V
or
21:0 M
M==21:0to.V ==2xB,B5x 10-12x5
(j 10-8
==B.B5x10-3m ==8.85 mm.
Example 18.A spark passes in air when the potential
gradient at the surface of a charged conductor is
3x106Vm-1. What must be the radius of an insulated
metalsphere which can be charged to a potential of3x106V
before sparking into air ?
Solution. Potential gradient,
dV==3x106Vm-1
dr
dV==3x106dr
V==3x106r
or
or
2.13
or
But V-==3x106V
..3x106r==3x106
r==1m.
Example 19.A uniform electric field Eof300NC1 is
directed along negative X-axis. A, BandCarethreepoints
in the field, having x and y coordinates (in metre), asshown
in Fig.2,21.Find the potential differences to.VBA'to.VCBand
to.VCA·
y
B(4, 4)C(-3,4)
.~----~--------~--
A(4, 1)
E'~----~--------~--
------------~o+---------------~~x
Fig. 2.21
Solution. (i)No work is done in moving a unit
positive charge from AtoBbecause the displacement
ofthe charge isperpendicular to the electric field. Thus
the pointsAandBare at the same potential.
.. to.VBA==0
(ii)Work is done by theelectricfield as the positive
~
charge moves from BtoCii.e.,in the direction ofE).
Thus thepointCisat a lowerpotential than the pointR
As E==_to.V
Sx
to.VCB==-Ef'o.x==-300 Ne1 x7m
==-2100 V.
(iii)Points AandBlie on an equipotential surface.
So VB==VA
to.VCA==VC-VA ==VC -VB==to.VCB
==-2100 V.
Example 20.Three points A,BandClie in a uniform
electric field(E)of5x103NC1 as shown in the figure. Find
the potential difference between A and C. [CBSE F 09]
~
,
,
5cm',
I
13cm
~E
I
'IC
Fig. 2.22···3wx«n£m~r±n3lxv

105. 2.14
Solutio
n. Points B and C lie on an equipotential
surface, soVc=VB"
P'D.betweenAand C=P'D,between AandB
=-Eill
=-Sx103NC1 x4x10-2m [.: E =-~J
=-200 V. [ill=AB=~S2_32= 4 em]
Example 21.Ifthepotential in the region of space around
the point (-1m,2 m,3 m) is given by
V=(10x2 +Sy2-3z2) volt,calculate thethree components
ofelectricfield at thispoint.
Solution. Here x=-1 m, y=2. m, z =3m
As V=lOx2+Sl-3z2
ava 2 2 2
E=--=--(10x +Sy -3z)
xax ax
= -20x =-20x (-1)=20 Vm-1.
av a 2 2 2
E=-=--(lOx +Sy -3z)=-10y
Y8y dy
=-10x2 =-20Vm-1
E=_av=-~(10x2+Sy2-3z2)=6z
Zazdz
=6x3=18 Vm-1.
j2)roblems ForPractice
1.A uniform electric field of 20NC-1 existsin the
vertically downward direction. Determine the
incree in the electric potential one goes up
through a height of 50 cm. (Ans.10 V)
2.A uniform electric fieldof 30 NC-1 exists along the
X-axis. Calculate the potential difference VB-VA
between the points A(4m,2m)and B(10m,5m).
(Ans.-180 V)
----t 1\ 1'1 1
3.An electric fieldE=20i+30jNC-exists in free
space. If the potential at the origin is ten zero,
determine the potential at point(2m,2m).
(Ans.-100V)
4.The electric field in a region is given by E= ~I,
x
Write the 51unit forA.Write an expression for the
potential in the region suming the potential at
infinity to be zero. ( Ans.Nm3C-1,;:2)
5.Figure 2.23 shows some equipotential surfaces. What
can you s about the mnitude and the direction of
the electricfield?
(Ans.E=~Vm-1,radially outward)
r
PHYSICS-XII
30V
20cm:9'",
,'
60~~~~
20V
Fig.2.23
HINTS
1.L'1V= -EL'1r=-20x(-~) =10V.
100
2.~V= -SoL'1x=-30(10 - 4)=-180 V.
3.~V= -SoL'1x-E,.L'1Y= -20x2 - 30x2=-100 V.
4..:51unit of electric field=NC-1
:.51unit ofA=NC-1 xm3=Nm3 C-1
(x,y,z) Ad A
Potential, V= -f-;=-2 .
x 2x
00
5.For the equipotential surface of 60V,
60V=kq=--'.:L
r0.10m
or kq=60Vx0.10m=6Vm
E-kq_£V-1
•. - 2- 2 m
r r
Clearly, Edecrees withr.The direction of electric
field will be radially outward because Vdecrees
withr.
2.8EQUIPOTENTIAL SURFACES AND THEIR
PROPERTIES
13.Whatisan equipotential surface?Give an
example.
Equipotential surface.Any surface that has same
electric potential at every point onitiscalled anequipo-
tential surface. The surface may be surface of a body or a
surface in space. For example, we shall see later on,
the surface of a charged conductor is an equipotential
surface. By joining points of constant potential, we can
dr equipotential surfaces throughout the region in
which an electric field exists.
14.State and prove theimportant properties of
equipotential surfaces.
Propertiesof equipotential surfaces: 1.Nowork is
done in moving a test charge over an equipotential surface.~~~1styjxiwn}j1htr

106. ELECTROSTATIC POTE
NTIALAND CAPACITANCE
Equipotential
surface
Fig. 2.24An equipotential surface.
LetAand Bbe two points over an equipotential surface,
shown in Fig.2.24.If the test chargeqois moved from
Ato B,the work done will be
WAB=Chargexpotential difference
=qo(VB-VA)
Asthe surface is equipotential, so VB-VA=0
Hence WAB=0.
2.Electric field isalways normal tothe equipo-
tential surface atevery point. Ifthe field were not normal
to the equipotential surface, it would have a non-zero
component along the surface. So to move a test charge
ainst this component, a work would have to be done.
Butthereisno potential difference between any two
points on an equipotential surface and consequently
no workis required to move a test charge on the
surface. Hence the electric field must be normal to the
equipotential surface at every point.
3.Equipotential surfaces are closer together in the
regions of strong field and farther apart in the regions
of weak field.We know that electric field at any point
is equal to the negative of potential gradient at that
point.
i.e.,
E=- dV
dr
dV
dr=--
E
or
Forthesame change in the value of dVi.e., when
dV=constant, we have
1
dr«:-
E
Thusthe spacing between the equipotential
surfaces will be smaller in the regions, where the
electric field is stronger and vice versa.
4.No two equipotential surfaces can intersect each
other. Ifthey interesect, then there will be two values
of electric potential at the point of intersection, which
is impossible.
2.9EQUIPOTENTIAL SURFACES OF
VARIOUS CHARGE SYSTEMS
15.Sketch and explain the equipotential surfaces for:
(i)a point charge,(ii)two point charges+q and - q,
2.15
separated by a small distance,(iii)two point charges +q and
+q separated by a small distance and (iv) auniform electric
field.
Equipotential surfaces of various charge systems.
For the various charge systems, we represent equipo-
tential surfaces bydashed curves and lines of force by
full line curves. Between any two adjacent equipotential
surfaces, we sume a constant potential difference.
(i)Equipotential surfaces of a positive point
charge. The electric potential duetoa point charge qat
distancerfrom it. is given by
V=_l_.!l.
41t EOr
This showsthatVisconstant if risconstant. Thus,
the equipotential surfaces of a single point charge are
concentric spherical shells with their centres at the
point charge, shown in Fig.2.25. As the lines of force
point radially outwards, sotheyare perpendicular to
the equipotential surfaces at allpoints.
Fig. 2.25Equipotential surface ofa+ve point charge.
(ii)Equipotential surfaces of twoequal and opposite
pointcharges:Electric dipole. Fig.2.26 shows the
equipotential surfaces of two equal and opposite charges,
+qand-q,separated by a small distance. They are close
together in the region in between the two charges.
Fig. 2.26Equipotential surfaces for two equal and
opposite charges.···3wx«n£m~r±n3lxv

107. 2.16
(iii)Equipotential surfaces of two equal positive
charges.Fig. 2.27 shows
the equipotential surfaces of
two equal and positive charges, each equal to+q,sepa-
rated by a small distance. The equipotentialsurfaces
are far apartin the regions in between the two charges,
indicating a wefieldinsuchregions.
Fig. 2.27
(iv) Equipotential surfaces for a uniform electric
field.Fig. 2.28 shows the equipotential surfaces for a
uniform electric field. The lines of force are parallel
straight lines and equipotential surfaces are equidis-
tant parallel planes perpendicular to thelines of force.
Equipotential
surfaces
.r>:
:---<.....r---r---
~
~
'-~
'-~
f-~
-- -- --
Fig.2.28 Equipotential surfaces for a uniform
electric field.
16. Give theimportance of equipotential surfaces.
Importance of equipotential surfaces. Like the lines
of force, the equipotential surfaces give a visual picture
~
of both the direction and the magnitude of fieldEina
region of space. If we dr equipotential surfaces at
regular intervals ofV,wefind that equipotential
surfaces arecloser together in the regions ofstrong
field and farther apart in the regions of we field.
~
Moreover,Eis normal to the equipotential surface at
every point.
2.10ELECTRIC POTENTIAL ENERGY
17. What is meant by electricpotential energy of a
charge system ?
Electric potential energy. It is the energy possessed
by a system of charges by virtue of their positions.
When two like charges lie infinite distance apart, their
potential energy is zero because nowork h to be
PHYSICS-XII
done in moving onecharge at infinite distance from
theother. Butwhen they arebroughtcloser to one
another, work has to be done against the force of
repulsion. As electrostatic force isa conservative force,
this work gets stored as the potential energy of the two
charges.
Theelectric potentialenergy of a system of point charges
may be defined as the amount of work done in assembling the
charges at theirlocations by bringing them in,frominfinity.
18. Deduce expressions for the potential energy of a
system of two pointcharges andthree point chargesandhence
generalise the result for a system ofNpoint charges.
Potential energy of a system of two
point charges.Suppose a point charge ql
isat rest at a pointPIin space, shownin
Fig. 2.29. It tes nowork to bring the first
chargeqlbecause there isnofieldyet to
work ainst.
,
r
,
,
,
Fig.2.29P.E.of two point charges.
Electric potential due to charge qlata point P2at
distance r12fromPIwillbe
V=_1_ !1l
I .
41t EO 'i2
Ifchargeq2is moved in from infinity to pointP2,the
work required is
W2=Potential xcharge
=VIxq2=_1_. qlq2
41tEO'12
As theworkdoneisstored the potential energy
U ofthe system (ql+q2)'so
U=WI+W2=_1_. ql q2
41t EOr12
Potential energy of
a system of three point
charges. As shown in
Fig. 2.30, now we bring
in the chargeq3from
infinity to the pointP3.
Work has tobe done
ainst the forces
exerted by qlandq2·
Fig.2.30P.E. ofthree
point charges.~~~1styjxiwn}j1htr

108. ELECTROSTATIC POTENTIAL AND
CAPACITANCE
Therefore
W3 = Potential at point P3 due to q1andq2
xchargeq3
or W3=_1_ [!!.L +31..]xq3=_1_[q1q3+q2q3]
41t EOr13123 41t EO r13 r23
Hence the electrostatic potential energy of the
systemq1+q2+q3is
U=Totalwork done to semble the three charges
=W1+W2+W3
or U=_1_ [q1q2+q1q3+q2q3]
41t EO r12 r13 r23
Potential energy of a system ofNpoint charges.
Theexpression for the potential energy of N point
charges can be written
U=_l_Lqiqj=.!.._1_f ffJiqj
41t EOall pairs'ij241t EOi=1j=1'i;
i7'j
Asdoublesummation counts every pair twice, to
avoid this the factor 1/2h been introduced.
NOT E Thepotential at jth charge due to all other
chargescan be written
V.=I3£
}k=1r'k
k7'j }
Theexpression for P.E. ofNpoint charges can be written
U=1.~q,[_1_Iqk]=1.Iq.Y.
2j=1 } 41t EOk=1rl 2j=1 } }
k7' } }
For Your Knowledge
~Electric potential energy isascalar quantity. While
finding its value, the value of various charges must be
substituted with their proper signs.
~Thepotential energy of two likecharges(qlq2>0) ispositive.
As the electrostatic force is repulsive, so a positive
amount of work h to be done ainst this force to
bring the charges from infinity to a finite separation.
~The potentialenergy oftwo unlike charges (ql q2< 0) is
negative. As the electrostatic force is attractive, soa
positive amount of work h to be done ainst this
force to te the charges from thegiven locations to
infinity. Conversely, a negative amount of work is
needed to bring the charges from infinity to the present
locations, so the potential energyisnegative.
~As electrostatic force is a conservativeforce, sothe
potential energy of a charge configurationisinde-
pendent of the manner in which the charges are
sembled to the present locations. The potential energy
is a characteristic of the present state of configuration,
not on how this state is attained.
2.17
~Positive potential energy implies thatworkcan be
obtained by releing the charges, whilenegative potential
energy indicates that an external ency will have to do
work to separate the charges infinite distance apart.
~Electric potential is a characteristic of an electric field, it
doesnot matterwhetheracharged object is placed in
thatfield ornot.It ismeasured in JC-1 orvolt.On the
otherhand, electric potentialenergy isthe energy of a
charged object in an external electric field. More
precisely, it is theenergy of the system consisting of the
charged object and the external electricfield (or charges
producing that field). It ismeuredin joule.
2.11POTENTIAL ENERGY IN AN
EXTERNAL FIELD
19.Write an expression for the potential energy of a
single charge in an external field. Hence define electric
potential.
Potential energy of a single charge.We wish to
determine the potential energy of a charge qinan
~
external electric fieldEat a point P where the corres-
ponding external potential is V.Bydefinition, Vata
point Pistheamount of work done in bringing a unit
positive charge from infinity tothepoint P.Thus,the
work done in bringing a charge qfrom infinity tothe
point P will beqV,i.e.,W =qV
This work done is stored the potential energy of
~
the chargeq.Ifris theposition vector of point P
relative to some origin,then
~ .~
U(r)=qV(r)
P.E.of a charge in an external field
= Chargexexternal electric potential
As V=U
q
So we can define electric potential at a given pointin
an external fieldas the potential energy of a unit positive
charge at that point.
20.Write an expression for the potential energy of
two point chargesqlandQ2'separated by distance rin an
electric fieldE.
Potential energy of a system of two point charges in
~ ~
an external field. LetV(r1)andV(r2)be the electric
~
potentials of the field E at the points having position
~ ~
vectorsr1andr2 shown in Fig. 2.31.
~
Work done inbringing q1from 00tor1ainst the
external field···3wx«n£m~r±n3lxv

109. 2.18
0--
'I \
ISource
, charges
-='=
-- ofE Q2VCr;)
Fig.2.31P.E. of two charges in an external field.
~
Work done in bringingq2fromOC!to'2ainst the
external field
~
=q2V( '2)
Work done on q2ainst the force exerted byq1
_1_q1q2
4m:o' '12
where'12is the distance betweenq1andq2'
Total potential energy of the system =Thework
done in sembling the twocharges
~ ~ 1q1q2
U=q1V( 1J)+q2V('2)+--'-
41t1:o '12
or
21.Defineelectronvolt. Express it in joule.
Units of electrostatic potential energy.Suppose an
electron (q=1.6x10-19C)is moved through a potential
difference of 1volt, then the change in its P.E. would be
I:J.U=qI:J.V=1.6 x10-19 ex1v=1.6 x10-19J
This is a commonly used unit of energy in atomic
physics and wecallitelectronvolt(eV).
Thus electron volt is the potential energy gained or lost by
an electron in moving through a potential difference of1volt.
1eV= 1.6 x10-19J
Multiples and submultiples of eV
1meV (milli electron volt)
= 10-3eV=1.6x10-22J
1keY (kilo electron volt)
= 103 eV= 1.6 x10-16J
1MeV (million electron volt)
=106eV= 1.6 x10-13J
1GeV (giga electron volt)
=109eV=1.6 x10-toJ
1TeV (tera electron volt)
=1012eV=1.6x10-7J.
PHYSICS-XII
2.12POTENTIAL ENERGY OF A DIPOLE IN
A UNIFORM ELECTRIC FIELD
22.Derive an expression for the potential energy of a
dipole in a uniform electric field. Discuss the conditions of
stable andunstable equilibrium.
Potential energy of a dipole placed in a uniform
electric field.As shown in Fig. 2.32, consider an electric
~
dipole placed in a uniform electric fieldEwith its
~
dipole momentpmaking an angle 8with the field.
Two equal and opposite forces+qEand-qEact on
itstwo ends. The two forces form a couple. The torque
exerted by the couple will be
't=qEx2asin 8=pEsin8
whereqx2a=p,isthe dipole moment.
+q ->
.0....:.... ...•. +qE
->
-----------r~~~--~----~E
: 2asine
->
-qE.•.••.--a:
-q
Fig. 2.32 Torque on a dipole in a uniform electric field.
1£the dipole is rotated through a small angled8
ainst the torque acting on it, then thesmall work
done is
dW='td8=pEsin 8d8
The total work done in rotating the dipole from its
orientation ming an angle 81, with the direction of
the field to 82 will be
82
W=fdW=fpEsin8d8
~
= pE[-cos 8]~ =pE(cos 81 -cos 82)
This work done is stored the potential energy U
of the dipole.
.. U=pE(cos81-cos82)
1£initially the dipole is oriented perpendicular to
the direction of the field(81= 90°) and then brought to
some orientation ming an angle8with the field
(82=8),then potential energy ofthe dipole willbe
U =pE(cos90°-cos8) =pE(0 -cos8)
~ ~
or U= -pEcos8=-P .E~~~1styjxiwn}j1htr

110. ELECTROSTATIC POTENTIAL AND
CAPACITANCE
Special Ces
1.Positionof stable equilibrium.Whene=0°,
u=-pEcos 0° = -pE
Thusthepotential energy of a dipole isminimum
when its dipole momentisparalleltothe external
field.This isthe position of stable equilibrium.
2.Position of zero energy.Whene= 90°,
U= -pEcos 90° =0.
Thus thepotential energy of a dipole iszero
whenitis held perpendicular tothe external
field. This canbe explained follows. If we
holdthedipole perpendicular tothe electric
field and bring it from infinity into the field,
then thework done on charge +qbythe
externalent is equal to the work done on
charge - q.Thenet work done on the dipole will
be zero and hence its potential energy is zero.
3.Position of unstable equilibrium.Whene= 180°,
U= -pEcos 180° =+pE
Thus the potential energy of a dipoleismaximum
whenitsdipole momentisantiparallel to the external
field.Thisis the position of unstable equilibrium.
Examples based on
Electric Potential Energy
Formulae Used
1.Electric potential energy of a system of two point
charges,
U=_I_. q1q2
41teo'i2
2. Electric potential energy of a system of N point
charges,
U=_l_ L qjqk
41teoall pairsrjk
3. Potential energy of an electric dipole in auniform
electric field,
u=-pE(cos 82-cos~)
Ifinitially the dipole is perpendicular to the field
E,~=90°and 82 =8 (s ), then
-4 -4
U= -pEcos8= -P .E
If initially the dipole is parallel tothe fieldE,
~ =0° and 82=8(s ),then
U= -pE(cos 8-1)=pE(I-cos 8)
Units Used
Charges arein coulomb, distances in metre,
energy in joule or in electron volt (eV) and dipole
moment incoulomb metre(Cm).
leV=1.6x10-19 C, 1 MeV =1.6x10-13 C.
2.19
Example 22
(a) Determine the electrostatic potential energy of a
system consisting of two charges 7 ~Cand-2 ~C
(andwithno external field) placed at(-9em,O,O)
and(9em,a,0)respectively.
(b) How much workisrequired toseparate thetwo
charges infinitely away from each other ?
(c) Suppose the same system of charges isnow placed in
anexternal electric field E=A(1/1);
A=9x105Cm-2. What would the electrostatic
energy of the configuration be ? [NCERT]
Solution. (a) q1= 7 ~C = 7x10-6c,.q2= - 2x10-6C,
r=18em =0.18 m
Electrostatic potential energy of the two charges is
U=_I_. q1 q2
41teo r
9x109x7x10-6 x(-2)x10-6
--------'-----'--- =-0.7J.
0.18
(b)Work required to separate two charges infinitely
from each other,
W=U2-U1=0 -U=-(-0.7)=0.7 J.
(c)Energy of the two charges in the external electric field
= Energy of interaction of two charges with the
external electric field
+Mutual interaction energy of the two charges
=q1V(r1)+q2V (r2) +_1_ql~2
4m;0r:
=q1A+q2A+_1_q1q2
1. r24m;01
= [7~C+-2~C]x9x105Cm-2 -0.7J
0.09 m 0.09m
= (70-20)-0.7 = 50 - 0.7 = 49.3 J.
Example 23.Three charges - q, +Qand - q are placed at
equal distances on astraight line. If the potential energy of
thesystem of three charges iszero,find the ratioQ /q.
Solution. As shown in Fig. 2.33, suppose the three
charges are placed at pointsA,Band C respectively on
a straight line, such that AB= BC =r.
-q +Q
• •
-q
•
A B
I--- r '1'
C
r------I
Fig.2.33
As the total P.E. of the system is zero, so
_1_ [- qQ+ (-q)(-q) +Q(-q)] =0
41t EO r 2r r
or -Q+!t.-Q=aor 2Q=!t.orQ=.!.= 1 : 4.
2 2 q4···3wx«n£m~r±n3lxv

111. 2.20
Example24.Two
positive point charges of0.2/lC and
0.01 /lC are placed10em apart. Calculate the work done in
redung the distance to 5em.
Solution.Hereq1= 0.2x1O-6C, q2= 0.01x10-6C
Initial separation(r)= 10 em = 0.10 m
Final separation(rf)= 5 em = 0.05 m
Work done = Change inpotential energy
= Final P. E. - Initial P. E.
__1_q1q2__1_q1q2 - q1q2[~-.!l
- 4m,0· rf 4m,o '; - 41tEo rf';
= 0.2x10-6x0.01x10-6x9x109[_1l_J
0.05 0.10
=1.8x10-4J.
Example25.Two electrons, each moving with a veloty of
106ms-l, are released towards each other. What will be the
closest distance of approach between them?
Solution.Letrobe the distance of closest approach
of the two electrons. At this distance, the entire K.E. of
the electrons changes into their P.E. Therefore,
1 21 2 1ee
-mv +-mv =---
2 2 41tEo ~
1 e2 9x109x(1.6x10-19)2
r,=--. -- =----'-,;-;---~;O-'---
o 41tEO mv2 9.1x10-31 x(106)2
=2.53x10-tO m.
Example26.Two particles have equal masses of5.0 g each
and opposite charges of+4x10-5Cand-4.0x10-5C.They
are released from rest with a separation of1.0m between
them.Find the speeds of the particles when the separationis
reduced to50 em.
Solution. Here m= 5.0 g = 5x10-3kg.
q=±4x 1O-5C, r1=1.0 m, r2=50 em =0.50 m
Letv= speed of each particle at the separation of
50 cm.
From energy conservation principle,
K.E. ofthetwo particles at 50 em separation
+ P.E. of the two particles at 50 em separation
= P.E. of the two particles at 1.0 m separation
.!mv2+.!mv2+_1_.q1 q2=_1_. q1q2
2 2 47tEO r2 41tEO r1
mv2=q1q2 [~_~Jlorv2=q1q2 [r2 -r1]
41tEOr1r2 41tEOm r1 r2
v2= 4x10-5x(-4x10-5) x9x109[0.50-1.0]
5x10-3 1.0x0.50
=2880 orv=53.67ms-t.
PHYSICS-XII
Example27.Four charges /~q -.q D
arearranged at the corners ofa
square ABCD of side d as
shown inFig.2.34.(i)Find the
work required to put together
thisarrangement. (ii)Acharge
qoisbrought tothecentre EofB__._----- .•C
the square, thefourcharges-q +q
beingheld fixed at its corners.
Fig.2.34
How much extra workis
needed to do this ? [NCERT;CBSE F 15]
Solution. (i)GivenAB=BC=CD=AD=d
.. AC= BD=~d2 +d2=..fid
d
,
,
Work required to put the four charges together
= Total electrostatic P.E. ofthe four charges
=_I_[qAqB +qAqC+qAqO+qBqC+qBqO+qcqo]
47t EOAB AC AD BC BD CD
_ 1[q2 q2 q2 q2 q2 q2]
-41tEO-d+J2d-d -d+J2d -d
=-L(4-J2).
47tEo
(ii)Extra work needed to bring chargeqoto centre E
W=qoxElectrostatic potential atEdue to the
four charges
-q[ q + -q
- 0 41tEo (dI..fi)41tEo(dI..fi)
+ q + -q]-0
41tEo(dl..fi) 41tEo(dl..fi) -.
Example28.Three point
charges, +Q,+2Q and -3Q A(+Q)
are placed at the vertices of an
equilateral triangle ABt of sideI
(Fig.2.35).If these charges are
displaced to the midpoints ~,
1\ andC1respectively, find the
amount of the work done inB(+2Q) B1 q-3Q)
shifting the charges to thenew
locations. [CBSE OD 2015] Fig.2.34
Solution. ~ 1\=1\ C1= ~C1=AB=.i
2 2
Initial P.E. of thesystem is
u.=_1_[QX2Q +2Qx(-3Q)+QX(-3Q)]
I41tEo I I I
__1_7Q2
41tEO' Iwww4±·tustryvu4s·~

112. ELECTROSTATIC P
OTENTIAL AND CAPACITANCE
FinalP.E.of the system is
U==_1_[QX2Q +2Qx(-3Q) +QX(-3Q)]
f41tEo 1/2 1/2 1/2
1 14Q2
-41tEo·-1-
Work done==Uf -U,
1 14Q2 17Q2 17Q2
==---.-- + ---- ==---.-- .
41tEo I 41tEoI 41tEo I
Example29.An electricdipole oflength2emisplaced
with its axis making an angle of 60°to auniformelectric
field of105uc:'.Ifitexperiences atorque of 8J3Nm,
callate the
(i)magnitude of the charge on the dipole, and
(ii)potential energlJ of the dipole. [CBSEOD2000]
Solution. Here 2a==2em==0.02 m,8==60°,
E==105 NC\ 1:==8J3Nm
(i) T==pEsin8==qx2axEsin8
8J3==qx0.02x105xsin60°
==8J3x2 ==8x10-3C.
q0.02x105xJ3
(ii)P.E.of the dipole is
U==-pEcose==-qx2axEcos8
==- 8x10-3 x0.02x105xcos60°==-8J.
Example30.An electric dipole oflength4em,when placed
withitsaxis making an angle of 60°with auniform electric
fieldexperiences atorqueoj4J3Nm.Calculate the(i) magni-
tudeof theelectric field, (ii)potential energy of the dipole, if
thedipole has charges of ±8nC. [CBSEOD04;D06C,14]
Solution. Here 2a==4cm==0.04m,8==60°,
T==4J3Nm,q==8nC==8x10-9C
Dipole moment,
p==qx2a==8x10-9 x0.04 ==0.32 x10-9Cm.
(i)As1:==pEsin 8
E==__1:_
Psin 80.32x10-9xsin60°
==4J3x109x2 ==2.5x1010NCl.
0.32xJ3
(ii)U==-pEcose
==-0.32 x10-9 x2.5x1010xcos60° ==-4J.
Example 31. A molele of a substance haspermanent
electric dipole moment equal to 10-29 Cm.Amole of this
substanceispolarized (at low temperature) t:yapplying a
strong electrostatic field of magnitude (10Vm-l). The
direction of the field issuddenly changed by an angle of60°.
or
2.21
Estimate theheat released by the substance in aligning its
dipolesalong the new direction of the field. For simplity
assume100% polarization of the sample. [NCERT]
Solution. Here p==10-29 Cm, E==106 Vm-1,
8==60°, N==6x1023
Work required to bring one dipole from position
8==0°to positioneis
W==pE-pEcos8==pE(1-cos 8)
==10-29 x106(1- cos60°)J==0.5 x10-23J
Work required for one mole of dipoles
==WxN==0.5x10-23 x6x1023==3.0J
Heat releed=Loss inP.E.= Work done= 3.0J.
jOrOblems For Practice
1.Two pointcharges+10~C and -10~C are
separated by a distance of2.0em inair.(i)Calculate
thepotential energy ofthesystem, suming the
zero of the potential energy tobe at infinity.
(ii)Draw an equipotential surface of thesystem.
[CBSE D 04] (Ans. -45J)
2.Two point chargesAandBof values +15IlC and
+9IlC are kept18em apart in air. Calculate the
work done whenchargeBis moved by3cm
towardsA. [CBSE OD2000] (Ans. 1.35 J)
3.Two point charges 20x1O-6C and -4x1O-6C are
separated by a distance of 50cm in air. (i)Find the
point on the line joining the charges, where the
electric potential iszero.(ii)Also find the electro-
static potential energy ofthesystem.[CBSE ODOS]
[Ans.(i)41em from the charge of20x10-6C
(ii)-144Jl
4.Twocharges, of mnitude 5nC and -2nC,are
placed at points (2 em, 0,0)and(xem,0,0)in a
region of space, where there is no other external
field. If the electrostatic potential energy of the
system is -O.5IlJ, what is the value ofx?
[CBSE D OSC] (Ans.x=4em)
5.Three point charges are arranged as shown in
Fig.2.36.What is their mutual potential energy ?
Teq==1.0x10--4C anda==10em. (Ans. 0.27J)
Dq a q
Fig.2.36 Fig. 2.37
6.Determine potential energy of thecharge configu-
rationshown in Fig. 2.37. ( q2 r;J
Ans.--(-.,,2)
41tEoa···3wx«n£m~r±n3lxv

113. 2.22
7.Find theamountof work
done
in arranging the
three pointcharges, on
the vertices of an equi-
lateral triangle ABC,of
side 10em,as shown in
the adjacent figure.
[CBSE Sample Paper 2011]
8e------,e C
6~C -6~C
(Ans. - 3.24J)
8.Calculate the work done to dissociate the system of
three charges placed on the vertices ofa triangle
shown in Fig. 2.38.Hereq= 1.6x10-10C.
[CBSE D 08; OD 13] (Ans.2.304 x10-8J)
q
ql~:4
q,~q,
Fig. 2.39
- 4q~----- __+2q
lOan
Fig. 2.38
9.Whatis the electrostatic potential energy of the
charge configuration shown in Fig. 2.39 ? Te
ql=+1.0x10-8C,q2=- 2.0x10-8C,
q3=+3.0x10-8C,q4=+2.0x10-8C
anda= 1.0 metre. (Ans.- 6.36x10-7J)
10.Three pointcharges+q,+2qand Qare placedat the
threevertices ofan equilateral triangle. Find thevalue
of chargeQ (in terms of q),sothatelectric potential
energy of the system is zero. (Ans. Q=-2q /3)
11.An electron (charge =-e)isplaced ateachofthe
eight comers of acubeofsideaandana-particle
(charge=+2e)at the centre of the cube. Calculate
the potential energy of thesystem.
(Ans.3.89x10lOe2/ajoule)
12.Two identical particles, each having acharge of
2.0x10-4C and mass of 10 g, are keptat a
separation of 10 em and then releed. Whatwould
be the speeds of the particles when the separation
becomes large? (Ans.600ms ")
13.Find the amount ofwork done in rotating an electric
dipole, of dipole moment 3.2 x10-8em,from its
position ofstable equilibrium, tothe position of
unstable equilibrium, in a uniform electric field of
-intensity 104 N / C. [CBSE Sample Paper 2011]
(Ans.6.4x10-4J)
14.An electric dipole consists of two opposite charges
each of mnitude11lCseparated by 2 em. The
dipole is placedin an external electric field of
1cPNC-1. Find(i)the maximum torque exerted by
PHYSICS-XII
thefieldon the dipole(ii)the work which the
.external agent will have to do in turning the dipole
through 180° starting from the position e= 0°.
[Ans. (i)2x10-3Nm(ii)4x10-3JJ
HINTS
1.U=_l_. qlq2
41tEOr
910x10-6 x(-10) x10-6
=9xlO x 2 =-45J.
2.0x10-
For equipotential surface, see Fig. 2.26 on pe 2.15.
2.W = Final P.E. -Initial P.E.
=4~q:J~-{]
= 9x109x15x10-6x9x10-6[100 _ 100]
15 18
=1.35J.
3.(i)Suppose the point ofzero potential is located at
distance xmetrefrom the charge of 20 x10-6C.
Then, V= _1_ [20 x 10-6 _4x10-6]= 0
41tEo x 0.50 -x
Thisgivesx= 0.41 m = 41 em.
(ii)U=_1_. qlq2
41tEo r
9x109x20x10-6x(-4)x10-6
-------'--'--- =-1.44J.
0.50
4.U=_1_qlq2
41tEo r
-69x109x5x10-9x(-2)x10-9
:. - 0.5 x10= 2
(x-2)xl0
•
Onsolving, x=4em
3q2
5.U=---
41tEo a
3x9x109x(1.0 x10-4)3= 0.27 J.
0.10
7.W=_l_[qAqB +qAqC+qBqC]
41tEo AB AC BC
=_1_[!Li+ q(-q)+q(-q)]
41tEo r r r
1l 9x109x(6x10-6)2
= - -- =- J= - 3.24 J.
41tEor 0.10 ..
8.Initial P.E. of the three charges,
U.=_1_[Ihq2+q2q3+qlq3]
I41tf.:o r r r~~~1styjxiwn}j1htr

114. ELECTROSTATIC
POTENTIAL AND CAPACITANCE
=_1_[q(-4q) +(-4q)x2q +qX2q]
41tEor r r
110q2 9x109x10x(1.6xlO-10)2
41tEo-r-= - 0.10 J
=- 2.304x10-8J
Final P.E., U f= 0
Work required todissociate the system of three
charges,
W=Uf-u,=2.304x10-8J.
9.U=_1_[q1q2 +q~3+%q4+q2q3+q~4+q3q4]
41tEo av2a a av2a a
=;9x109[(1)(-2) +(1)f]) +(1)(2)+(-2)(3)
1.0 -n
+(-2) x(2)+(3)(2)] x 10-16J
12
9x109x10-16
12 J=-6.36x10-7J.
10.Suppose the charges +q,+2qand Q are placed at
the comers A,Band C of anequilateralMBCof side
a.Then
_1_[qx2q+qxQ+2qxQ]=0
41tEo r r r
or 2q+Q+2Q=0 or Q = -'lq/3.
11.U=9 x 109 [12(-:)(-e)+12 (-jte)
+ 4(-e)(-e)+8(-e)(2e)]
.s; ../3a/2
9 x109x4xe2[3+2.+ ~_~lJ
a 12J3J3
36x109e2 e2
---- [3+2.12-4.04] = 3.89x1010-joule.
a a
12.Hereq= 2.0x10-4C,m= 10 g = 10-2 kgI
r= lOcm =0.10 m
Letvbethespeedofeachparticle at infinite
separation. By conservation of energy,
P.E.of two particles at theseparation of 10 em
=K.E.ofthe two particles at infinite separation
_1_ .ql q2=.!.mv2+.!.mv2
41t£o r 2 2
2 1 qlq)
orv=--.---
41tEo rm
9x109x2.0x10-4x2.0x10-4 4.
0.10 x10-2 = 36x10
v=600 ms-1•
2.23
13.Here '\=0°,82= 180°, P= 3.2x10-8Cm,
E=104N/C
W=pE(cos,\-cos82)
=3.2xlO-8 x104(cosOO-cos180°)
= 3.2x10-4x(l+1)= 6.4x10-4J.
14.P=qx2a= 10-6 xO.02 = 2x1O-8Cm
(i)"max=pEsin90°= 2x10-8x1ef xl
=2x10-3 Nm.
(ii) W=pE(cos '\-cos 82)
=2x10-8x105(cos OO-cos 180°)
=2x10-3(1+ 1) =4xlO-3J.
2.13CONDUCTORS AND INSULATORS
23.What are conductors and insulators?Whywere
insulators called dielectrics and conductors non-electrics ?
Conductors and insulators.On the bis of their
behaviour in an external electric field, most of the
materials can be broadly clsified intotwocategories:
1.Conductors. These are the substances which allow
largescale physical movement of electric chargesthrough
themwhenanexternal electricfield is applied. For example,
silver, copper, aluminium, graphite, human body, acids,
alkalies, etc.
2.Insulators.These arethe substances which do not
allow physical movement of electric chargesthrough them when
an external electricfield isapplied. Forexample, diamond,
gls,wood, mica, wax, distilled water, ebonite, etc.
The rubbed insulators were able to retaincharges
placed on them, so they were calleddielectrics. The
rubbed conductors (metals) could not retain charges
placed on them but immediately drained the
charges, so they were callednon-electrics.
2.14FREE AND BOUND CHARGES
24.Discuss the various free and bound charges
present in conductors and insulators.
Free and bound charges.The difference between
theelectrical behaviour of conductors and insulators
can be understood on the basis of free and bound
charges.
Inmetallic conductors, the electrons of the outer
shellsof the atoms are looselybound to the nucleus.
They get detached from the atoms and move almost freely
inside the metal. In an external electric field, these free
electrons drift in the opposite direction of the electric
field. Thepositive ionswhich consist of nuclei and
electrons of inner shells remain held in their fixed posi-
tions. These immobile charges constitute theboundcharges.···3wx«n£m~r±n3lxv

115. 2.24
Inelectrolyticconductors
,both positive and
negative ions act charge carriers. However,their
movements arerestricted by the external electric field
and the electrostatic forces between them.
Ininsulators, the electrons are tightly bound to the
nuclei and cannot be detached from the atoms, i.e.,
charges ininsulators arebound charges. Due to the
absence of freecharges, insulators are poor conductors of
electricity.
For Your Knowledge
~A third important category of materials is the
semiconductors whichwe shalldiscuss inchapter 14.
~In metallic conductors, electrons of outer shells of the
atoms are the free charges while the immobile
positive ions are the boundcharges.
~In electrolytic conductors, both positive and negative
ions are the free charges.
~In insulators, both electrons and the positive ions are
the bound charges.
~There is noclear cut distinction between conductors
and insulators - theirelectrical properties vary
continuously within avery large range. Forexample,
the ratio of the electrical properties between ametal
and gls m be high 1020.
2.15BEHAVIOUR OF CONDUCTORS IN
ELECTROSTATIC FIELDS
25. State andprovethe variouselectrostatic properties
shownbyconductors placed in electrostatic fields.
Electrostatic properties of a conductor.When
placed in electrostatic fields, the conductors show the
following properties :
1.Net electrostatic field iszero in the interior of a
conductor.As shown in Fig.2.40,when aconductor is
~
placed inan electric field Eext'itsfree electrons begin to
~
move in the opposite direction of Eext' Negative
charges are induced on the leftend and positive
+
+
..• +
Eind
-+-+
+
->
E=0+
Conductor
Fig. 2.40Electric field inside a conductor is zero.
PHYSICS-XII
charges are induced ontherightendofthe conductor.
~
Theprocesscontinues till the electric field Eind set up
bythe induced charges becomes equal and opposite to
~ ~ ~ ~
the field Eext'The net field E(=Eext- Eind) inside the
conductor will be zero.
2.Just outside the surface of a charged conductor,
electric fieldisnormal tothe surface. Iftheelectric
field isnot normal to the surface, it will have a
component tangential to the surface which will
immediately cause the flow of charges, producing
surface currents. But no such currents can exist under
static conditions. Hence electric field is normal to the
surface of the conductor at every point.
3. The net charge in the interior of a conductor is
zero and any excess charge resides at its surface.As
shown in Fig. 2.41, consider a conductor carrying an
excess chargeqwith no currents flowing in it.Choose a
Gaussian surface inside the conductor just near its
~
outer boundary. As the field E =0 at allpoints inside
the conductor, the flux<Itthrough the Gaussian surface
must be zero. According toGauss's theorem,
I.~ ~ q
<It=jE.dS=-
EO
As <It=0,soq=0
->
E
+ + + +
+ +- --- - --:q+
+/ ",+
-> + I -> \+
E+-----1, E=0 '
+\ I+
+ " ",,/+
+
Gaussian
surface
+
Fig. 2.41
Hence there can be nocharge in the interior of the
conductor because theGaussian surface lies just near
the outer boundary. The entire excesschargeqmust
reside at the surface of the conductor.
4. Potential is constant within and on the surface
of a conductor. Electric field at any point is equal tothe
negative of the potential gradient,
. E=_dV
t.e.,
dr
But inside a conductor E =0and moreover, Ehas
no tangential component on thesurface, so
dV=0 or V=constant
dr~~~1styjxiwn}j1htr

116. ELECTROSTATI
C POTENTIAL AND CAPACITANCE
Hence electric potential is constant throughout the
volume of a conductor and has the samevalue (
inside) on its surface.Thus the surface of a conductor isan
equipoteniialsurface.
Ifa cond uctor is charged, there exists an electric
field normal to its surface. This indicates that the
potential on thesurface will be different from the
potential at a point just outside the surface.
5.Electric field at the surface of a charged conductor
isproportionaltothe surface chargedensitq.Consider
a charged conductor of irregular shape. Let cbe the
surface charge density at any point of its surface.To
2.25
inside the cavity. Imine a Gaussian surface inside the
conductorquiteclose to the cavity. Everywhere inside
theconductor, E =0.By Gauss's theorem, charge
enclosed by this Gaussian surface is zero (E=0=>
q=0).Consequently, the electric field must be zero at
every point inside the cavity(q=0=>E=0).The entire
excess charge+qlies on its surface.
E
+ +
E'---(
+ +
+
E=O
+
+ +
Surface of
conductor E
Fig.2.42A small pill box as a Gaussian surface
of a charged conductor.
determineEat this point, we choose a short cylinder
(pill box) the Gaussian surface about this point. The
pill box lies partly inside andpartly outside the con-
ductor. It h a cross-sectional area L'lSand negligible
height.
Electric field is zero inside the conductor and just
outside, it is normal to the surface. The contribution to
the total flux through the pill boxcomes only from its
outer cross-section.
~ =E»s
Charge enclosed by pill box,
ByGauss'stheorem,
~=l..
EO
EL'lS=crL'lS or
EO
q=cL'lS
~
As E points normally outward, so we write
~cr"
E=-n
EO
where;; is aunit vector normal to the surface in the
outer direction.
6.Electric fieldiszero in the' cavity of a hollow
charged conductor. As shown in Fig:'2.43, consider a
chargedconductor having a cavityi-with no charges
Fig. 2.43Electric field vanishes in the cavity
of a conductor.
2.16 ELECTROSTATIC SHIELDING
26.Whatiselectrostatic shielding? Mention its few
applications,
Electrostatic shielding.Consider a conductor with a
cavity, with no charges placed inside the cavity.
Whatever be the size and shape of the cavity and
whatever be the charge on the conductor and the
external fields in which it might be placed, the electric
field inside thecavity iszero,i.e.,the cavity inside the
conductor remains shielded from outside electric
influence. This is known electrostaticshielding.
Such a field free region is called aFaraday cage.
The phenomenon of making a region free from any
electric fieldiscalled electrostatic shielding. Itisbased on
the fact that electric field vanishes inside the cavity of a
hollow conductor.
Applicationsof electrostatic shielding
1.In a thunderstorm accompanied by lightning, it
is safest tosit inside' a car, rather than near a tree
or on the open ground. The metallic body of the
car becomes an electrostatic shielding from
lightning.
2. Sensitive components of electronic devices are pro-
tected or shielded from external electric distur-
bances byplacing metal shields around them.
3.In a coaxial cable, the outer conductor connec-
ted to ground provides an electrical shield to
thesignals carried by the central conductor. •···3wx«n£m~r±n3lxv

117. 2.26
For Your Knowledge
~In the interior of a conductor,the electric f
ield and the
volume charge density both vanish. Therefore,
charges in aconductor can only be at the surface.
~Electric field at the surface of a chargedconductor
must be normal to the surface at every point.
~For a conductor without any surface charge, electric
field is zero even at the surface.
~The entire body of each conductor, including its
surface, is at a constant potential.
~If we have conductors of arbitrary size, shape and
charge configuration, then each conductor will have a
characteristic value of constant potential which may
differ from one conductor to another.
~A cavity inside a conductor is shielded from outside
electrical disturbances. However, the electrostatic
shielding does not work the other w round. That is,
if we place charges inside the cavity, the exterior of
the conductor cannot be shielded from the electric
fields of the inside charges.
2.17 ELECTRICAL CAPACITANCE OF A
CONDUCTOR
27. Defineelectrical capacitance of a conductor. On
which factors does it depend?
Electrical capacitance of a conductor. The electrical
capacitance of a conductor isthe measure ofitsability to hold
electriccharge. When an insulated conductor is given
some charge, it acquires a certain potential. If we
incree the charge on a conductor, its potential also
increes. If a chargeQput on an insulated conductor
increes its potential byV,then
QocV or Q=CV
The proportionality constantCis called the
capacitance of the conductor. Thus
. Charge
Capacitance= --"""-
Potential
Hence thecapatance of a conductor may be defined
as the charge required to increase the potential of the conductor
by unit amount.
The capacitance of a conductor is the meure of its
capacity to hold a large amount of charge without running
a high potential. It depends upon the following factors:
1.Size and shape of the conductor.
2.Nature (permittivity) of the surrounding medium.
3.Presence of the other conductors in its neigh-
bourhood.
or
It is worth-noting that the capacitance of a con-
ductor does not depend on the nature of its material
and the amount of charge existing on the conductor.
PHYSICS-XII
28. Define the unit of capacitance fora conductor.
Give its dimensions.
Units of capacitance. The51unit of capacitance is
farad(F),named in the honour ofMichael Faraday.
The capacitance of conductor is1faradifthe addition
of acharge ofl coulomb toit,increases its potential by1volt.
:.1farad =1coulomb or 1F=1C=1cv'
1volt 1 Y
One farad is a verylarge unit of capacitance. For
practical purposes, we use its following submultiples:
1millifarad=1mF=10-3F
1microfarad=IIlF=10-6F
1picofarad=1pF=10-12 F
Dimensions of capacitance.The unit of capacitance is
1F_1C _1C_1C2_1(As)2
------------
1Y 1J /C 1J1Nm
..Dimensions of capacitance
2 2
A T =[~lL-2y4A2]
MLr2.L
2.18CAPACITANCE OF AN ISOLATED
SPHERICAL CAPACITOR
29. Obtain anexpression for the capatance of an
isolated spherical conductor of radius R.
Capacitance of an iso-
lated spherical conductor.
Consider an isolated sphe-
rical conductor of radiusR.
The chargeQis uniformly
distributed over its entire
surface. It can be sumed
to be concentrated at the
centre of the sphere. The
potential at any point on the
Fig. 2.44Capacitance of a
surface of the spherical
spherical conductor.
conductor will be
V=_I_. Q
41t EO R
Capacitance of the spherical conductor situated
in vacuum is
Clearly, thecapacitance ofaspherical conductor is
proportional to its radius.~~~1styjxiwn}j1htr

118. ELECTROSTATIC POTENTIAL AND CAPACITANCE
Let us calculate the
radius of the spherical
conductor of capacitance 1 F.
R=_1_. C=9x109mF-1.1 F
4rcSo
= 9x109m =9 x106km
This radius is about 1500 times the radius of the
earth(-6x 103 km).So we conclude:
1.One farad is a very large unit of capacitance.
2.It is not possible to have a single isolated
conductor of very large capacitance.
For Your Knowledge
~The formula: C=41t EORis valid for both hollow and
solid spherical conductors.
C
As E=--
o41tR
So the51unit ofEOcanbe written farad per metre
(Fm-1). From Coulomb's l, the 51unit ofEOcomes
out to beC2N-1m-2. Both ofthese units are equivalent.
~The farad(1F =1Cy-l) is an enormously large unit
of capacitance because the coulomb is a very big unit
ofcharge while the volt istheunitof potential having
reonable size.
Formulae Used
1.Capacitance of a spherical conductor of radiusR,
C=4rcso R
. Charge
2. Capacitance=-----'~
Potential
or
Units Used
Charge is in coulomb, potential in volt and
capacitance in farad (F).
Example32.An isolated sphere has a capacitance 50pF.
(i) Callate its radius. (ii)How much charge should be
placed011it to raise its potential to104V?
Solution. Here C= 50 pF = 50x10-12F,V= 104 V
(i)R=_1_. C=9x109mF-1 xSOx10-12F
4rcSo
=45x1O-2m =45em.
(ii)q=CV= 50 x 10-12 x104= 5 x 10-7 C =0.5 1lC.
Example 33. Twenty seven spherical drops of radius3mm
and carrying10-12Cof charge are combined toform a single
drop.Find the capatance and the potential of the bigger
drop. [Haryana 01]
2.27
Solution. Let rand R be the radii of the small and
bigger drops, respectively.
Volume of the bigger drop
= 27xVolume of a small drop
. ArcR3=27xArcr3
t.e., 3 3
or R=3r=3x3mm=9x10-3m
.'. Capacitance of the bigger dropis
C= 4rcso R=_1-9.9 x10-3F
9x10
=10-12 F=1pF
Charge on bigger drop
q= 27xCharge on a small drop
=27x 10-12 C
.'. Potential of bigger drop is
q27x10-12
V=-= =27V.
C 10-12
Example34.Eight identical spherical drops, each carrying
a charge1nC are at a potential of900 Veach. All these drops
combine together to form a single large drop. Callate the
potential of this large drop. (Assume no wastage of any kind
and take the capatance of a sphere of radius r as
proportional to r). [eBSESample Paper15]
Solution.
Capacitance of each small drop, Cex:r=>C=kr
Charge on each small drop, q=CV=(krx900)C
Charge on large drop, q=8q=7200krC
Volume of a large drop = Volume of 8 small drops
ArcR3=8xArcr3 => R=81/3r=2r
3 3
Capacitance of large drop, C' =kR= 2kr
Hence, the potential of the large drop is
V' =!L=7200kr=3600V.
C' 2kr
Example3S.A charged spherical conductor has a surface
charge density of0.07 Cem-2.When the charge is increased
by4.4C,the surfacecharge density changes by
0.084 Cem-2.Find the initial charge and capacitance of the
spherical conductor.
Solution. Let qbe the charge on the spherical con-
ductor andrits radius. Its surface charge density is
-q- = 0.07 Cern-2 (i)
47t? ...
When the charge is increed by 4.4 C, the surface
charge density becomes
q+4.4 = 0.084 C cm-2 ( ..)
2 •••II
4rcr:···3wx«n£m~r±n3lxv

119. 2.28
Dividi
ng equation(ii)by(I),we get
q+4.4 0.084 C
--=-- orq=22
q 0.07
From equation(i),we get
~ 22x7
r- - =5cm=0.05 m
- 41tx0.07-4x22x0.07
Capacitance,
C= 41t EO r=__1-9x0.05=5.56x10-12 F.
9x10
j2)roblems ForPractice
1.Find the capacitance of a conducting sphere of
radius10cmsituatedinair.How much charge is
required to raise it to a potential of1000volt?
(Ans.11pP,1.1x10-8 C)
2.Assuming the earth to be a spherical conductor of
radius 6400km,calculate its capacitance.
[Himachal 98C; Haryana 98C]
(Ans.711IlF)
3.Ndrops of mercury of equal radii and possessing
equal charges combine to form a big drop. Compare
thecharge, capacitance and potentialof bigger drop
withthe corresponding quantities ofindividual drops.
[Punjab01]
(Ans.N,N1/3, N2/3)
HINTS
1.C = 41t EoR = _1-9 x 0.10=11x10-12 F=11pF.
9xlO
q=CV= 11x10-12p xl00G V = 1.1x10-8 C.
1 6
2.C=41tEoR=--9 x6.4xl0
9x10
=0.711x1O-3p =711 IlF.
3.Letqbe the charge on each small drop andrits
radius.
Capacitance of each small drop, C=41tEOr
Potential of each small drop, V=_1_ !1
41tEor
IfRis the radius of the big drop, then
4 3 4 3 1/3
-nrxN=-1tR orR=N r
3 3
Chargeon the big drop,q'=Nqor
Capacitance of thebigdrop.
C'=41tEOR=41t Eo N1/3 r=NI/3 C
e=N1/3.
C
or
PHYSICS-XII
Potential of the bigdrop,
V'-1q' _1 Nq
- 41tEO•R-41tEo.NI/3 r
=N2/3 .__~.!1=N2/3 V
47tEor
or
2.19CONCEPT OF A CAPACITOR AND
ITS PRINCIPLE
30.Anisolated conductorcannot have alarge
capacitance, why?
The capacitance of an isolated conductor is small.
When a conductor holds a large amount ofcharge, its
potential isalsohigh. If the sociated electric field
(£=a/Eo) becomes high enough, the atoms or
molecules of the surrounding airget ionised. A
breakdown occurs in the insulation of the surrounding
medium and the charge put on the conductor gets
neutralised or les . Forair,the bredown
point occurs at fields ofthe order of3x106Vm-1.This
putsthelimit on the capacitance of a conductor.
Moreover, if wetend to have a single conductor of
large capacitance, it will have practically inconvenient
large size.
31.Why does the capacitance of a conductor increase,
when an earthed connected conductor is placed near it ?
Briefly explain.
Principle of a capacitor.Consider a positively
charged metal plateAandplacean uncharged plate B
closeto it,as shown in Fig. 2.45.Due to induction, the
closer face of plate B acquires negative charge and its
farther face acquires apositive charge. The negative
charge on plate Btends to reduce the potential on plate
A,while thepositive charge on plate B tends to
incree thepotential on AAsthe negative charge of
plate Biscloser toplateAthan its positive charge, so
the net effect is thatthe potentialofAdecrees by a
smallamount and hence its capacitance increases by a
smallamount.
A B A B
+ + -+ ++
+ + -+ ++
++ -+ + +
++ -+ + +
++ -+ ++
+ + -+ + +
++ -+ ++
-i-
+ -+ ++
++ -+ ++
++ - + + +
+ + -+ ++
+"+ - + ++ -
++ -+ ++
++ -+ + +
++ -+ + +
+ + -+ + +
+ + - + + +
Fig. 2.45Principle of a capacitor.···3wx«n£m~r±n3lxv

120. ELECTROSTATIC POTENTIAL AND
CAPACITANCE
Now if the positive face of plate B is earthed, its
positive charge gets neutralised due to the flow of
electrons from the earth to the plateB.The negative
charge on Bisheld in position due to the positive
charge on A.The negative charge on Breduces the
potential of Aconsiderably and hence increes its
capacitance by a large amount.
Hence we see thatthe capacitance of an insulated
conductorisconsiderably increased when weplace an
earthed connected conductor near it. Such a system of two
conductors is called a capator.
32.Whatisa capator?Define capatance of a
capator. On what factors does itdepend ?
Capacitor. A capatorisan arrangement of two
conductors separated by an insulating medium that isused
to store electric charge and electric energy.
A capacitor, in general, consists of two conductors
of any size and shape carrying different potentials and
charges, and placed closed together in some definite
positions relative to one another.
Pictorial representation of a capacitor. The pictorial
symbol for a capacitor with fixed capacitance is
shown in Fig.2.46(a)and for that with a variable capa-
citance is shown in Fig. 2.46(b).
--111---
(a) (b)
Fig. 2.46Symbols for a capacitor with
(a)fixed,(b)variable capacitance.
Capacitance of a capacitor. As shown in Fig. 2.47,
usually a capacitor consists of two conductors having
charges+Q and - Q. The potential difference between
themisV=V+- V_. Here Q is called the charge on the
capacitor. Note thatthecharge on capatordoes not
mean the total charge given to the capacitor which is
+ Q-Q=O.
Fig. 2.47 Two conductors separated by an
insulator form a capacitor.
For"agiven capacitor, the charge Q on the capacitor
isproportional to the potential differenceVbetween
the two conductors. Thus,
QocV or Q=CV
2.29
The proportionality constant Ciscalled thecapa-
citance of the capator. Clearly,
C=Q
V
C. Charge on either conductor
or apatance= ---~---------
P.O.between the two conductors
Thecapacitanceof a capatormay be defined as the charge
required to besupplied to either of the conductors of the
capacitor so as to increase the potentialdifferencebetween
them by unit amount.
The capacitance of a given capacitor is a constant
and depends on thegeometric factors,such the
shapes, sizes and relative positions of the two cond-
uctors, and the nature of the medium between them.
SI unit of capacitance is farad (F). Acapator has a
capatance of 1faradif1coulomb of chargeistransferred
fromits one conductor to another on applying a potential
difference of1volt across the two conductors.
2.20 PARALLEL PLATE CAPACITOR
33.Whatisa parallel plate capacitor ?Drive an
expression for itscapatance. On what factors doesthe
capacitance of a parallel plate capator depend?
Parallel plate capacitor.The simplest and the most
widely used capacitor is the parallel plate capacitor. It
consists of two large plane parallel conducting plates,
separated by a small distance.
LetA=area of each plate,
d=distance betweenthetwo plates
±c=uniform surface charge densities on the
two plates
±Q=±oA=total charge on each plate.
Area=A E=O
~I~~_~_~~~ ~~I
I+1+1+ftl+r=~~·
'-- --:::--:::-- ---'1density -cr
E=O
Fig. 2.48 Parallel plate capacitor.
In the outerregions above the upper plate and
below the lower plate, the electric fields duetothe two
charged plates cancel out. The net fieldiszero.
E=~-~=O
2Eo 2Eo
In the inner regionbetween the two capacitor
plates, the electric fields due to the two charged plates
add up. The net field is
E=~+~=~
2Eo2Eo EO···3wx«n£m~r±n3lxv

121. 2.30
The
direction of the electric field is from the
positive to the negative plate and the field is uniform
throughout. For plates with finite area
bend at the edges. This effect is calledfringingof the
field.But for large plates separated by small distance
(A» d2), the field is almost uniform in the regions far
from the edges. For a uniform electric field,
P'D.between the plates
=Electric fieldxdistance between the plates
V=Ed=ad
EO
Capacitance of the parallel plate capacitor is
C=Q= ~ or C=EoA
V ad /EO d
Factors on which the capacitance of a parallel
plate capacitor depends
1.Area of the plates(CocA).
2.Distance between the plates(Coc1 /d).
3.Permittivity of the medium between the plates
(COCE).
or
2.21SPHERICAL CAPACITOR*
34. Whatisa spherical capator?Derive an
expression for its capatance.
Spherical capacitor. A spherical capator consists of
two concentric spherical shells of inner and outer radii a and
b.The two shells carry charges - Qand+Q
respectively. Since the electric field inside a hollow
-t .
conductor is zero, soE=0forr<a.Also the field is
-t
zero outside the outer shell,i.e.,E=0forr>b.A radial
~
fieldEexists in the region between the two shells due
to the charge on the inner shell only.
To determine the electric field at any point P at dis-
tancerfrom the centre, consider a concentric sphere of
radiusr the Gaussian surface. Using Gauss's theorem,
4t=E.4n? =QorE=~
EO 4nEor
£=0 Charge+Q
+
Charge-Q
Gaussian
surface
Fig. 2.49Spherical capacitor.
PHYSICS-XII
The potential difference (caused by the inner
sphere alone) between the two shells will be
b b b
V= -fE.;t=fEdr=f ~dr
a a a4nEO'
Qb-2 Q[l]bQ[1 1]
=,4nEO ~r dr=4nEO -; a=4nEO -;; -b
-t -t
[.: Epoints radially inward anddrpoints
-t -t
outward soE . d r=Edr1800= -Edr]
The capacitance of the spherical capacitor is
Q Q orC=4nEOab .
C=V= Q[1 1] b-a
4nEOab
2.22CYLINDRICAL CAPACITOR*
35. What is a lindrical capator?Derive an
expression for its capatance.
C}lind.ica capacitor. Alindrical capator consists
of two coaxial conducting linders of inner and outer radii a
and b.Let the two cylinders have uniform linear charge
densi ties of±A.Cm-1.The lengthLof the capacitor is
so large(L»radiiaorb)that the edge effect can be
neglected. The electric field in the region between the
two cylinders comes only from the inner cylinder, the
outer cylinder does not contribute due to shielding. To
calculate the electric fieldEat any pointPin between
the two cylinders at a distancerfrom the central axis,
we consider a coaxial Gaussian cylinder of radiusr.
Using Gauss's theorem, the flux through Gaussian
surface must be
b a
Charge
density + A.
+1-
1
+1-
1
+1
1
1
+1
1
+1-
1
+1--J..-I---L-
Charge
density - A.
Gaussian
cylinder
Fig.2.50«««2uv{lzkyp~l2jvt

122. ELECTROSTATIC POTENTIAL AND CAPACITANCE
or E .21trL=')...
L
eo
E=_J..._
21teo r
:. Potential difference between the two cylinders is
b-+-+ b
V=-fE.dr=fEdr
aa
[.: Eandd-;are in
opposite directions]
or
b J... J...b1
= f --dr=-f -dr
a21teo r21teoar
J... b J...
=-[lnr] =-[lnb-lna]
21teo a2 1teo
V=~ln~
21teoa
Total charge on each cylinder is Q= LA
.. Capacitance of cylindrical capacitor is
C=Q= LA or C=21teoL
V_J..._ln~ ln~
21teoa a
Exam /es basedon
-.--. .•.......
Formulae Used
1.Capacitance,C=3..
V
2.Capacitance of a parallel plate capacitor,C=BodA
3. P.D. between the two plates of a capacitor having
charges%andq2'
V=q1-q2
2C
4. Capacitance of a spherical capacitor,C=41tSo..!!!!.....
b-a
Hereaandbare the radii of inner and outer shells
of the spherical capacitor.
S.Capacitance of a cylindrical capacitor,
L L
C = 21tSo--b =21tSo b
loge- 2303 log10-
a a
Hereaandbare the radii of inner and outer coaxial
cylinders and L is the length of the capacitor.
Units Used
Capacitance C is in farad, chargeqin coulomb,
potential differenceVin volt, thicknessesdandt
in metre.
Constant Used
Permittivity constant,EO= 8.85 x 10-12 C2N-1m-2
2.31
Example 36.When1.0x1012electrons are transferred
from one conductor to another of a capator, a potential
difference of10V develops between the two conductors.
Callate the capatance of the capator.
Solution.Hereq=ne=1.0 x 1012.x 1.6 x 10-19
=1.6 x 10-7 C
V=lOV
..C=..i=1.6 x 10-7 =1.6x10-8F.
V 10
Example 37. Acapator of unknown capatance is
connected across a battery of V volts. The charge stored in it
is360~c.When potential across the capator is reduced by
120V, the charge stored in it becomes120~c.Callate:
(i) The potentialyand the unknown capatanceC
(ii) What will be the charge stored in the capator,if
the voltage applied had increasedby 120 V?
[CBSE D 13]
Solution.(i)LetCbe the capacitance of the capa-
citor andVthe potential drop across the plates. Then
q= CV=360~C
When the potential difference is reduced by 120 V,
if=C(V-120)=120 ~C
.._V_=36O =3
V-120 120
C =..i= 360 ~C = 2 F.
V180V ~
V=180V
(it)When the volte is increed by 120 V,
l'=C(V+120) =2~Fx (180+120) =600~C
Example 38.Aparallel plate capator has plate area of
25.0artand a separation of2.0 mm between its plates. The
capator is connected to12V battery.(j)Find the charge on
the capator. (ii) If the plate separation is decreased by
1.0 mm what extra charge is given by the battery to the
positive plate?
Solution.A=25.0 cm2 =25 x 10-4 m2,
d= 2.0mm= 2x10-3m,V= 12 V
C=eoA= 8.85 x 10-12 x 25 x 10-4 =1.1 x 10-11 F
d 2x10-3
(i)q=CV= 1.1 x 10-11 x 12 =1.32x10-10C
(ii)Hered'= 2.0 -1.0 = 1.0 mm= 1 x 10-3 m
C'= 8.85 x 10-12 x 25 x 10-4 =2.2 x lO-11F
.. Ix10-3
if=cv=2.2 x 10-11 x 12 =2.64 x 10-10 C
Extra charge given by the battery to the positive
plate is
q' - q=(2.64-1.32)x 10-10 =1.32x10-10C.···3wx«n£m~r±n3lxv

123. 2.32
Example 39
.Two parallel plate aircapacitors have their
plateareas100and500err?-respectively. If theyhave the
samechargeandpotential and the distance between the
plates of the first capacitor is0.5mm whatisthedistance
between the plates of the second capacitor ?[Punjab97C]
Solution. As capacitance, C=q /Vand the two
capacitors have the same chargeqand potentialV,so
they have the equal capacitances, i.e.,
C1=C2
EO~= EO~
d1. d2
d=~d
2 ~ 1
But ~ =100 cm2,'~ =500 crn2,
d1= 0.5 mm = 0.05 ~m
d-500x0.05 -025 - 2 5
. . 2 - 100 -. em -.nun.
or
or
Example40.Asphere of radius 0.03mis suspended within
a hollowsphere of radius 0.05m.If the inner sphereis
charged to a potential of1500 volt and outer sphere isearthed,
findthe capacitance and the charge on the inner sphere.
Solution. Here a= 0.03 m,b=0.05m,V= 1500 V
The capacitance of the air-filled spherical capacitor is
41tEOab 0.03x0.05
C=--"--
b-a9x109x (0.05 -0.03)
=8.33x.10-12 F=8.33pF.
Charge, q=CV=;8.33x10-12x1500
=1.25x10-8C.
Example 41.Thethickness of air layer between thetwo
coatings of a spherical capator is2em The capator has
the same capatance as the sphere ofl.2 mdiameter. Find
the radii0/its surfaces.
41tEab
Solution.Here 0=41t EO R
b-a
~=R
b-a
Nowb -a=2 cm and R=.!2 m =60 ern
2
",~..
·or
ab =60
2
or ab·=!120
(b+a)2=(b - a)2+4ab
= 22+4x 120 =484
or b+a=22
or 2+a+a=22 [':b-a=2crn]
a=10emandb=12 em.
PHYSICS-XII
Example 42. Thenegativeplate ofaparallel plate capator
isgiven a charge of - 20x10-8C.Find the charges
appearing on thefour surfaces of the capator plates.
Solution. As shown in Fig. 2.51, let the charge
appearing on the inner surface of the negative plate be
- Q.Then the charge onits outer surface will be
Q-20 x10-8C
12 3 4
•p
-8
Q -20x10 C-Q +Q -Q
Fig. 2.51
The induced charge on the inner surface of the
positive platewill be +Qand that on the outer surface
willbe-Q,asthe positive plate is electrically neutral.
To find Q,we consider the electric field at apointP
inside the negative plate.
Field due to surface 1= ~, towards left
2EoA
Field due tosurface 2= ~, towards right
2EoA
Field due to surface3=~, towards left
2EoA
. 20x10-8C
Field due to surface 4= ,towards left
2EoA
Asthe pointPlies inside the conductor, thefield
here must be zero.
~_~+~+ Q-20x10-8 =0
2EoA2EoA2EoA 2EoA
2Q-20x10-8 =0'
Q=+10x1O-8C
:. Charge on surface1=-10x10-8C
Charge onsurface 2=+10x10-8C
Charge on surface3==:10x10-8C
Charge on surface4=-10x10-8C.
problems For Practice
or
1.Acapacitor of20J.1Fis charged to a potential of
10 kV. Find the charge accumulated on each plate
of the capacitor. (Ans,0.2 C)···3wx«n£m~r±n3lxv

124. ELECTRO
STATIC POTENTIAL AND CAPACITANCE
2.Calculate the capacitance of a parallel plate capa-
citor having circular discs ofradii 5.0 cm each. The
separation between the discs is 1.0 mm.
(Ans.0.69 x10-10F)
3.A parallel plateair capacitor consists of two circular
plates ofdiameter 8cm. At what distance should
theplatesbeheld soasto have the same capa-
citance thatof a sphere of diameter 20 cm?
(Ans.4mm)
4.A parallel-plate capacitor h plates of area200 em2
andseparation between the plates 1.0 mm. (i)What
potential difference will bedeveloped if a charge of
1.0 nCis given to the capacitor? (ii)Ifthe plate
separation isnow increed to 2.0 mm, what will be
the new potential difference? (Ans. 5.65 V, 11.3 V)
5.Two metallic conductors havenetcharges of
+70 pCand - 70 pC, which result ina potential
difference of 20 V between them.What isthe
capacitance of the system? (Ans. 3.5pF)
6.A spherical capacitor h an inner sphere of radius
9 emandan outer phere of radius 10em. Theouter
sphere is earthed and the inner sphere is charged.
What is the capacitance of the capacitor?
(Ans.0.1 nF)
7.Thestratosphere acts a conducting l er for the
earth. If the stratosphere extends beyond 50krn
from the surface of the earth, then calculate the
capacitance of the spherical capacitor formed
between stratosphere and earth's surface. Te radius
of theearth 6400krn. (An.0.092F)
8.Acharge of+2.0 x 10-8 C is placed on the positive
plate and a charge of -1.0 x10-Con the negative
plateofaparallel plate capacitor of capacitance
1.2x10-3~F.Calculate the potential difference
developed between the plates. (Ans. 12.5 V)
HINTS
1.C=20 ~F=20xlO-6F, V=10 kV=104V
Charge, q=CV=20 x 10-6 x 104C =0.2C
2.Here r=5.0em=0.05 m,d=1.0mm=10-3m
Capacitance,
3.
EoA eonr2
C=-=--
d d
nx(0.05)2 -10 .
= 9 3 =0.69x10 F.
4nx9x10x10
eA Eonri
_0_=4nl'.~Ror --- =4ne R
d -u 4d 0
d=.s:=(0.08)2 =4x10-3m=4mm.
16 R 16x0.10
or
2.33
c-EoA _8.85 x10-12x 200x10-4
-d - Lx10-3
=0.177 x10-9F=0.177 nF
. q 1nC
(I)V=-= =5.65 V.
C0.177nF
(ii)When the plate separation increes from
1.0mm to 2.0 mm, the capacitance decrees by
a factor of2.Forthesame charge, the potential
difference will increase by a factor of 2.
:. V'=2V=2x5.65=11.3 V.
5.Charge on the capacitor,
q=70 pC=70x10-12C
C=.i=70x10-12C = 3.5pF.
V 20V
4.
6.Herea=9em =0.09 m, b=10cm=0.10 m
C=41tEoab=_1_. 0.09x0.10 F
b - a9 x 109 (0.10-0.09)
=0.01xO.10xlO-9F =0.lx10-9F =0.1 nF.
0.01
7.Here
a=radius ofthe earth=6.4 x106m
b=distance of the stratosphere layer from the
centre oftheearth
=6400+50=6450km=6.45x106m
ab 1 6.4 x 106x6.45x106
C=4nEoa _ b=9x109x (6.45 _6.4)x106
=0.092F.
8.V= ~ -q2=20x10-8 +1.0x10-8 =12.5V.
2C 2 x1.2x109
2.23 COMBINATION OF CAPACITORS IN
SERIES AND IN PARALLEL
36.Anumber ofcapacitors are connected inseries.
Deriveanexpression for the equivalent capatance of the
series combination.
Capacitors in series.When thenegative plate of one
capator is connectedtothepositive plate of the second, and
the negative of the second tothepositive of third and soon,
the capacitors are said to be connected inseries.
Figure252shows three capacitors of capacitances
C1, C2and C3connected inseries. Apotential
difference Visapplied across the combination. This
setsup charges±Qonthetwo plates of each capacitor.
What actually happens is, a charge +Qis given to the
left plate of capacitorC1during thecharging process.
Thecharge+Qinduces a charge-Qonthe right plate
of C] and a charge-Qon the left plate ofC2,etc.···3wx«n£m~r±n3lxv

125. 2.34
+Q-Q +
Q -Q +Q -Q
:=H:=H:
+- + - +
+ - + - +
CI Cz C3
~ VI ~I~ V2 ~I~ V3--+I
L..--------o Vo-------~
(+)H
Fig. 2.52Capacitors in series.
The potential differences across the various
capacitors are
Q Q Q
VI=-,V2=-,V=-
CI C23C3
For the series circuit, the sum of these potential diffe-
rences must be equal to the applied potential difference.
V=V +V +V =Q+Q+Q
1 2 3Cl C2 C3
VII 1
or -=-+-+- ...(1)
QCl C2 C3
Clearly, the combination can be regarded an
effective capacitor with charge Q and potential dif-
ferenceV.If Cs is the equivalent capacitance of the
series combination, then
C=Q
sV
1 V
CsQ
or
From equations(1)and(2),we get
111 1
-=-+-+-
c,C1 C2 C3
For a series combination ofncapacitors, we can
write
111 1
-=-+-+ .....+-
c,Cl C2 c,
For series combination of capacitors
1.The reciprocal of equivalent capacitance is equal
to the sum of the reciprocals of the individual
capacitances.
2.The equivalent capacitance is smaller than the
smallest individual capacitance.
3.The charge on each capacitors is same.
4. The potential difference across any capacitor is
inversely proportional to its capacitance.
PHYSICS-XII
37. A number of capacitors are connected in parallel.
Derive an expression for the equivalent capatance of the
parallel combination.
Capacitors in parallel. When the positive plates of all
capacitors areconnected to one common point and the
negative plates to another common point, thecapators are
said to be connected in parallel.
Figure 2.53 shows three capacitors of capacitances
Cl'C2and C3 connected in parallel. A potential
differenceVis applied across the combination. All the
capacitors have a common potential differenceVbut
different charges given by
Ql=c,V,Q2= C2V,Q3= C3V
+ -
V
(+) (-)
Fig. 2.53Capacitors in parallel.
...(2)
Total charge stored in the combination is
Q = Q1 +Q2+Q3=(C1 +C2o+C3)V ...(1)
If Cis the equivalent capacitance of the parallel
bir, th
com mation, en °
Q=CpV ...(2)
From equations (1)and(2),we get
CpV":(C1 +C2+C3)V
or C p= Cl+ C2+ C3
For a parallel combination ofncapacitors, we can write
Cp=Cl-+C2+ .....+Cn
For parallel combination of capacitors
1.The equivalent capacitance is equal to the sum
of the individual capacitances.
2.The equivalent capacitance is larger than the
largest individual capacitance.
3.The potential difference across each capacitor is
same.
4. The charge on each capacitor is proportional to
its capacitance.···3wx«n£m~r±n3lxv

126. ELECTROSTATIC POTENTIAL AND CAPACITANCE
iliiiiiiiiiiiiiiiiiii~LE~
xampleS_b_a.s __e_d_O_n -----'
Formulae Used
1.In series combination,
1 1 1 1
-=-+-+-+ ...
Csc,.CZC;
2.In parallel combination, Cp=c,.+CZ+ C; +...
3.Inseries combination, charge on each capacitor is
same (equal to thecharge supplied by battery) but
potential differences across the capacitors may be
different.
4.Inparallel combination, potential difference on
each capacitor is same but the charges on the capa-
citors m be different.
Units Used
Capacitances are in farad, potential differences in
volt and charges in coulomb.
Example 43.Two capators ofcapatance of6!IFand
12 !IFare connected inserieswith a battery. Thevoltage
across the 6!IFcapatoris2V.Compute the total battery
voltage. [CBSE 0006]
Solution. As the two capacitors are connected in
series, thecharge on each capacitor must be same.
Chargeon 6!lF = Charge on 12 !IF
capacitor capacitor
or 6 !IF x2 volt=12 !IFxVvolt
6x2
:.PD.across 12 !IF capacitor =-- =1volt
12
Battery volte =VI+V2=2 V + 1 V = 3V.
Example44.Two capators ofcapatances3!lFand
6 !IF,arecharged to potentials of2 V and 5V respectively.
These two chargedcapacitors are connected inseries. Find
the potential across each of the two capators now.
[CBSE SamplePaper 04]
Solution. Total charge onthe two capacitors
=CIVI +C2V2 =(3x2 +6x 5)!lC=36!lC
Inseries combination, charge is conserved.
..Charge on either capacitor,
q= 36!lC
Potential on 3 !IF capacitor=!L= 36 !lC = 12 V
CI 3!lF
Potential on 6 !IF capacitor=!L= 36 !lC =6V.
C2 6!lF
Example 45.Two capators have a capatance of5!lF
when connectedinparallel and1.2 !IFwhen connectedin
series. Callate their capatances.
2.35
Solution. Letthe two capacitances be CI !IF and C2 !IF.
Inparallel, C p= CI+ C2 = 5 !IF
CC
Cs= 1 2= 1.2 !IF
CI+C2
CI(5 - CI) =1.2
5
2
CI- 5CI + 6 = 0
Hence, CI= 2 or 3!lF
:.The capacitances are of2!lFand31lF.
Example46.Three capators ofequalcapacitance, when
connectedinseries have netcapacitance CI,and when
connectedinparallelhquenet capatanceC2.Whatisthe
value ofCI/C2?
Solution. Let C = capacitance of each capacitor.
For series combination,
In series,
or
or
1 1 1 1 3
-=-+-+-=- or
CIC C C C
For parallel combination,
CI_C 1_1
C2-3"'3C-9
Example 47. In Fig. 2.54,each of the uncharged capators
has a capacitance of25 !IF.What charge will flow through
themeterMwhen the switch5isclosed?
C2= C+C + C=3C
r~
4200 V
l~_I JIe
Fig. 2.54
Solution. As the three capacitors are connected in
parallel, their equivalent capacitance is
Cp= C + C + C =3C =3x25!lF =75!lF
V=4200 V
Charge,q=CpV= 75x10-6x4200
= 315x10-3C =315 me
Example 48. Callate the charge supplied by the battery
inthe arrangement showninFig. 2.55.
Fig. 2.55
10V···3wx«n£m~r±n3lxv

127. 2.36
Solution.T
he given arrangement is equivalent to
the arrangement shown in Fig. 2.56.
c1=5J.1f
II
+11-
II
+11
Cz=6J.1f
+I-
I
10V
Fig.2.56
Clearly, thetwocapacitors are connected inparallel.
Their equivalent capacitance is
C= C1+ C2=5 + 6 = 11 ~F
Charge supplied bythebattery is
q=CV=11~Fx10 V =110J..Ic.
Example 49.Three capators C1, C2andC3areconnected
toa6V battery, as showninFig.2.57.Find the charges on
the three capators.
I
Fig. 2.57
Solution. The given arrangement is equivalent to
thearrangementshown in Fig. 2.58(a).
6V
+ -
(a)
C1=10 IlF C'=10 IlF
C:J
6V
(b)
Fig. 2.58
PHYSICS-XII
Clearly, C2 and C3 areinparallel. Their equivalent
capacitance is
C'= C2+ C3= 5 + 5 = 10 ~F
NowC1andC'form a series combination,
shown in Fig.2.58(b). Their equivalent capacitance is
C=C1C= 10x10 =5 ~F
C1+C10 + 10
Charge drn from the battery,
q=CV=5 ~Fx6V=30~C
Charge onthe capacitor C1=q=30~C
Charge on the parallel combination of C2and
C3=q=30 ~C
AsC2andC3are equal, so the charge is shared
equally by the two capacitors.
30
Charge on C2 = charge on C3 =-= 15~C
2
Example50.Find the equivalent capatance of the
combination of capators between the points A andBas
shownin Fig.2.59.Also callate the total charge that flows
in the rit when a 100 V batteryisconnected between the
points A andB. [CBSE D02]
40J.1f 60J.1f
Ao-------i ~
I ~J.1f
lOp!' ~ 6Op!'
L---~----------~---oB
Fig.2.59
Solution.Here three capacitors of 60 ~F each are
connected inseries. Their equivalent capacitance C1is
given by
1 1 1 1 3 1
-=-+-+-=-=-
•C160 60 60 60 20
or C =20~F
The given arrangement now reduces
equivalent circuit shown in Fig. 2.60(a)
40J.1f
(,)Ao-i::-Lo,I1,0,'
TT T
to the
(b)
40 IlF 40 IlF
AOI----IIf---IIf---O B
Fig.2.60~~~1styjxiwn}j1htr

128. ELECTROSTATICPOTENTIAL
AND CAPACITANCE
Clearly, the three capacitors of IOIlF, IOIlF and
20 IlF are in parallel. Their equivalent capacitance is
C2=10 + 10 + 20 = 40 IlF
Now the circuit reduces to the equivalent circuit
shown in Fig. 2.60(b). We have two capacitors of 40 IlF
each connected in series. The equivalent capacitance
betweenAandBis
40x 40
C= =20IlF.
40 + 40
GivenV=100V
.'.Charge, q=CV=20~Fx100V
= 2000 ~C =2 mC
Example51.IfC1=3pFandC;=2pF,callate the
equivalent capatance of the given network between points
A andB.
Fig. 2.61
Solution. Clearly, capacitors 2,3and 4 form a
series combination. Their totalcapacitance C' is
given by
11111117
-=-+-+-=-+-+-=-
C'C1 C2 C13236
C'=~pF
7
The capacitance C'formsa parallel combination
with capacitor5,so their equivalent capacitance is
C" C' C 6 20F
= + 2=-+2=-p
7 7
The capacitance C"forms a series combination with
capacitors 1 and 6. The equivalent capacitance Cof the
entire network is given by
1 1 1 1 7 1 1 61
-=-+-+-=-+-+-=-
C C" C1 C120 3 3 60
C=60 F.
61 P
Example52.From the network shown in Fig. 2.62,find
the value of the capacitance Ciftheequivalent capatance
between points A andBisto be1~F.All the capacitances are
in~F.
2.37
L-------~-----------+----~B
Fig. 2.62
Solution. Capacitors C2andC3form a parallel com-
bination of equivalent capacitance,
Cs=C2+C3= 2 + 2 = 4~F
CapacitorsC4andCsform a series combination of
capacitanceC9given by
1111131
--= --+-- =-+ - =-=-
C9 C4c,12 6 12 4
C9=41lF
The equivalent circuit can be shown in Fig. 2.63(a)
Fig. 2.63(a)
CapacitorsC1andCsform a series combination of
capacitanceCIOgiven by
C_ C1 Cs_8x4 _32 _ ~F
10- - - - ~
C1+Cs8+412. 3
CapacitorsC6andC9form a parallel combination
of capacitance.
Cn=C6+ C9=4+4 =8 ~F
The given network reduces to the equivalent circuit
Fig.2.63(b).
Fig.2.63(b)~~~1styjxiwn}j1htr

129. 2.38
Again, capacitors
C7and Cn form a series combi-
nation of capacitanceC12given by
C_C7xCn_1x8 _ 8 \IF
12- - -- r
C7+Cn 1+89
Now ClO andC12form a parallel combination of
capacitanceC13 shown in Fig.2.63(c).
8832
C13=ClO+C12=3"+9'=9JlF
C
A~~
clOLL~C12 ~3 9
B
Fig. 2.63(c)
Finally, the capacitors C andC13form a series com- or
bination of capacitance 1 JlF shown in Fig. 2.63(d).
C C13
A~~~B
32
9
Fig.2.63(d)
1 1 9
-=-+-
1 C 32
or
32
C=-JlF.
23
Example 53. Connect three capators of3JlF,3 JlF
and6JlFsuch that theirequivalent capatance is5JlF.
Solution. Capacitors connected in parallel have
maximum equivalent capacitance.
Cmax= 3 + 3 + 6=12 JlF
Capacitors connected in series have minimum
equivalent capacitance.
1 1 1 1 5
--=-+-+-=-
Cmin 3 3 6 6
6
or Cmin =-= 1.2 JlF
5
The required equivalent capacitance of5JlF lies
between Cmax and·Cmin. So
3x6
5 JlF = 3 JlF + 2 JlF = 3 JlF+-JlF
3+6
So we should connect the series combination of3 JlF
and 6 JlF capacitors in parallel with the third capacitor
of 3 JlF.
Example54.Seven capacitors, each of capacitance 2JlFare
tobe connected inaconfiguration to obtain an effective
capacitance of10 / 11JlF.Suggest a suitable combination to
achievethedesired result. [lIT 90]
PHYSICS-XII
Solution. Suppose a parallel combination of n
capacitors is connected in serieswitha series
combination of (7 -n)capacitors.
Capacitance of parallel combination, C1 =2nJlF
2
Capacitance ofseries combination, C2= --JlF
7-n
Asthese two combinations are in series, so
C=10 F
511Jl
1 1 1 11 1 7-n
But -=-+- -=-+--
~ ~ ~ 10 2n 2
Multiplying both sides by 10n,we get
11n=5+3Sn -5n2
Sn2-24n-S =0or
(n - S)(5n+1)=0
n=5[Rejecting-vevalue]
Hence parallel combination of5capacitors must be
connected in series with the other 2 capacitors.
Examp e 55 Find the equivalent capatance between the
pointsPandQas showninFig.2.64.GivenC=18JlFand
C1=12JlF. REC 97]
:&lliDc
c E C B C
Fig 2.64
Equivalent capacitance between points FandBis
18x18+ 18= 27 F
18 + 18 Jl
Equivalent capacitance between pointsAandBis
18x27
12+-- =12 + 10.8 =22.8 "'-23 JlF
18 +27
Equivalent capacitance between points AandEis
23x18 +18= 28 F
23+18 Jl
Equivalent capacitance between points0andEis
28x18 + 12 =23F
28 +18 Jl
Equivalent capacitance between points 0and Qis
23x18 +18= 28 F
23+18 Jl
Equivalent capacitance between points PandQis
28x18= 11 F.
28 +18 Jl~~~1styjxiwn}j1htr

130. ELECTROSTATIC POTENTIAL
AND CAPACITANCE
Example56.Fourcapacitors are connected as shownin
the Fig. 2.65. Calculate the equivalent capacitance between
thepointsXandY. [CBSE D 2000]
xe1,1!~~hj;i~Y
Fig. 2.65
Solution. Clearly, the first plate of 2IlFcapacitor,
the second plate of 3 IlFcapacitor and the first plate of
5 IlFcapacitor are connected to the point A On the
other hand, the second plate of 2 IlFcapacitor, the first
plate of 3IlFcapacitor and the second plate of 5 IlF
capacitor are connected to the point B.Thus the capa-
citors of 2IlF,3IlFand 5IlFare connected in parallel
between pointsAand B, shown in the equivalent
circuit diram of Fig. 2.66.
21lF
Xo---A+----II----+-B---I~y
10JlF
Fig. 2.66
Total capacitance of the parallel combination of
capacitances21lF,3IlFand 5IlFis
C=2+3+5=IOIlF
As shown in Fig.2.67, this parallel combinationis
inseries with capacitance of 10 1lF.
xo>-----II--~II--Oy
Fig. 2.67
Equivalent capacitance between X and Y
=10x10=5 F.
10+10 Il
Example 57 .Five capacitors ofcapacitance 10IlFeach are
connected with each other, as shown in Fig. 2.68. Calculate
the total capacitance between the points A andC.
C4
A$~~c
Fig. 2.68
2.39
Solution. The given circuit canbe redrn in the
form of a wheatstone bridge shown in Fig. 2.69.
B
c'4~
A~~rc
C
Fig. 2.69
As C1 =C2=C4=Cs'
C C
Therefore, -.l=--.! .
C2c,
Thus the given circuit is a balanced wheatstone
bridge. So the potential difference across the ends of
capacitor C3 is zero. Capacitance C3 is ineffective. The
given circuit reduces to the equivalent circuit shown in
Fig.2.70(a).
Fig. 2.70(a)
Capacitors C1 and C2 form a series combination of
equivalent capacitance C6 given by
C=C1xC2=10x10=5IlF
6C1+C210+10
Similarly, C4and Cs form a series combination of
equivalent capacitance C7 given by
C=C4xCs=10x10=5IlF
7C4+Cs10+10
As shown in Fig. 2.70(b), C6and C7 form a parallel
combination. Hence the equivalent capacitance of the
network is given by
C=C6+C7=5+5=10IlF.
C6
A~~C
Fig. 2.70(b)···3wx«n£m~r±n3lxv

131. 2.40
Example58.The
re are infinite number of capators, each
of capatanceI/-1F.Thelj are connected in rows, such that
the number of capators inthe first row, second row, third
row,fourthrow,are respectively 1, 2, 4, 8, ..... The rows of
thesecapators are thenconnected between points A and B,
as shown in Fig. 2.71.Determinetheequivalentcapatance
of thenetwork between the points A and B.
1IlF
A B
Fig.2.71
Solution.LetCl,C2,CyC4,.....bethe effective or
capacitances of the capacitors of first row, second row,
third row, fourth row, ..... respectively. Then
Cl=IIJF
1xII
C2=-- =-/-IF
1+1 2
11111
-=-+-+-+-=4
C3 1 11 1
1
C3=4"/-IF
Similarly, C4=.!./-IF,and so on.
8
Astheserows are connected in parallel between
points Aand B, sothe equivalent capacitance between
points AandBis
1 1 1
C=Cl+ C2+ C3+ C4+ ..... =1 +- + -+-+ .....
2 4 8
This is an infinite geometric progression with first
terma=1andcommon ratio r=1 /2. Hence
a 1
C=-=--=2/-1F.
l-r 1-1/2
Example 59. Find the equivalent capacitor of the ladder
(Fig.2.72)between points AandB.
21lF 21lF
A~T,~T,~~-l"~
T T TIhh_
Bo
Fig.2.72
PHYSICS-XII
Solution. Let C be the equivalent capacitance of the
infinite network. It consists of repeating units of two
capacitors of I/-1Fand2/-IF.The addition of one such
more unit will not affect the equivalent capacitance.
But then the network would appear shown in
Fig.2.73.
Fig.2.73
The equivalent capacitance of the new arrangement
must be equal toC.
C=I+2xC
2+C
or C2-C -2=0
C = 2 /-IF or - 1 /-IF
Asthe capacitance cannot be negative, sothe
equivalent capacitance of the ladder is 2/-1F.
Example60.IfCl=20/-IF,C2=30 /-IF andC3=15 /-IF
andtheinsulated plate of Clbe at a potential of90 V,one
plate ofC3being earthed. What isthepotential difference
between the plates of C2' three capacitors being connected in
series? [CBSE OD 15]
Solution. Here Cl = 20/-lF, C2 =30 ~lF, C3= 15 /-IF,
V=90V
Fig. 2.74
The equivalentcapacitance C of the series com-
bination is given by
11111113
-=-+-+-=-+-+-=-
C Cl C2 C320 30 15 20
C=20 /-IF
3
Total potential difference =90 -0 =90 V
:.Totalcharge,
q=CV= 20x10-6.90 =600x10-6C
3
PD.between the plates of capacitor C2 is
V=..!L= 600x10-6C =20 V.
2C2 30x10-6F«««2uv{lzkyp~l2jvt

132. ELECTROSTATIC POTE
NTIAL AND CAPACITANCE
Example 61. In the rit shown in Fig. 2.75,ifthe point'
Cisearthedand point A isgiven a potential of + 1200V,find'
the charge on each capator andthe potential at the pointB.
C2
cb-[J
12000VI 4~F
A 3~F B 3
2 ~F
v=O
c
Fig. 2.75
Solution. Capacitors C;and C3 form a parallel com-
bination. Their equivalent capacitance is
C' =C2+C3= (4 + 2) J.lF= 6 J.lF
Now Cl andC'form a series combination, there-
fore,the equivalent capacitance of the entire network is
CC' 3x6
C=C + C' = 3+6 =2J.lF
Thecharge on the equivalent capacitor is
q=CV=2x10-6x1200 C =2.4 x10-3 C
This must be equal to the charge on C,and also the
sum of the charges on C2 andC3•Thus
V _v:=-.:L= 2.4x10-3=800V
A BCl 3x10-6
VA=1200V
.. VB =1200-800=400 V
HenceVe - VB= 400-0 =400V
q2=C2(Ve - VB)=4x10-6x400C ==1.6x10-3C
q3==C3(Ve-VB)=2x10-6 x400C = 0.8x10-3C
ql ==q=2.4x10-3C.
Example 62. Anetwork of four10 J.lFcapators is
connected to a500V supply as shown in Fig. 2.76.Deter-
mine (a) the equivalent capatance of the network, (b)the
chargeon each capator. [NCERT]
+q -q
Brlhc
_q.L.+c,-" ~+q
CI C3
+q++++ -----q
+q'_q'
+
A
+
D
+
+
C4
SOOV
Fig. 2.76
2.41
Solution. (a)In the given network, Cl' C2 and C3 are
connected in series. Their equivalent capacitance CIis
given by
11111113
-==-+-+-==-+-+-=-
C' Cl C2 C310 10 10 10
C'==10 J.lF
3
Now C' and C4 form a parallel combination. There-
fore,the equivalent capacitance of the whole network is
I 10 40
C=C + C4 ==-+10 =-J.lF =13.3 J.lF.
3 3
or
(b)It isclearfrom Fig. 2.76 that the charge on each
of the capacitors Cl, C2 and C3 is same. Let it beq.Let
the charge on C4 beq'.
•.P.D.across AB,
or
P.D. across BC,
P.D.acrossCD,
But
or
==1.7x 10-3 C
Also, P.D. across AD='£ =500 V
C4
•.q'=500xC4==500xl0J.lC
= 5000x10-6C==5x10-3C.
Example 63.Fourcapators Cl'C2'C3andC4are con-
nected to a battery of12 V, as shown in Fig. 2.77.Find the
potential difference between the points A andB.
CI=8~F C2=4~F
I1I
+-
12 V
Fig. 2.77···3wx«n£m~r±n3lxv

133. 2.42
Solution.
LetVAbe the potential at pointAandVB
that at B.Then
P.D.acrossC1=12 -VA
P.D.acrossC2=VA-0=VA
P.D.acrossC3=12-VB
P.D.acrossC4=VB-0=VB
As the capacitors C1andC2are connected in series, so
ql=q2
or C1(12 -VA)=C2VA
or 8(12 -VA)=4VA
or VA=8 V
Again, the capacitorsC3andC4are connected in
series, so
q3=q4
or C3(12 -VB)=C4VB
or 3(12-VB)=6VB
or VB=4V
The potential difference between the pointsAandBis
VA-VB=8 - 4=4V.
Example 64.Five identical capator plates, each of areaA
are arranged such that the adjacent plates are at distance d
apart. The plates are connected to a source of emf V, as
shown in Fig.2.78.Find the charges on the various plates.
[lIT 84]
1 2 3 4 5
..;;;;..v
+
A
Fig.2.78
Solution. As shown in Fig.2.79,the given network
isequivalent tothree parallel-plate capacitors con-
nected in parallel.
Their capacitances are
EOA2EOAandEOA
-d-' d d
A
The p.d. across each V
capacitor is V. +-
As
Fig. 2.79
Charge =Capacitance xp.d.
PHYSICS-XII
So charges on various plates are
_ +EOA V __ 2EOA V
ql - d'q2 - d
_2EOAV _ 2EOAV _EOAV
q3-+d'q4 --d 'qs-+-d- .
Example 65. For the network shown in Fig. 2.80,find the
potential difference between points A and B,and that bet-
ween BandCin the steady state.
3flF IJlF
3~~
Ion
20n
Ac>----'V\J'\r--+-:-I1111 r---<lC
100V
Fig. 280
Solution. The two capacitors of 3 flFand 3flFon the
left side of the network are in parallel,their equivalent
capacitance=6flF
The two capacitors of1flFand1flFon the other side
of the network are also in parallel, their equivalent
capacitance=2flF.So the given network reduces to the
equivalent circuit shown inFig.2.81.
~
6flF 2flF
Ion
.20 n 100V
A<r--'vvv--+-:-11111 _
C
Fig. 2.81
Inthe steady state, when all the capacitors are
charged,there isno current in the circuit. So there is no
potential drop across any resistance. Hence
p.d. across1ufcapacitor
=p.d. between pointsAandC=100 V
As6flFand2flFcapacitances are in series,the p.d.
of100 Vis divided between them in the inverse ratio of
their capacitancesi.e.,inthe ratio2 : 6or1 : 3.
1
VAB=p.d. across6flF=4"x100=2SV
3
VBC=p.d.across 2flF=4"x100=7SV.···3wx«n£m~r±n3lxv

134. ELECTROSTATIC POTENTIAL
AND CAPACITANCE
~rOblems For Practice
1.Two capacitors have a capacitance of 51lF when
connected in parallel and 1.2IlFwhen connected in
series.Calculate their capacitances.
(Ans. 21lF,31lF)
2.Two capacitors of equal capacitance when
connected in series have net capacitance C;,and
when connected in parallel have net capacitance c;..
Whatisthe value ofC; /c;.? [CBSE D93CI
(Ans.C;/c;.=1/4)
3.Three capacitors of capacity 1, 2 and 31lF are
connected such that second and third are in series
andthe first one in parallel. Calculate the resultant
capacity. (Ans. 2.2 IlF)
4.The capacities of three capacitors are in the ratio
1:2 : 3. Their equivalent capacity in parallel is
greater than the equivalent capacity in series by
60/11 pF. Calculate the individual capacitances.
(Ans.1 pF, 2 pF, 3pF)
5.The equivalent capacitance of the combination
betweenAandBin Fig. 2.82 is 41lF.
20f.1F
~----~II---H--;
C
Fig.2.82
(i)Calculate capacitance of the capacitorC.
(ii)Calculate charge on each capacitor if a 12V
battery is connected across terminals AandB.
(iii)What will bethe potential drop across each
capacitor? [CBSE D 091
[Ans.(i)51lF(ii)48 IlC(iii)2.4V,9.6Vj
6.How would you connect 8, 12 and 241lF capacitors
toobtain(i)minimum capacitance (ii)maximum
capacitance? If a potential difference of 100 volt is
applied across the system, what would be the
charges on the capacitors in each ce?
[Ans.(i)In series,Cmin=41lF,q=41lC,
(ii)In parallel, Crnax=44IlF, ql=8001lC,
q2=1200IlC, q3=24001lC]
7.Calculate the capacitance of the capacitor in Fig. 2.83,
ifthe equivalent capacitance ofthe combination bet-
weenAandBis151lF. [CBSE D 941
(Ans.601lF)
Fig.2.83
2.43
8.In the combination of four identical capacitors shown
inFig.2.84, theequivalent capacitance between
points P andQis 1IlF.Find the value of each
separate capacitance. (Ans.41lF)
Fig.2.84
9.Findthe equivalent capacitance of the combination
shown in Fig. 2.85between the points AandB.
L l (Ans.c;+~J
A~~B ~+~
C1-r= C2=r-
Fig.2.85
10For the network shown in Fig. 2.86,calculate the
equivalent capacitance between points AandB.
(Ans.6IlF)
Fig.2.86
11.Calculate thecapacitance of the capacitor Cin
Fig. 2.87. The equivalent capacitance of the combi-
nation between PandQis 30 1lF. [CBSE OD95]
(Ans. 60IlF)
C 20 f.1F
P~~~--4~--~S--~Q
Fig.2.87
12.Calculate theequivalent capacitance between points
AandBofthe combination shown in Fig. 2.88.
A~~~"~:F)
IIlF IIlF
Fig.2.88···3wx«n£m~r±n3lxv

135. 2.44
13.Fi
ndthe equivalent capacitance between points A
andBfor thenetwork shown in Fig. 2.89.
C1 Cz (Ans.~ JlF)
"T"
A C3 1 J!F B
Fig.2.89
14.Calculate the equivalent capacitance between the
pointsAandBof the circuit givenbelow.
C C [CBSE F 95]
1 z 28
Ao----1~~ (Ans.-JlF)
4J!Fr~ •..:J!Fr~ ~
C6I'"'C'W'
C,1'"'"lB
Fig.2.90
15.A network of six identical capacitors, each of value
Cis made, as shown in Fig.2.91.Findthe equivalent
capacitance between thepoints AandB.
(Ans.4C/3)
A0---,:--1
I----'--<l B
Fig.2.91
16.Find the equivalent capacitance between the points
AandB of the network of capacitors shown in
Fig.2.92. (Ans. 1JlF)
Fig.2.92
17.Find the capacitance between the points AandB
of the assembly shown in Fig. 2.93.
(Ans.2.25JlF)
PHYSICS-XII
Fig.2.93
18.Findthe resultant capacitance between the points X
andYof the combination of capacitors shown in
Fig.2.94. [Haryana 01]
(Ans. 2.5 JlF)
5J!F
Fig.2.94
19.The outer cylinders of two cylindrical capacitors of
capacitance 2.2 JlFeach are kept in contact and the
inner cylinders are connected through a wire. A
battery of emf 10 Vis connected, shown in
Fig. 2.95. Findthecharge supplied bythe battery to
the inner cylinders. (Ans.44 JlC)
lOV
Fig.2.95
20.In Fig. 2.96, C1=1JlF,C2=2JlFandc,=3JlF.Find
the equivalent capacitance between points AandB.
(Ans.6JlF)
A0---< >---QB
Fig.2.96
21.Four capacitors of equal capacitances are con-
nected inserieswith a battery of 10 V, shown in~~~1styjxiwn}j1htr

136. ELECTROSTATIC POTENTIAL AND CAPACITANCE
Fig.2.97
.The middle point B is connected to the
earth. What will be the potentials of the pointsA
and C? (AnS.VA= + 5V,Vc= - 5V)
lOV
J;~:l
"
Fig. 2.97
22.Determine the potential difference across the plates
of each capacitor of the network shown in Fig. 2.98.
Take Ez>fl·
(Ans.v:=(Ez-fl)c;rv:=(Ez-fl)c;.J
1c;.+c; 2c;.+c;
Fig. 2.98
23.Find the potential difference between the points A
andBof the arrangement shown in Fig. 2.99.
(Ans. - 8V)
Fig.2.99
24.Determine the potential difference VA - VBbet-
ween points AandBof the circuit shown in
Fig.2.100.Under what condition is it equal to zero?
Po--f---l 1-----4t--- I---t----o Q
v
Fig. 2.100
2.45
25.A variable capacitor hnplates andthedistance
between two successive plates isd..Determine its
capacitance. (Ans.C =(n-~EoAJ
26.A network of four capacitors each of 12 JlF
capacitance is connected to a 500 Vsupply as
shown in Fig. 2.101. Determine
(a)equivalent capacitance of the network, and
(b)charge on each capacitor. [eBSE OD 10]
[Ans.(a)16 JlF(b)ql=q2=q3=2000JlC,
q4=6000flq
Fig. 2.101
27.For the network shown in Fig. 2.102, compute
3J.lF 3J.lF 3J.lF
'~~,~~
bo--1~~~
3J.lF 3J.lF 3J.lF
Fig. 2.102
(i)the equivalent capacitancebetween points a
andb.
(ii)the charge on each of the capacitorsnearest to a
andbwhenVab= 900V.
(iii)V cdrwhenVab= 900V.
[Ans.(i)1 JlF(ii)900 JlC (iii)100Vj
HINTS
1.Proceed in Example 45 on pe 2.35.
2.Proceed in Example 46 on pe 2.35.
2x3
3.C = 1+ -- =2.2JlF.
2+3
4.Let the capacitances be C, 2 C and 3C.Then
Cp= C + 2 C + 3 C = 6 C
1 1 1 1 11 6C
-=-+-+-=- or C=-
CsC 2C 3C 6C s11
60 6C 60
GivenCp-Cs=11pF or 6 C-11=11 pF
or C= 1pF
Sotheindividual capacitances are 1 pF, 2pF and
3pF.···3wx«n£m~r±n3lxv

137. 2.46
5.(i)As 20
/IF capacitor and capacitor C are in series,
their equivalent capacitance is
C =Cx20
ABC+ 20
20C
or 4/lF=--
C+20
or 4C+80=20C
or C=5/lF.
(ii)Charge on each capacitor,
q=CAB V= 4/lFx12V=48 /lC
(...) P D F . q48/lC V
III . .on 20/l capacitor =-- =-- =2.4
20/lF 20/lF
q48/lC
P.D.on capacitor C = - = -- =9.6V.
C5/lF
6.(i)For minimum capacitance, the three capacitors
must be connected inseries. Then
1 1 11 1
--= - + - + - =-or Cmin =4/lF.
Cmin 81224 4
(ii)For maximum capacitance, thethree capacitors
must be connected in parallel. Then
Cmax=8+12+ 24 =44/lF.
(iii)In seriescombination, chargeis same on all
capacitors.
q=CV=4/lFx100V=4/lC.
In parallel combination, charges on the
capacitors are
1ft=C1V=8/lFx100V= 800/lC
q2= C;V=12/lFx100V= 1200/lC
q3=C;V=24/lF x100V= 2400/lc.
7.The combined capacitance of the parallel com-
bination of two 10 /IF capacitors is20 /IF. This com-
bination is connected in series with capacitance C.
1 1 1 1 1 14-3 1
..-+-=- or-=---=--=-
20 C 15 C 15 20 60 60
or C = 60/lF.
8.All capacitors are in series.
4 1
--- orC=4/IF.
Cl/lF
9.C=~+~+~=~+ 2C1C;.
C1+C; C1+C; C1+C;
1 1 1 1 1
10.C ="'9+ "'9+ "'9= '3'C =3/lF
C =3/lF+ C =3/lF+3/lF= 6/lF.
1 1 1
11.-+-=- :. C = 60/lF.
C60 30
1 1 112
12.-=--+ - + --=-..C = 0.5 /IF.
C 1+1 1 1+1 1
PHYSICS-XII
13.C1and~ are in parallel between points AandD. So
the equivalent capacitance between AandDis
C'=C1+~=1+1=2/lF
The given network now reduces to theequivalent
circuitshown inFig.2.103.Between points AandB,
now C and C; are inseries andC4in parallel. Hence
theequivalent capacitance between AandBis
CZ
H~
C'~C~B
2!iF
CC; 2x1 8
C =-- +C= -- + 2 =-/IF.
C+C; 42+1 3
Fig. 2.103
14Capacitors C;, ~and C4 areconnected in series,
theirequivalent capacitance ~ is given by
11111111
-=-+-+-=-+-+-=-
c,C;c,C4 4 2 4 1
~ =l/lF
Also, CSand C6 are in series,theequivalent capa-
citance is
CSxC6 2x4 4
Cs=--=-=-/IF
CS+C6 2+43
~ and Csform a parallel combination of capacitance,
4 7
~=~+Cs=I+-=-/lF
3 3
NowC1and ~ form a series combination. Theequi-
valentcapacitance C between AandBis given by
1 1 1 1 319
-=-+-=-+-=-or
CC1 ~4728
28
C =-/IF.
19
15.The equivalent network is shown in Fig. 2.104.
Fig. 2.104
Clearly, the equivalent capacitance
= [2 Cand C inseries]II[Cand 2 C inseries]
2CxC Cx2C 4C
=---+---=-.
2C+C C+2C 3
16.Two2/lFcapacitors at the leftsideofthenetwork
are in series. Their equivalent capacitance is
2x2
C=-- = l/lF
s2+2~~~1styjxiwn}j1htr

138. ELECTROSTATIC POTENTIAL AND CAPACITANCE
The
capacitance Cs and the next capacitor of 1J..lFare
in parallel. Theirequivalentcapacitance is
Cp= 1+ 1= 2J..lF
Proceedinginthis w , wefinally get two 2 J..lF
capacitors connected in-series.
·.Equivalent capacitance betweenAandB
2x2
=--=lJ..lF.
2+2
17.The given arrangement is a balanced wheatstone
bridge. Proceed inExample 57 on pe 2.39.
18.The arrangement between the pointsAand B is a
balanced wheatstone bridge. Proceeding as in
Example 57, we find that the equivalent capacitance
between Aand Bis
C'= 5J..lF
Nowthe capacitorC'and the left out capacitor of
5J..lFare in series. The equivalent capacitance
between points XandYwill be
C'x5 5x5
C=-,- =--=2.5IlF.
C+5 5+5
19.The two capacitors are connected in parallel
· . C= 2.2+2.2 = 4.4 J..lF
Charge,q=CV= 4.4 J..lF x10 V =44J..lc.
20.Thethree capacitors are connected inparallel
between points Aand B.
·. C = ~ +Cz+ ~ = 1+ 2 + 3 =6J..lF.
21.HereVB=O.As the capacitances are equal on the
two sides of point B,
.. VA - VB=VB - Vc
or VA+VC=2VB=0
But VA - Vc=10V
.. VA=+5 VandVc=-5 V.
22.Letchargeqflow across the capacitor platesuntil
thecurrent stops.Ina closed circuit,
L 6V=0
f1+!L-E2+!L=0
~ Cz
q [ \+CzCzJ=E2 -f1
(E2-f1)~Cz
q= ~+Cz
P.D.across plates of ~=!L=(E2-f1)Cz
~ ~+Cz
PD.acrossplates ofCz=!L =(E2-f1) ~ .
Cz ~+Cz
or
or
or
23.The given arrangement is equivalent to the circuit
shown in Fig. 2.105.
2.47
Fig.2.105
Proceeding in the above problem22,weget
_(E2 -f1)~Cz
q- ~+Cz
P.D.across the plates of ~,
v=!L=(E2-f1)Cz=(12V-24V)4J..lF =-SV
1~ ~+Cz 2J..lF+4J..lF
24.Suppose the chargeqlflows in the upper branch
and'hin the lower branch. Then
V=ql[~+ ~Jor
Also,V=q2[.2.+.2.J
c,C4
V~C4
or q2= ~+C4
:.VA-VB=(VQ-VB) -(VQ -VA)=i£._!!L
C4Cz
Putting the values of ~ and q2'weget
V-V -~-~
A B-~+C4 ~+Cz
_ V[Cz~ -~C4 ]
(~+Cz)(~+C4)
ForVA- VB= 0,we have
£L= ~
CzC4
25.The given arrangement is equivalent to (n-1)
capacitors joined inparallel.
. . C=(n-1)EoA.
d
12J..lF
26.(a)~23= -3- =4 J..lF
Ceq=~23+C4=4+ 12 = 16 J..lF.
(b)ql=q2=q3=~23V= 4 J..lFx500V =2000 J..lC
q4= C4V=12 J..lF x500 V = 6000 uc
27.(i)Three 3J..lFcapacitors in serieshave equivalent
capacitance = 1J..lF. The combination is in
parallel with 2JlFcapacitor.
•. Equivalent capacitance between.c andd
=1+2=3J..lF
or···3wx«n£m~r±n3lxv

139. 2.48
The situation
is repeated for points eandf
Hence there are three 31lF capacitors in series
between pointsaandb.Equivalent capacitance
between aandb=11lF.
(ii)Potential drop of 900Vacrossaandbis equally
shared by three 31lF capacitors.
Hence charge on each capacitor nearest to a
andb
=300x3=900IlC
(iij)Potential drop of 300Vacrosseandfis equally
shared by 31lF capacitors.
Hence Vcd=100V.
2.24ENERGY STORED IN A CAPACITOR
38.How does a capatorstoreenergy?Derive an
expression for theenergy stored in a capator.
Energy stored in a capacitor. A capacitor is a device
to store energy. The process of charging up a capacitor
involves thetransferring of electric charges from its
one plate to another. Thework doneincharging the
capatorisstored as itselectrical potential energy.This
energy is supplied by the battery at theexpense of its
stored chemical energy and can be recovered by
allowing the capacitor to discharge.
Expression for the energy stored in a capacitor.
Consider a capacitor of capacitanceC.Initially, its two
plates are uncharged. Suppose the positive charge is or
transferred from plate 2 to plate 1 bit by bit.In this
process, external work h to be done because at any
stage plate 1 is at higher potential than the plate 2.
Suppose at any instant the plates 1and2have charges
Qand -Qrespectively, shown in Fig.2.106(a).Then
the potential difference between the two plates will be
V,=Q
C
Q' -Q'-dQ' Q -Q
+
•
+
+
+
+ dQ'
+
+~ -
+
+
+
+
+ •
+ •
+ -
+-> ~ -
+E~
+ •
•
+ •
+ •
+ •
1 2 1 (b) 2(a)
Fig.2.106(a)Work donein transferringchargedQ'from plate 2
to plate~.(b)Total work done in charging the capacitor m be
considered the energy stored in the electric field between
the plates.
PHYSICS-XII
Suppose now a small additional charge dQbe trans-
ferred from plate 2 toplate 1. The work done will be
dW '" V' . dQ=Q . dQ
, C
Thetotalworkdonein transferring a charge Q from
plate 2 to plate 1 [Fig.2.105(b)]will be
QQ [Q,2]Q
W=f dW=f -.dQ= -
oC 2C 0
1Q2
2'C
This work done is stored as electrical potential
energy Uofthe capacitor.
1Q21 21
U=-.-=-.CV=-QV [.: Q=CV]
2 C 2 2
39.If several capators are connected in series or
parallel, show that the energy stored would be additive in
either case.
Energy stored in a series combination of capacitors.
For a series combination, Q=constant
Total energy,
U=Q2 ..!.=Q2 . [~+~ +~+...]
2 C ZC1 C2 C3
Q2 Q2 Q2
=--+-+--+ ...
2C1 2C2 2C3
U=U1+U2+U3+ ...
Energy stored in a parallel combination of
capacitors. For a parallel combination, V=constant
Total energy,
1 21 2
U=-CV=-[C1+C2+C3+...]V
2 2
1 21 21 2
=-C1V+ -C2V+-C3V+ ...
222
U=U1+U2+U3+...or
Hencetotalenergyisadditive both inseries and parallel
combinations of capators.
2.25ENERGY DENSITY OF
AN ELECTRICFIELD
40. Whereistheenergystoredin acapator? Derive
an expression for the energy density of an electric field.
Energy density of an electric field. When a
capacitor is charged, an electric field is set up in the
region between its two plates. We can s that the
work done inthecharging process h been used in
creating the electric field. Thus the presence ofan
electric field impliesstoredenergy orthe energyisstored
in the electric field.···3wx«n£m~r±n3lxv

140. ELECTROSTATIC POTENTIAL AND
CAPACITANCE
Consider a parallel plate capacitor, having plate
areaAand plate separationd.Capacitance of the
parallel plate capacitor is given by
EA
C=_o_
d
Ifcis the surface charge density on the capacitor
plates, then electric field between the capacitor plates
will be
E=~
EO
Charge on either plate of capacitor is
Q=cA= EoEA
Energy stored in the capacitor is
U=Q2 =(EoEA)2=..!.E E2Ad
2C 2.EoA20
d
ButAd=volume of the capacitor between its two
plates. Therefore, theenergy stored per unit volume or the
energy densityof the electric field is given by
U 1 2
U=Ad=2EOE
Although we have derived the above equation for a
parallel plate capacitor, it is true for electric field due to
any charge configuration. In general, we can s that
an electric field E can be regarded as a seat of energy with
energy density equal to.3.. EOE2.Similarly, energy is also
2
sociated with a mnetic field.
2.26REDISTRIBUTION OF CHARGES
41. If two charged conductors are touched mutually
and then separated, prove that the charges on them will be
dividedinthe ratio of their capatances.
Redistribution of charges.Consider two insulated
conductorsAand B of capacitances C1 and C2, and
carrying charges Q1 andQ2respectively. LetVIand
V2be their respective potentials. Then
Q1=C1VIand Q2 =C2V2
Fig. 2.107 Redistribution of charges.
Now, if the two conductors are joined by a thin
conducting wire, then the positive charge will flow
2.49
from the conductor at higher potential to that at lower
potential till their potentials become equal. Thus the
charges are redistributed. But the total charge still
remains Q1 +Q2'
If the capacitance of the thin connecting wire is
negligible and the conductors are a sufficient distance
apart so that do not exert mutual electric forces, then
their combined capacitance will be C1 +C2.
. Total charge
Common potential=---------'=------
Total capacitance
V=gl+Q2=C1VI+C2V2
C1+C2 C1+C2
or
If after redistribution charges onAand BareQ'1and
Q'2respectively, then
Q'1=C1V
Q'1=C1
Q'2C2
Thus, after redistribution, the charges on the two
conductors are in the ratio of their capacitances.
42. When two charged conductors having different
capaties and different potentials are joined together,
show that thereisalways a loss of energy.
Loss of energy in redistribution of charges.Let C1
and C2 be the capacitances andVIandV2be the poten-
tials of the two conductors before they are connected
together. Potential energy before connection is
1 21 r2
v,=2C1VI+2C2"2
After connection, letVbe their common potential.
Then
V=Total charge=Q1+Q2= C1VI+C2V2
Total capacitanceC1+C2 C1+C2
Potential energy after connection is
1 21 21 2
Uf= 2 C1 V+2 C2V= 2(C1+C2)V
=..!.(C +C )[C1V1 +C2V2]2
21 2 C +C
1 2
=..!.(C1VI+C2V2)2
2 (C1+C2)
Loss in energy,
U=u, -Uf
=..!.CV2+..!.CV2 _..!..(C1VI+C2v2l
21 122 22 (C1+C2)···3wx«n£m~r±n3lxv

141. 2.50
1 2 2 2 2
---
- [C1VI+C1C2VI+C1C2V2
2 (C1 +C2)
22 2 2 2 2
+C2V2 -C1VI-C2V2 -2 C1C2VIV2]
.!C1C2 [V2+V2-2V V]
2 (C1 +C2) 1 2 1 2
_ 1C1C2(VI - V2)2
-2·c1+C2
This isalw s positive whether VI>V2orVI<V2.
So when two charged conductors are connected,
charges flow from higher potential side to lower
potential side till the potentials of the two conductors
get equalised. In doing so, there isalways some loss of
potential energy in the form of heat due to the flow of
charges in connecting wires.
Examples based on
"iff'Energ StoreCiin Ca acitors
Formulae Used
1.Energy stored in a capacitor,
2
U=].cv2 =]...L=].qV
2 2 C 2
2.Energy stored per unit volume or the energy
density of the electric field of a capacitor,
1 2
u=-EOE
2
3. Electric field between capacitor plates, E =~
EO
Units Used
Capacitance is in farad, charge in coulomb,
electric field in NC-1 or Vm-1,energyin joule and
energy density in Jm-3.
Example 66. How much work must be doneto charge a
24 JlFcapator when the potential difference between the
plates is 500V? [Haryana 02]
Solution. Here C = 24 JlF = 24 x10-6F,V= 500 V
Work done,
W =.! CV2=.!x24x10-6x(500)2 =3J.
2 2
Example 67. Acapacitor is charged through a potential
difference of200 V, when0.1 Ccharge is stored in it.How
much energy will it release, when it is discharged ?
[ISCE 98]
Solution. HereV=200 V,q=0.1 C
1 1
U=-qV =-x0.1x200=10J
2 2
Energystored,
When the capacitor is discharged, it releases the
same amount of energyi.e.,10J.
PHYSICS-XII
Example 68. Two parallel plates, separated by2mm of air,
haveacapatance of3x10-14Fandare charged to a
potential of200V.Then without touching theplates, they
aremoved apart till the separation is6mm.(i)Whatis the
potential difference between the plates?(ii)What is the
change in energy?
Solution. Charge, q=CV=3x10-14x200 =6x10-12 C
When theseparation increes from 2 mm to6 mm,
the capacitance becomes
C' =!£. C=~x3x 10-14 =10-14 F
d' 6
(i)P.D.between the plates becomes
610-12
V'=~ = x = 600 V.
C 10-14
(ii)Initial energy stored in the capacitor,
U=.!CV2=.!x3x10-14 X(200)2 = 6x10-10J
2 2
Final energy stored in the capacitor
U'=.!CV,2 =.!x10-14x(600l = 18 x10-10J
2 2
Incree in energy =U'- U=12x10-10J.
Example 69.Twocapators ofcapacitances C1=3JlFand
C2=6 JlFarranged inseriesareconnected in parallel with a
third capatorC3=4JlFThe arrangement isconnected to a
6.0V battery. Callate thetotalenergy stored in the
capacitors, [CBSE Sample Paper 98]
Solution. Equivalent capacitance of the series
combination of C1 and C2 isgiven by
C=C1C2=3x6= 2 JlF
C1+C2 3+6
Combination C' is in parallel withC3.
:. Totalcapacitance,
• ., -6
C = C+C3=2+4=6 JlF =6x10 F
Energy stored,
U=.!cv2=.!x6x10-6x62=1.08x10-4J.
2 2
Example 70. Threeidentical capators C1' C2andC3of
capacitance 6 JlFeachare-connected to a12Vbattery as
shown. Find:
(i) charge oneach capator.
(ii) equivalent
capatance of the
network.
(iii)energy stored in
the network of
capators.
[CBSE D09]
Fig. 2.108«««2uv{lzkyp~l2jvt

142. ELECTROSTATI
C POTENTIAL AND CAPACITANCE
Solution.(i)C1and C2 are connected in series across
12battery whileC3is inparallel with this combination.
Equivalent capacitance ofC1andC2is
C C 6x6
C= 1 2=--= 3/IF
12C+C6+6
1 2
Charge on either of the capacitors C1andC2issame.
q1=q2=C12V=3/lFx 12 V = 36/lC
Charge onC3'
q3= 6/lFx 12 V =n/lC
(ii)Equivalent capacitance of the network,
C=~2+ ~= 3 IlF+6/lF =9/IF.
(iii)Energy stored in the network,
U=.!CV2=.!x9xlO-6 x(12)2 = 6.48x10-4 J.
2 2
Example 71. In Fig.2.109, theenergy stored illC4is27J.
Callate thetotal energy stored in the system.
2!!F
~IlI-F_Qo--+-_3_IlF--I ~
C1 61lF ~
C4
Fig.2.109
Solution. Energy storedinC4is
1 2
U4=2"C4V=27J
.!x6x 10-6 x V2 =27
2
V2= 27x2 = 9x106
6x10-6
or
or
Energy stored in C2 '
U2=.!x2x10-6 x9x 106 =9J
2
Energy stored in C3'
U3=.!x3x10-6x9x106= 13.5J
2
Energy stored inC2'C3and C4
= U2 + U3 + U4 =9+13.5 + 27 =49.5J
Equivalent capacitance of C2'C3and C4 connected
in parallel
=2 +3+5=l1JlF
q2 = 49.5J[u=2qC2]
2x11x10-6
2.51
EnergystoredinC1'
U=L= 49.5x2x11x10-6= 544.5 J
12C1 2x1x10-6
Total energy stored in the arrangement
=544.5 + 49.5 = 594.0 J.
Example 72.In acamera-flash cirit(Fig.2.110), a
2000 /IF capator ischarged bya1.5Vcell. When aflashis
required, the energy stored inthecapacitor isdischarged by
means of atrigger Tthrough a discharge tubein0.1
millisecond. Find the energy stored inthecapacitor and the
power oftheflash. [ISCE 97]
Electronic trigger
Discharge
tube
Fig. 2.110
Solution. HereC= 2000 /IF = 2x10-3F,V= 1.5 V
Energy stored in the capacitor,
U=.!cv2 =.!x2x10-3x(1.5)2 = 2.25 x10-3J
2 2
Time during which capacitor is discharged for
producing flh,
t=0.1millisecond =0.1x10-3s = 10-4 s
Power of flh, P=U = 2.25 x 10-3= 22.5 W.
t 10-4
Example 73.A800pF capacitor ischarged by a 100 V
battery. After sometime thebatten)isdisconnected. The
capator isthen connected to another 800pF capator.
Whatistheelectrostatic energystored? [CBSE F09]
Solution. Here C1 = C2= 800 pF =8x10-10F,
VI= 100 V, V2=a
Common potential,
V=~V1+c;V2= 8xlO-10xlOO+0 =50V
~ +c;8x10-10+8xlO-10
1 2
Uf=2(~+C;)V
=.!(8xlO-10 + 8xlO-10) x(50)2 = 2 x10-6J
2
Example 74
(i)A900pF capacitor ischarged by a 100V battery.
How much electrostatic energy isstored by the
capator?···3wx«n£m~r±n3lxv

143. 2.52
(ii) Thec
apatorisdisconnectedfrom the battery and
connected to another900pF capacitor. Whatisthe
electrostatic energy stored by thesystem?
(iii)Where hastheremainder of the energy gone?
[NCERT;CBSE OD 90]
Solution.(i)The chargeon the capacitor is
q=CV=900x10-]2 Fx100 V =9x 10-8 C
The energy stored by the capacitor is
U= ~CV2=~qV= ~x9x1O-8C x100V
222
=4.5x10-6J.
+q-q -.i.+ s.
Co~
2+-2
+
+
.!L+
q
2+ -2'
+
+
Fig. 2.111
(ii) Inthesteadysituation, thetwo capacitors have
their positive plates at thesame potential, and their
negative plates at the same potential. Let the common
potential difference beV'.The charge on eachcapacitor
is thenq=CV'.By charge conservation,q'=q/2.
.', TotaIenergy of the system
=2x~qV'=qV'=q .£
2 C
=±.q; =±.qv=~x~qv [-:q'=~and~=V]
= ~x4.5x10-6J=2.25x10-6J.
2
PHYSICS-XII
or
V==Total ch!~ = _Cf.L!"_!lL =CIV]+C2V2
Total capacitance C] +C2 C]+C2
V= C]V]
C1+C2
(ii)Energy stored in the capacitors before connection,
1 2
u,=2C]V]
Total energy after connection,
1 2
Uf=2(C1+C2)V
22
=~ (C +C) C1 V]
2 i 2(C1+C2)2
=~CiV]2=(C]Ju.
2 C]+C2 C]+C2 I
Clearly, Uf<Ui
Hence total energy of the combination is less than
the sum of the energy stored in the capacitors before
they are connected.
Example 76.Two capators of unknown capatances C]
andC2are connected first in series and then inparallel
across a battery of 100V.Ifthe energy stored in the two
combinations is0.045Jand0.25Jrespectively, determine
the values of C]andC2•Alsocallate the charge on each
capacitorin parallel combination. [CBSE D 15]
Solution. For series combination, we have
U= 1C]C2 V2
2 C]+C2
0.045 = ~ C]C2 x (100l
2 C]+C2
For parallel combination, we have
1 2
U=2(C]+C2)V
0.25 =~(C]+C2)x(100)2
2
orC]+C2=0.5x10-4
... (i)
(iii)There is a transient period before the system
settles to the situation(ii).During this period, a
transient current flows from the first capacitor to the
second. Energy is lost during this time in the form of
heat and electromnetic radiation.
Example 75. A capacitor ischargedtopotential V]'The C C
power supplyisdisconnected and the capacitor isconnected From(i), 0.045 =~x ] 2 x (100)2
20.5x10-4
in parallel to another uncharged capacitor.
(i)Derive theexpression for the common potential of theor C]C2= 0.045x10-8
combination of capators. Now (C]-C2)2 =(C]+C2)2-4C]C2
(ii)Show that total energy of the combination is less than =(0.5x10-4)2 _4x0.045x10-8
the sum of the energy stored in them before theyare = (0.25 -0.180) x10-8= 0.07x10-8
connected. [CBSE OD 15]
SIti (.)LCd C b th it f .. C] -C2=.J0.07x10-4=026x 10-4 ...(iii)
o uIOn. Iet]an 2 e e capac! ances 0
the two capacitors andVbe their common potential. On solving (ii)and(ii.i),we get
Then C] = 0.38x10-4 F and C2 = 0.12x10-4 F
...(ii)···3wx«n£m~r±n3lxv

144. ELECTROSTATIC POTENTIAL AND CA
PACITANCE
Charges on capacitors CIandC2in parallel
combination are:
QI=C1V=0.38x 10-4x100C=O.38x 10-2 C
Q2=C2V=0.12x10-4x100C=0.12x 10-2C.
Example77.Acapacitor ofcapacitance 6/IFischarged to
a potential of150YItspotential fallsto90V,when another
capacitor isconnected toit.Find the capacitance of the
second capacitor and the amount ofenergy lost due to the
connection.
Solution. Here CI=6/IF, VI=150V,V2=0,
V=90V,C2=?
Common potential,
V=CIVI+C2V2
C1+C2
90V= 6x10-6x150 + 0
6x10-6+C2
or
or
C2+6x10-6= 6x10-6x150=10x10-6
90
C2=4x10-6F=4/IF.
Initial energy stored,
1 21 -6 2
Ui=U1=2"CIVI=2"x6x10x(150)
= 6.75x10-2J
or
Final energy stored,
1 2
Uf=2"(CI+C2)V
=.!(6+4)x10-6x (90)2 = 4.05x10-2J
2
Theloss of energy on connecting the two capacitors,
flU=Ui-Uf= (6.75-4.05)x10-2
= 2.7x10-2J= 0.027J.
Example78.Abattery of10 Visconnected to a capator
ofcapacity 0.1F.The battery isnow removed and this
capacitorisconnected to a second uncharged capator. If the
charge distributes equally on these two capacitors, find the
total energtj stored inthe two capacitors. Further, compare
thisenergy with the initialenergy stored in the first
capator. [Roorkee 96]
Solution.Initial energy stored in the first capacitor is
U=.!CV2=.!x0.1x(10)2= 5.0J
'22
When the first capacitor is connected to the second
uncharged capacitor, the charge distributes equally.
This implies that the capacitance of second capacitor is
2.53
also C. The volte across each capacitor is nowV/2.
The final total energy stored in thetwo capacitors is
U=.!C(V)2+.!C(V)2=.!cv2
f2 2 2 2 4
=2.5J
Uf_2.5 _1 _ 1. 2
u-:-5.0 -2" -..
~rOblems for Practice
1.Acapacitor charged from a 50Vd.c. supply is
found to have charge of 10/lc. What is the
capacitance of the capacitor andhow much energy
isstored in it ? [ISeE 93]
(Ans.0.2/lF, 2.5x10-4J)
2.For flh pictures, a photographer usesacapacitor
of 30/IFand a charger that supplies 3x103V.Find
thecharge and energy expended in joule foreach
flash. (Ans.9x10-2C,135J)
3.An electronicflh lamp has 10capacitors, each 10 /IF,
connected in parallel. The lamp is operated at
100 volt. How much energy will be radiated in the
flash? (Ans.0.5J)
4.Three capacitors of capacitances 10/lF, 20/lF and
30/IFare connected in parallel to a 100 Vbattery
shown in Fig. 2.112. Calculate the energy stored in
thecapacitors. [ISeE 94]
(Ans.0.3J)
+-
'------11------'
lOOV
Fig. 2.112
5.Avariable capacitor is kept connected to a 10 V
battery. Ifthecapacitance of the capacitor is
changed from 7J.!Fto3/IF,what is the change in the
energy? What happens to this energy?[ISeE 96]
(Ans.2x10-4J,decree in energy)
6.The plates of a parallel plate capacitor have an area
of 100em2each and are separated by 2.5 mm.The
capacitor is charged to 200 V. Calculate the energy
stored in the capacitor. [Punjab 96]
(Ans.7.08x10-7J)
7.A80 /IF capacitoris charged by a 50 V battery. The
capacitor is disconnected from the battery and then···3wx«n£m~r±n3lxv

145. 2.54
across another
unchanged 320I-lF
Calculate the charge on thesecond
.[CBSE D 94 C]
(Ans. 3.2xlO-3q
8.Find the total energy stored in the capacitors in the
network shown below. [CBSE D 04]
(Ans. 3.6x10-5J)
connected
capacitor.
capacitor.
Fig. 2.113
9.A 10 I-lFcapacitor is charged by a 30 V d.c. supply
andthen connected across an uncharged 50 I-lF
capacitor. Calculate(i)thefinal potential difference
across thecombination, and(ii)theinitial and final
energies. How will you account forthedifference in
energy? [CBSE OD 04]
[Ans.(i)5V,(ii)Uj= 4.5x10-3J,
Uf= 0.75x10-3Jl
10.Netcapacitance of three identical capacitors in
series is 1 I-lF. What willbe their net capacitance if
connected in parallel ?
Findtheratio of energy stored in thetwo configu-
rations iftheyare bothconnected tothe same source.
[CBSE OD11] (Ans.9 I-lF, 1: 9)
11.Two capacitors of capacitances 251-lFand 100I-lFare
connected in series and are charged by a battery of
120 V. The battery is then removed. The capacitors
are nowseparated and connected inparallel. Find
(i)p.d. across each capacitor (ii)energy-loss inthe
process. (Ans.38.4 V, 0.05184 J)
12.Figure 2.114shews a network of five capacitors
connected to a 100V supply. Calculate the total
charge and energy stored in the network.
[CBSE Sample Paper 08]
(Ans.4xlO-4C, 0.02 J)
Fig. 2.114
'------IIIr---~
lDDV
PHYSICS-XII
13.Twocapacitors arein parallel and the energy stored
is 45J,when the combination is raised to potential
of 3000 Y.With the same two capacitors inseries,
the energy stored is 4.05 Jfor the same potential.
Whatare their individual capacitances?
(Ans.91-lF,11-lF)
14.Find the ratio ofthepotential differences that must
beappliedacrosstheparallel andtheseries
combination of two capacitors C1.andc;withtheir
capacitances in theratio1:3 sothattheenergy
stored inthe two cases, becomes the same.
[CBSE F 10]
(Ans ..f3:4)
HINTS
1.C =!L= 10 I-lC= 0.2 1-lF.
Y 50V
Energystored,
U =1 Cy2 =1x0.2x10-6x(50l= 2.5x10-4J.
2.HereC=30I-lF=3xlO-5F, Y=3x103V
Charge, q=CY = 3 x 10-5 x 3 x103C = 9x10-2C
Energy, U=1CY 2=1x3xlo-5x9xl06
= 135J.
3.Total equivalent capacitance,
C = 10x10 f.lF= 100 I-lF= 1O-4F
Energyradiated
= 1 CY 2 = 1x10-4x(100)2 = 0.5J.
4.C =C1.+c;+C;= 10+20+30 = 60I-lF
=60 x 10-6 F
U=1 Cy2 = 1x60x10-6x(100)2= 0.3J.
5Here Cj =7f.lF= 7x1O-6F,Y = 10 V
u,= 12=1 x7 x10-6x(10)2 = 3.5x10-4J
Again, Cf=31-lF=3x10-6 F, Y =10 V
Uf=1Cfy2=1x3 x10-6x(1O)2= 1.5xlo-4J
Decrease in energy= Uj -Uf= 2.0x10-4J.
Energy islost heat and electromnetic
radiation.
6.HereA= 100cm2 = 10-2m2,
d= 2.5 mm = 2.5 x10-3m, Y = 200V
U =..! CY 2 =..!. EOA.Y2
2 2 d
=..!x8.85x10-12x10-2(200)2
2 2.5 x10-3
=7.08x10-7J.···3wx«n£m~r±n3lxv

146. ELECTROSTA
TIC POTENTIAL AND CAPACITANCE
8.Thetwo 2!-1Fcapacitors on the right sidearein
2x2
series, their equivalent capacitance =--=1!-IF
2+ 2
This 1!-IFcapacitance is in parallel with the central
1 !-IFcapacitor. Their equivalent capacitance
=1+1=2!-1F
This 2!-1F capacitance is in series with the 2 !-IF
capacitor at the bottom. Their equivalent capacitance
2x2
=-- = l!-1F
2+2
Finally, 1!-IFcapacitance is in parallel with the left
out 1!-IFcapacitor. The equivalent capacitance is
C =1 +1=2 !-IF= 2x10-6F
V=6V
U=.!CV2 =.! x2x10-6x(6)2
2 2
=3.6x10-5J.
9.HereC1.= 10 !-IF= 10 x10-6F,
~ =50 !-IF =50x10-6F,
(i)Common potential,
C1.VI+ ~V210x10-6x30 + 0
V= = =5 V.
C1.+~ (10 +50)x10-6
~=30V,
V2=0
(ii)Initial electrostatic energy of 10 !-IFcapacitor,
1 2
Ui=2"C1.VI
=.! x10 x10-6x(30)2= 4.5x10-3J
2
Final electrostatic energy of the combination,
Uf=~(10 +50)x10-6x(5)2= 0.75x]0-3J
Loss in energy=Ui-Uf= 3.75x10-3J
Thedifference inenergy islost in the form of heat
and electromagnetic radiation as thecharge flows
from first capacitor to second capacitor.
C
10.Here Cs ="3 = 1!-IF
C=3 !-IF
Cp=3C=9!-1F
U 1c,V2 C 1
_s=_2__ =~=_=1: 9
UIcV2 C 9
p2p p
11.(i)Equivalent capacitance in series,
C = 25x100=20 F
25+100 !-I
Charge on each capacitor in series,
q=CV=20 !-IFx120 V=2400 !-IC
2.55
Equivalent capacitance in parallel,
C' = 25 + 100 = 125 !-IF
Total charge,
q'= 2400+2400=4800 !-IC
P.D.across each capacitor,
V'=1..= 4800 !-IC = 38.4V.
C'125 !-IF
(ii)In series,
U=.!cv2=.!x20x10-6 x(120)2
2 2
= 0.144J
In parallel,
U'=.!C'V,2=.!x125x10-6x(38.4)2 = 0.09216J
2 2
:.Energy loss
=U - U' = 0.144-0.09216 = 0.05184J.
12.The equivalent circuit diram for the given
network is shown below:
100 V
'-------iI It-------'
Fig. 2.115
Two 3!-1F capacitors in paraUel. The equivalent
capacitance,
C1.= 3 + 3 = 6 !-IF
The l!-1F capacitor and a 2!-1F capacitor are in
parallel. Their equivalent capacitance,
~ = 1+ 2=3!-1F
ThenC1.and ~ form a series combination of equi-
valent capacitance,
This combination is in parallel with the fifth capa-
citor of 2 !-IF.
:.Net capacitance, C=2 + 2 = 4 !-IF
Total charge,
q=CV= 4x10-6x100 = 4x10-4C
Total energy stored,
U =.!CV2=.!x4x10-6 x(100)2 = 0.02J.
2 2···3wx«n£m~r±n3lxv

147. 2.56
14
.Given
Now
or
or
2.27DiElECTRICSAND THEIR POLARIZATION
43. What are dielectrics ?Explain the difference inthe
behaviour of a conductor and a dielectric in the presence
anexternalelectric field. Distinguish between polar and
non-polar dielectrics.
Dielectrics.In insulators, the electrons remain
attached to the individual atoms or molecules.
However, theseelectrons cansuffer small movements
within the atoms or molecules underthe influence of
an external electric field. The net effect of these micro-
scopic movements gives rise to some important electric
properties tosuch materials. In viewof theseelectrical
properties, insulators are called dielectrics.
Adielectricisa substance which doesnot allow theflow
of charges through it but permits them to exert electrostatic
forces ononeanother through it. A dielectricisessentially
an insulator which can be polarised through small localised
displacements of itscharges.
Examples. Gls, wax, water, air, wood, rubber,
stone, pltic, etc.
Difference in the behaviour of a conductor and a
dielectric in the presence of an external electric field.
Dielectrics have negligibly small number of charge
carriers compared to conductors.
In a conductor, theexternal fieldEomoves thefree
charge carriers inducing field Eind in theopposite
direction ofEo.The process continues until the two
fields cancel each other and the net electric field in the
conductor becomes zero.
PHYSICS-XII
Eo
Eo
Eind
Conductor
Eo
Eo
Dielectric
Fig. 2.116Difference in the behaviour of a conductor and
a dielectric in an external electric field.
In a dielectric, theexternal field Eoinduces dipole
moment by stretching or re-orienting the molecules of
the dielectric. Theinduced dipole moment sets up an
electricfieldEindwhich opposes Eobut does not
exactly cancel this field. It only reduces it.
Polar and non-polar dielectrics.A dielectric m
consist ofeither polar or non-polar molecules. A
molele in which the centre of mass of positive charges
(protons) does not coinde with the centre of mass of
negative charges (electrons) iscalled apolarmolele.
The dielectrics made of polar molecules are called
polar dielectrics. The polar molecules have unsym-
metrical shapes. They have permanent dipole mo-
mentsof theorder of 10- 30Cm.For example, a water
molecule h a bent shape withits two 0- H bonds
inclined at anangle of105c shown in Fig. 2.117. It has
a verylarge dipole moment of 6.1 x 10- 30Cm. Some
otherpolar molecules areHCI,N~,CO,Cf\0H, etc.
Non-polar
CO,
Polar p
HCl
/
p
Fig. 2.117 Some polar and non-polar molecules.
Amolele inwhich thecentre of mass of positive
charges coincides with the centre of mass of negative charges
is called a non-polar molecule. The dielectrics made of
non-polar molecules are callednon-polar dielectrics.
Non-polar molecules have symmetrical shapes. They
have normally zero dipole moment. Examples of
non-polar molecules are ~, N2, 02' CO2, CH4, etc.···3wx«n£m~r±n3lxv

148. Polarization of a polar dielectric in an external
electric field.The mo
lecules of a polar dielectric have
permanent dipole moments. In the absence ofany
external electric field,the dipole moments ofdifferent
molecules are randomly oriented due to thermal
itation in the material, shown in Fig.2.11S(b)(i). So
the total dipole moment is zero. When an external fieldFig.2.119(a) Polarization of a dielectric.
is applied, the dipole moments of different molecules -CJp +CJp
ELECTROSTATIC POTENTIAL AND CAPACITANCE
44. How does a dielectric develop a net dipole moment
in an external electric field when it has (i) non-polar
moleles and(ii)polar moleles?
Polarization of a non-polar dielectric in an external
electric field.In the absence of any electric field, the
centres of positive and negative charges of the
molecules of a non-polar dielectric coincide, shown
in Fig. 2.llS(a)(i). The dipole moment of each molecule
is zero. In the presence of an external electric field Eo '
the centres of positive charges are displaced in the
direction of external field while the centres of negative
charges are displaced in the opposite direction. The
displacement of the charges stops when the force
exerted on them by the external field is balanced by the
restoring force due to the internal fields in the mole-
cules. This induces dipole moment in each molecule
i.e.,each non-polar molecule becomes an induced
dipole. The induced dipole moments of different mole-
cules add up giving a net dipole moment to the dielec-
tric in the direction of the external field, shown in
Fig.2.11S(a)(ii).
Eo=°
Eo,,0
~ ~ 8) 8)
~
8)
~
~ ~ 8) 8) 8)
~
8)
--Eo
(i) (ii)
Fig.2.118(a)Polarization of anon-polar dielectric
in an external electric field.
-r-+E;
(ii)
Fig.2.118(b)Polarization of a polar dielectric
in anexternal electric field.
2.57
tend to align with the field. As a result, there is anet
dipole moment in the direction of the field, shown in
Fig. 2.llS(b)(ii). The extent of polarisation depends on
relative values of two opposing energies :
1.The potential energy ofthe dipole in theexternal
field which tendsto align the dipole with thefield.
2. Thermal energy of itation which tends to
randomise the alignment of the dipole.
Hence bothpolar and non-polar dielectrics develop a
net dipole moment in the presence of an external electric
field. This factiscalled polarization of the dielectric.
~
Thepolarization P isdefined asthe dipole moment
per unit volumeand its mnitude is usually referred to
~
thepolarization density. The direction of P is same
~
that of the external field E o'
45. Explain why the polarization of dielectric reduces the
electric field inside the dielectric. Hence define dielectric
constant.
Reduction of electric field by the polarization o~ a
dielectric.Consider a rectangular dielectric slab placed
~
in a uniform electric field Eo acting parallel totwo of its
faces, shown in Fig.2.119(a).Its molecular dipoles
Dielectricslab
+
+$<:::3) <:::3) $
+
8<:::3)<:::3)e+
:
+
@<:::3) <:::3)@
+ Eo
.-
+
@<:::3) <:::3)@
+
..-----~-------- ----------'
Region of zero
charge density
+
+
+
+
+
+
+
+
I~----~------------~I
+
+
+
Fig.2.119(b) Reducedfield inadielectric, E=Eo-Ep'~~~1styjxiwn}j1htr

149. 2.58
->
align
themselves in the direction ofEo.This results in
uniform polarization of the dielectric, i.e., every small
volume of theslab has a dipole moment in the
->
direction ofEo.The positive charges of the dipoles of
first vertical column cancel the negative charges of the
dipoles of second column andsoon.Thusthevolume
charge density in the interior oftheslab is zero.
However, there is a net uncancelled negative charge on
the left face and uncancelled positive charge on the
right face of the slab.
The uncancelled charges are the induced surface
->
charges due to the external field Eo.Sincethe slab as a
whole remains electrically neutral, the mnitudeof
the positive induced surface charge is equal to thatof
the negative induced surface charge.
Thusthepolarized dielectric isequivalent totwo
charged surfaces with induced surface chargedensities ±(Jp'
Reduced field inside a dielectric and dielectric
constant. In ce of a homogeneous and isotropic
dielectric, the inducedsurface charges set up an
->
electric fieldEp(field duetopolarization) inside the
dielectric in a direction opposite to that of external
->
fieldEo'thus tending to reduce the originalfieldin the
->
dielectric. The resultant field Einthedielectric will be
-> -> ->
equalto Eo -Epanddirected inthe direction of Eo.
->
The ratio of the original fieldEoandthereduced
-> ->
fieldEo -Epin the dielectric is called dielectric constant
(K)orrelative permittivity (Er). Thus
-> ->
K=Eo= Eo
-> -> ->
EE-E
op
46. Define polarisation density. How isit related to
the induced surfacecharge density?
Polarisation density.Theinduced dipole moment
developed per unit volume of adielectric when placed inan
external electric fieldiscalled polarisation density. Itis
denoted byP.Suppose a dielectricslabof surface area A
and thicknessdacquires a surface charge density±(Jpor
due to its polarisation in the electric field and its two
faces acquire charges±Qp.Then
Q
(J=---E.
pA
We can consider the whole dielectric slab asalarge
dipole having dipole moment equal to Qpd.Thedipole or
PHYSICS-XII
moment per unit volume or the polarisation density
will be
P=dipole moment of dielectric
volume of dielectric
o,do,
=--=-=(J
Ad A p
Thusthepolarisation density may be defined as the charge
inducedper unit surface area.
Obviously, a uniformly polarised dielectric with
uniform polarisation density Pcan be replaced by two
->
surface l ers (perpendicular toP)of surface charge
densities±(JP,andzerocharge density in the interior.
47.Define electric susceptibility. Deduce the relation
between dielectric constant and electric susceptibility.
->
Electric susceptibility. If the fieldEisnot large,
->
thenthe polarisation Pisproportional to theresultant
->
fieldEexisting in the dielectric, i.e.,
-> ->
or P=EOXE
whereX(chi) is a proportionality constant called
electric susceptibility. The multiplicative factor EOisused
to keepXdimensionless. Clearly,
->
P
x=--
->
EOE
Thustheratio of the polarisation toEOtimesthe electric
fieldiscalled the electric susceptibility of the dielectric. Like
P,it also describes the electrical behaviour of a dielec-
tric.The dielectrics with constant Xare calledlinear
dielectrics.
Relation betweenKandx.Thenet electric field in a
polarised dielectric is
But
-> -> ->
E=Eo -Ep
E=(Jp=~
PEO EO
-> ->P
E=Eo--
EO
->
Dividing bothsidesbyE,we get
E
1=---.1l. -X
->
E
l=K-X or K=1+X···3wx«n£m~r±n3lxv

150. ELECTROS
TATIC POTENTIAL AND CAPACITANCE
2.28DiElECTRIC STRENGTH
48. What do youmean by dielectric strength of a
dielectric ?
Dielectric strength.Whenadielectric is placedina
very highelectric field, the outer electrons m get
detached from their parent atoms. The dielectric then
behaves like a conductor. This phenomenonis called
dielectric breakdown.
Themaximum electric field thatcanexistinadielectric
without causing thebreakdown of its insulating property is
calleddielectric strength of the material.
The unit of dielectric strength is same that of
electric field i.e.,Vm-1.But the more common practical
unit is kV mm-1.
Table 2.1Dielectric constants and dielectric
strengths of some common dielectrics.
Dielectric
Dielectric Dielectric strength
constant inkVmm1
Vacuum 1.00000 00
Air 1.00054 0.8
Water 81 -
Paper 3.5 14
Pyrex gls 4.5 13
Mica 5.4 160
Porcelain 6.5 4
2.29CAPACITANCE OF A PARALLEl PLATE
CAPACITOR WITH A DIELECTRIC SLAB
49.Deduce the expression for the capacitance of a
parallel plate capator when a dielectric slab isinserted
between its plates. Assume the slab thickness less than the
plateseparation.
Capacitance of a parallel plate capacitor with a
dielectric slab.Thecapacitance of a parallel plate
capacitor ofplate area Aand plate separationdwith
vacuum between its plates is given by
_ EoA
CO-d
Suppose initially the charges on the capacitor plates
are±Q.Then the uniform electric field set up between
thecapacitor plates is
(JQ
Eo=-=--
EOAEo
When a dielectric slab of thickness t<dis placed
between the plates, the field Eo polarises the dielectric.
This induces charge -Qpon the upper surface and
2.59
Fig. 2.120A dielectric slab placed in a
parallel plate capacitor.
+Qpon the lower surface of the dielectric. These
induced charges set up a field Epinside the dielectric in
-+
the opposite direction of Eo.The induced field is given by
E=(JP=R[(J=Q=P,polarisation density]
pEO EO PA
The net field inside the dielectric is
where Kisthedielectric constant of the slab. So
between the capacitor plates, the field E exists over a
distancetand field Eo exists over the remaining
distance(d-t).Hence the potential difference
between the capacitor plates is
V=Eo(d -t)+Et=Eo(d-t)+~o t[.: ~=K]
=Eo(d-t+.!)= ~(d-t+.!)
K EOA K
The capacitance of the capacitor on introduction of
dielectric slab becomes
C=Q= EO A
Vd-t+.!
K
Special Ce If the dielectric fills the entire space
between the plates, thent=d,and we get
EA
C=-O-.K=KCO
d
Thusthe capacitanceof a parallelplate capator increases
Ktimes when its entire spaceisfilled with a dielectricmaterial.
C
Clearly, K= -
Co
Dielectric constant
Capacitance with dielectric between two plates:
Capacitance with vacuum between two plates
Thus the dielectric constant of a dielectric material
may be defined as the ratio of thecapatance of a capator
completely filled with that material to the capatance of the
same capator with vacuum between itsplates.···3wx«n£m~r±n3lxv

151. 2.60
2.30CAPACITANCE OF A PARALLEL PLATE
CAPACITOR WITH A CONDUCTING SLAB
50.Deduce- the
expression for the capacitance ofa
parallel plate capator when a conducting slab is inserted
between itsplates.Assume the slab thickness less than the
plateseparation.
Capacitance of a parallel plate capacitor with a
conducting slab. Consider aparallel plate capacitor of
plate area Aand plateseparation d.If the space bet-
ween the plates is vacuum, its capacitance is given by
_EoA
CO-d
Suppose initially the charges on the capacitor plates
are±Q.Then the uniform electric field set up between
thecapacitor plates is
crQ
Eo=-=--
EOAEo
where cris thesurface charge density. The potential
difference between the capacitor plates will be
Qd
Vo= Eod=-
~AEO
When aconductingslabof thickness t<dis placed
between thecapacitor plates, free electrons flow inside
itso to reduce the field to zero inside the slab,
shown in Fig.2.121.Charges -Qand+Qappear on
the upper and lower facesof theslab. Now the electric
field exists only in the vacuum regions between the
plates of the capacitor on the either sideofthe slab, i.e.,
the field exists only in thickness d -t,therefore, potential
difference between the plates of the capacitor is
V=Eo(d - t)=---.fL(d-t)
AEo
I
1
+~+~+~Eo~
+~+-+Q
I1+
~ Conductingd £=0
1
+ + ++++Q slab
-~-~
_~£~~
-~---Q
I
Fiq, 2.121 A conducting slab placed in a
parallel plate capacitor.
:. Capacitance of the capacitor in the presence of
conducting slab becomes
_Q _EoA_EoA d C_[_d_) C
C-------or -
- V - (d-t)-d.d - t d - t.0
Clearly, C>Co' Thus the introduction of a conducting
slab of thickness t in a parallel plate capator increases its
capatance by afactor of_d_.
d -t
PHYSICS-XII
2.31USES OF CAPACITORS
51. Mention some important uses of capators.
Uses of capacitors.Capacitors are very useful
circuit elements in any of the electric andelectronic
circuits. Some of their usesare
1.To produce electric fields of desired patterns,
e.g.,for Millikan's experiment.
2.Inradio circuits for tuning.
3.In power supplies for smoothing therectified
current.
4.Forproducing rotating mnetic fields in
induction motors.
5.In the tank circuit of oscillators.
6.They store not onlycharge, but also energy in
the electric field between their plates.
2.32EFFECT OF DIELECTRIC ON
VARIOUS PARAMETERS
52.Aparallel-plate capacitor is charged by a battery
which is then disconnected. A dielectric slab is then
inserted to fill thespace between the plates. Explain the
changes,ifany,that or in the values of (i) charge on
the plates, (ii)electric field between the plates, (iii)p.d.
between the plates, (io)capatance and (v) energy stored
inthe capacitor.
Effect of dielectric when the battery is kept
disconnected from the capacitor.LetC!o'Co 'Vo'Eo
andUobe thecharge, capacitance, potential difference,
electric field and energy stored respectively before the
dielectric slab is inserted. Then
Vo 1 2
C!o=CoVo' Eo=d 'Uo=:2CoVo
(i)Charge. The charge on the capacitor plates
remains C!obecausethe battery hasbeen disconnected
before the insertionofthe dielectric slab.
(ii)Electric field.When the dielectric slab is
inserted between the plates, the induced surface
charge on thedielectric reduces the field to a new value
given by
(iii)Potential difference. The reduction in the
electric field results in the decreasein potential difference.
EdV
V= Ed=_o-=~
K K
(iv)Capacitance. Asa result ofthe decree in
potential difference, thecapatance increases Ktimes.
C=C!o=---.9L=KC!o=KCo
V VolK Vo«««2uv{lzkyp~l2jvt

152. ELECTROSTATIC POTENTIAL AND CAPACITANCE
(v)Energ
y stored. The energy stored decrees by a
factor ofK.
U=..!CV2=..! (KC )(VO)2=..!...! C 11: 2=Uo .
2 2 0 K K 200 K
53.A parallel plate capator is charged by a battery.
When battery remains connected, a dielectric slab is inserted
between the plates. Explain what changes, ifany,or in
the values ofti)p.d. between the plates, iii)electric field
between the plates, (iii) capatance, (iu) charge on the
plates and (v) energy stored in the capator?
Effect of dielectric when battery remains con-
nected across the capacitor. Let 00rCo' VO' Eoand Uo
be the charge, capacitance, potential difference, electric
field and energy stored respectively, before the
introduction of the dielectric slab. Then
Vo 1 2
00 = Co Vo 'Eo=-,Uo=-CoVo
d 2
(i)Potential difference. As the battery remains
"connected across the capacitor, so the potential difference
"<remains constantatVoeven after the introduction of
dielectric slab.
(ii)Electric field. As the potential difference
remains unchanged, so the electric field Eobetween the
capacitor plates remains unchanged.
V11:
E=-=~=Eo
d d
(iiz)Capacitance. The capacitance increases from Co to C.
C=K Co
(iv)Charge.Thechargeon the capacitor platesincreases
from 00 toQ.
Q=CV=KCO·VO=KOo·
(v)Energy stored. The energystored in the capacitor
increasesKtimes.
1 21 2 1 2
U="2CV ="2(KCO)VO =K'"2COVO =KUO'
Table 2.2 Effect of dielectric on various parameters.
Battery disconnected Battery kept connected
from the capacitor across the capacitor
Q= (1(constant) Q=K(1
11:
V=Vo(constant)
V=~
K
E= fu E=fu(constant)
K
C=Kc:;, C=Kc:;,
U
U=KUOU =-.!l.
K
2.61
For Your Knowledge
~Capacitance of a parallel plate capacitor with
compound dielectric.
A. Series typt. ~rang-nenIfa capacitor is filled with
ndielectric slabs of thicknesses t1,t2, .....,tn' shown
inFig.2.122(a),thenthisarrangement is equivalent ton
capacitors connected inseries.
With a single dielectric slab,
eoA
C= t
d-t+-
K
Capacitance with ndielectric slabs will be
eoA
C=d _ (~+t2+... +tn)+[!L+!L+... +~J
KrK2 Kn
Butd=~+t2+t3+.....+tn
eoA
C= ~ t2 tn
-+-+ ..... +-
Kr K2 Kn
...-- I_A _
~_::_t
1<3 it3
___ K,_,It.
L
ig.2.122(a) Fig.2.122(b)
B.Para'! Is',;earr;'1em'1The arrangement shown
inFig.2.122(b)consists of ncapacitors in parallel,
having plate are ~,~, ..... rAn ' and plate
separation d.
The equivalent capacitance of the parallel arrange-
mentwill be
or
C=C;+C2+ .....+Cn
Kreo ~ K2eo ~ KneoAn
= d+ d + ...+ d
eo
C =d(Kr ~+K2 ~+ ...+KnAn)
A
~ =~= ..... =An=-;;-,then
eoA
C=---;t;;(Kr +1<2+.....+Kn)
If···3wx«n£m~r±n3lxv

153. 2.62
Formulae Used
1. Capacitance of a parallel plate capa
citorfilled with a
dielectric of dielectric constant K.
EoKA
C:=KCo=--
d
2.Capacitance of a parallel plate capacitor with a dielec-
tric slab of thicknesst«d)in between its plates,
C= EoA
d-t(1-~)
3.Capacitance of a parallel plate capacitor with a con-
ducting slab of thicknesst«d)in between its plates,
C=EOA
d-t
4.Capacitance of spherical capacitor filled with a
dielectric,
ab
C=41t Eo K. b_a
5. Capacitance of a cylindrical capacitor filled with a
dielectric,
C= 21t EO KI
b
2303log10 -
a
6. Effect of dielectric with battery disconnected from
the capacitor,
Q=QyrV =Vo ,E= Eo ,C= K Co'U=Uo
- KKK
7.Effect of dielectric with battery connected across the
capacitor,
Q= K(;b ,V=Vo 'E= Eo'C= K Co 'U= KUo
Units Used
CapacitanceCis in farad, charge qin coulomb,
potential differenceVinvolt, areaAin m2,
thicknessesdandtin metre.
Constant Used
Permittivity constant, EO=8.85x1O-12C2N-1m -2.
Example 79. In a parallel plate capator, the capatance
increases from4~F to SO~F,onintrodung a dielectric
medium between the plates. Whatisthe dielectric constant of
the medium?
Solution.
K = Capacitance with dielectric = SO~F = 20.
Capacitance without dielectric 4~F
ExampleBO.A parallel plate capator with air between
the plates has a capacitance of 8~F. The separation between
the plates isnowreduced by half and the space between them
PHYSICS-XII
isfilled with a medium of dielectric constant 5.Callate the
value ofcapacitance of the capator in the second case.
[CBSEOD 06]
Solution. Capacitance of thecapacitor with air
between its plates,
EA
Co=_0_ =SpF
d
When the capacitor is filled with dielectric (K =5)
between its plates and the distance between the plates
isreduced by haltcapacitance becomes
EOKA EO x5xA
C=d/2= d/2 =10Co
or C=10xS=80 pF.
ExampleB1 .Figure2.123showstuio-identical capators,
C1andC2,eachof1~Fcapatance connected to a battery of
6V.Initially switch'5'is closed. Aftersome time '5'isleft
Fig.2.123
open and dielectric slabs of dielectric constant K=3are
inserted tofill completely the space between the plates of the
twocapators. How will the (i) charge and (ii) potential
difference between theplates of the capators beaffected
after the slabs are inserted? [CBSE D III
Solution.Withswitch 5closed, VI=V2= 6V
:.q1=q2=1~Fx6V=6~C
When dielectric slabs (K =3) are inserted, capaci-
tance of each capacitor becomes 3 ~F.
P.D. across Cl, V{= 6 V
Charge, eft=3 ~Fx 6V =18~C
Withswitch5open, the p.d. on C2attains a new
value but charge q2isstill 6~C
V'=6~C=2 V.
23 ~F
ExampleB2.Anebonite plate (K=3), 6mmthick, is
introduced between the parallel plates of a capator of plate
area2x1O-2~and plate separation 0.01mFind the
capatance.
Solution. Heret=6 mm =6x 1O-3m,
A=2x10-2m2, d=0.01 m, K=3
C= EOA = 8.S5x10-12 x2x10-2
d-t(1- ~) 0.01-6x10-3 (1- ~)
17.7x10:14= 29.5 x10-12F= 29.5 pF.
6x10-~~~1styjxiwn}j1htr

154. ELECTROST
ATICPOTENTIAL AND CAPACITANCE
x y
Example 83. Two parallel PlateL:J
capators, XandY, have the same
area of plates and same separation
between them. Xhas air between + -
theplates while Ycontains a 12 V
dielectric medium of s,=4. Fig. 2.124
(i)Calculate capatance of each capator ifequivalent
capatance of the combination is4IlF.
(ii)Calculate the potential difference betweenthe plates
ofXandY.
(iii)What is the ratio of electrostatic energy stored in X
andY? [CBSE D 04, 09]
Solution. (i)Let Cx =C.Thenc,=e,c=4C
Now X and Yare connected inseries.
.. C =CXCy = C.4C
eqCx+Cy C+4C
4 IlF =iC or C = 5 IlF
5
Hence Cx =C=5IlF andc,=4C=4x5=20IlF.
(ii)LetVbethe p.d. acrossX.Then p.d. across Y
will be V/4.
.. V+V=12or
4
Hence Vx=V=9.6VandVy=V /4=2.4V.
(iii)Energy stored in X= ~ C(9.6 )2=i=4 :1.
Energy stored inY~4C(2.4)2 1
/Example 84. An electric field Eo =3 x104Vm-1 is
established between the plates, 0.05m apart, of a parallel
platecapacitor. After removing the charging battery, an
uncharged metal plate of thickness t=0.01misinserted
betweenthe capator plates. Find the p.d. acrossthe
capacitor(i)before,(ii)after the introduction of the plate.
(iii)Whatwould bethep.d.ifadielectric slab (K=2)were
introduced inplace of metal plate? [Roorkee 91]
Solution.(i)The p.d. across the capacitor plates
before metal plate is inserted,
Vo=Eod=3x104x0.05 = 1500 V.
(ii)Asnoelectric field exists inmetal plate, so the
p.d.aftertheintroduction of metal plate is
V=Eo (d -t)=3x104x(0.05-0.01) =1200 V.
(iii)When dielectric slab(K=2)isintroduced, the
p.d. becomes
V=Eo(d -t)+ Eo t=1200+ 3x 104x0.01 =1350 V.
. K 2
or
V=9.6V
Example 85. A parallel plate capator ischarged to a
certain potential difference. When a3.0mm thick slab is
slipped between the capator plates,then to maintain the
2.63
same p.d. between the plates, the plate separation is tobe
increased by2.4mm Find the dielectric constant of the slab.
Solution. LetEobethe electric field between the
capacitor plates before the introduction of the slab.
Then,thep.d. between the plates is
Vo=Eod
Suppose theseparation between the plates is
increed by d'tomaintain the same p.d. after the
introduction of the slab of thickness t.Then
Vo=Eo (d+d'- t)+Eo . t
K
E
Eo (d+d' -t)+-.!1..t=Eo d
K
K=_t_= 3.0mm =5.
t-d'3.0mm-2.4 mm
or
Example 86. The area ofparallel plates of an air-filled
capatoris0.20~and the distance between them isO.Olm
The p.d. acrossthe plates is 3000 V.Whena0.01m thick
dielectric sheetisplaced between the plates,thep.d.decreases
to1000V.Determine (i)capatance of the capacitor before
plangthesheet (ii)charge on each plate (iii)dielectric
constant ofthe material tio)capatance ofthe capator after
plangthedielectric (v) permittivity of the dielectric. Given
Co=8.85 x10-12Fm-1.
Solution. (I)Capacitance of air-filled capacitor is
Co = CoA=8.85x10-12x0.20 = 1.77 x10-10 F.
d 0.01
(ii)Chargeon each plate,
q= CoVo=1.77x10-10x3000
= 5.31 x10-7C.
(iii)Dielectric constant of the material is
K-C_q/V_Vo_3000_3
-Co -q/Vo-V-1000 - .
(iv)Capacitance after thedielectric sheet is introduced,
C =KCo=3x1.77x10-10 = 5.31x10-10F.
(v)Permittivity of thedielectric is
C=KEO= 3x8.85x10-12= 2.65x10-11Fm -1.
Example 87. The capacitance of a parallel plate capatoris
50pF and the distance between theplatesis4mm Itis
charged to 200Vand then the charging battery isremoved.
Now adielectric slab (K=4)of thickness 2mm is placed.
Determine (i) final charge on each plate (ii)final potential
difference between theplates (iii)final energy in the
capacitor and (io) energy loss.
Solution.Capacitance of air-filled capacitor,
_ CoA
Co--d- ...(1)···3wx«n£m~r±n3lxv

155. 2.64
Capacitance with dielectric
slabof thickness
t«d)is
C= goA
d-t+t/K
(i)The charge on capacitor plates, when 200 V p.d.
is applied, becomes
q=CoVo=50x10-12x200=10-8C
Even after thebattery is removed, the charge of
10-8Con the capacitor plates remains the same.
(ii)Onplacing the dielectric slab, suppose the capa-
citance becomes Cand potential differenceV.Then
q=Co Vo=CV
V-CoV_d-t+t/K V
-C 0- d 0
or
[Using (1) and (2)]
=4- 2+2 / 4x200=125V.
4
(iii)Final energy in the capacitor is
U=.!qV=.!x10-8 x125=6.25x10-7J.
2 2
(iv)Energy loss
1
=uo-u=2q (Vo - V)
=.!x10-8 x(200 -125)
2
=3.75x10-7J.
Example 88. A parallel plate capacitor isformed by two
plates, each of area100cd,separated by a distance of1mm
A dielectric of dielectric constant5and dielectric strength
1.9x107Vm-l isfilled between the plates. Find the
maximum charge that can be stored on the capacitor without
causing any dielectric breakdown.
Solution.Electric field between capacitor plates is
given by
E=~=-q-
KgoKgoA
Asthe electric field should notexceed 1.9 x107Vm-I,
so the maximum charge thatcan be stored is
q=KgoAE
=5x8.85x10-12 x100x10-4x1.9x107
=8.4x10-6C.
Example 89.A slab of material of dielectric constant Khas
the same area as the plates' of a parallel plate capator but
has a thickness 3d /4,where disthe separation of the plates.
Howis the capatance changed when the slab is inserted
between the plates? [NCERT]
PHYSICS-XII
...(2)
Solution. IfVoisthe potential difference when
there is nodielectric, then the electric field between the
capacitor plates will be
V
E-~
0-d
After the dielectric is inserted, the electric field in
the dielectric reduces to
E
E=~
K
Now the potential difference between the plates
will be
Thus the potential differencedecrees by a factor
of(K+3) /4K,while the free charge qoon the plates
remains same. The capacitance increes to a new
value given by
C-qo_4Kqo_4KC
-Y--K+3'Vo -K+3 o·
Example 90
(a) Find the ratio of thecapacitances of a capator filled
with two dielectrics of same dimensions but of dielectric
constants KIandK2'respectively.
(b) A capacitor isfilled with two dielectrics of the same
dimensions but of dielectric constants KI=2and
K2=3.Find the ratio of capacities intwo possible
arrangements. [MNREC 85]
Solution.(a)The two possible arrangements of the
two dielectrics are shown in Figs.2.125(a)and(b).
I
-=E2J=-[J
I
K2
Fig.2.125 (a) (b)
(i)The arrangement (a)can be supposed to be a
parallel combination of two capacitors, each with plate
areaA/2and separationd.Therefore, the total capa-
citance is···3wx«n£m~r±n3lxv

156. ELECT
ROSTATIC POTENTIAL AND CAPACITANCE
(ii)The arrangement (b)can besupposedto bea
series combinationof two capacitors, each with plate
totalareaAand separationd/2.Therefore, the
capacitance C ' isgiven by
1 1 1 1 1
-=-+ -=--,-- +---;--
cC'C 6AK 60AK2
1 2 _0__ 1
d/2 d/2
d(11)
=260A-;s+K2
C'=260A(KlK2)
d Kl+K2
or
Ratio of the capacitances in thetwoarrangements is
2
C _60A(K1 +K2) d(K1 +K2) _ (K1 +K2)
C,- 2d '260AK1K2- 4K1K2
(b)Here Kl =2, K2=3
C(2+3)2 25
C4x2x 324
~roblems For Practice
1.A parallel-plate capacitor having plate area 100cm2
andseparation 1.0 mm holds acharge of 0.12~C
when connected to a 120 Vbattery. Find the
dielectric constant of the material filling thegap.
(Ans.11.3)
~. Find the length of the paper used in a capacitor of
capacitance 2 ~F, if the dielectric constant ofthe
paperis 2.5 andits width and thickness are 50 mm
and 0.05 mm, respectively. (Ans.90m)
3.A parallel-plate capacitorconsists of 26metal strips,
each of 3 emx4em, separated by mica sheets of
dielectric constant 6 and uniform thickness 0.2 mm.
Find the capacitance. (Ans. 7.97x10-9F)
4.A parallel-plate capacitor of capacity 0.5~F isto be
constructed using paper sheets of thickness 0.04mm
dielectric. Find how many circular metalfoils
of diameter 0.1 m will have to be used. Take the
dielectric constant of paperused as 4. (Ans. 73)
5.When a slab of insulating material 4mm thick is
introduced between the plates of aparallel plate
capacitor, it is found that the distance between the
plates h to be increed by 3.2 mm to restore the
capacitance to the original value. Calculate the
dielectric constant of the material. (Ans.5)
6.The two plates of a parallel plate capacitor are
.4 mm apart. A slab of dielectric constant 3and
thickness 3 mm is introduced between the plates
with its faces parallel to them. The distance
between the plates is so adjusted that the capa-
2.65
citance ofthe capacitor becomes2/3rd of its original
value. What is the newdistance between the
plates? [CBSEODOSC](Ans.8 mm)
7.The distance between the parallel plates of a charged
capacitor is 5 cm and the intensity of electric field is
300Vem-1.A slab ofdielectric constant 5 andthick-
ness 1 em is inserted parallel to the plates. Determine
the potential differencebetween the plates, before and
afterthe slab isinserted ? (Ans. 1500V,1260V)
8.A parallel plate capacitor with plate separation
5mm is charged by a battery. It is foundthaton
introducing a mica sheet 2mmthick, while keeping
thebattery connections intact, the capacitor draws
25%more energy from the battery thanbefore. Find
thedielectric constant ofmica. (Ans.2)
9.Figure 2.126 shows a parallel plate capacitor of
plate areaAandplate separation d.Its entire space
is filled with three different dielectric slabs of same
thickness. Find the equivalent capacitance of the
arrangement. [Ans. C = 3eoAKlK2K3 ]
d(K1K2 +K2K3+K3K1)
r
iT
P
d/3
+
d/3 Kj K2 K3
+
d/3
.i.
bB
Fig.2.127
r- I_A__ ---"l
Fig.2.126
10.The space between the plates of a parallel plate
capacitor of capacitance C is filled with three
dielectric slabs of equal thickness, shown in
Fig. 2.127. Ifthedielectric constants of thethree
slabs are K1, K2and K3, findthenewcapacitance.
[Ans. C=~(KI +K2+K3)]
3
11.A slab of material ofdielectric constant Kh the
same area the plates of a parallel plate capacitor
but h thickness d/2,wheredisthe separation
between the plates. Find the expression forthe
capacitance when the slab isinserted between the
plates. [CBSE F 10;OD13] (Ans.~C )
K+l0
HINTS
1.Capacitance,
IfKisthedielectric constant, then
C=KeoA= Kx8.85 x10-12xlOOx10-4 = 1O-9F
d 1.0x10-3
:. K=11.3.···3wx«n£m~r±n3lxv

157. 2.66
3.Arrang
ement ofnmetal plates separated by dielectric
acts a parallel combination of (n-1) capacitors.
C=(n-1) ICEOA
d
25x6x8.85x10-12x3x4x10-4
0.2x10-3
=7.97x10-9F.
C=(n -1)ICEOA
4.As
d
.. 0.5 xlO-6 F
(n-1) x4x8.85x10-12x3.14 x(0.05)2
0.04x10-3
0.5x0.04x103 7 97
orn -1= =1.=:72
4x8.85x3.14x(0.05)2
or n=73.
EOA
5.Capacitancewithout dielectric, C =--
d
When dielectric is introduced,
As the capacitance remains same in both ces, so
EoA EoA
-d- =d'-t (1-;)
or d=d'-t (1-;) ord'-d=t(1-;)
Butd'- d=3.2mm,t=4 mm
. . 3.2=4(1- ;)
or 1-~ = 3.2 = 0.8 or ~= 0.2or IC=5.
K 4 IC
6.~xCapacitance with air = Capacitance with dielectric
2EoA eaA
3d=d'-t+~
K
or ~(d' -3 +~)=d= 4 mm or d'=8mm
or
7.P.D.before the dielectric slab is inserted,
Vo=EUd= 300Vem-1x5 em =1500 V.
EaA EA
Cu=--=_0_ farad
d 0.05
or EOA= 0.05 Co
Capacitance with dielectric slab,
EOA 0.05 Co 25 Co
C= t 0.01 -
d -t+-0.05 - 0.01 + _ 21
K 5
PHYSICS-XII
Forcharge to remain constant,
CuVo=CV
Cux1500=25CuxVorV=1260 V.
21
8.Asthe battery connections are intact(V= constant)
and the capacitor drs 25%more charge, so the
capacitance also increes by 25%. That is
C=125 C =~r
10004'1J
eaA 5EoA
d-t(1_ ~) =4.-d-
d -t(1-;)= ~ort(1-;)=~
1-~ = ~ = _5_ = ..!or IC=2
K5t5x2 2
or
or
or
9:The given arrangement is equivalent to three
capacitors connected in series. Each such capacitor
hasplate areaAand plate separation d.
K1EOA 31C1EOA
Cr=~= d
Cz=31C2EOAandc,= 3 K3 EOA
d d
The equivalent capacity C ofthegiven arrangement
isgiven by
1 1 1 1 d(1 1 1J
C=Cr +Cz +c; =3EoA K1 +1(2 +1C3
or C = 3 EOAKl K2 K3
d(K1K2+K2 ~ +K3K1)
10.Original capacitance, C =EOA
d
The new arrangement is equivalent to three capaci-
tors connected in parallel. Each such capacitor h
plate areaA/3and plate separationd.The new
capacitance is
C'=Cr+Cz+c;
1C1EOA/3 1C2EaA/3 K3EOA/3
= + +~--"----
d d d
EOA
=3dCK1+K2 +K3)
C'=~(ICI+Kz+ ~).
3
v:
11.Without dielectric, EU=-.l!
d
E=EU
K
V=EU.!!+E.!!=fu.!!+Eu.!!
2 2 2 1C2
EUd(K+l) VoCK+1)
=T'-K-= 2K
C_qo _2Kqo _ 21C
- V-Vo(K+1)- IC+1Cu
or
With dielectric,
or~~~1styjxiwn}j1htr

158. ELECTROSTA
TICPOTENTIAL AND CAPACITANCE
2.33DISCHARGING ACTION OF SHARP
POINTS : CORONA DISCHARGE
54.Brieflyexplain discharging action of sharp points
or corona discharge.
Discharging action of sharp points : Corona
discharge.When a spherical conductor of radius r
carries a chargeq,its surface charge density is
G=!L.:s:
A.4nr2
Electric field on the surfaceis
LowE+
Fig. 2.128Corona discharge.
The pointedend ofa conductor is highly curved
and its radiusof curvature r very small. Ifthecon-
ductor is given a charge q,then the charge density Gat
the pointedendwill be very high. Consequently, the
electric field near the pointed endwillbe veryhigh
which m cause the ionisation or electrical bre-
down of the surrounding air. Theoppositely charged
ions neutralise the pointed end while the similarly
charged ions are repelled ay. Fresh air molecules
come near the pointed end and te ay itscharge,
setting up a kind ofelectric wind. Thisprocess by which
the charge at the pointed end of aconductor gets discharged
iscalled corona discharge. The dischargeis often
accompanied by a visible glow near the pointed end.
2.34COLLECTING ACTION OF A
HOLLOW CONDUCTOR
55. A smallsphere ofradius r and charge qisenclosed
by a spherical shell of radius Rand charge Q.Show thatif
qispositive, charge qwill necessarily flow from the
sphere to the shell (when the two are connected by a wire)
no matter what the chargeQonthe shell is. [NCERT]
Collecting action of a hollow sphere.Consider a
small sphere of radiusrplaced inside a largespherical
shell of radius R. Let the spheres carry chargesqandQ,
respectively.
2.67
Total potential on the outer sphere,
VR=Potential due to its own charge Q
+Potential due to the chargeqon
theinner sphere
=4n\J~ + ~]
Potential on the inner
sphere due to its own charge is
1q
V=---
14neo.r
As the potential at every
point inside a chargedsphere
is the same as that on its
surface, so potential on the
inner sphere due to charge Q
onoutersphere is
V=_1_ Q
24neo· R
Total potential on inner sphere
Vr=4:eJ;+ ~]
Insulating
suspension
Fig. 2.129Small charged
sphere suspended inside a
charged spherical shell.
Hence the potential difference is
V - V=-q-[!-!]
r R4neo rR
Soifqispositive, the potential of the inner sphere
willalways be higher than that of the outer sphere.
Now if the two spheres are connected by a conducting
wire,thechargeqwill flowentirely to the outer sphere,
irrespective of the charge Q already present on the
outer sphere. In fact this istrue for conductors of any
shape.
2.35
*
VAN DE GRAAFF GENERATOR
56. Explain the basic prinple, construction and
working of Van de Graaffgenerator.
Van de Graff generator. Itisanelectrostatic
generatorcapable of building up high potential differences of
the order of107 volts.
Principle.The working of a Van de Graff gene-
rator is bed on following two electrostatic phenomena:
(i)Discharging action ofsharp points (corona
discharge) i.e.,electricdischarge takes place in air or
gasesreadily at the pointed ends of conductors.
(ii)If a charged conductor isbrought intointernal
contact with a hollow conductor, all ofitscharge
transfers to the hollow conductor, howsoever high the
potential of the latter may be.···3wx«n£m~r±n3lxv

159. 2.68
Construction. A largespherical
conducting shell (of
few metres radius) is supported at a height several
metres above the ground on an insulating column. A
long narrow belt of insulating material, like rubberor
silk,iswound around two pulleys, P1at ground level
andP2at the centre of the shell.
This belt is kept continuously moving by an electric
motor attached to the lower pulley Pl'Near the bottom
and the top of its run, ·sp mpw· {±p± nwz±p ·z ·rz
sharply pointed brs combs Bland ~, pointing
towards the belt. The comb ~, calledspray combis
given a positive potential of 10 kV with respect to the
earth by means of a battery; while the comb ~,called
collecting comb, is connected to thespherical shellS.
Working. Due to the high electric field at the
pointed ends of comb ~, the air of the neighbourhood
gets ionised and its positive charge repelled or spr ed
on to the belt, which moves up into the shell S. Asit
{±p± nwz±e to comb ~, it induces a negative charge at
the pointed ends of comb~and a positive charge on
the shell S. The positive charge spreads uniformly on
the outer surface of theshell S.The high electric fieldat
the pointed ends of comb~ionises the air there and
repels the negative charges on to the beltwhich
neutraliseits positive charge. This process continues.
Asmore and more positive charge is given to the shell,
its potential continues to rise. In this w , a high
potential of 6 to 8 million volts can be built upon the
sphere.
A discharge tube is placed with its upper end
inside the hollow sphere and lower end earthed. The
ion source is placed at the upper end of the tube. The
high potential on the sphere repels the charged
particles downwardwith large acceleration, where
theyhit the target atoms to bring about the nuclear
disintegration.
PHYSICS-XII
+ + +
Metal
shell,S
+
+
+
CoJlector
comb
4-~'--- Ion
source
Insulating belt to carry
and deliver charge
Insulating pillar
Discharge tube
Grounded metal
be
..:c..
T
H.T.B. ,
-'- Target
Fig. 2.130
Use.The high potential difference set up in a
Van de Graff generator is used to accelerate charged
particles like protons, deutrons, a-particles, etc. to high
energies of about 10 MeV, needed for experiments to
probe the small scale structureof matter.~~~1styjxiwn}j1htr

160. GI DELINES To NCERT
EXERCISES
2.1.Two charges 5x1O-8C and-3x10-8Carelocated
16cm apart. At what point on the line joining the two charges is
theelectric poteial zero ?Takethepoteialat infinity to be
zero.
Ans. Zero of electric potential for two charges. As
shown in Fig. 2.186, suppose the two charges are placed on
X-axis with the positive charge located at the origin0.
-8 -8
ql=5x10 C q2= -3x10 C
·.-----------+I--------------~·
o p A
I--- x----0·01+1_ --- 0.16- x-----I
Fig. 2.186
Let the potential be zero at the point P and OP =x.For
x<0(i.e.,to the left of0),the potentials of the two charges
or
cannot add up to zero. Clearly, xmust be positive. If xlies
between0andA,then
Vl+V2=0
_1_ [ql+q2 ]-0
4TCEO x0.16-x
9 [5x10-83x10-8]
or9x10 - = 0
x 0.16 -x
5 3 =0
x0.16 -x
or
or x= 0.10 m =10 em,
The other possibility is thatxmayalso lie onOA
produced, asshown in Fig. 2.187.
-8 -8
ql=5xl0 C q2=-3xlO C
• • I
o A P
11+-: 0_.1_6_-_-_-~-x--'I'-- x -0.16=1
Fig. 2.187
As V;+V2=0
_1_[ 5x10-8 _3x10-8 ]=0
4TCEO X X -0.16
5 3 =0
x x -0.16
or x=0.40 m = 40 em.
Thus the electric potential is zero at 10 em and 40 em
away from the positive charge on the side ofthe negative
charge.
2.2.Aregular hexagon of side 10em has acharge of 5J.1Cat
each ofitsvertices. Calculate thepoteial at the cere of the
hexagon.
Ans.Clearly, distance of each charge from the centre
ois
r= IDem =D.1Dmwww5notes•rive5~om

161. 2.92
Fig.2
.188
Magnitude of each charge is
q=5IlC=5xio+c
:.Potential at the centre 0is
1q6x9x109x5x10-6
V=6.--.-=-------
41tEOr 0.10
=2.7x106 V.
2.3.Two charges +2IlC and -2IlC are placed at pois A
andB,6em apart. (i)Ideify anequipoteial surface ofthe
system(ii)Whatisthe direction of the electric field at every or
pointonthe surface?
Ans.(i) The equipotential surface will be a plane
normal toABand passing through its midpoint0,as
shown in Fig. 2.189. Ithas zero potential everywhere.
A o
Equipotential
surface
Fig. 2.189
(ii)Thedirection of electric field is normal to the plane
in the direction ABi.e.,from positive to negative charge.
2.4.Aspherical conductorofradius 12cm has acharge of
1.6x10-7Cdistributed uniformly on its surface. What isthe
electric field
(a)inside the sphere (b) just outside the sphere
(c) at a point 18cm from the cere of the sphere?
Ans.Refer to the solution of Example 75 on page 1.60.
2.5.A parallel plate capacitor with air between the plates
has a capacitance of8pF (lpF=10-12F).Whatwill be the
capacitance if the distance between the plates be reduced by half,
the space between them isfilled with a substance of dielectric
consta, lC=6?
Ans.Capacitance of the capacitor withairbetween its
plates,
EA
= -0-=8pF
d
PHYSICS-XII
When the capacitor is filled with dielectric(lC=6)
between its plates and the distance between the plates is
reduced by half, capacitance becomes,
EOlCA EOx6xA EoA
C'=--= =12-
d' d / 2 d
C'=12x8=96pF.
2.6.Threecapacitors each of capacitance 9pF are conected
inseries. (a)Whatisthetotal capacitance of the combination?
(b)What is thepoteial difference across each capacitor when
the combination isconected to a120Vsupply ?
Ans.(a)IfCisthe equivalent capacitance of theseries
combination, then
111111131
-=-+-+-=-+-+-=-=-
CC1, <;c;99993
or
C=3pF.
(b)As all the capacitors have equal capacitance, so
potential dropt1Vwould be same across each capacitor.
.. V =t1VI+t1V2+t1V3
=t1V+t1V+t1V=3t1V
t1V=V= 120 = 40V.
3 3
or
2.7.Three capacitors of capacitances 2pF,3pF and4pF are
conected in parallel. (a) What is thetotalcapacitance of the
combination? (b)Determine the charge on eachcapacitor ifthe
combinationisconected to a100Vsupply.
Ans.(a)For the parallel combination, total capacitance
is given by
C=C1,+<;+<;=2+3 + 4=9pF.
(b)When the combination is connected to 100V
supply, charges on the capacitors will be
ql=C1,V=2x10-12 x100=2x10-10C
q2=<;V= 3x10-12x100=3x1().-lO C
q2=c;V=4x10-12 x100=4x10-10C.
2.8.Ina parallel plate capacitor with air between the plates,
each plate has an area of6x10-3m2and the distance between
the plates is3mm.Calculate the capacitance of the capacitor. If
the capacitor isconected to a100V supply, whatisthe charge
on each plate of the capacitor ?
Ans.Capacitance of capacitorwithair between its
plates is
Co=EoA= 8.85x10-12 x6x10-3
d 3x10-3
=1.8x10-11F=18 pF.
Charge,
q=CoV= 1.8x10-11x100=1.8x10-9 C.
2.9.Explain what would happen if in the capacitor givenin
Exercise 2.8,a3mm thick mica sheet(of dielectric consta=6)
were inserted between the plates, (i) while the voltage supply
remains conected (ii)after the supply was disconnected.
[CBSE Sample Paper 98]www3xy·o¥n~s•o3myw

162. ELECTROSTATIC P
OTENTIAL AND CAPACITANCE
Ans.From the above question, we have
CD=1.8x10-11F=18pF,qo=1.8x10-9C
Also, K=6
(i)When the voltage supply remains conected, the
potential diference between capacitor plates remains
same i.e.,100V.
Thecapacitance increases Ktimes.
.. C=KCD= 6x18 =108 pF.
The charge on the capacitor plates will be
q=CV=108x10-12x100 =1.08x10-8 C.
(ii)Afterthe supply isdisconnected, thecharge on the
capacitor plates remains same i.e.,
qo=18x10-9C
The capacitance increases Ktimes.
C=KCD=108 pF.
The potential diference between the capacitor plates
becomes
v=Va= 100 =16.6 V.
K 6
2.10.A12pFcapacitor isconected to a50Vbattery. How
much electrostatic energy is stored in the capacitor?
Ans.Here C=12 pF=12x10-12F,V=50 V
Energy stored,
U=-.!CV2 =-.!x12x10-12 x(50)2=1.5x10-8J.
2 2
2.11.A600pFcapacitor ischarged by a200V supply. It is
then disconected from the supply and isconnected to another
uncharged 600pFcapacitor.How much electrostatic energy is
lostin the process?
Ans.Here C; =600 pF, VI=200 V,
<;= 600 pF, V2= 0
Common potential,
V= C;VI+<;V2= 600x 10-12 x200+ 0= 100 V
C;+<; (600+ 600) x10-12
Initial energy stored,
121 -12 2
Ui= U1="2C;VI="2x600x10 x(200)
= 12x10-6J
Final energy stored,
1 2
Uf="2(C;+<;)V
=-.!(600+600)x10-12x(100)2 =6x10-6J
2
Electrostatic energy lost,
6.U=U,-Uf=12x10--{; -6x10-6
= 6x10-6J
2.93
2.12.A charge of8mC is located at the origin. Calculate
the work done in taking a small charge of-2x10-9Cfrom
apoiP(0,0,3em)toa poiQ(0,4em,0)via a poiR(0,
6em,9em).
Ans. As the work done in taking a charge from one
point to another is independent of the path followed,
therefore
W=qo[VQ-VpJ=qo[-q- - -q-]
41t EOr2 41t Eo '1
=4~:J~-I]
z
R(O,6ern,9ern)
P(O,0, 3em)
q=2meQ(O,4 ern, 0)
x
Fig. 2.190
Here q= 8 mC = 8x1O-3C,qo=-2x10-9C
'1=3em =3x10-2m,r2=4em =4x10-2m
..W= - 2x10-9x8x10-3x9x109
x[4x~o-2 -3x:0-2 ]
=1.2J.
2.13.A cube of side b has acharge q at each of itsvertices.
Determine thepoteial and electric field due tothischarge
array at the cere of the cube.
Ans.Length of longest diagonal of the cube
=~b2+b2+b2=J3b
Distance of each charge (placed at vertex) from the
centre of the cube is
r=J3b
2
:. Potential at the centre of the cube is
V= 8._l_.!i=8x_1_.~
41t Eo r 41t EOJ3b
4qwww5notes•rive5~om

163. 2.94
Electric fields
at the centre due to any pair of charges
at the opposite comers will be equal and opposite thus
cancelling out in pairs. Hence resultant electric field at the
centre will be zero.
2.14.Two tiny spheres carrying charges 1.5JlCand2.5JlC
are located30em apart. Find the poteial (a) at the midpoi of
the line joining the two charges, and (b) at a poi10em from
this midpoi in a plane normaltothe line and passing through
the midpoi.
Ans.The two situations are shown in Fig. 2.191.
C
§f:;&;
6~¥
\~~/
A~------...J...------->eB
15em 0 15 em
q]=1.5 JlC q2=2.5 JlC
~
IDem ~
Fig.2.191
(a)Here1.=r2= 15em = 0.15 m
:.Potential at the midpoint 0 of the line joining the
two charges is
v:-_1_[ql+q2]
o - 4m;0 1.r2
= 9x109[ 1.5 x10-6+2.5x10-6] V
0.15 0.15
=9x109x10-6[ 10+5;]V
= 9x103x80 V =2.4xl0sV.
3
(b)Here1.=r2= ~102 +152
=55:::.18em = 0.18 m
.. Potential at point C due to the two charges is
v: -_1_[ql+q2]
c- 41tEo1.r2
9 [ 1.5 x10-6 2.5x10-6]
= 9xl0 + ----
0.18 0.18
9x109x4x10-6 5
------=2xl0 V.
0.18
2.15.A spherical conducting shell of iner radius 1.and
outer radius r2 has a chargeQ.
(a) A charge qisplaced at the cere of the shell. Whatisthe
surface charge density on the iner and outer surfaces of
the shell?
(b)Isthe electric field inside acavity(with nocharge) zero
evenifthe shell isnot spherical, but has any irregular
shape?Explain.
PHYSICS-XII
Ans.(a)The chargeqplaced at the centre of the shell
induces a charge -qon the inner surface of the shell and
charge+qon its outer surface.
.. Surface charge density on the inner surface of the
shell
charge q
---2
surface area41t1.
Surface charge density on the outer surface of the shell
Q+q
4m:2 .
2
(b)Even if the shell is not spherical, the entire charge
resides on its outer surface. The net charge on the inner
surface enclosing the cavity is zero. From Gauss's theorem,
electric field vanishes at all points inside the cavity. For a
cavity of arbitrary shape, this is not enough to claim that
electric field inside must be zero. The cavity surface may
have positive and negative charges with total charge zero.
S
P@_R
+ -
Q
Fig. 2.192Electric field vanishes inside a cavity ofanyshape.
To overrule this possibility, consider a closed loop
PQRSP, such that partPQRis inside the cavity along a
line of force and the partRSPis inside the conductor.
Since the field inside a conductor is zero, this gives a
network done by the field (in partRSP)in carrying a test
charge over a closed loop. But this is not possible for a
conservative field like the electrostatic field. Hence there
are no lines of force(i.e.,no field), and no charge on the
inner surface of the conductor, whatever be its shape.
2.16.(a) Show that the normal compone of electrostatic
field has a discoinuity from one side of a charged surfaceto
another given by
4 ~ "cr
(f2-f1).n =-
Eo
wherenisaunitvector normal tothe surface at a poi andcris
the surface charge density at that poi.(The direction ofnis
fromside1toside 2)
Hence show that just outside a conductor, the electric field
is cr~ /EO'
(b)Showthat the tangeial compone of electrostatic field
iscoinuous from one side of a charged surface toanother.
[Hint:For(a),Use Gauss's law. For(b),use the fact
that work done by electrostatic field on a closed loop is
zero.]www3xy·o¥n~s•o3myw

164. ELECTROSTATIC POTENTIALAND CAPACITANCE
Ans. (a)Elec
tric field near a plane sheet of charge is
given by
E=~
2Eo
Ifnis a unit vector normal tothe sheet from side 1 to
side 2, then electric field on side 2
~ cr"
~=-n
2Eo
in the direction of the outward normal to the side 2.
Similarly, electric field on side 1 is
~ cr"
E,.=--n
2Eo
in the direction of the outward normal to the side 1.
(E; - ~).n=.z.-(-~J= ~
2Eo 2Eo EO
~ ~
AsE,.and ~ act in opposite directions,there mustbe
discontinuity at thesheet of charge. ow electric field
vanishes inside a conductor, therefore
Hence outside the conductor, theelectric field is
~
(b)LetXYbe thecharged surface of a dielectric and E,.
~
and ~ be the electric fields on the two sides of the charged
surface as shown in Fig. 2.193.
~~
d~:
A E;=E1CDS91 D
----------------
1+ ++++++++ +1
X---..."-----------...,,,---- Y
1 ---------1
B C
~' E2 ,
__~2+__J
Ei.=E2CDS92
Dielectric
Fig. 2.193
Consider a rectangular loop ABCDwith length Iand
negligibly small breadth. Line integral along the closed
path ABCDwillbe
JE.di=F;.l-~.l =0
or E,.Icos 91 -~Icos 92 =0
(E,.cosf\-~cos 92) I=0
2.95
~
whereE{andEiare the tangential components of E,.and
~
~' respectively. Thus,
E{=Ei (',: I.*- 0)
Hence thetangentialcomponentof theelectrostatic
fieldis continuous acrossthe surface.
2.17.A longcharged cylinder of linear charged density A.is
surrounded by a hollow co-axial conducting cylinder. Whatis
the electric field in the space between the twocylinders ?
Ans.Refer answer toQ.35 on page 2.30.
2.18.In a hydrogen atom, the electron and proton are bound
at a distance of about053A.
(i)Estimate the poteial energy of the system in eY,
taking the zero of poteial energy at infinite
separation ofthe electron from proton.
(ii)Whatisthe minimum work required tofreethe elec-
tron,given that its kinetic energy in the orbit ishalf
the magnitude of poteial energy obtained in (i)?
(iii)What arethe answers to (i)and(ii)aboveifthe zero
of poteial energyistaken at1.06Aseparation ?
Ans.(i) ifI=-1.6x10-19C,
q2=+1.6x10-19 C,
r=0.53A=0.53 x10-10m
P.E. ofthe electron-proton systemwill be
U=_1_. qlq2
41tEor
(-1.6 x 10-19) x1.6x10-19
=9x109x J
0.53x10-10
9x1.6x1.6xlO-19 J
0.53
9x1.6 x1.6x10-19eV
0.53x1.6x10-19
=-27.2 eV.
(ii)K.E.of the electron in the orbit
1 1
= -P.E.= -x27.2eV =13.6eV
2 2
:. Total energy of the electron
=r.E.+K.E.
= (-27.2+13.6)eV
=-13.6eV.
As minimum energy of the free electron iszero, so
minimum work required to free the electron
=0 -( -13.6)
=13.6 eV.www5notes•rive5~om

165. 2.96
(iii) When
the zero of potential energy is not taken at
infinity, the potential energy of thesystem is
U-ql q2[]._ ~]
4m,01.r2
= 9x109x(-1.6x10-19) x1.6x10-19
[1 1] J
x0.53x10-10 -1.06x10-10
9x109x1.6x10-19x1.6 x10-19 [1]
-- 1-- eV
-.1.6x10-19 x0.53x10-10 2
9x1.6
= ----eV =-13.6eV
0.53x2
This indicates that the K.E.of 13.6 eV of case (i) is used
up in increasing theP.E.from-27.2eV to -13.6eV as the
electron is carried from 0.53 Ato 1.06Aposition. K.E.in
thissituation should be zero. As the total energy in this
case is zero, therefore, minimumwork required to free the
electron
= 0 - (-13.6eV) = 13.6 eV.
2.19.If one of the two electrons of a H2molecule isremoved,
we get a hydrogen molecular ion(H2+). Inthe groundstate of a
H;ion,the two protons are separated by roughly 15Ii,and the
electronisroughly 1Afrom each proton. Determine the
poteial energy of the system. Specify yourchoice of the zero of
poteial energy.
Ans.The system of charges is shown in Fig. 2.194.
qz=+qfl--------:---- ...•q3=+q
Proton rz=1.5A Proton
Fig.2.194
Charge on an electron,
%=-e= -1.6x10-19C
Charge on each proton,
q2=q3=+e=+1.6x10-19C
If the zero of potential energy is taken at infinity, then
potential energy of the system is
U=U+U+U=_1_ [ql q2+q2q3+ql q3 ]
12 23 1341ts r. t: t:
o 1 2 3
_ 1 [ (-q)q+ q.q+(-q)q]
-41tso 1x10101.5x10101x1010
_ e2[-1+2.-1] [q=e]
- 41tso x1010 1.5
PHYSICS-XII
(1.6x10-19)2 x9x109x (-4)
10-10 3J
(1.6x10-19)2 x9x109x4
=- 19 eV=-19.2eV
1.6x10-x3
[.: 1eV=1.6x10-19JJ
2.20.Twocharged conducting spheres of radii a and bare
conected to each other by a wire. What isthe ratio of electric
fields atthe surfaces ofthetwo spheres? Usethe result obtained
to explain why charge density on the sharp and poied ends of a
conductor ishigher than thaton its flatter portions.
Ans.The charges will ow between the two spheres
till their potentials become equal. Then the chargeson the
two sphereswould be
Q1=c;V=.S.
Q2 CzV Cz
.S.=~
Cz b
But
The ratioofthe electric fields at the surface of the two
spheres will be
Also,
1Q1
11 ~ .7qb2a b2 b
Ez=_1_ Q2=Q2.a2=b'a2=-;;.
41tSo .b2
!i.=0'1
Ez 0'2
~=~
0'2 a
Thus the surfacecharge densities are inversely
proportional to the radii of the spheres. Since the at
portion may be considered as a spherical surface of large
radius and a pointed portion as that of small radius, that is
why,the surface charge density on the sharp and pointed
endsof a conductor is much higher than that on its atter
portion.
2.21.Two charges -q and+q arelocatedatpois (O,O,-a)
and (O,O,a) respectively.
(i)What isthe electrostatic poteial at the pois
(O,O,z)and (x,y,O) ?
(ii)Obtain thedependence of poteial on the distance r
of apoifromthe origin when r/a»1
(iii)How much work isdone in moving a small test
charge from the poi (5,0,0)to(-7,0,0) alongthe
x-axis?
Doesthe answer change ifthepath of the testcharge
betweenthe same pois isnot along the x-axis ?www3xy·o¥n~s•o3myw

166. ELECTROSTATIC POTENTIAL AND
CAPACITANCE
Ans.(i)When the point P lies closer to the charge +q
as shown in Fig. 2.195(a),the potential at this point P
will be
V=4n1eJ ~-!]=4n1eJ z~a -z-~-a) ]
q 2a
=4ne'T7
o-
V=_l p_
4neo .z2-a2.
or [.: p=qx2a]
z
P (0, 0,z)
+q(0,0,a)
y(a)
a
-q(0,0,-a)
X
Z
+q(0, 0,a)
y
(b)
a
-q(0,0,-a)
X
P (0,0, z)
Fig.2.195
When the point P lies closer to charge-q,as shown in
Fig.2.195(b),it can be easily seen that
V=__l p_
4ne.z2 -a2
o
Again, any point (x,y,0) lies in XY-plane which is
perpendicular bisector of Z-axis. Such a point will be at
equal distances from the charges - qand+q.Hence
potential at point (x,y,0) will bezero.
(ii)If the distance of point P from the or-igin 0isr,then
from the results of part (i),we get
V=+_1_ __P_ [Putz=r]
-4nE •r2-a2
o
Ifr»a,we Canneglect a2compared to r2,so
V=+_l_ E.
- 4nea.r2
:.Forr»a,thedependence of potential Vonris1/r2
type.
2.97
(iii)(5,0, 0) and (- 7,0,0) are the points on theX-axis
i.e.,thesepoints lie on the perpendicular bisector of the
dipole. Each point is atthe same distance from the two
charges. Hence electric potential at each ofthese points is
zero.
Workdone in moving the testchargeqofromthepoint
(5,0, 0)to (- 7,0, 0) is
W=q(1)-V2)=q(0 - 0)=O.
No, the answer will not change if the path of the test
charge between the same two points isnotalong X-axis.
This is because the work done by the electrostatic field
between two points is independent of the path connecting
the two points.
2.22.Figure 2.196below shows a charge array known as an
electric quadrupole. For a poi on the axis of the quadrupole,
obtainthe dependence ojpoteial on rfor r» a.Corast your
result with that due to an electric dipole and anelectric
monopole (i.e.a single charge).
I--a-----..jI--a------l
• •• • •
+q -q-q +q
I,
P
'I
Fig.2.196
Ans.Potential at point P is
V=4:eJr ~ a-;-;+r1a ]
=_1_.q[r(r+a)-2(r -a)(r+a)+ r(r-a)]
4nea r(r- a)(r+a)
=_1_. q [r2+ar-2r2+2a2+r2- ar]
4nea r(? - a2)
1 2qa2 1 Q
=4neo.r(r2- a2) =4neo.r(r2- a2)
whereQ=2q a2 isthequadrupole momeof the given
chargedistribution. Asr»a,so we can write
V__1_Q
-4neo. r3
Hence for larger,quadrupole potential varies as 1/ r3,
whereas dipole potential varies as 1/ r2and monopole
potential varies as 1/r.
2.23.An electrical technician requires a capacitance of2IlF
in a circuit across a poteial difference of1 kV. A large number
of1.IlFcapacitors are available to him each of which can
withstand a poteial difference of not more than400V.
Suggest a possible arrangeme that requires a minimum
number ofcapacitors.
Ans.Let this arrangement require ncapacitors of 11lF
each in series andmsuch series combinations tobe
connected in parallel.www5notes•rive5~om

167. 2.98
P.O. ac
ross each capacitor of a series combination
=1000=400 or n=1000=2.5
n 400
But number of capacitors cannot be a fraction,
n= 3
Equivalent capacitance of the combination is
1
-. m=2 or m=2n=6
n
:.Total number of capacitors required
=3x6=18
So six series combinations, each of three Lufcapacitors,
shouldbe connected in parallel as shown in Fig. 2.197.
IIlF IIlF IIlF
HH
IIlF IIlF IIlF
~HH
IIlF IIlF IIlF
~HH
IIlF IIlF IIlF
~HH
IIlF IIlF Illf
~HH
IIlF IIlF IIlF
~HH
0------ 1kV---.0
Fig.2.197
2.24.Whatisthe area of the plates of a2F parallel plate
capacitor? Given that the separation between the platesis05em
Ans.Here C=2F,d=0.5cm = 5 x1O-3.m
As C=EoA
d
Cd2x5x10-3 2
A=-= m
EO8.85 x 10 12
.:::1130 x 106 m2=1130 km2.
2.25.Obtain the equivale capacitance of the network
shown in Fig.2.198.For a300V supply, determine the charge
and voltage across each capacitor. [eBSE OD081
Fig. 2.198
100pF
L-----li
C4
PHYSICS-XII
Ans.Asc;andc;are in series, their equivalent capa-
citance C;C;200x200
=c;+c;= 200+200 = 100 pF
Series combination of C2 and C3 is in parallel withC1'
their equivalent capacitance
=100pF+100pF=200pF
The combination ofC;,C;andc;is in series withC4'
equivalent capacitance of the network
200x100 F _ 200 F
200+100 p -3p
Total charge on the network is
q=CV= 200x10-12x300 = 2x10-8C
3
This must be equal to charge on C4 and also to the sum
ofthe charges onthe combination of C;,C;andC;.
.. q4=q=2x10-8C
q4 2x10-8
V4=-= 12 V=200V
C4 100x10
P.O. between pointsAand B
=V - V4 =(300 - 200)V= 100V
V;=100 V
%=C;VI= 100x10-12x100 = 10-8 C
Also theP'D,across the series combination of C2 and C3
= 100V
owsinceC;=c;
100
V2=V3=""2=50 V
and q2=q3=200x10-12x50 = 10-8 C.
2.26.The plates of a parallel plate capacitor have an area of
90 cm2 each and are separated by2.5mm The capacitor is
charged by conecting it to a400V supply.
(i)How much energyisstored by the capacitor?
(ii)View this energy stored in the electrostatic field
between the plates and obtain the energy per unit
volume u. Hence arrive at a relation between u and
the magnitude of electric fieldEbetween the plates.
Ans.(i)HereA= 90 cm2 =90x10-4 m 2 = 9x1O-3m2
d= 2.5 mm=2.5x10-3m,
EO=8.85xlO-12Fm -1,
V=400V'
Capacitance of the parallel plate capacitor is
C=EoA= 8.85 x10-12x9x10-3 F
d 2.5x10-3
= 31.86 x10-12 F = 31.86 pF.
Electrostatic energy stored by the capacitor,
U=1.CV2 =1.x31.86x10-12x(400)2J
2 2
= 254.88 x 10-8 J=2.55x10-6J.www5notes•rive5~om

168. ELECTROSTATIC POTENTIAL AND CAPACITANCE
(ii)Energy stored per unit volume or energy density of
the capacitor is
U 2.55 x10-
6 -3
U=Ad=9x10-3 x2.5x10-3 Jm
=0.113 Jm-3.
The relation betweenuandEcan be arrived at as
follows:
1 2
U"2CV 1EoA V2 1(V)2
U=Ad= ~ ="2-d-' Ad ="2Eod
u= ~EoE2.
2
2.27.A4flF capacitorischarged by a200V supply. Itis
then disconected from the supply andisconected to another
uncharged2flF capacitor. How much electrostatic energy of the
first capacitorislost in the form of heat and electromagnetic
radiation? [eBSE 00 OSI
Ans.Initial electrostatic energy of the 4 F capacitor is
U.= ~CV2= ~x4x10-6x(200)2=8x10-2J
I2 2
Charge on 4 F capacitor
=CV=4x10-6 x200=8xlO-4C
or
When the 4 F and 2 F capacitors are connected
together, both attain a common potentialV.Thus
V=Total charge=8x10-4 C =400V
Total capacitance (4+2)x10-6F 3
Final electrostatic energy of the combination,
Uf= ~x(4+2)x10-6 x(4~r J=~6x1O-2J
=5.33x10-2J
Electrostatic energy of the first capacitor lost in the
form of heat and electromagnetic radiation is
t.U=Uj-Uf=(8-5.33)x10-2J
=2.67x10-2J.
2.28.Show that the force on each plate of a parallel plate
capacitor has a magnitude equal to ~ qE, where qisthe charge
2
on the capacitor, and Eisthe magnitude of electric field between
the plates. Explain the origin of the factor ~ .
2
Ans.LetAbe the plate area andcr,the surface charge
density of the capacitor. Then
q=crA
E=~
So
Suppose we increase the separation of the capacitor
plates by small distancet.xagainst the forceF.Then work
done by the external agency=F. Sx
2.99
Ifube the energy stored per unit volume or the energy
density of the capacitor, then increase in potential energy
of the capacitor
=uxincrease in volume=u.A.t.x
F.t.x=u. A.t.x
F=uA=1soE2. A=1(soE)A E
=lcrA. E=1qE
The physical origin of the factor1in the force formula
lies in the fact that just inside the capacitor, field isE,and
outside it is zero.Sothe average valueE/2 contributes to
the force.
2.29.A spherical capacitor consists of two conceric
spherical conductors, held in position by suitable insulating
supports (Fig.2.199).Show that the capacitance of a spherical
capacitorisgiven by
C=_4_1tS-,0,-t:.•...17:-,,-2
1.-r2
or
where1.andr2are the radii of outer and iner spheres, respectively.
ChargeQ
Fig.2.199
Ans.Refer answer toQ.34 on page 2.30.
2.30.A spherical capacitor has an iner sphere of radius
12em and an outer sphere of radius13em. The outer sphereis
earthed and the iner sphereisgivenacharge of25C.The
space between the co-ceric spheresisfilled withaliquid of
dielectric consta32.(a) Determine the capacitance of the
capacitor. (b) Whatisthe poteial of the iner sphere?
(c) Compare the capacitance of this capacitor with that of an
isolated sphere of radius12em. Explain why the latterismuch
smaller.
Ans.Herea=12 em=12x1O-2m ,
b=13 em=13x1O-2em ,
q=2.5 C=2.5x10-6C, K=32
(a)Capacitance of the spherical capacitor is
ab
C=41tso K.--
b-a
32 12x10-2 x13x10-2
---- F
9x109' (13 - 12)x10 2
32x12x13
= x10-11F=5.5x10-9F.
9www5notes•rive5~om

169. 2.100
(b)Potential of the inner sphere is
q2.5 x
10-6 3 2
V=-= 9V = 0.45 x 10 V = 4.5 x10V.
C 5.5 x 10-
(c)Capacitance of the isolated sphere of radius 12 em is
12x10-2 -11
C =41tEOR = 9F = 1.3x10F.
9x10
When an earthed conductor is placed near a charged
conductor, the capacitance of the latter increases. The two
conductors form a capacitor. But the capacitance of an
isolated conductor is always small.
2.31.Answer carefully:
(i) Two large conducting spheres carrying chargesQ1and
Q2are brought close to each other. Is the magnitude of
electrostatic force between themexactly given by
qQ22rwhere r is the distance between their ceres?
41t EOr
(ii) If Coulomb's law involved 1Ir3 dependence (instead of
1Ir2), would Gauss' law be still true?
(iii) A small test charge is released at rest at a poi in an
electrostatic field configuration. Will it travel along the
line of force passing through that poi?
(iv) What is the work done by the field of a nucleus in a
complete circular orbit of the electron? What if the orbit
is elliptical?
(v) We know that electric field is discoinuous across the
surface of a charged conductor. Iselectric poteial also
discoinuous there?
(vi) What meaning would you give to the capacity of a single
conductor?
(vii) Guess a possible reason why water has a much greater
dielectric consta(=80)than say, mica(=6).
Ans.(i)No. When the two spheres are brought close to
each other, their charge distributions do not remain
uniform and they will not act as point charges.
(ii)No. Gauss's law will not hold if Coulomb's law
involved 1/r3or any other power ofr(except 2). In
that case the electric ux will depend uponralso.
(iii)Not necessarily. The small test charge will move
along the line of force only if it is a straight line. The
line of force gives the direction of acceleration, and not
that of velocity.
(iv)Zero. But when the orbit is elliptical, work is done in
moving the electron from one point to the other. How-
ever, net work done over a complete cycle is zero.
(v)No, potential is everywhere constant as it is a
scalar quantity.
(vi)A single conductor is a capacitor with one plate at
infinity. It also possesses capacitance.
(vii)Because of its bent shape and the presence of
two highly polara -H bonds, a water molecule
possesses a large permanent dipole moment about
0.6 x 10-29Cm. Hence water has a large dielectric
constant.
PHYSICS-XII
2.32.A cylindrical capacitor has two co-axial cylinders of
length15em and radii1.5em and1.4em.The outer cylinder is
earthed and the iner cylinder is given a charge of35/lC
Determine the capacitance of the system and the poteial of the
iner cylinder. Neglect end effects(i.e.,bending of field lines at
the ends).
Ans.Here L=15 em =0.15 m,
q=3.5 liC =3.5 x10-6Ca= 1.4 cm =0.014 rn,
b=1.5 em =0.015 rn
Capacitance of a cylindrical capacitor is given by
21tEoL L
C=-b-= 1 b
In- 2~-2.303 log-
a 41tEo a
0.15 F
2 x 9 x 109 x 2.303 log 0.015
0.014
0.15 x 10-9 F
18x2.303 x 0.03
= 0.1206 x 10-9 F = 1.2 x 10-10 F
Potential,
_ q_3.5x10-6 _ 4
V --- 10V - 2.9x10 V.
C 1.2 x10-
2.33.A parallel plate capacitoristo be designed with a
voltage rating1kV,using a material of dielectric consta3
and dielectric strength about107Vm-1. For safety, we would
like the field never to exceed say10%of the dielectric strength.
What minimum area of the plates is required to have a
capacitance of50pF? [CBSEOD 05]
Ans.Maximum permissible voltage
= lkV = 103 V
Maximum permissible electric field
= 10%of 107 Vrn-1 = 106 Vm-1
:. Minimum separationdrequired between the plates
is given by
E =Vord=V= 10: = 10- 3m
d E 10
Capacitance of a parallel plate capacitor is
C=KEoA
d
Cd50x10-12x10-3 2
A=--= m
KEO3x8.85x10-12
= 18.8x10-4m2 .:::19cm2.
2.34.Describeschematically the equipoteial surfaces corres-
ponding to
(i)a consta electric field in the Z-direction.
(ii) a field that uniformly increases in magnitude but
remains in a consta (say, Z)directions.
(iii)a single positive charge at the origin.
(iv) a uniform grid consisting of long equally spaced
parallel charged wires in a plane.www5notwsvr{vw5uom

170. ELECTROSTA
TIC POTENTIAL AND CAPACITANCE
Ans.(i)For a constant electric field in Z-direction,
equipotential surfaceswill be planes parallel to XY-planes,
as shown in Fig. 2.200.
iZ
I 7
E1I
7) Equipotentials
I
/.
~-----------------.y
.: a
,
,/
,
,
,
X~
Fig. 2.200
(ii)In this case also, the equipotential surfaceswill be
planes parallel to XY-plane. However, as field increases,
such planes will get closer.
(iii)For asingle positive charge at the origin, the
equipotential surfaces will be concentric spheres having
origin astheircommon centre, as shown in Fig.2.25.The
separation between the equipotentials difering by a
constant potential increases with increase in distance
from the origin.
(iv)Near the grid the equipotentialsurfaces will have
varying shapes. At far of distances, the equipotential
surfaces will be planes parallel to the grid.
2.35.In a Van deGraafftype generator, a spherical metal shell
is to be a15x106V electrode. The dielectric strength of the gas
surrounding the electrode is5x107Vm-1. What is themini-
mum radius of the sphericalshell required? [CBSE OD08]
Ans.Maximum permissible potential, V=1.5x106V
Forsafety, the maximum permissible electric field is
E=10%of dielectric strength
=10%of 5x107Vm-1=5x106Vm-1
Nowfor a spherical shell,
V=_1_.3.
47tEor
E=_l_ 3...=V
47tEO .r2r
Minimum radius required is
V 1.5x106V 1
r=-= =3x10-m=30em.
E 5x106Vm1
2.36.A small sphere of radius '1and charge q1isenclosed by
aspherical shellofradius r2 andchargeq2'Showthatif q1 is
positive, charge will necessarily flow fromthe sphere tothe shell
(when the two are conected by a wire) nomatter what the
chargeq2 on the shell is.
Ans.Refer answer toQ.55 on page 2.67.
2.101
2.37.Answer the following:
(i) The top of the atmosphere isat about 400kVwith respect
to the surface of the earth, corresponding toan electric field that
decreases with altitude. Near the surface of the earth, thefieldis
about100Vm-1. Why do thenwenotget anelectric shock as we
step outof our house io the open? (Assume the house to be a
steel cage sothereisno field inside.)
(ii) A man fixes outside his house one evening a two metre
highinsulating slab carrying on itstop a large aluminium sheet
of area 1m2.Will he get an electric shock ifhetouches the metal
sheet next morning?
(iii) The discharging curre intheatmosphere due to the
smallconductivity of airisknown to be1800Aonan average
overthe globe. Whythendoes theatmosphere not discharge
itselfcompletely indue course and become electrically neutral ?
In other words, what keeps theatmosphere charged?
(iv)What arethe forms ofenergy io which the electric
energy oftheatmosphere isdissipated during a lightning?
Ans.(i)Normally theequipotential surfaces are
parallel to the surface of the earth asshown in Fig.2.201.
Now our body is a good conductor. So as we step out into
the open, the original equipotential surfaces of open air
getmodified, but keeping our head and the ground at the
same potential andwedonot get any electric shock.
300 V 300V
200 V 200 V
lOOV
77777777777777
Ground
(a)
77777777777777
Ground
(b)
Fig. 2.201
(ii)Yes.Thealuminium sheet and the ground form a
capacitor with insulating slab as dielectric. The dis-
charging current in the atmosphere will charge the capa-
citor steadily and raise its voltage. Next morning, if the
man touches the metal sheet, he will receive shock to the
extent depending upon the capacitance of the capacitor
formed.
(iii)The atmosphere is continuously being charged by
thunder storms and lightning bolts all over globe and
maintains an equilibrium with the discharge of the
atmosphere in ordinary weather conditions.
(iv)The electrical energy is lost as (i)light energy
involved in lightning(ii)heat and sound energy in the
accompanying thunder.www5notes•rive5~om

171. 2.102 PHYSICS-XII
Text Based Exercises
~YPE A :VERY SHOR
TANSWER QUESTIONS (1 mark each)
1.Define electric potential. Is it a scalar or a vector
quantity? [Punjab01;CBSEaD 06]
2.Define the unit of electric potential. [Punjab02]
3.Write down the relation between electric field and
electric potential at a point.
4.Name the physical quantity whose 51,unit is JC-I.
Is it a scalar or a vector quantity?[CBSEaD 2010]
5.Write the 51unit of potential gradient.
6.Define electric potential diference between two
points. Is it scalar or vector? [Punjab01]
7.What do you mean by a potential diference of 1volt?
8.Write the dimensional formula of potential
diference.
9.5 J of work is done in moving a positive charge of
0.5 C between two points. What is the potential
diference between these two points?[ISCE9S]
10.A charge of 2 C moves between two points main-
tained at a potential diference of 1 volt. What is the
energy acquired by the charge ?
[!SCE94;CBSED10C]
11.In a conductor, a point P is at a higher potential
than another pointQ.In which direction do the
electrons move?
12.Give two examples of conservative forces.
[Himachal93]
13.How much is the electric potential of a charge at a
point at infinity ?
14.What is the nature of symmetry of the potential of a
point charge ?
15.What are the points at which electric potential of a
dipole has maximum value?
16.What are the points at which electric potential of a
dipole has a minimum value?
17.What is the nature of symmetry of a dipole
potential?
18.What is electrostatic potential energy? Where does
it reside?
19.What is the value of the angle between the vectors
pandEfor which the potential energy of an
electric dipole of dipole momentp,kept in an
-4
external electric field Erhas maximum value?
[CBSESPIS]
-4
20.Write an expression for potential at point P(r )
due to two charges'handq2located at positions
-4 d-4 .I
rIanr2respective y.
21.Define electron volt. How is it related to joule?
22.How many electron volts make up one joule?
[Himachal 93]
23.Will there be any efect on the potential at a point if
the medium around this point is changed ?
24.What work must be done in carrying an a-particle
across a potential diference of 1 volt ?
25.What is an equipotential surface? Give an example.
[Punjab2000, 02;CBSED 03]
26.Why are electric field lines perpendicular at a point
on an equipotential surface of a conductor?
[CBSEcoISC]
27.Can you say that the earth is an equipotential
surface?
28.What is the geometrical shape of equipotential
surfaces due to a single isolated charge?
[CBSED13]
29.What is the shape of the equipotential surfaces for a
uniform electric field? [Punjab01]
30.How much work is done in moving a 500 IlC charge
between two points on an equipotential surface?
[CBSED 02]
31.A charge of+1Cis placed at the centre of a spherical
shell of radius 10 cm. What will be the work done in
moving a charge of+11lCon its surface through a
distance of 5 cm ?
32.What is the optical analogue of an equipotential
surface?
33.The middle point of a conductor is earthed and its
ends are maintained at a potential diference of
220 V. What is the potential at the ends and at the
middle point?
34.Define capacitance of a conductor.
35.Can there be a potential diference between two
conductors of same volume carrying equal positive
charges?
36.The capacitance of a conductor is 1 farad. What do
you mean by this statement?
37.What is a capacitor? [Punjab96C]www6notesdrive6com

172. ELECTROSTATIC POTENTIAL AND CAPACITANCE
38.Write the physical quantity that has its unit
coulomb volt-1.Is it a
vector or scalar quantity ?
[CBSE093C, 98]
39.Define capacitance. Give its51unit.
[CBSE093C;ISCE 98]
40.Define51unit of capacitance. [CBSE0094]
41.Write the dimensions of capacitance.
42.What is the net charge on a charged capacitor?
43.On what factors does the capacitance of a capacitor
depend?
44.Write two applications of capacitors in electrical
circuits.
45.. In what form is the energy stored in a charged
capacitor?
46.What is the basic purpose of using a capacitor?
47.Write diferent expressions for the energy stored in
a capacitor.
48.Write down the expression for the capacitance of a
spherical capacitor.
49.The diference between the radii of the two spheres
of a spherical capacitor is increased. Will the capa-
citance increase or decrease? [Punjab 2000]
50.What is a dielectric ?
51.Define dielectric constant in terms of the capa-
citance of a capacitor. [CBSE006]
52.Write down the relation between dielectric constant
and electric susceptibility.
53.
~
Write a relation for polarisationPof a dielectric
material in the presence of an external electric field
~
E.
54.
[CBSEOO 15]
Define dielectric strength of a medium. What is its
value for vacuum ?
55.Where is the knowledge of dielectric strength
helpful?
What is the efect of temperature on dielectric
constant?
An air capacitor is given a charge of 2 J.lCraising its
potential to 200V.If on inserting a dielectric
medium, its potential falls to 50V,what is the
dielectric constant of the medium ?
An uncharged insulated conductorAis brought
near a charged insulated conductor B. What
happens to the charge and potential of B?
[CBSE00OlC]
For a given potential diference, does a capacitor
store more or less charge with a dielectric than it
does without a dielectric ?
56.
57.
58.
59.
2.103
60.Can we place a parallel plate capacitor of 1 F
capacity in our house?
61.What is the basic diference between a capacitor
and an electric cell ?
62.Two capacitors of capacitancesc;andc;are
connected in parallel. A chargeqis given to the
combination. What will be the potential diference
across each capacitor?
63.What is the order of capacitances used in a radio
receiver?
64.Is there any conductor which can take unlimited
charge?
65.A parallel plate capacitor with air between the
plates has a capacitance of 8 pF. What will be the
capacitance if the distance between the plates be
reduced by half and the space between them is
filled with a substance of dielectric constantK=6?
[CBSE005]
66.A 500J.lCcharge is at the centre of a square of side
10 cm. Find the work done in moving a charge of
10J.lCbetween two diagonally opposite points on
the square. [CBSE008]
67.The graph of Fig. 2.202, shows the variation of the
total energy(E)stored in a capacitor against the
value of the capacitance (C)itself
the charge on the capacitor or the potential used to
charge it is kept constant for this graph?
[CBSE Sample Paper 08]
Fig. 2.202
68.Define the term 'potential energy' of charge'ifat a
distance'r'in an external electric field.
[CBSEOO 09]
69.What is the work done in moving a test chargeq
through a distance of 1 cm along the equatorial axis
of an electric dipole? [CBSE0009]
70.What is the electrostatic potential due to an electric
dipole at an equatorial point? [CBSE0009]
71.A metal plate is introduced between the plates of a
charged parallel plate capacitor. What is its efect
on the capacitance of the capacitor?[CBSE F 09]www7notesdrive7com

173. 72.A hollow metal
sphere of radius 5 em is charged
such that the potential on its surface is 10V: What is
the potential at the centre of the sphere?
[CBSEOD11]
73.In which orientation, a dipole placed in a uniform
electric field is in(i)stable,(ii)unstable equili-
brium ? [CBSEOD 08;D 10]
74.Write the expression for the work done on an
electric dipole of dipole momentpin turning it
from its positionof stable equilibrium to a position
....
of unstable equilibrium in a uniform electric fieldE.
[CBSED13C]
75.Two charges21lCand-21lCare placed at pointsA
andB,5 cm apart. Depict an equipotential surface
of the system. [CBSED13C]Fig. 2.203
2.104
Answers
PHYSICS-XII
76.What is the amount of work done in moving a
charge around a circular arc of radiusrat the centre
of which another point charge is located ?
[CBSEOD13C]
77.What is the equivalent capacitance,C,of the five
capacitors, connected as shown in Fig. 2.203 ?
[CBSESamplePaper 2011]
v
1.The electric potential at any point in an electric
field is defined as the amount ofwork done in
moving a unit positive charge from infinity to that
point against the electrostatic force. It is a scalar
quantity.
2.The SI unit of electric potential is volt. The electric
potential at a point is said to be 1 volt, if 1 joule of
work is done in moving a positive charge of 1
coulomb from infinity to that point against the
electrostatic force.
E= _dV.
3.
dr
4.Electric potential or potential diference. It is a
scalar quantity.
5.SI unit of potential gradient = Vm-1.
6.The potential diference between two points in an
electric field may be defined as the amount of work
done in moving a unit positivecharge from one
point to the other against the electrostatic force. It is
a scalar.
7.The potential diference between two points is said
to be 1 volt if 1 joule of work is done in moving a
positive charge of 1 coulomb from one point to the
other against the electrostatic force.
.. work doneML2T-2
8.Potential diference = = ---
charge C
ML2T-2
=0 =[ML2r3 A-I]
AT
9.V= W = ~ =10 V.
qO.5C
••
10.Energy acquired by thecharge =qV= 2Cx1V =2J.
11.FromQtoP.
12.(i)Electrostatic force, (ii)Gravitational force.
13.Zero.
14.The potential of a point charge is spherically
symmetric.
15.At axial points, the electric potential of a dipole has
a maximum positive or negative value.
16.At equatorial points, the electric potential of a
dipole is zero.
17.The dipole potential is cylindrically symmetric.
18.The electrostatic potential energy of a system of
charges may be defined as the work required to be
done to bring the various charges to their respective
positions from infinity.
19.P.E.=-pEcos9.Clearly, P.E. is maximum when
cos9=-lor9= 180°.
V(17)=_1_[ql+q21
4m:o I17-171I I17-172I
Electron volt is the potential energy gained or lost
by an electron in moving through a potential
diference of one volt.
1electron volt=leV=1.6x10-19J
1J=6.25x1018eV.
Yes.If the dielectric constant of the medium is
increased, the electric potential will decrease.
24.W=qLW =2e t1V
= 3.2x10-19 Cx1V = 3.2x10-19J.
20.
21.
22.
23.www3xyto°n~svo3myw

174. ELECTROSTATIC POTENTIAL AND CAPACITANCE
25.Any surface which has same electric potential at
every point is called an equipotential surface. The
surface of a charged conductor is an equipotential
surface.
26.If it were not so, the presence of a component of the
field along the surface would destroy its equi-
potential nature.
27.Yes. Earth is a conductor, so its surface is equi-
potential.
28.For a point
charge, the equipotential surfaces are
concentric spherical shells with their centre at the
point charge.
29.For a uniform electric field, the equipotential
surfaces are parallel planes perpendicular to the
direction of the electric field.
30.Zero.
W=q!lV=500f,lCx0=0 .
31.Zero. This is because the surface of the spherical
shell will be an equipotential surface.
32.Wavefront.
33.The potential at the middle point of the conductor
is zero and that at the ends+110V and -110V,so
that the p.d. at ends=110 - (- 110)=220V.
34.The capacitance of a conductor may be defined as
the charge required to raise its potential by unit
amount.
35.Yes. Two conductors of same volume but of
diferent shapes will have diferent capacitances.
A conductor issaid to have a capacitance of 1 farad,
if 1 coulomb of charge increases itselectric potential
through 1 volt.
A capacitor is a device to store electric charge. It
consists of two conducting plates separated by an
insulating medium.
Capacitance has its unit coulomb volt-1.It is a
scalar quantity.
The capacitance of a capacitor may be defined as
the charge required to be supplied to either of the
conductors so as to increase the potential diference
between them by unit amount.
The S1unit of capacitance is farad (F). A capacitor
has a capacitance of 1 F if 1 coulomb of charge is
transferred from its one plate to another on
applying a potential diference of 1 volt across the
two plates.
As 1F=lC= ~=lC2=1(As)2
. 1V IJ/C IJ INm
A2T2 [M-1L-2T4A2)
[Capacitance)=MLT-2L = .
36.
37.
38.
39.
40.
41.
2.105
Zero, because the two plates have equal and
opposite charges.
The capacitance of a capacitor depends on the
geometry of the plates, distance between them and
the nature of the dielectric medium between them.
(i)Capacitors are used in radio circuits for tuning
purposes.
(ii)Capacitors are used in power supplies for
smoothening the rectified current.
In a charged capacitor, energy is stored in the form
of electrostatic potential energy in the electric field
between its plates.
To store charge and electric energy.
1 21d1
U=-CV =- -=>-QC
2 2 V2
C=4m:o . ~ , whereaandbare the radii of the
b-a
inner and outer spheres respectively.
The capacitance will increase.
A dielectric is essentially an insulator which allows
electric induction to take place through it but does
not permit the ow of charges through it.
The ratio of the capacitance(Cd)of the capacitor
completely filled with the dielectric material to the
capacitance(Cv)of the same capacitor with vacuum
between its plates is called dielectric constant.
Cd
K=-
CV
52.K=1+X,whereKis dielectric constant and X is
electric susceptibility.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
53.
54.
--> -->
P=EOXeE.
The maximum value of electric field that can exist
inside a dielectric without causing its electrical
breakdown is called its dielectric strength. The
dielectric strength for vacuum is infinity.
The knowledge of dielectric strength helps in
designing a capacitor by determining the
maximum potential that can be applied across the
capacitor without causing its electrical breakdown.
The value of dielectric constant decreases with the
increase of temperature.
K=Vvacuum =200=4.
Vdielectric 50
The charge on the conductor Bremains unchanged
but its potential gets lowered.
A capacitor with a dielectric has a higher capaci-
tance and hence stores more charge.
55.
56.
57.
58.
59.www7notesdrive7com

175. 2.106
60.No.Ifd=1em=10-2m,then
area of such a capa-
citor would be
A=Cd= 1Fx10-2 m =109 m2
Eo8.85x10-12C2N-1m-2
This is a plate about30kmin length and breadth.
61.A capacitor provides electrical energy stored in it.
A cell provides electrical energy by converting
chemical energy into electrical energy.
62.In parallel combination, potential diference is same
across each capacitor.
Net capacitance, C=c;+ ~
:.P.D. across each capacitor, V=!L=--q- .
Cc;+ ~
63.In the powersupply, it is1- 10/-IFand for tuning
purposes, it is100/-lF.
64.Yes, the earth because of itslarge capacitance can
take unlimitedcharge.
65.With air between the capacitor plates,
EA
Co=-"T-=8pF
With dielectric between the capacitor plates,
C=K EoA=2KCo=2x6x8 =96pF.
d/2
66.The work done in moving a charge of 10 uC
between two diagonally opposite points on the
square will be zero because these two points will be
equipotential.
67.Energy stored, E=-.!CV2 =-.!Ii
2 2 C
1
When Q is constant, Eex:-,and we get a graph of
C
the type given in thequestion.
Hence the chargeQon the capacitor is kept constant.
PHYSICS-XII
68.The potential energy of a charge qisthe work done
in bringing chargeqfrom infinity to thepositionr
inthe external electric field.
U(r)=qV(r)
69.As potential at any point on the equatorial axis of
anelectric dipole is zero, so
W=q ~V=q(0 - 0)=o.
70.Zero.
71.The introduction of a metal sheet of thickness tin a
parallel plate capacitor increases its capacitance by
a factor of.s:rwheredistheplate separation of
d-t
the capacitor.
72.Potential at the centre
=Potential at the surface=10V.
73.(i)When the dipole momentpis parallel to the the
-+
electric fieldE(8=0°),the dipole is in stable
equilibrium.
(ii)When the dipole momentpis antiparallel to the
-+
electric fieldE(8=180°),thedipole is in unstable
equilibrium.
74.W=pE(cos~-cos82) =pE(cosOO-cos1800)
=pE(l+1)=2pE.
75.See Fig.2.26on page2.15.
76.Zero, because all points of the circular arc will be at
the same potential.
77.C=~,becausethecombinations of C;and ~ as
well asC4andCshave been shorted.
"YPE B:SHORT ANSWER QU ESTIONS (2or 3marks each)
1.Distinguish between electric potential and poten-
tial energy andwrite the relation between them.
[Punjab96C]
2.Define electric potential. Derive an expression for
the electric potential at a distance r from a chargeq.
[Punjab99C]
3.Drawaplotshowing the variation of (i)electric
field (E) and(ii)electric potential (V)with distance r
due to a point chargeQ. [CBSE D 12]
4.Derive an expression for the electric potential due
to an electric dipole. [Haryana01]
5.Derive an expression for the electric potential at a
point along the axis line of the dipole.
[CBSE D 2000, 08 ; OD OlC, 02,13C]
6.Show mathematically thatthe electric potential at
any equatorial point of an electric dipole is zero.
[CBSEOD01]
7.Give three diferences between the nature of electric
potentials of a single point charge and an electric
dipole.
8.Obtain an expression for the electric potential at a
point due to group ofNpoint charges.www6notesdrive6com

176. ELECTROSTA
TICPOTENTIAL AND CAPACITANCE
9.Obtain an expression for the potential at a point due
toacontinuous charge distribution. [CBSE aD 92C]
10.Show that the electric field at any pointisequal to
the negative of the potential gradient at that point.
11.Describe howcanwe determine the electric potential
at a point from the knowledge of electric field.
12.Twoclosely spaced equipotential surfacesAand B
with potentials VandV+BV,(where 8Visthe
changeinV),are kept81distance apart as shown in
the figure. Deduce the relation between the electric
B v+ov
~--7V
Fig.2.204
field and the potential gradient between them. Write
the two importantconclusions concerning the
relation between the electric field andelectric
potentials. [CBSE D14C]
13.Show that the amount of work done in moving a
test charge over an equipotential surface is zero.
[Haryana97]
14.Show that the direction of the electric field is
normal to theequipotential surface atevery point.
15.Show thatthe equipotential surfaces are closed
together in the regions of strong field and far apart
in the region of weak field.
16.Sketch equipotential surfaces for
(i)a positive point charge. [CBSE D2000]
(ii)anegative point charge. [CBSE DOl]
(iii)twoequal and oppositecharges separated by a
small distance.
(iv)twoequal and positive charges separated by a
smalldistance.
a uniform electric field.
Draw equipotential surfaces
charge Q>O.
(b)Are these surfaces equidistant from each
other? Ifnot,explain why. [CBSE D HC]
18.Draw the equipotential surfaces due to an electric
dipole. Locate the points where the potential due
to the dipole is zero. [CBSE aD 13]
19.Twopoint charges qlandq2arelocated atTiand
r;respectively in anexternal electric fieldE.
Obtain the expression for the total work done in
assembling this configuration. [CBSE D 14C]
(v)
17.(a)
[CBSE D 2000, 01]
due to a point
2.107
20.(a)Depict the equipotential surfaces for a system
oftwo identical positive point charges placed a
distance'd' apart.
(b)Deduce the expression for the potential energy
ofasystem of two point charges qlandq2
brought from infinity to the points?and ~
respectively in the presence of external electric
->
fieldE.
21.
[CBSE D 10;on 15]
Derive anexpression for the potentialenergy of an
->
electric dipole ofdipole momentpin an electric
->
field E.
22.
[Himachal02;CBSEDOS]
An electric dipole isheldinauniform electric field
->
E.
(a)Showthat the net force acting on itiszero.
(b)The dipole is aligned with itsdipole momentp
->
parallel tothe field E . Find:
(i)the work done in turning the dipole till its
dipole moment points in the direction
->
opposite to E .
(ii)the orientation of the dipole for which the
torque acting on it becomes maximum.
[CBSE oo 12, 14C]
23.UsingGauss's law, show thatelectric field inside a
conductor is zero. [CBSE D 2000]
24.Just outside a conductor electric field isperpen-
diculartothe surface. Give reason.
25.Showthat the excesscharge on a conductor resides
onlyonits surface.
26.Show that the electric field at thesurface of a
charged conductor is given byE= ~~,where o is
EO
the surface charge density and~ isa unit vector
normal to thesurface in the outward direction.
[CBSEOD 10]
Or
Derive anexpression for the electric field at the
surface ofa charged conductor. [CBSE aD 09]
27.Show thatelectric field is zero in the cavity of
hollow charged conductor.
28.What is electrostatic shielding ?Mention its two
applications.
29.Define electrical capacitance of a conductor. On
what factors does itdepend ?
30.Show that the capacitance ofaspherical conductor
isproportional to its radius. Hence justify that farad
is alarge unit ofcapacitance. [Himachal96]www7notesdrive7com

177. 2.108
31.Anisolated
conductor cannot have a large capa-
citance. Why ?
32.Why does the capacitance of aconductor increase,
when an earth connected conductor is brought near
it ? Briey explain.
33.What is a capacitor? Explain its principle.
[Punjab2000, 02,03]
34.Derive an expression for thecapacitance of a parallel
plate capacitor. [CBSE0OSC,14;0003]
35.(a)Obtain the expression for the energy stored per
unitvolume in acharged parallel plate capacitor.
(b)The electric field inside a + +++ ++ + + + + + +
parallel plate capacitor is aOb
E.Find the amount of
work done in moving a
d c
chargeqover aclosed _
rectangular loopabed a.
[CBSE0 14] Fig.2.205
36.Distinguish between polar and non-polar dielec-
trics.Give one example of each.
37.Three capacitors of capacitances ;, C;andC3are
connected in series. Find theirequivalent capa-
citance. [CBSE0 92,93; Haryana 94;Himachal 97]
38.Three capacitors of capacitances q.C;andC3are
connected inparallel. Find their equivalent
capacitance. [CBSE092,94;Himachal99]
39.Deduce the expression for the electrostatic energy
stored ina capacitor of capacitance 'C' and having
charge'Q'.
How will the(i)energy stored and(ii)the electric
field inside the capacitor be afected when it is
completely filled with a dielectric material of
dielectric constant'K? [CBSE0008,12]
40.Iftwo charged conductors are touched mutually
and then separated, prove that the charges on them
willbe divided in the ratiooftheir capacitances.
41.Two capacitors with capacity; and C;arecharged
topotentialV;andV2respectively and then
connected in parallel. Calculate the common
potential. acrossthe combination, thecharge on
eachcapacitor, the electrostaticenergy stored in the
systemandthe change in the electrostatic energy
fromits initial value. [CBSESamplePaper08]
42.Explain why the polarization ofa dielectric reduces
the electric field inside the dielectric. Hence define
dielectric constant. [CBSE099]
43.Define' dielectric .constant' of a medium. Briey
explain why the capacitance of a parallel plate capa-
citor increases, on introducing a dielectric medium
between the plates. [CBSE0006C]
PHYSICS-XII
44.Whatismeant by dielectric polarisation ? Hence
establish the relation: K=1+X [Haryana01]
45.Twodielectric slabs ofdielectric constants KlandK2
are filled in between thetwoplates, each of areaA,
ofthe parallel plate capacitor as shown in Fig. 2.206.
Find the net capacitance of the capacitor.
ICBSE0005]
DOl
I-- 1/2------II-- 1/2------I
Fig.2.206
46.Aparallel plate capacitor of capacitance C is
charged to a potential V.Itis then connected to
another uncharged capacitor having the same
capacitance. Find out the ratio of the energy stored
inthe combined system to that stored initially in
thesinglecapacitor. ICBSE0014]
47.Iftwo similar plates, eachof areaAhaving surface
chargedensities +cr and-crare separated by a
distance dinair,write expressions for: (I)The
electric field between the two plates (iI)The
potential difference between the plates (iil)The
capacitance of the capacitor so formed.
ICBSEOD07]
48.Explain, using suitable diagrams, thedifference in
thebehaviour of a(i)conductor and (ir)dielectric in
thepresence of external electric field. Define the
terms polarization of adielectric and write its
relation withsusceptibility. ICBSE015]
49.A capacitor is chargedwith a batteryand then its
plate separation isincreasedwithout disconnecting
thebattery. What will be the change in
(a)charge stored in the capacitor?
(b)energy stored in the capacitor?
(c) potential difference across the plates ofthe
capacitor?
(d)electric field between the plates of the
capacitor? [CBSESamplePaper2011]
50.The charges ql=3flF,
q2=4 F andq3=-7flF
are placed on the
circumference of a circle of
radius 1.0 m, as shown in
Fig. 2.207. What is the
value of charge q4placed
on the same circle if the
potential at centre, Vc=O? Fig.2.207
IISCE03]www6notesdrive6com

178. ELECTROSTATIC P
OTENTIAL AND CAPACITANCE
51.Two thin concentric shells of radii 1. andr2(r2>1.)
have charges qlandq2.Write the expression for the
potential at the surface of innerand outershells.
[CBSE 00 13 C]
52.(a)Acharge+Qisplacedon a large spherical
conducting shell of radius RAnother small con-
ducting sphere of radius rcarrying charge 'ifis
introduced inside the large shell and is placedat its
centre. Find the potential diference between two
points, one lying on the sphere and the other on the
shell.
Answers
2.109
(b)Howwould the charge between the two ow, if
theyare connected by a conducting wire? Name
the device which works on this fact. [CBSE 0009]
53.Briefly describe discharging action of sharp points
(orcorona discharge).
54.Draw a labelled schematic diagram of a Van-de-
Graaf generator. State its working principle.
Describe briey how it is used to generate high
voltages. [CBSE 0 13 C]
••
1.Refer to points 3and11ofGlimpses.
2.Refer answer toQ.2on page2.2.
3.See Fig.2.3on page2.2.
4.Refer answer to Q.5onpage2.3.
5.Refer answer to Q.3 onpage2.2.
6.Refer answer toQ.4 onpage2.3.
7.Refer answer toQ.6onpage2.3.
8.Refer answer toQ.7 on page 2.4.
9.Refer answer to Q.8 on page2.4.
10.Refer answer to Q.10on page2.11.
11.Refer answer to Q. 11onpage2.12.
12.Work donein moving a unit positive charge
through distanceol,
Exol =VA -VB=V -(V+OV) =-OV
E=- OV
01
For conclusions, refer answer toQ.10onpage2.12.
13.Refer answer to Q. 14on page2.14.
14.Refer answer toQ.14on page 2.14.
15.Refer answer toQ.14on page 2.14.
16.Refer answer toQ.15onpage2.15.
17.(a)See Fig.2.25on page 2.15.
dV dV
(b)AsE=-dr ordr=-r:
1
:.For constantdV, droc- ocr2
E
Hence thespacing between the equipotential
surface will increasewith theincrease in distance
from the point charge.
18.See Fig. 2.26on page2.15.The electric potential is
.zero at the equatorial points of the dipole.
19.Refer answer toQ.20on page2.17.
20.(a)See Fig. 2.27on page 2.16.
(b)Refer answer. toQ.20on page2.17.
21.Refer answer to Q. 22on page2.18.
22.(a)Refer answer to Q.40on page1.41of chapter1.
n n
(b)(i)W=J,d8=JpEsin8=pE[-cosera=-2pE.
a a
(ii)As,=pEsin8,so , is maximum when8=90°.
23.Refer answer to Q.25on page2.24.
24.Refer answer toQ.25on page2.24.
25.Referanswer to Q. 25on page2.24.
26.Refer answer to Q.25on page2.24.
27.Refer answer toQ.25on page2.24.
28.Refer answer to Q.26on page2.25.
29.Refer answer to Q.27on page2.26.
30.Referanswer to Q. 29on page2.26.
31.Refer answer to Q. 30on page2.28.
32.Refer answer to Q. 31on page2.28.
33.Refer answer toQ.31on page2.28.
34.Refer answer to Q. 33on page2.29.
35.(a)Refer answe~ to Q.40on page2.49.
(b)E.Laband E.L dc,soWab=0andWed=O.
Also, Wbc=-Wda
Total work done in movingchargeqover the closed
loopabcda,
W=Wab+Wbc:+-Wcd+Wda
=0-Wda+ 0+Wda=O.
36.Refer answer to Q.43on page2.56.
37.Refer answer to Q. 36on page2.33.
38.Refer answer to Q.37on page2.33.
.39.Refer answer to Q.38page2.48.
When the capacitor is completely filled with a
dielectric material and for constant charge Q,
C=KCUandV=Va /Kwww7notesdrive7com

179. 2.110
(i)U=!:C
V2 =!:(KCu)(VO]2=.!..!CuV02 =Uo
2 2 K K2 K
(ii) E=fu
~
40.Refer answer to Q.41on page2.49.
41.Refer answer toQ.42on page 2.49.
42.Refer answer toQ.45on page2.57.
43.Refer to point35of Glimpses on page 2.116.
44.Refer answer to Q.47on page2.58.
45.The given arrangement is equivalent to a parallel
combination of two capacitors eachwith areaAI2
and plate separationd.Hence the net capacitance of
the resulting capacitor is
C=c;+c;
EO(AI2)K} EO(AI2)K2
= +~---..=.
d d
46.Initialenergy stored in thesinglecapacitor
= !:CV2 =!:q2
2 2C
Capacitance of the combined (parallel) system
=C+C=2C
Asthe totalchargeqremains the same, so
Final energy storedin the combined system =!:i....
22C
47.
Final energy =!: =1 : 2.
Initial energy 2
(i)Electric field at points between the two plates,
E=~+~=~.
2ea2Eo EO
(ii)Potential diference between the plates,
ad
V= Ed=-.
EO
48.
(iii)Capacitance of the capacitorso formed,
C=..i=~= EoA.
V adIEO d
Refer answer to Q. 43on page 2.56.
Polarisation of a dielectric. The induced dipole
moment set up per unit volume of a dielectric when
PHYSICS-XII
placed inanexternal electric field is called
polarisation. Forlinear isotropic dielectrics,
~ ~
P=EOXe E
where Xeis the electric susceptibility of the
dielectric medium.
49.C=EoA.
d
Whendisincreased, C decreases.
(a) q=CV decreasesdueto the decrease in the
value of C.
(b)U=!:CV2 decreasesdue to the decrease in the
2
value of C.
(c)Vremainsunchanged because' the battery
remains connected.
(d)E=VIddecreases due to the increase in the
value ofd.
SO.AsVc=0
:. _1_[!!1.+q2+q3+q4 ]= 0
4m,0 r r r r
or Ih+q2+q3+q4=0
or 3+4-7+q4=0
or q4=O.
51.Potential at the surface of inner shell,
"t=Potential due to its own charge q}
+Potential due to chargeq2on outer shell
-4:EO( ::+::J
Potentialat thesurface of outer shell
V2=Potential due to charge q}on inner shell
_ 4:J~n:~:rtochargeq,onoutershell
52.Refer answer to Q. 55on page 2.67.
53.Refer answer to Q.54on page2.67.
54.Referanswer to Q.56on page 2.67.
~YPE C:LONG ANSWER QUESTIONS (5 marks each)
1.Find the expression for the electric field intensity
and the electric potential, due to a dipole at a point
on theequatorial line. Would theelectric field be
necessarily zero at a point where the electric
potential is zero?Give an example to illustrate
your answer. ICBSESample Paper2011]
2.Define electrostatic potential energy of a charge
system. Derive an expression for the potential energy
of asystem of three point charges. Generalise the
result for a system of N pointcharges.www5notesdrive5com

180. ELECTROSTATIC POTENTIAL
AND CAPACITANCE
3.Define the term electric dipole moment. Derive an
expression for the total work done in rotating the
dipolethrough an angle ein uniform electric field
.....
E.
[CBSEOD93, 95;D96C)
4.Derive an expression for the potential energy of an
electric dipole placed in a uniform electric field.
Hencediscussthe conditions of its stable and
unstable equilibrium.
5.Explain the principle of a capacitor. Derive an
expression for the capacitance of a parallel plate
capacitor. [CBSED92, 94]
6.Obtain the expression for the capacitance of a
parallel plate capacitor.
Three capacitors of capacitancesC1,Czandc;are
connected (i)in series,(ii)in parallel. Show that the
energy stored in the series combination is the same
as that in the parallel combination. [CBSEOD03]
7.Deduce an expression for the total energy stored in
a parallel plate capacitor and relate it to the electric
field. [CBSEF94C)
8.(a)Derive the expression for the energy stored in a
parallel plate capacitor. Hence obtain the
expression for the energy density of the electric
field.
(b)A fully charged parallel plate capacitor is
connected across an uncharged identical
capacitor. Show that the energy stored inthe
combination is less than that stored initially in
the single capacitor. [CBSEOD15]
9.Define the terms(i)capacitance of a capacitor
.(ii)dielectric strength of a dielectric. When a dielectric
is inserted between the plates of a charged parallel
plate capacitor, fully occupying the intervening
region, how does the polarization of the dielectric
medium afect the net electric field ? For linear
dielectrics, show that the introduction of a
dielectric increases its capacitance by a factor K.
characteristic of the dielectric. [CBSEDOSC)
Answers
2.111
10.Find the expression for the capacitance ofaparallel
plate capacitor of areaAand plate separation dif
(i)a dielectric slab of thickness tand(ii)a metallic
slab of thicknesst,where(t<d)areintroduced one
by one between the plates ofthe capacitor. In which
case would the capacitancebemore andwhy?
[CBSE Sample Paper 2011]
11.What is a dielectric? Adielectric slab of thicknesst
is kept between the plates of a parallel plate
capacitor separated by distance d.Derive the
expression for the capacitance of the capacitor for
t«d. [Himachal 02;CBSED93 ;OD0lC)
12.(a)Show that in a parallel plate capacitor, if the
medium between the plates of a capacitor is filled
with an insulating substance of dielectric constant
K.its capacitance increases. (b)Deduce the
expression for the energy stored in a capacitor of
capacitance C with charge Q. [CBSED09C)
13.(a)A small sphere, of radius'o',carrying a positive
chargeq,is placed concentrically inside a larger
hollow conducting shell of radius b(b>a).This
outer shell has a chargeQonit.Show thatifthese
spheres are connected by aconducting wire, charge
will always ow from the inner sphere to the outer
sphere, irrespective of themagnitude of the two
charges. [CBSEF15]
(b)Namethe machine which makes use of this
principle. Draw a simple labelled line diagram of
this machine. What 'practical dificulty' puts on
upper limit on themaximum potential diference
which this machine can built up ?
[CBSED09C ;OD14]
14.Explainthe principle of a device that can build up
high voltages of theorderof afew million volts.
Draw a schematic diagram and explain the working
of this device.
Is there any restriction on theupper limit of the
high voltages set up in thismachine? Explain.
[CBSE D 12]
•
1.For derivation of electric field intensity at
equatorial point of a dipole, refer answer to Q.38
on page 1.40 of chapter 1.
1 P
El,qua=41tEo(r2+a2 )3/2
For derivation of electric potential at an equatorial
pointof adipole, refer answer toQ.4 on page 2.3 of
chapter 2.
Vequa=0
No,the electric field maynot be necessarily zero at
a point wherethe electric potential is zero. For
example,the electric potential at anequatorial point of
a dipole is zero, while electric fieldis not zero.
2.Refer answer toQ.18 on page 2.16.
3.Refer answer toQ.22onpage2.18.
4.Refer answer to Q. 22 on page 2.18.
5.Refer answer to Q. 31 on page 2.28and Q. 33 on
page 2.29.www7notesdrive7com

181. 2.112
6.Refera
nswer toQ. 33 on page 2.29 and Q. 39 on
page 2.48.
7.Refer answer to Q.38 on page 2.48 and Q.40 on
page 2.48.
8.(a)Refer answer to Q. 38on page 2.48 and Q. 40 on
page 2.48.
(b)Refer to the solution of Example 75 on page 2.52.
9.Refer answer toQ.45 on page 2.57 and Q.49 on
page 2.59.
10.For derivation, refer answers toQ.49 on page2.59
andQ.50 on page 2.60.
EoA
Cdielectric = t
d=t+-
K
C =EOA
metald=t
Clearly, Cmetal >Cdielectric
[For metal, K=001
PHYSICS-XII
11.Refer answer to Q. 49 on page 2.59.
12.(a)Refer answer to Q.49 on page 2.59.
(b)Refer answer toQ.38 onpage 2.48.
13. (a)Refer answer to Q.55 on page 2.67
(b)A Van-de-Graaf generator works on this prin-
ciple.SeeFig.2.130. The potential on the outer
surface of its metallic shell cannot exceed the
breakdown field of air (=::3x106Ym-l)
because thenthe charges begin toleak into air.
This puts the limit on the potential difference
which themachine can built up.
14.Thedevice is Van de Graf generator. Foritsprinciple
andworking, refer answertoQ.56 on page 2.67.
Yes,there is arestriction ontheupper limit of the
highvoltages set up in the Van de Graaff generator.
The high voltages can be built uponly upto the
breakdown field of thesurrounding medium .
.l'rYPE 0 :VALUE BASED QUESTIONS (4 marks each)
1.Immediately after school hour, as Birnla with her
friends carne out, they noticed that there was a
sudden thunderstorm accompanied by the
lightning. They could not findany suitable place for
shelter. Dr. Kapoor whowaspassing thereby in his
car noticed these children and ofered them to corne
in his car. He even tookcare to drop them to the
locality where they werestaying. Birnla's parents,
who werewaiting, saw this and expressed their
gratitude to Dr. Kapoor. [CBSE00 lSC]
(a)Whatvalues didDr.Kapoor andBirnla's
parents display?
(b)Why isit considered safe tobeinside a car
especially during lightning and thunder-
storm?
(c) Define the term' dielectric strength'. What does
this term signify ?
Answers
2.One evening, Pankaj outside his house fixed atwo
metre high insulating slab and attached a large
aluminium sheet of area 1m 2over its top. Tohis
surprise, next morning when he incidently touched
thealuminium sheet, he received an electric shock.
Hegot afraid. Henarrated the incident to his
Physics teacher in the school who explained him
the reason behind it.
Answer thefollowing questions based on the above
information:
(a)What arethe values being displayed by
Pankaj?
(b)What may bethe reason behind the electric
shock received by Pankaj ?
•
1.(a)Dr.Kapoor displayed helpfulness, empathy and
scientific temper.
Bimla parents displayed gratefulness and
indebtedness.
(b)It is safer to sit inside a car during a thunder-
storm because the metallic body of the car
becomes an electrostatic shielding from
lightning.
(c) The maximum electric field that adielectric
medium canwithstandwithout break-down of
itsinsulating property is called itsdielectric
strength. Itsignifies the maximum electric field
upto which the dielectriccansafely play its role.
2.(a)Keen observer and curiosity.
(b)Thealuminium sheet and the ground form a
capacitor alongwith the insulating slab. The
discharging current of theatmospherecharges
the capacitor steadily and raises its voltage. So,
when Pankaj touches thealuminiumsheet, he
receives an electric shock.www3xyto°n~svo3myw

182. Electrostatic Potential and Capacitance
GLIMPSES
1.Potential
difference.The potential difference
between two points isdefined as the amount of
workdone in bringing a unit positive charge
from one point to another against the electro-
static forces.
.. Work done
Potential difference= ----
Charge
_ _WAB
VAB-VB-VA ---
q
or
2.51 unit of potential difference is volt (V). The
potential difference between two points in an
electric field is said to be1volt if1joule of work
has to be done in moving apositive charge of1
coulomb from one point to the other against the
electrostatic forces.
1 V=1 JCI =1 Nm CI
3.Electric potential. It is defined as the amount of
work done in bringing a unit positive charge
from infinity to the observation point against
the electrostatic forces.
Work done
Electric potential
Charge
or V=W
q
Electric potential is a scalar quantity.
4.51 unit of electric potential is volt. The electric
potential at a point in an electric field is said to be
1 volt if one joule of work has to be done in
moving a positive charge of 1 coulomb from
infinity tothat point against the electrostatic
forces.
5.Electric potential due to a point charge.The
electric potential of a point chargeqat distancer
from it is given by
1q
V=--.-
41t 1:0.r
i.e -.,
1
Voc-
r
It is spherically symmetric.
6.Electric potential due to a dipole.Electric potential
at a point having position vector r,due to a
dipole of momentpat the origin is given by
--+--+ 8
V-1 p.r_ 1pcos
-41tl:o·7- 41tl:o -,:;.--
At points on the axial line of the dipole
(8 =Ooor 180°),
V.=+_1_ E
axial - 41t I:.,:;.
a
Atpoints on the equatorial line of the dipole
(8 =90°),
Vequa =0.
7.Electric potential due to a group of N point
charges.Ifrl,r2, r3... rNare the distances ofN
point charges from the observation point, then
V=_1_ [qi+q2+q3+ ....+qN ]
41tl:o 't ~ ~ rN
8.Determination of electric field from electric
potential. The rate of change of potential with
distance is called potential gradient. Electric
(2.113)wwwnnotesdrivencom

183. 2.114
field at any point is equal to the negativ
e of the
potential gradient at that point
E=- dV
dr
SI unit of electric field=Vm-1
-7
The direction ofEis in the direction ofsteepest
decrease of potential.
9.Determination of electric potential from electric
field. The electric potential atapoint having
position vectortis given by
r-7 -7
V=-fE .dr
00
10.Equipotential surface.Any surface that has
same electric potential at every point on it is
called an equipotentialsurface. The surface of a
charged conductor is an equipotential surface.
Some of the important properties of equi-
potential surface are as follows :
(i)No work is done in moving a test charge
over an equipotential surface.
(ii)Electric field is always normal to the
equipotential surface at every point.
(iii)Equipotential surfaces are close togetherin
the regions of strong field and farther apart
in the regions of weak field.
(iv)No two equipotential surfaces canintersect
each other.
11.Electric potential energy. The electric potential
energy of a system of point charges is defined as
the amount of work done in assembling the
charges at their locations by bringing them in,
from infinity.
P.E.of acharge=ChargexElectric potential
at the given point
It is measured in jouleG)or electronvolt (eV).
1eV=1.6x10-19J
12.Potential energy of a system of two point
charges.If two point chargesq1andq2are
separated by distancer12'then their potential
energy is
U=_l_. q1 q2
41t Sor12
PHYSICS-XII
13.Potential energy of a system of three point
charges.Itisgiven by
U=_1_ [q1 q2+q2 q3+q3 q1 ]
41t So '12 r23 r31
14.Potential energy of N point charges. It is given
by
U=_l_ Lqjqj
41tSo Allpairs';j
15.Potential energy of a dipole in a uniform electric
field.It is equal to the amount of work done in
turning the dipole from orientation91to92in
the field E.
U= -pE(cos92-cos91)
If initially the dipoleisperpendicular to the
field E,91=90° and92=9(say),then
-7 -7
U=-pEcos9=-P.E
When 9=0°,U=-pEi.e.,thepotential energy
of the dipole is minimum. The dipole is instable
equilibrium.
When9 =90°, U =0
When9 =180°, U =+pE
i.e.,the potential energy of the dipole is
maximum. The dipole is inunstable equilibrium.
16.Conductors and insulators.Conductors are the
substances which allow large scale physical
movement ofelectric charges through them
when anexternal electric field is aplied. They
contain a large number of free electrons.
Insulators are the substances which do not
allowphysical movement of electric charges
through themwhen an external electric field is
aplied. They contain a negligibly small
number of free charge carriers.
17.Electrostatic properties of a conductor.When
placed in anelectrostatic field, a conductor
shows the following properties:
(i)Net electrostatic field is zero in the interior
of a conductor.
(ii)Just outside the surface of a conductor,
electric field is normal to the surface.www5notysxr·vy5wom

184. ELECTROSTATIC POTENT
IALAND CAPACITANCE (Competition Section) 2.115
(iii)The netchargein the interior of a conductor
is zero and any excess charge resides on its
surface.
(iv)Potential is constant within and on the
surface of a conductor.
(v)Electric field at the surface of a charged
conductor is proportional to thesurface
charge density.
(vi)Electric field is zero in the cavity of a hollow
charged conductor.
18.Electrostatic shielding. The phenomenon of
making a region free from any electric field is
called electrostatic shielding. It is based on the
fact that electric field vanishes inside the cavity
of a hollow conductor.
19.Capacitance of a conductor.It is thecharge
requiredto increase the potential of a conductor
byunit amount.
. Charge
Capacitance =---=-
Potential
or
C=.!i
V
20.Capacitance of a spherical conductor.It is pro-
portional to the radius R of the spherical
conductor.
C=41t1,0 R
21.Capacitor.Itisan arrangement of two con-
ductors separated by aninsulating medium
that is used to store electriccharge and electric
energy.
22.Capacitance of a capacitor. The capacitance of a
capacitor is the charge required to be suplied
to one of its conductors so as to increase the
potential difference between two conductors by
unit amount.
q
c=-
V
23.Farad. It is the SI unit of capacitance. The
capacitance of a capacitor is 1 farad (F) if 1
coulomb of charge is transferred from its one
plate to another on aplying a potential dif-
ference of 1 volt across the two plates. .
1farad=1coulomb or 1F=1C
1 volt 1 V
1mF=10-3F,1J.lF=10-6F,1pF=10-12F.
24.Parallel plate capacitor.It consists of two large
parallel conducting plates, eachof area A,and
separated by asmall distance d.Itscapacitance
is
25.Spherical capacitor.It consists of two concentric
spherical conducting shells of inner andouter
radiiaandb.
41tEOab
C=--"--
b-a
26.Cylindrical capacitor.It consists of two coaxial
conducting cylinders of inner and outer radii
aandband of common length 1.
I 1
C =21tEO--b=2rtEO b
loge- 2303loglO -
a a
27.Capacitors in series.The equivalentcapa-
citanceC5ofnumber of capacitors connected in
series is given by
1 1 1 1
-=-+-+-+ ...
C5 C1 C2 C3
In a series combination of capacitors, the charge
on each capacitor is same but the potential
difference across anycapacitor is inversely
proportional to its capacitance.
28.Capacitors in parallel. The equivalent capa-
citance of a number of capacitors connected in
parallelisgiven by
Cp=C1+C2+C3+...
In a parallel combination of capacitors, the
potential difference across each capacitor is
same but the charge on each capacitor is
proportional to its capacitance.
29.Energy stored in a capacitor. Theenergy stored
in a capacitor of capacitance C and charge qwith
voltage Vis
U =.! CV2=.!.Q2=.!QV
2 2 C 2
30.Energy density. The electrical energy stored per
unit volume or energy density in a region with
electric fieldEis
1 2
U=2" EOEwww<notesdrive<com

185. 2.116
31.Common potential.If a number of
conductors of
capacitances CI, C2,C3, .... ,at potentialsVI' V2,
V3, ..... , having chargesql'q2' Q3' .... respectively
are placed in contact, their common potentialVis
given by
V=Total charge=Ql+Q2+Q3+.
Total capacitanceCI+ C2+ C3+.
=CIVI+C2V2+ C3V3+····
CI+C2+C3+····
32.Loss of energy on sharing charges. If two
conductors of capacitances CIand C2 at
potentialsVIandV2respectively are connected
together, a loss of energy takes place which is
given by
su=.!. CIC2 (V _ V)2.
2C+C I 2
I 2
33.Dielectric.A dielectricis asubstance which does
not allow the flow of charges through it but
permits them to exert electrostatic forces on one
another. It isessentially an insulatorwhich can
be polarised through small localised displace-
ments of its charges.
34.Polar and non-polar dielectrics.The dielectrics
made of polar molecules (such as HCl, N~,
~O,C~OH, etc.) are called polar dielectrics. In
a polar molecule, the centre of mass of positive
charges does not coincide with the centre of
mass of negative charges.
The dielectrics made of non-polar molecules are
called non-polar dielectrics. In a non-polar
molecule, the centre of mass of positive charges
coincides with the centre of mass of negative
chargese.g., ~, 02'CO2, CH4, etc.
35.Polarisation of dielectric.If the medium
between the plates of a capacitor is filled with a
dielectric, the electric field due to the charged
plates induces a net dipole moment in the
dielectric. This effect is called polarisation which
induces a field in the oposite direction. The net
electric field inside the dielectric and hence the
potential difference between the plates are
reduced. Consequently, the capacitance C increases
from its valueCowhen there is vacuum.
C='KCO'
36.Dielectric constant. It is the ratio of the
capacitance(C)of the capacitor with the
PHYSICS-XII
dielectric as the medium to its capacitance (Co)
when conductors are invacuum.
C
K=-
CO
It is also equal to the ratio of the aplied electric
field(Eo)to the reduced value of electric field
(E)on inserting the dielectric slab between the
plates of the capacitor.
E E
K=~= 0
E Eo - E'
Here E' is the field set up due to polarisation of
the dielectric in the oposite direction ofEo.
37.Capacitance of a parallel plate capacitor filled
with a dielectric.
C =K C =EOKA
o d
38.Capacitance of a parallel plate capacitor with a
dielectric slab between its plates.Iftis the
thickness ofthe dielectric slab andt<d,then
_ EoA
C-d-t(l-~r
39.Capacitance of a parallel plate capacitor with
conducting slab between its plates.Fort<d,
c-1_d )EOA =(_d )C
l.d-t d d-t o·
40.Capacitance of a spherical capacitor filled with a
dielectric.
ab
C=41t EO K --.
b-a
41.Capacitance of a cylindrical capacitor filled with
a dielectric
C=21t EO KI
b·
2.30310gl0 -
a
42.Van de Graaff generator.It is an electrostatic
generator capable of building up high potential
differences of the order of 107 volts,Itis based
on the principle that when a charged conductor
is brought into internal contact with a hollow
conductor, it transfers whole of its charge to the
hollow conductor, howsoever high the potential
of the latter may be.Also, it uses discharging
action of sharp points. It is used for accelerating
charged particles.www5notysxr·vy5wom

186. C H A PT E R
CuRRENT EL
ECTRICITY
3.1
CURRENT ELECTRICITY
1.Whatiscurrent electricity ?
Current electrity.In chapters 1 and 2, we studied
the phenomena assoated with the electric charges at
rest. The physics of charges at rest is called
electrostatics orstatic electricity. We shall now study
the motion or dynamicsof charges. As the term rrent
implies some sort ofmotion, so the motion of electric
chargesconstitutes an electric current.
The study of electric charges inmotion iscalledcurrent
electrici ty.
3.2ELECTRIC CURRENT
2.Define electric current.
El~tric rrent. If two bodies charged to different
poten .alsare connected together by means of a
conduc ing wire, charges begin to flow fromone body
to another. The charges continue to flow till the
potentials of the two bodies become equaL
Theflow of electriccharges through aconductor constitutes
anelectric current. Quantitatively, electric current in a
conductor across an area held perpendicular to the direction
or
offlow of chargeisdefined as theamountof charge flowing
across that area per unit time.
If a charget.Qpasses through an area in timetto
t+M,thenthe rrent Iat time tis givenby
1=lim
Ilt-7a
t.Q=dQ
M dt
If the rrent issteady i.e.,the rate of flow of charge
does not change with time, then
1=Q
t
EI. Electric charge
or ectnc rrent = ------"'-
Time
where Qisthecharge that flows across the given area
in time t.
Lightning, which is the flow of electric charge
between two clouds or from a cloud to the earth, is an
example of a transient rrent (a rrent of short
duration). But the charges flow in a steady manner in
devices like a torch, cell-driven clock, transistor radios,
hearing aids, etc.
3.Give the51unit of current.
SI unit of rrent is ampere.If onecoulomb of charge
crosses an area in one second, then the current through that
area is one ampere (A).
1 1coulomb
ampere=----
Isecond
Ampere is one basic SI unit. We shall formally define it
inchapter4in terms of magnetic effect of rrent.
Smaller rrents are expressed in following units:
1milliampere=1mA=10-3A
1microampere =1!iA=10-6A
(3.1)···2xy§n~m}r±n2lyv

187. 3.2
The orders of magnitude ofs
ome electric rrents
we come across in daily life are as follows:
Current in a domestic appliance ~ 1A
Current carried by a lightning ~104A
Current in our nerves =-10-6 A=1~A.
4..Distinguish between conventional and electronic
currents.
Conventional and electronic rrents. By con-
vention, the direction of motion of positive charges is
taken as the direction of electric rrent. However, a
negativecharge moving in one direction is equivalent to
an equal positive charge moving in the opposite
direction, as shown in Fig. 3.1.Asthe electrons are
negatively charged particles, sothe direction of electronic
rrent (i.e.,the rrent constituted by the flow of
electrons) is opposite to that of the conventional
rrent.
Conventional current
~
Electronic rrent
••
~I
Fig. 3.1Flow of negative charge is equivalent to the flow of
positive charge in the opposite direction.
5.Is electric current a scalaror vector quantity?
Electric rrent is a scalar quantity. Although
electric rrent has both magnitude and direction, yet
it is a scalar quantity. This isbecause the laws of
ordinary algebraare used to add electric rrents and
the laws ofvector additionare not applicable to
J.ne additionof electric rrents. For example, in
(Fig.3.2,two different rrents of 3A and4A flowing
in two mutually perpendilar wiresAOand BO meet
at the junction0and then flow along wire Oc.
The rrent in wire OC is 7A which is the scalar
addition of3A and4A and not 5A as required by
vector addition.
A
3A
90°<'>-0--1--- C
7A
4A
B
Fig. 3.2Addition of electric rrents is scalar.
PHYSICS-XII
Formulae Used
1.Electric rrent=ChargeorI=!i
Time t
ne
2.Asq=ne,soI=-
t
3.In case of anelectron revolving in a rcle of radius
rwith speed v,period of revolution ofthe electron is
T=21tr
v
Frequen of revolution, v=2=~
TZrtr
Current at any point of the orbit is
I=Charge flowing in1revolution
xNo. of revolutions per second
orI=ev=3!!.-.
21tr
Units Used
Electric charge is in coulomb (C),time in second
(s),andcurrent in ampere (A)
Constant Used
Charge on an electron, e=1.6x1O-19c.
Example 1.1020electrons, each having a charge of
1.6x10-19 C,passfrom a point A towards another point Bin
0.1s.Whatisthe current in ampere? What isits direction?
Solution. Heren=1020, e=1.6x10-19C,t=0.1s
Current,
The direction of rrent is from BtoA.
Example 2. Show that one ampereisequivalent to aflow of
6.25x1018elementary chargespersecond. [CaSE D92C]
Solution. Here 1=1A,t=1s,e=1.6x10-19 C
As[=!i=ne
t t
umber of electrons,
li 1x1 u
n=-= 19=6.25x10.
e1.6x10-
Example 3.How many electrons passthrough a lamp in
one minute, ifthe current is300mA?
[Himachal 95;Punjab 02]
Solution.I=300mA=300x10-3A,
t=1minute=60s,e=1.6x10-19C
As [=!i=ne
t twww9notesdrive9com

188. CURREN
T ELECTRICITY
.',Number of electrons,
.n=It= 300x10-3x60 = 1.125 x1020
e 1.6x10-19
Example4.How many electrons per second flowthrough a
filament of a120Vand60 Welectric bulb? Given electric
power isthe product ofvoltage and current.
Solution. Here V= 120 V, P = 60 W, t= 1 s
P60
AsP=VI,therefore, I=-=-= 0.5A
V120
Number of electrons,
It0.5xl 18
n=-= 19 = 3.125x10.
e1.6x10-
Example5.In the Bohrmodel ofhydrogenatom, the
electron revolves around the nucleus in a circular pathof
radius5.1x10-11m at afrequency of6.8 x1015revolutions
per second. Calculate the equivalent current.
Solution. Here r= 5.1x10-11m,
v=6.8x1015rps,e=1.6x10-19C
Current,
1=ev= 1.6x10-19x 6.8x1015=1.088x10-3A.
Example6.Inahydrogen atom, an electron movesinan
orbit of radius 5.0x10-11m with a speed of 2.2x106ms-1.
Find the equivalent current. (Electronic charge =1.6x10-19
coulomb). [Roorkee 84]
Solution. Herer= 5.0x10-11m,
v=2.2x106ms-1, e=1.6xlO-19C
Period of revolution of electron,
T=2rrr=2rrx5.0x10-11 s
v 2.2x106
1 2.2 x106
Frequency, v=-=-----,.,
T2rrx5.0x1011
17
2.2x7x10 =7x1015s-1
2x22x5
Current, 1=ev = 1.6 x10-19 x7x1015
=1.12x10-3A.
Example7.Figure3.3shows a plot ofcurrent Ithrough
thecross-section of awireoveratimeinterval of10 s.Find
the amount of I(A)
charge that ows
through the wire 5
during this time
period.
[CBSE00 lSC]
5 10
Fig. 3.3
3.3
t(s)
Solution. Amount of charge that flows in 10 s
=Area under the 1-tgraph
= ~x5x5+(10-5) 5 = 37.5 C
Example8.Theamountof chargepassingthrough cross-
sectionof a wire isq(t)=at2+bt+c .
(i)Write the dimensional formulae for a,bandc.
(ii)If the values ofa, band cinSIunitsare5,3and1
respectively, findthe value of current at t =5second.
Solution. (i) Given q (t)=at2+bt+c
Dimension ofa=[t~]= ~;=Ar1
Dimension ofb=[7]=~T=A
Dimension ofc=[q]=AT
(ii)Current, 1=dq=~(at2+bt+c)=2at +b
dt dt
Att=5s,I= 2x5x5+3= 53 A.
rproblems For Practice
1.One billion electrons passfrom a point P towards
another point Qin10-3 S .What is the current in
ampere? What is itsdirection?
(Ans. 1.6x10-7A,directionof
rrent is fromQtoP)
2.If 2.25 x1020electrons pass through a wireinone
minute, find the magnitude of the current flowing
through the wire. [Punjab 02](Ans.0.6 A)
3.A solution of sodiumchloride discharges
6.1xHy6N a+ions and 4.6 x1016Cl"ions in 2 s. Find
therrent passing through thesolution.
(Ans. 8.56x10-3A)
4.An electric rrentof 2.0I!A existsinadischarge tube.
How much charge flows across a cross-section of
the tube in 5 minutes? (Ans.6.0x10-4C)
5.In a hydrogen atom, the electron makes about
0.6xHy6revolutions persecondaround thenucleus.
Determine the average rrentatanypoint on the
orbit of the electron. (Ans.0.96 mA)
6.An electron moves in acirlar orbit of radius 10 em
with a constantspeed of 4.0x106ms-1. Determine
the electric rrent at a point on the orbit.
(Ans.1.02x10-12 A)
7.In a hydrogen discharge tube, the number of
protons drifting across a cross-section per second is
1.1x1018,while the number of electrons drifting in
the opposite direction across anothercross-section
is3.1x1018persecond. Find the rrent flowingin
the tube. (Ans. 0.672 A)···2xy§n~m}r±n2lyv

189. 3.4
HINTS
1.[=ne=
109x1.6x10-19 =1.6X10-7A
t 10-3
ne2.25x1020x1.6x10-19
2.[=-= = 0.6 A
t 60
(n++n-)e
3.[=[cations+[anions = t
(6.1x1016 +4.6x1016)x1.6 x10-19 -3
-'---------'------ =8.56x10 A
2
4.q=It=2.0x10-6 x5x60 = 6.0x10-4 C.
5.[=ve= 0.6 X1016 X1.6x10-19
=0.96X10-3A=0.96 mA
6.T=21tr= 21txO.!O s :.V= ~ = 4x106s-1
V 4x10 T21tX0.10
l__ 4x106 x 1.6 x10-19-102 10-12A
=ve - -. X •
21t xO.10
(n+ne)e
7.[=l+I=--,-P---
P" t
(1.1 X1018+ 3.1X1018) x1.6X10-19
= =0.672A
1
3.3 MAINTENANCE OF STEADY CURRENT
IN A CIRCUIT
6. With the help of a mechanical analogy, explain how
~theow of electric currentismaintainedinan electric circuit.
Maintenance of steady rrent in an electric rit.
The flow of electric rrent in a rit is analogous to
the flow of water in a pipe. As shown in Fig.3.4,
suppose we wish to maintain a steady flow of water in
a horizontal pipe fromAto B. As pressure atAis
higher than that at B, so water flows spontaneously
from the upper tank to the lower tank. To maintain a
steady flow, a water pump must do work at a steady
rate to pump water back from the lower tank to the
upper tank. Obviously, the water pump makes water
flow from lower to higher pressure. It helps to maintain
the pressure difference betweenAand B.
-11\
'n\
I
h
l~A======~ ..'
Water
pump
Fig. 3.4A closed water flow rit.
PHYSICS-XII
A steady flow of electric rrent in a conductor is
maintained in a similar way. As shown in Fig.3.5,
positive charge flows spontaneously in a conductor
from higher potential(A)to lower potential(B)i.e.,in
the direction of the electric field. To maintain steady
rrent through the conductor, some external device
must do work at a steady rate to take positive charge
from lower potential(B)to the higher potential(A).
Such a device is the source ofelectromotive force(emf)
which may be an electrochemical cell or an electric
generator. A source of emf transfers positive charge
form lower potential to higher potentiali.e.,in the
opposite direction of the electric field. Clearly, a charge
flow rit is analogous to the water flow rit.
R
A-:~Source of emf
B-(Charge pump)
Fig. 3.5A closed charge flow rit.
3.4ELECTROMOTIVE FORCE :EMF
7. Define emf of a battery. Is itreally aforce? When is
theemf of a battery equal to the potential diference
between itsterminals? Define emf of1volt.
Electromotive force.A battery is a device which
maintains a potential difference between its two ter-
minalsAand B.
~-------------I
---B·F,•f+-
:A Fe BI
Fig.3.6A schematic diagram of a battery.
Figure3.6shows a schematic diagram of a battery.
Due to certain chemical reactions, a force (of non-
electrostatic origin) is exerted on the charges of the
electrolyte. This force drives positive charges towards
terminalAand negative charges towards terminal B.
-+
Suppose the force on a positive chargeqisF".As the
charges build up on the two terminalsAand B, a
potential difference is set up between them. An electric
-+
field E is set up in the electrolyte fromAto B.This field
-+ -+
exerts a forceFe=qE on the chargeq,in the opposite
-+
direction ofFn.In the steady state, the charges stop
amulating further and F"=Fe .
The work done by the non-electrostatic force during
the displacement of a chargeqfrom BtoAis
W=Fn d
wheredis the distance between the terminalsAand B.···2xy§n~m}r±n2lyv

190. CURRENT ELECTRICI
TY
The work done per unit charge is
e=w=Fnd
q q
Thequantity e=W /qiscalled the electromotive force
oremfof the battery or any other source.
Theelectromotive forceof a source may be definedas
the work doneby the source intaking aunit positive charge
from lower tothe higher potential.
If the two terminals of the battery are not connected
externally, then
Fn=Fe=qE
Fnd=Fed=q Ed=qV
whereV=Edis the p.d. betweenthe two terminals. Thus,
e=Fnd=qV=V
q q
Hencethe emf of a sourceisequal to the maximum potential
diference between its terminals when itis inthe open circuit
i.e.,when it is not sending any rrent in the rit.
Basically, an electrochemical cell consists of two
electrodes P andNimmersed in an electrolyte, as
shown in Fig. 3.7
C I R D
AA
YYY
IP NI
~ r--
A B
'-- '--
+-f--
R
CJ
Cellsymbol
Electrolyte
Fig. 3.7An electrochemical cell connected to an
externalresistance andthesymbolic representation. Here
Vp - VA=V+>0andVN-VB=-V_<o.
Thetwo electrodes exchange charges with the
electrolyte. Consequently, the positive electrode P
develops a positive potential V+(V+>0) with respect to
its adjacent electrolyte marked AThe negative
electrode N develops a negative potential- V_(V_ >0)
withrespect tothe adjacent electrolyte B. When. no
current flows through the cell, the electrolyte has the
same potential throughout, so that the potential dif-
ference between the· two electrodes P and Nis
V+-(-V_)=V++V_=e,the emf.
Obviously, V++V_>o.
3.5
In case of a closed rit, we can define emf in
another way as follows:
The emf of asource may be defined asthe energy supplied
bythe sourceintakinga unitpositive charge once round the
complete circuit. Again, we note that
emf= Work done or e=W ~
Charge q
Literally, emf means the force which causes the
flow of charges in a rit. However, the term emf is a
misnomer. The emf is not a force at all. It is aspeal
case of potential difference, so it has the nature of work
done per unit charge.
SI unit of emf is volt.If anelectrochemical cellsupplies
an energy of1joulefor the ow of 1coulomb of charge
through the whole circuit (including the cell), then itsemfis
saidto be one volt.
3.5EMFVS.POTENTIAL DIFFERENCE
8.Giveimportant points of diferences between
electromotive force and potential diference.
Differences between electromotive force and
potential difference.
Electromotive force Potential difference
1.
Itis the workdonebyaIt istheamountof work
source in taking a unit done in taking a unit
charge once round the charge from onepoint
complete rcuit. of a rcuit to another.
2.Itis equaltothemaxi- Potential difference
mum potential diffe- may exist between any
rence between the two twopoints of aclosed
terminals of a source whencircuit.
it isinanopenrit.
3.Itexists even when the It exists only when the
cirit is notclosed. ciritisclosed.
4.It has non-electrostaticIt originates from the
origin. electrostatic field set up
bythe charges amu-
lated on the two termi-
nals of the source.
5.Itisa cause. When emf Itis an effect.
isapplied in a rit,
potential difference is
caused.
6.Itisequal to the sum of Every cirit
potential differences component has its own
across all the compo-potential difference
nents of a rit inclu-across its ends.
ding the p.d. required
tosend rrent through
the cell itself.
7.It is larger than the p.d. It is always less than the
across anyrit emf.
element.
8.It is independent of theIt isalways less than the
external resistance in emf.
the cirit.···2xy§n~m}r±n2lyv

191. 3.6
3.6OHM'S LAW :RESISTANCE
9.
State Ohm's law. Define resistance and state itsS1unit.
Ohm's law.On the basis of his experimental
observations, a German physistGeorge Simon Ohm
derived a relationship between electric rrent and
potential difference in 1828. This relationship is known
asOhm's law and can be stated as follows:
Thecurrent owing through aconductorisdirectly
proportional to the potential diference applied across its
ends, provided the temperature and other physical
conditions remain unchanged.
Thus, Potential differenceex:Current
Vex:I
or V=RI
The proportionality constantRis called theresis-
tanceof the conductor. Its value is independent ofV
andIbut depends on the nature of the conductor, its
length and area of cross-section and physical con-
ditions like temperature, etc. Ohm's law may also be
expressed as
V=R
T~raph ~etween the
potential differenceV
applied across a conductor
to the rrentIflowing
through it is astraight line, Fig.3.8V-Igraph for an
as shown in Fig. 3.8. ohmic conductor.
Resistance. Theresistance of aconductoristhe property
byvirtue of which it opposes the ow of charges through it.
The more the resistance, the less is the rrentIfor a
given potential difference.It is equal to the ratioof the poten-
tial diference applied across the conductor to the current
owing through it.Thus
R=V
I
r
v~
51unit of resistance is ohm (Q). If the potential
difference(V)is1volt and rrent(I)is1ampere, then
the resistance(R)is1ohm. .
hI volt
10m=----
1ampere
or
Thus,the resistance of a conductorissaid to be 1ohm if
a current of1ampere ows through it on applying a
potential diference of1uolt across its ends.
Any material that has some resistance iscalled a resistor.
Pictorial symbols for resistors and meters are given in
Fig.3.9.
PHYSICS-XII
Fixed ~or~
resistor
Variable ~ or-+-
resistor~
Potential ~ or
divider I
Meters--0---0-
Voltmeter Ammeter
~
t
-@--<D-
Galvanometer
Fig.3.9Symbolsfor resistors and meters.
10. Briefly explain how can we measure theresistance
of awire.
Measurement of resistance.Fig.3.10 shows a simple
rit for measuring the resistance of a wire. Here the
battery and ammeter are connected in series with the
wire and the voltmeter in parallel with it. The ratio of
the voltmeter reading(V)and the ammeter reading (/)
gives the resistance(R)of the wire.
Battery
+,\I-----{
R
Voltmeter
Fig.3.10To measure resistance of wire.
3.7FACTORS AFFECTING THE
RESISTANCE:RESISTIVITY
11.What are the factors on which the resistance of a
conductor depends? Define resistivity and state its SI unit.
Factors affecting the resistance. At a constant
temperature, the resistance of a conductor depends on
the following factors :
1.Length. The resistanceRof a conductorisdirectly
proportional to its length i.e.,
Rex:I
2.Area of cross-section. The resistance Rof a uniform
conductorisinversely proportional toitsarea ofcross-
section A, i.e.,
I
Rex:-
A
3. Natureof the material. The resistance of a
conductor also depends on the nature of its material.
For example, the resistance of a nichrome wire is
60 times that of a copper wire of equal length and area
of cross-section.···2xy§n~m}r±n2lyv

192. CURRENT EL
ECTRICITY
Combining the above factors, we get
I I
Rex: - or R =p-
A A
where pistheconstant of proportionality called resistivittj
orspecific resistance ofthematerial ofthe conductor. It
depends on the nature of the material of the conductor
andon the physical conditions like temperature and
pressure but it is independent of its size or shape.
Resistivity or spefic resistance. If in the above
equation, we take
I=1unitandA=1square unit
then R=p
Thus,the resistivity or specific resistance of a material
may be defined as the resistance of a conductor of that
material,having unit length and unit area of cross-section.
Or,itisthe resistance ofered bythe unit cube of the material
ofaconductor.
51unit of resistivity. We can write
RxA
p=--
I
51.f ohm x metre/
unit0p=------
metre
=ohm meter (Qm)
Thus,the51unitof resistivity isohm metre(Qm).
3.8 CURRENTDENSITY, CONDUCTANCE
AND CONDUCTIVITY
12.Define the termscurrent density, conductance
and conductivity. Write their51units. Express Ohm's
lawinvector form.
Current density.Thecurrent density at any point
insideaconductor is defined as theamount ofchargeowing
per second through a unitareaheldnormal to the direction of
theow of charge at that point. It is a vector quantity
having the same direction as that of the motion ofthe
positive charge. It is a characteristic property of any
->
pointinside the conductor and isdenoted by j .
As shown in Fig.3.11(a), if acurrent 1is flowing
uniformly and normally through anarea ofcross-
section Aof aconductor, then the magnitude of current
density at anypoint of this cross-section willbe
. q /tI
t=r+=r:
A A
Ifthe area Aisnot perpendilar to the direction of
rrent andnormal tothis area makes angle 8 with the
direction ofcurrent as shown in Fig.3.11(b), then the
component ofAnormal tothedirection ofcurrent
flow willbe
~=Acos8
3.7
Area =A
--+
A
(a)
--+
A
(b)
Fig. 3.11Current density.
Current density,
.I I
}=-=
An Acos 8
or
->->
I=jAcos 8=j .A
This equation again shows thatelectric rrent,
being scalar product oftwo vectors, is a scalar quantity.
The51unit of current density isampere per square
metre (Am-2)anditsdimensions are [AL-2].
NOT E The current Ithrough a partilar surface 5 in
->
aconductor isthe flux of jthrough that surface and is
givenbythe surface integral
I=ff.as
5
->
wheredSis a small element ofthe given surface area.
Conductance.The conductance of aconductor isthe
ease with which electric chargesflow through it. Itisequalto
the reciprocal of its resistance andisdenoted by G.
Thus,
or
1
Conductance =----
Resistance
1
G=-
R
The51unitof conductance isohm-lor mho or
siemens (S)
Conductivity.The reciprocal of the resistivity of a
materialis called its conductivity and is denoted by 0.
Thus,
C d .. 1
onuctivity =----
Resistivity
1
or 0=-
p
The51unitof conductivity isohm-1m-1or mho m-1
orSm-1.
Vector form of Ohm's Law. IfEisthe -nagnitude of
electric field ina conductor of lengti I,then the
potential difference across its ends is
V=EI···2xy§n~m}r±n2lyv

193. 3.8
I
~
Also from Ohm's law, we can write
V=IR=Ipl
A
I
EI=
-pl
A
or E=jp
~
As the direction of rrent densityjis same as
~
that of electric fieldE,we can write the above
equation as
~ ~
E=pj
or
~ ~
j=CJE
The above equation is thevector[ormof Ohm's
law.It is equivalent to the scalar formV=RI.
3.9CLASSIFICATION OF MATERIALS IN
TERMS OF RESISTIVITY
13. How can weclassify solids on the basis of their
resistivity values?
Classification of solids on the basis of their
resistivity values. The electrical resistivity of sub-
stances varies over a very wide range, as shown in
Table 3.1. Various substances can be classified into
threecategories:
1. Conductors.The materials which conduct electric
current fairly well are called conductors.Metals are good
conductors. They have low resistivities in the range of
10-8nm to 10- 6nm. Copper and aluminium have
the lowest resistivities of all the metals, so their wires
are used for transporting electric rrent over large
distances without the appreable loss of energy. On
the other hand nichrome has a resistivity of about 60
times that of copper. It is used in the elements of
electric heater and electric iron.
2. Insulators.The materials which do not conduct
electric current are called insulators. They have high
resistivity, more than 104 nm. Insulators like glass,
mica, bakelite and hard rubber have very high
resistivities in the range 1014 nm to 1016 nm. So they
are used for blocking electric rrent between two
points.
3. Semiconductors.These are the materials whose
resistivities lieinbetween those of conductors and insulators
i.e.,between 10-6 nm to 104 nm. Germanium and
silicon are typical semiconductors. For moderately high
resistances in the range of kn,resistors made of
carbon (graphite) or some semiconducting material are
used.
PHYSICS-XII
Table 3.1 Electrical resistivitiesofsome substances
...
A. Conductors
Silver 1.6x10-8
Copper 1.7x10-8
Aluminium 2.7x10-8
Tungsten 5.6x10-8
Iron 10x10-8
Platinum 11x10-8
Merry 98x10-8
Nichrome 100x10-8
(alloy of
Ni, Fe, Cr)
Manganin 48x10-8
(alloy of Cu.
Ni,Fe, Mn)
B. Semiconductors
Carbon 3.5x10-5
(graphite)
Germanium 0.46
Silicon 2300
C.Insulators
Pure water 2.5x105
Glass 1010 _1014
Hard Rubber 1013 _1016
NaCl _1014
Fused quartz _1016
0.0041 1
0.0068
0.0043 3
0.0045 6
0.0065 8
0.0039 10
0.0009 2
0.0004
0.002x10-3
-0.0005 4
-0.05
-0.07
4
4
8
14. What are the two common varieties of commercial
resistors?
Common commeral resistors. The commeral
resistors are of two major types :
1.Wire-bound resistors.These are made by winding
the wires of an alloy like manganin, constantan or
nichrome on an insulating base. The advantage of
using these alloys is that they are relatively insensitive
to temperature. But inconveniently large length is
required for making a high resistance.
2.Carbon resistors.They are made from mixture of
carbon black, clay and resin binder which are pressed
and then moulded into lindrical rods by heating. The
rods are enclosed in a ceramic or plastic jacket.°°°2uv{lzkyp£l2jvt

194. CURRENT ELECTRICITY
The carbon
resistors are widely used inelectronic
cirits of radio receivers, amplifiers, etc. They have
the followingadvantages :
(i)They can be made with resistance values rang-
ing from few ohms to several million ohms.
(ii)They are quite cheap and compact.
(iii)They are good enough for many purposes.
3.10COLOUR CODE FOR CARBON RESISTORS
15. Describe the colour code used for carbon resistors.
Colour code for resistors. A colour codeisused to
indicate theresistance value of a carbon resistor and its
percentage accuracy. The colour code used throughout
the world is shown in Table 3.2.
Table 3.2 Resistor colour code
Black B 0 10° Gold 5%
Brown B 1 101 Silver 10%
Red R 2 102 No fourth 20%
band
Orange 0 3 103
Yellow Y 4 104
Green G 5 105
Blue B 6 106
Violet V 7 107
Grey G 8 108
White W 9 109
How to remember colour code:
B B ROY of Great Britain had Very Good Wife
.J.- .J.- .J.- .J.-.J.- .J.-.J.- .J.-.J.-.J.-
012345 6 78 9
There are twosystems of marking the colour codes:
First system.Aset of coloured co-axial rings or
bands is printed on the resistor which reveals the
following facts :
1.The first band indicates thefirst significant figure.
2.Thesecond band indicates the second significant
figure.
3.The third band indicates the power of ten with
which the above two significant figures must be
multiplied to gettheresistance value in ohms.
4. The fourth band indicates the tolerance or possible
variation in percent of the indicated value. If the
fourth band is absent, it implies a tolerance of
±20%.
3.9
ifFirst significant figure
lrSecond significant figure
IDecimal multiplier
1ITolerance
4))))}-
Fig. 3.12Meanings of four bands.
Illustrations:1. In Fig.3.13,the colours of the four
bands are red, red, red and silver; the resistance value is
Red
.J.-
2
Silver
.J.-
±10%
Red
.J.-
2
Red
.J.-
2
Fig. 3.13
2. In Fig.3.14, the colours of the four bands are
yellow, violet, brown and gold; the resistance value is
Yellow
.J.-
4
Violet
.J.-
7
Brown
.J.-
1
Gold
.J.-
±5%
R=47xl01 Q±5%.
Violf1jet :wnYellow _ ~r=Gold
-t~~)--
Fig. 3.14
3. When there are only three coloured bands
printed on a resistor and there is no gold or silver
band, thetolerance is 20%. In Fig. 3.15,there are only
three bands of green, violet and red colours ; the
resistance value is
Green
.J.-
5
No 4th band
.J.-
±20%
Violet
.J.-
7
Red
.J.-
2
R=57x102Q±20%.
)--
Fig. 3.15···2xy§n~m}r±n2lyv

195. 3.10
Second System
:
1.The colourofthe body gives the firstsignificant
figure.
Fig.3.16
2. The colourof the end gives the second signi-
ficant figure.
3.The colour of the dot gives the number ofzeroes
to be placed after the second figure.
4.The colourof the ring gives the tolerance or
percent aracy of the indicatedvalue.
Illustration.Suppose for a given resistor, the body
colour is yellow, end colour is violet, dot colour is
orange and the ring colour is silver.
Body End IDot Ring
Yeiow Vi~letIOrrgeISilrer
4 7' 3 ±10%
:.R = 47x1030±10%= 47 kO±10%.
Conductance, ConductiviW, I--.
•• -.......
•
Formulae Used
V
1.Ohm's law,R=-orV=IR
I
2.Resistance of a uniform conductor, R=p~
A
RA
3.Resistivity or spefic resistance, p=-[-
1
4. Conductance=-
R
C d .. 1
5. onuctivity =----
Resistivity
. Current
6. Current density =---
Area
1 [
or(J=- =-
PRA
.I
or]=-
A
7. Colour code of carbon resistors. Refer to Table3.2.
Units Used
Potential differenceVis in volt (V), rrentIin
ampere (A), resistance R in ohm (0), resistivityp
in Om, conductance in ohm-lor mho orsiemens
(S), conductivity in 0-1m-lor Sm-1and rrent
density jin Am-2.
PHYSICS-XII
Example 9. In a discharge tube,the number of hydrogen
ions (i.e., protons) drifting across a cross-section per second
is1.0x1018,while the number of electrons driftinginthe
opposite direction across another cross-section is2.7x1018
per second. If the supply voltage is 230V,what isthe
effective resistance of the tube? [NCERT]
Solution. The rrent carried by a negatively charged
electron is equivalent to the rrent carried by a proton
in the opposite direction, therefore, total rrent in the
direction of protons is
I=Total charge flowing per second =(ne+np)e
= [2.7 x1018+1.0x1018] x1.6x10-19
= 3.7x1.6x10-1=0.592 A
Effective resistance,
R =V= 230 0 =388.50=-3.9x102O.
I 0.592
Example10.A10Vbattery of negligible internal
resistance isconnected across a 200Vbattery and a resis-
tance of 38 0as shown inthe figure. Find the value of the
current in circuit. [CBSE D13]
10V
~~
200 V
Solution.1=V= 200-10 = 5 A
R 38
Example11.Acopper wire of radius0.1mmand
resistance 1k0isconnected across a power supply of 20V.
(i)Howmanyelectrons are transferred per second between
the supply and the wire at one end? (ii)Write down the
current density inthe wire.
Solution. Herer=0.1 mm =0.1 x10-3m,
R =1 kO =1030, V=20 V
(i)Current, 1=V=203=0.02 A
R10
No. of electrons,
qIt
n=-=-
e e
0.02x1 17
---Cl~9 =1.25x10.
1.6x10-
(ii) Current density,
.I I 0.02
] =A=1tr2= 3.14 x(0.1x10-3)2
=6.37x105Am-2.www4not~s{riv~4zom

196. CURRENT ELECTRICITY
Example1
2.Current ows throughaconstricted con-
ductor, as shown inFig.3.17.The diameter 01=2.0mm
andthe current density to theleft of the constriction is
7= 1.27x106Am-2. (i) What current ows into the
constriction ?(ii)If the current density isdoubled as it
emerges from the right side of the constriction, what is
diameter 02?
,
,
,
r,
I
II,
,,
,
,,
,,, ,,,
,
,
Fig. 3.17
Solution.Here01= 2.0 mm, 71= 1.27 x106Am-2,
72=2 71
(i)Current flowing into the constriction,
II=71A=i,xn(~1J
= 1.27 x106x3.14x(1x10-3)2 = 3.987A
(ii)For a steady flowof rrent,
11=12
71~ =72x Az
71xn(~1r= 72xn(~2r
71xn(~1r=2jl xn(~2rr.72=271]
1
02=.fi01= 0.707 01
or
or
or
or
=0.707 x2.0 mm=1.414 mm.
Example 13. A current of2mAispassedthrough a
colour coded carbon resistor with first, second and third
rings of yellow, green and orange colours. What isthe
voltage drop across the resistor?
Solution.
Yellow
t
4
Orange
t
3
Green
t
5
R=45x103D
Given 1=2mA = 2x10-3A
V=RI= 45x103x2x10-3V = 90 V.
Example 14. An arc lamp operates at80V,10ASuggest
a method to use it with a240Vd.c.source. Calculate the
valueof the electric component required for this purpose.
[CBSE F 94]
3.11
Solution. Resistance of the arc lamp is
R=V=80 =8D
I 10
Inorder to use arc lampwitha source of 240V,a
resistance R' should be connectedinserieswith itso
that rrent through the rit does not exceed 10 A.
Then
I(R+R')=V or 1O(8+R')=240
or R'=24 - 8=16 D.
Example 15. Calculate the resistivity vf a material of a
wire10mlong,0.4mm in diameter and having a resistance
of2.0 D. [Haryana 02]
Solution. Here I= 10 m, r =0.2 mm=0.2x10-3m,
R=2D
Resistivity,
RARxnr2
P==-I-= I
= 2x3.14x(0.2x10-3)2=2.513x10-8Dm.
10
Example 16. The external diameter of a 5metre long
hollow tube is10em andthethickness of its wall is5mmIf
the specificresistance of copper be 1.7x10-5ohm-metre,
thendetermine its resistance.
Solution. The cross-sectional area of the tube is
2 2
A=n(r2-r1)
=3.14 x[(5x10-2)2 -(4.5 x10-2)2]
= 14.9x10-4 m2
Also, P= 1.7x10-8Dm,I=5 m
.. Resistance,
R-i_1.7x10-8x5
- PA-14.9x10-4
=5.7x10-5D.
Example 17. Findtheresistivity of a conductor inwhich a
current density of2.5Am-2 isfound to exist, when an
electricfieldof15Vm- 1isapplied onit. [ISCE 98]
Solution. Here 7=2.5 Am-2, E=15 Vm-1
RA V A
Resistivity, P=- = -. -
III
=VII=~=~=6Dm.
l/ A72.5
Example 18. Calculate the electrical conductivity of the
material of aconductor of length 3m, area ofcross-section
0.02m~having a resistance of2 D.
Solution. Here I= 3m,R= 2 D,
A=0.02mm'=0.02 x 10- 6m2···2xy§n~m}r±n2lyv

197. 3.12
Electrical conductivity = 1
Resistivity
1 I 3
0=
-=-=---------0-
P RA 2x0.02x10-6
=75x1060-lm-1.
Example 19. A wire of resistance 4 0 is usedtowind acoilor
ofradius 7em The wire has adiameter of1.4mm andthe
specific resistance of itsmaterial is2x10-7OmFindthe
number of turns inthecoil.
Solution. Letnbe the number of turnsin the coil.
Then total length ofwire used
=21tRxn=21tx7x 10-2xnmetre
Total resistance,
or
1
R=p-
A
or
n=70.
Example20.Awireof10ohm resistance isstretched to
thriceitsoriginal length. What will be its(i)new resistivity,
and(ii)new resistance? [CBSE D 98C]
Solution.(i)Resistivity premains unchanged
because it is the property of the material of the wire.
(ii)In both cases, volume of wire is same. So
V= A'l' =AI
A'I I 1
or -=-=- [.:l'=1+21=3ij
A l'31 3
l'
R' PA'l'A3 3
-=--=-x -=-x -=9
R I I A'11
p-
A
Hence R' =9R=9x10 =900.
Example 21. A wire hasa resistance of16O.Itismelted
and drawninto a wire of halfitslength. Calculate the
resistance of thenew wire. What isthepercentage change in
itsresistance ?
Solution. In both cases, volume of the wire is same.
V= A'l' =AI
A'I I 2
A=r=ll=l
2
or
or
l'
R' PA'l'All 1
-=--=-x -=-x -=-
R IIA'224
p-
A
R' =.!R =.! x16 = 40.
4 4
Change in resistance
R-R' 12
= -- x 100=-x100 = 75%.
R 16
PHYSICS-XII
Example22.The resistance of a wire isRohm.What will
beitsnewresistance ifit isstretched to n times itsoriginal
length?
Solution. In both cases, volume of the wire is same.
V= Al=A'l'
Al'
-=-=n
A' I
[.:I'=nil
or
l'
R' P=:l' A 2
-=-LL=-.-=n.n=n
R I IA'
p-
A
R'=n2R.
Example23.A cylindrical wire isstretched toincrease its
length by 10%. Calculate the percentage increase in
resistance.
or
Solution.New length, l'= 1+ 10% ofI
=1+0.11=1.11
£.=1.1
I
AI=A'l'
Al'
A'I
~=£.x~=(£.)2 =(1.1l=1.21
R I A' I
or
The percentage increase in resistance,
R'- R(R')
-R- x100 =R-1x100 =(1.21-1)x 100 =21%.
Example24.Two wires A andBof equal mass and of the
samemetalare taken. The diameter of the wire Aishalf the
diameter of wireB.If the resistance of wire Ais240,
calculate the resistance of wire B.
Solution. Mass of wire=volume xdensity
= area of cross- section xlengthx density
2 2
m=1trAIAd= 7trB IBd
or !JL=(rAJ2=(~)2 1
IA rB 1 4
:~0: ':~:0;~,(:.r0~'G)'01~
1trA
1 1
or RB=16RA= 16x240 = 1.5 O.
Example25.A piece of silver has a resistance of10.What
will be the resistance of aconstantan wire of one-third length
and one-half diameter, ifthe specific resistance ofconstantan
is30times that of silver ?°°°2uv{lzkyp£l2jvt

198. CURRENT. ELECTRICIT
Y
Solution. For silver,
R=4pI=10n
nd2
Forconstantan,
1
4 /l'4x30Px-
R'=-P-= 3
nd/2n(~J
40x4p1
= 2= 40 R = 40x1 = 40 n.
ttd
Example 26. On applying the same potential diference
between the ends of wires of iron andcopper of the same
length, the same current flows in them. Compare their radii.
Specific resistances of iron andcopper are respectively
1.0x10-7and1.6x10-8nmCan theircurrent-densities
be made equal by taking appropriate radii?
Solution. On applying same potential difference,
same rrent flows in the two wires. Hence the
resistances of the two wires should be equal.
1 1
But R=p-=p-
A nr2
For the two wires of same length 1,we have
1 1
Rl =PI-2 and ~ =P2 -2'
rtr1 rtr2
As Rl=~
PI P2
1=rf
or
riron = Piron = 1.0 x10-7=2.5.
~opper Pcopper 1.6x10-8
No,rrent densities cannot be equalbecause they
depend on nature of the metals.
cproblems ForPractice
1.A voltage of 30 V is applied across a colour coded
carbon resistor with first,second and third rings of
blue,black and yellow colours. What isthe rrent
flowing through the resistor? [CBSED05]
(Ans.0.5x10-4A)
2.Apotential difference of 10 V is applied across a
conductor ofresistance 1k n. Find the number of
electrons flowing through the conductor in
5minutes. (Ans.1.875x1019)
3.What length of a copper wire of cross-sectional area
0.01mm2wouldberequired to obtain a resistance
of 1k n ? Resistivity of copper =1.7x1O-8nm.
(Ans. 588.2m)
3.13
4.A metal wire of spefic resistance 64 x1O-8n m
and length 1.98m has a resistance of 7 n. Find its
radius. (Ans.2.4x10-4m)
5.Callate the resistance of a 2 m long nichrome wire
of radius 0.321 mm. Resistivity of nichrome is
15x10-6n m. If a potential difference of 10 V is
applied across this wire, whatwill be the rrent in
the wire? (Ans.9.26n, 1.08A)
6.An electron beam has an aperture of 1.0mm 2.A
total of 6x1016 electrons flow through any
perpendicular cross-section per second. Callate
(i)the rrent and (ii)the rrent density in the
electron beam.
[Ans. (i)9.6x10-3A(ii)9.6x103Am-2]
7.Calculate the electric field in a copper wire of
cross-sectionalarea 2.0 mm2 carrying a rrent of
1 A.The resistivity of copper = 1.7 x10-8nm.
(Ans.0.85x10-2Vm-1)
8.A given copper wire is stretched to reduce its
diameter to half its previous value. What would be
its new resistance? [CBSE D92C]
(Ans.R'= 16 R)
9.Whatwill bethe change in resistance of a
constantan wirewhen its radius is made half and
length reduced to one-fourth of its original length ?
(Ans.No change)
10.A wire of resistance 5n is uniformly stretched until
its newlength becomes 4times the originallength.
Find its new resistance. (Ans.80n)
11.A metallic wire of length 1 m is stretched to double
its length. Callate the ratio of its initial and final
resistances assuming that there is no change in its
density on stretching. [CBSED94]
(Ans.1 :4)
12.A wire of certain radius is stretched so that its
radius decreases by a factornCallate its new
resistance. (Ans.n4R)
13.A wire 1 m long and 0.13 mm in diameter has a
resistance of 4.2 n.Callate the resistance of
another wire ofthe samematerial whose length is
1.5m and diameter 0.155mm. (Ans.4.4n)
14.A rheostat has 100 turns of a wire of radius 0.4 mm
having resistivity 4.2 x10-7nm. The diameter of
eachturnis 3 . What isthe maximum value of
resistance that it can introduce? (Ans. 7.875n)
15.Giventhatresistivity of copper is 1.68 x10-8nm.
Callate the amount of copper required to draw a
wire 10km longhaving resistanceof 10n. Thedensity
of copper is8.9x103kgm -3. (Ans.1495.2kg)···2xy§n~m}r±n2lyv

199. 3.14
16.Thesize
of a carbon block is 1.0 emx1.0 x50 cm.
Find its resistance(i)between the opposite
square faces(ii)between the opposite rectangular
faces of the block. The resistivity of carbon is
3.5x10-50 . (Ans. 0.1750, 7.0x10-50)
17.Two wiresAandBof the same material have their
lengths in the ratio 1:5and diameters in the ratio
3:2.If the resistance of thewireBis1800, find the
resistance of the wireA. (Ans. 160)
18.A uniform wire is t into foursegments. Each
segment is twice as long as the earlier segment. If
the shortest segment has a resistance of 4 0, find the
resistance of the original wire. (Ans.60 0)
19.Callate the conductance and conductivity of a
wire of resistance 0.010, area of cross-section
10-4m2 and length 0.1 m. [Haryana2000]
(Ans.100 S, 105Sm-1)
HINTS
1.R = 60x1040,V= 30 V
V 30 4
I=-= 4= 0.5x10A.
R 60x10
2.I=V=10V = ~ = 10-2 A
RUO 10000
_3. _~ _10-2x5x60 _ 1 875 1019
n- e-e-1.6x10 19 -. x
4A R-pl_~
.s-A-n';
:. y2=~ = 64x10-8x1.98x7 = 5.76 x1O-8m2
nR 22x7
ory=2.4x10-4m.
I V
5.Use R=p~and I=-.
nr R
qne6x1(y.6x16x10-19
6.(i)I=-=-=------
tt 1
= 9.6xl0-3 A.
(ii)Current density,
.-i-9.6x10-3 _9 6103Am-2
] -A-1.0x10-6- .x .
VIR Ipl Ip 1x1.7x10-8
7.E=-=-=-=-=---..".-
I I IA A 20x10-6
=0.85xl0-2Vm-1.
8.When the diameter of the wire is reduced to its half
value, area of cross-section becomes one-fourth and
the length increases to four times the original
length. .
,I' 41 I
R =p A,=P'-1-=16p A =16R.
-A
4
PHYSICS-XII
I I
9.R=P-=P-2
A 1tY
R' = p'l/4 = p _1_ =R
1t(Y/2)2 ny2
I .
10.R=p- =50
A
R'=p ~ = 16p ~ = 16 R= 16x5=800.
A/4 A
I
11.R=p-
A
R' =p~=4p ~ = 4 R
A/2 A
..R: R'=1:4.
12. V=A'I'=Al
orV=1t(.;;YI'=1ty2IorI'= ~I
R' I ' n2I 4I 4R
=P--2 =p 2 =np-=n .
1tY' 1t(y / n) nr2
13.Rz=1)[t][~r
=4.2[1.5] [0.13x10-3 ]2= 4.4O.
1 0.155xlO-3
14.Length ofthe wireused, I= lOOnD
I 100nDlOOpD
R=p ~ =P'-;r=-,;-
100x4.2x10-7x3x10-2
-----"3 "2-- = 7.875O.
(O.4x10 )
I
15.As R=p-
A
pi 1.68 x10-8x10xl03 52
A=-= =1.68xl0- m
R 10
Mass of copper required,
m=Volume xdensity = Alxdensity
= 1.68x10-5x10x103x 8.9x103
=1495.2kg.
(i)R = ~ = 3.5 x10-5x50x10-2=0.175o.
PALO x10-2x1.0x10-2
(ii)R=~=3.5 x10-5 x1.0x10-2=7.0xl0-50.
PALO x 10-2x50 x10-2
IA
RA= P~ =IA(.'!JL]2 =.!x(~)2= ~
R8 _18_ 18dA 5 3 45
pnd~/4
4 4
RA= 45 R8 = 45xI80=160.
16.
17.°°°2uv{lzkyp£l2jvt

200. CURRE
NT ELECTRICITY
18.Letthelengths of the four segments be I,2/,41and
8/.Then their corresponding resistances will be R,
2R,4R and 8R
Given R=4Q
Resistance of the original wire
= R+2R+4R+8R =15R= 15x4 =60Q.
1 1
19.Conductance,G=-=-= 100 S.
R 0.01
Conductivity,
cr=..!=_I_= 0.1 =10SSm-1.
pRA0.01x10-4
3.11 CARRIERS OF CURRENT
16.Mention diferent types of charge carriersIn
solids, liquids and gases.
Carriers of rrent. The charged particles whichby
owing in a definite direction set up an electric current are
called current carriers.The different types of rrent
carriers are as follows:
1.Insolids. In metallic conductors, electrons are
the charge carriers. The electric rrent is due to the
drift of electrons from low to high potential regions. In
n-type semi-conductors, electrons are the majority
charge carriers while in p-type semiconductors, holes
are themajority charge carriers. A holeis a vacant state
from which an electron has been removed and itactsas
a positive charge carrier.
2.Inliquids. In electrolytic liquids, thecharge
carriers are positively and negatively charged ions. For
example, CuS04 solution has Cu2+ andSO~- ions,
which act as the charge carriers.
3.Ingases. In ionised gases, positive and negative
ions and electrons are the charge carriers.
4.Invaum tubes. In vaum tubes like radio
valves, cathode ray oslloscope, picture tube etc ; free
electrons emitted by the heated cathode act as charge
carriers.
17. Whyisit that electrons carry current in metals?
Metallic conduction.In metals, the atoms are
closely packed. The valence electrons of one atom are
closetotheneighbouring atoms and experience electrical
forcesdue tothem. So they do notremain attached to a
partilar atom, but can hop from one atom to another
andare freetomove throughout the l.')ttice.These free
electrons are responsible for conduction in metals.
The fact, that the negatively charged electrons carry
rrent in metals, was"first experimentally confirmed
by the American physistsTolman and Stewart in1917.
They measured the angular momentum of the charges
3.15
flowing steadily in a rlar loop. Their observations
indicated that
1.The sign of the charges isnegative.
2.Theratioe /mof the charges isequal to that mea-
suredfor the electronsin other experiments.
Itwasthus established directly that rrent in
metals is carried by negatively charged electrons.
3.12MECHANISM OF CURRENT FLOW IN A
CONDUCTOR : DRIFT VELOCITY AND
RELAXATION TIME
18. Explain the mechanism of the ow ofcurrent ina
metallic conductor. Hence define theterms driftvelocity
and relaxation time.Deduce a relation between them.
Mechanism of the flow of electric' charges in a
metallic conductor : Concepts of drift veloty and
relaxation time.Metals have a large number of free
electrons, nearly 1028 percubic metre. In the absence of
any electric field, these electrons are in a state of
continuous random motion due to thermal energy. At
room temperature, they move with veloties of the
order of105ms-1. However, these velocities are
distributed randomly in all directions. There is no
preferred direction of motion. On the average, the
number of electrons travelling in any direction will be
equal to number of electrons travelling in the opposite
~ ~ ~
direction. Ifu1'u2' .... ,UNare the randomveloties ofN
free electrons, thenaverage velocity of electrons will be
~ ~ ~
~ u1+u2+···+uN
u= =0
N
Thus, there is no net flow of charge in any direction.
~ .
Inthepresence of an external field E, each electron
~ ~
experiences a force -eE in the opposite direction of E
(since an electron has negative charge) and undergoes
~
an acceleration agiven by
~
~Force eE
a=--=--
Mass m
where mis the mass ofan electron. Asthe electrons
acelerate, theyfrequently collidewith the positive
metal ions or other electrons of the metal. Between two
successive collisions, an electrongains a veloty
component (in addition to its random veloty) in a
~
direction opposite to E. However, the gain in veloty
lasts for ashort time and is lost in the next collision. At
eachcollision, the electron starts afresh with a random
thermal velocity.···2xy§n~m}r±n2lyv

201. 3.16
~
If an
electron having random thermal veloty u1
acelerates for time'1(before it suffers next collision),
then itwill attain aveloty,
Similarly, the veloties of the other electrons will be
~ ~ ~
v3=u3+a '3'....y
~
The averagevelocity vdof all the N electrons will be
~~ ~ ~ ~~
=(u1+a '1)+(u2 +a'2)+···+(uN +a 'N)
N
~ ~ ~
=u1+ ~ + ...+UN+-;'1+'2+ ...+'N
N N
~
=0+a,
where'=('1 +'2+·····+'N)/N is the average time
between two sucessive collisions.The average time that
elapsesbetween two successive collisions of an electronis
called relaxation time. For most conductors, it is of the
order of10-14 s. Theveloty gained by an electron
during this time is
~
~ ~ e E,
vd=a,=---.
m
~
The parametervdis called drift velocity of
electrons. Itmaybe defined as the average velocity gained
bythefree electrons of a conductor in the opposite direction
of the externally applied electric field.
It may be noted that although the electricfield
acelerates an electron between two collisions, yet it
does not produce any net aceleration. Thisisbecause
the electron keeps colliding with the positive metal
ions.The veloty gained by it due to theelectricfield is
lost in next collision. As a result, it acquires a constant
~ ~
average veloty vdin the opposite direction of E. The
motion of the electron issimilar to that of asmall
spherical metal ball rolling down a long flight ofstairs.
As the ball falls from one stair to the next, it acquires
aceleration due to the force of gravity. The moment it
collides with the stair, it gets decelerated. The net effect
is that after falling through a number of steps, theball
begins to roll down the stairs with zero average
aceleration i.e., atconstant average speed. Moreover,
PHYSICS-XII
as the average time r between two successive collisions
is small, anelectronslowly and steadily drifts in the
~
opposite direction of E, as shown in Fig. 3.18.
Drift
E
B B'
tt
,
,
,
,
~
,
,
t
r
,
Fig. 3.18Slowandsteady drift of an electron in the opposite
->
direction ofE.Thesolidlines represent the path in the
-> ->
absence of Eand dashed lines in the presence of E.
3.13RELATION BETWEEN ELECTRIC CURRENT
AND DRIFT VELOCITY: DERIVATION OF
OHM'S LAW
19. Derive relation between electric current and drift
velocity. Hence deduce Ohm'slaw.Also write the
expression for resistivity interms of number density offree
electrons and relaxation time.
Relation between electric rrent and drift veloty.
Suppose a potential difference Visapplied across acon-
ductor of lengthIand of uniform cross-sectionA.The
electric field E set up inside the conductor is given by
E=V
I
~
Under theinfluence of field E, thefree electrons
~
begin to drift in the opposite direction E with an
average drift velocity vd'
Let the number of electrons perunitvolume or
electron density =n
Charge on an electron =e
14 I ~I
E
Free electron
----+
-e -e
-e -e
--e -e
• ••
Conventional Electronic
current rrent
11-+
Battery
Fig. 3.19Drift of electrons and electric field inside a conductor.°°°2uv{lzkyp£l2jvt

202. CURRE
NT ELECTRICITY
Numberofelectrons inlength Iof the conductor
=nx volume of the conductor =nAl
Totalcharge contained inlength Iofthe conductor is
q=en Al
All the electrons which entertheconductor at the
right end will passthrough the conductor at the left
end in time,
:.Current,
distance I
t= =-
velocity "«
I=:J. =enAI
t1/vd
I=enAvdor
This equation relates the rrent Iwith the drift
veloty vd.
The rrent density' j ,is given by
.I
]=-=envd
A
7 ~
Invector form ] =envd
Theabove equation is validforboth positive and
negative values of .
Deduction of Ohm's law. When a potential
difference Vis applied across a conductor of length I,
the drift veloty in terms ofVisgiven by
eE't eV't
vd=-=-
m mi
Ifthe area of cross-section of the conductor isAand
the number of electrons perunitvolume or the electron
density of the conductor is n,thenthe current through
theconductor willbe
eV't
I=enAvd =enA.-
mi
or
V
I
mi
-2--'
ne'tA
Ata fixed temperature, the quantities m,I,n, e,tand
A,all have constant values foragiven conductor.
Therefore,
V
-=a constant, R
I
ThisprovesOhm's law for aconductor and here
R=~
ni'tA
is the resistance ofthe conductor.
Resistivity in terms of electron density and relaxation
time. Theresistance Rof a conductor oflength I,area of
cross-section Aandresistivity p is given by
I
R=p-
A
3.17
R=~
ni'tA
where risthe relaxation time. Comparing the above
twoequations, we get
But
m
P=ne2't
Obviously, pisindependent ofthe dimensions of
the conductor but depends on its two parameters:
1.Number offree electrons perunit volume or
electron density of the conductor.
2.The relaxation time r,theaverage time between
twosucessive collisions of an electron.
~ ~
20.Write relation between quantitiesj r0andE.
~ ~
Relation betweenj,0andE.For an electron,
q=-e
~
~ eE«
and vd=---
m
~
7 ~ ( eE't)ne2't ~
] =nqvd=n(-e) ----;;;- =-;;;- E
ne2't 1 ..
But--=- =0,conductivity of the conductor
m p
~ ~ ~ ~
j=0EorE=pj
This is Ohm's law in terms of vector quantities like
~ ~
current density jandelectricfieldE.
21.What causes resistance inaconductor?
Cause of resistance.Collisions are the basic cause of
resistance. When a potential difference is applied
across a conductor, its free electrons get acelerated.
On their way,theyfrequently collide with the positive
metal ionsi.e.,their motion is opposed and this
opposition to theflow of electrons is called resistance.
Larger the number of collisions per second, smaller is
the relaxation time r, and larger will be the resistivity
(p=m/ne2t).
Thenumber of collisions that the electrons makewith
the atoms/ions dependson thearrangement of atoms
orionsinaconductor. So the resistance dependson thenature
of thematerial (copper, silver, etc.) of the conductor.
Theresistance of a conductor depends on its length. A
long wire offers more resistance thanshort wire
because there will be more collisions in the longer wire.
The resistance of conductor depends on its area ofcross-
section. A thickwire offers less resistance than a thin
wire because in athick wire, morearea of cross-section
is available for the flow of electrons.···2xy§n~m}r±n2lyv

203. 3.18
22. Alloys of me
talshavegreater resistivity than their
constituent metals. Why?
High resistivity of nichrome. In an alloy, e.g.,
nichrome (Ni - Cr alloy), Ni2+ and Cr3+ ions have
different charge and size. They opy random locations
relative to eachother, though their ionicsitesform a
regular crystalline lattice. An electron, therefore, passes
through a very random medium and is very frequently
deflected. So there is a small relaxation time and hence
large resistivity. In general, alloys have more resistivity
than that of their constituent metals.
23.Explain the cause of instantaneous current in an
electric circuit.
Cause of instantaneous rrent. Although the drift
speed of electrons is verysmall, typically 1 mm/s, yet
an electric bulb lights up as soon as we turn the switch
on. This is because electrons are present everywhere in
an electric rit. When a potential difference is
applied to the rcuit, an electric field is set up through-
out the rit, almost with the speed of light. Electrons
in every part of the rit begin to drift under the
influence of this electric field and a rrent begins to
flow in the rit almost immediately.
The above situation is analogous to the flow of
water in a long pipe. As soon as the pressure is applied
at one end of the water filled pipe, a pressure wave is
transmitted along the pipe with a speed of about
1400 ms-1.When this wave reaches the other end,
water starts flowing out. But water inside pipe moves
forward with a much smaller speed.
Formulae Used
1.Current in terms of drift veloty (vd)isI=enAvd
2.Current density, j =envd
3.No. of atoms in one gram atomic mass of an
element, N= Avogadro's number = 6.023 x1023.
4.In terms of relaxation time r,
R=~ and p=~
ne2'tA ne2't
5.Relation between rrent density andelectric field,
j=crE or E=pj
Units Used
Driftvelocity vdisin ms -1,free-electron density in
m-3,cross-sectional areaAin m2,rrent density
jin Am-2,allresistances in n.
Constants Used
e=1.6x10-19C andNA= 6.023 x 1023mol-1.
PHYSICS-XII
Example 27. Assuming that there isone free electron per
atom incopper, determine the number of free electrons in
1 metre3 volume of copper. Density of copperis
8.9x103kgm-3 and atomic weight63.5. (Avogadro's
number, N= 6.02 x1026per kg-atom).
Solution. If the atomic weight of a material isMkg
and the density isdkgm-3,then the volume of its
1 kg-atom will be(Mid) m3.
Acording to Avogadro's hypothesis, there are
6.02x1026atoms in 1 kg-atom of the material. This
number is called Avogadro's number(N).Thus
Number of atoms in(Mid)m3volume of a material
=N
.. Number of atoms in 1 m3 volume
N dxN
=--=--
Mid M
Assuming 1 free electron per atom in copper, the
number of free electrons in 1 m3volume of copper will be
dxN
n=--
M
Nowd=8.9 x103kg m-3, N=6.02 x1026,
M=63.5 kg
8.9x103x6.02x1026 28 3
n= =8.4x10 m-
63.5
Example 28. Acopper wirehasa resistanceof10 nand an
area of cross-section 1m~.A potential diference of 10V
existsacross the wire. Calculate the drift speed of electrons if
thenumber of electrons per cubic metreincopperis
8x1028electrons. [CBSE D 96]
Solution. Here R= lOn, A=lmm2=10-6m2,
V= 10V,n=8x1028electrons Im3
Now I=enAvd
V
-=enAvd
R
V 10
v - -- -------::-:0:-------::-;;:----,--
d -enAR-1.6x10-19 x8x1028x10-6 x10
= 0.078x10-3ms-1 =0.078mmS-l.
Example 29. (a)Estimate the average drift speed of
conduction electronsina copper wire of cross-sectional area
1.0x10-7 ~,carrying a current of1.5 AAssume that each
copper atom contributes roughly one conduction electron.
Thedensity ofcopperis 9.0x103kg m-3, anditsatomic
massis 63.5 u.Take Avogadro's number =6.0x1023mol-I.
(b) Compare the driftspeed obtained above with
(i)thermal speeds of copper atoms at ordinary temperatures,
(ii)speeds of electrons carrying the current and(iii)speed of
propagation of electric field along the conductor which
causes the drift motion. [NCERT]
or°°°2uv{lzkyp£l2jvt

204. CURRENT ELECTRICITY
Solution.Massof 1
m3of Cu
= 9.0x103kg =9x106g
Since Avogadro's number is 6.0 x1023and atomic
massof Cu is 63.5 u, therefore, 63.5g ofCu contains
6.0x1023atoms.
So 9x106g of Cu contains
60x1023
. x9x106atoms =8.21028atoms
63.5
Number of conduction electrons,
n= number of Cu atoms =8.5 x1028
Now1=1.5 A,A=10-7m2, e=1.6xlO-19C
I 1.5
v-------::-;,-----:::.,,-----:=-
d-enA -1.6x10-19 x8.5x1028x10-7
15 =11 10-3 -1
•x ms.
16x85x10
(b)(i)Atanytemperature T,the thermal speed of a
copper atom of mass M is given by
_tkBT
vrms - M
Butordinary temperature, T"'-300 K,
Boltzmann constant, kB= 1.38x10-23JK-I,
Mass of a copper atom,
M _ 63.5
-6.0x1023g
63.5x10-3k
6.0x1023 g
3x1.38x10-23x300x6.0x1023
63.5x10-3
=.J117354.33 =342.57 ms-1
From part (a),drift speed of electrons,
vd= 1.1x1O-3ms-1
vd(electrons) 1.1x10-3 6
---'''------- = =3.21x10- .
vrms(Cu atoms) 342.57
(ii) The maximum kinetic energy ..!mv~of electron
2
in copper corresponds to a temperature,
To=105K
1 2
-mVF =kBT
2
or
VF=)2kBT= 2x1.38x10-23 x105
m 9.1x10-31
=1.74x106ms-1.
vd(electron) = 1.1x 10-3"'-10-9.
VF(electron) 1.74 x106
3.19
(iii)An electric field propagates along a conductor
with the speed of an electromagnetic wave i.e.,
3x108ms-1.
vd(electron) 1.1x10-3
speed of propagation ofelectric field 3 x108
"'-10-11•
Example 30. Calculate the electric fieldina copper wire of
cross-sectional area2.0m~carrying acurrent of1AThe
conductivity of copper =6.25x107Sm-1.
Solution. HereA=2.0 mm2 =2.0x 1O-6m2,
I=1 A,o=6.25x107Sm-1
As j=~=crE
A
E=_1_= 1
Acr2.0x10-6x6.25x107
=8x10-3Vm-1.
Example 31.A potential diference of100Visapplied to
the ends of acopper wire one metre long.Calculate the
averagedriftvelocity of the electrons. Compare it with the
thermalvelocity at27°C. Givenconductivity of copper,
c= 5.81x107~r1m-1and number density ofconduction
electrons, n=8.5x1028m-3. [NCERT]
Solution. Electric field,
E=V=100V =100Vm-1
I1m
As j=crE=envd
:.Drift speed,
crE 5.81x 107x100
v=-=
de n1.6 x10-19 x8.5x1028
=0.43 ms-1.
kB= 1.38x1O-23JK-1,
T= 27+273 = 300 K
me= 9.1x10-31kg
Thermal veloty of electron at 27°C,
3x1.38x10-23 x300
9.1x10-31
Now,
~ 0.43=3.67x10-6•
vrms 1.17x105
Example 32. Find the timeof relaxation between collision
and free path ofelectrons in copper at room temperature.
Given resistivity of copper = 1.7x10-8Om,number density
of electrons incopper=8.5x1028m-3, chargeon electron
= 1.6x10-19 C,mass of electron= 9.1x10-31kg and drift
velocity of free electrons = 1.6x10-4ms-1.···2xy§n~m}r±n2lyv

205. 3.20
Solution. Here
p = 1.7 x10-8nm,n= 8.5x1028m -3,
e=1.6x 1O-19c, me=9.1x 10-31 kg,vd=1.6x10-4 ms-l.
m
As resistivity, p=_e_
ne21"
:.Relaxation time,
m 9.1x10-31
1" -__e_---------:;;c-;;,-------;;;;-------;;-
-e2np - (1.6 x10-19)2 x8.5x1028x1.7x10-8
=2.5x10-14s
Mean free path of electron
=vit= 1.6x10-4x2.5x10-14
= 4.0x10-l8m.
Example33.An aluminiumwireofdiameter 0.24emis
connectedinseries to a copper wire of diameter 0.16em The
wires carry an electric current of10ampere. Find
(i) current-density inthealuminium wire (ii)drift velocity
of electronsinthe copper wire. Given: Number of electrons
per cubic metre volume of copper =8.4x1028.
Solution. (i)Radius of Al wire,
r=0.24 =0.12 em =0.12 xlO-2m
2
Area of cross-section,
A=1t?=3.14x(0.12x10-2)2 = 4.5x10-6m2
:.Current density,
.=.i= 10 =2.2x106Am-2.
] A 4.5x10-6
(ii)Area of cross-section of Cu wire is
A=1tX(0.08x10-2)2 =2.0x10-6m2
Also,
n= 8.4x1028m-3, e=1.6x 10-19C,I=10 A
I 10
v ----------:;-;::----~_;:_---__,_
..d -enA -1.6x10-19x8.4x1028x2.0x10-6
=3.7x10-4ms-l.
Example34.A current of 1.0 ampereisowing through a
copper wire of length 0.1 metre and cross-section
1.0x1O-6~. (i) If the specific resistance of copper be
1.7x10-8nmcalculate the potential diference across the
ends of the wire. (ii)Determine current density in the wire.
(iii)If there be one free electron per atom in copper, then
determine the drift velocit¥ of electrons. Given : density of
copper =8.9x103kgm-, atomic weight ='63.5,
N= 6.02x1026per kg-atom.
Solution. HereI=1.0A,1 =0.1 m,A=1.0x10-6m2,
p=l.7x 1O-8nm, d=8.9x103kg m-3
(i)Resistance of wire is
R=pl 1.7x10-8xO.1=1.7x10-3n
A LOx 10-6
PHYSICS-XII
:. Potential difference,
V=IR= 1.0x1.7x10-3=1.7x10-3V.
(ii)Current density,
. =.i= 1.0 = 1.0 x106Am-2.
] A 1.0x10-6
(iii)Free-electron density,
dxN8.9x103x6.02x1026
n=-- =---------
M 63.5
=8.4x1028m-3
:.Drift veloty,
j 1.0x106
V=-=-----;r;------::;o
den1.6x10-19x8.4x1028
=7.4x10-5ms-l•
~roblems ForPractice
1.The free electrons of a copperwire of cross-
sectional area 10-6 m2acquire a driftveloty of
10-4m/s when a certain potential difference is
applied across the wire. Find the rrent flowing in
the wireifthe density of free electrons in copper is
8.5x1028electrons/m '. (Ans.1.36 A)
2.Estimate the average drift speed of conduction
electrons in acopper wire of cross-sectional area
2.5x10-7m2carrying a rrent of 2.7 A. Assume
the density of conduction electrons to be
9x1028m-3. [CBSE OD 141
(Ans. 0.75 mms ")
3.A rrent of 1.8 A flows through a wire of cross-
sectional area 0.5 mm 2. Find the rrent density in
the wire. If the number density of conduction
electrons in the wire is 8.8x1028m-3,find the drift
speed of electrons.
(Ans.3.6x106Am-2,2.56 x10-4ms-l)
4..The resistivity of copper at room temperature is
1.7x1O-80m. If the free electron density of copper
•is 8.4 x1028m-3,find the relaxation time for the
free electrons of copper. Givenme= 9.11x10-31kg
ande= 1.6x10-19C. (Ans. 2.49x1O-14s)
5.A copper wire of diameter 1.0 mm carries a'rrent
of 0.2 A. Copper has 8.4 x1028atoms per bic
metre. Find the drift veloty of electrons, assuming
that one charge carrier of 1.6 x 1O-19C is assoated
with each atom of the metal. [ISCE 971
(Ans.1.895x10-5rns")
6.A rrent of 2 A is flowing through a wire of length
4 m and cross-sectional area 1 mm2. If each bic
metre of the wire contains 1029 free electrons, find
the averagetimetaken by an electron to cross the
.length of the wire. (Ans.3.2x104s)www3xy°o«n¥s•o3myw

206. CURRENT ELECTRICITY
7.A10
Cof charge flows through a wire in 5minutes.
The radius of the wire is 1 mm. Itcontains 5 x1022
electrons percentimetre '.Calculate therrent
and driftvelocity.
(Ans. 3.33x10-2 A,1.326 x10- 6ms-1)
8.Acopperwire ofdiameter 0.16em is connected in
series to an aluminium wire of diameter 0.25 . A
rrent of 10 A is passed through them. Find
(i) rrent density in the copp wire (ii) drift
velocity offreeelectrons in thealuminium wire.
The number of free electrons perunit volume of
aluminiumwire is1029m- 3.
(Ans. 4.976x106 Am-2, 1.28x10-4ms-l)
9.Acurrent of30ampere is flowing through a wire of
cross-sectional area 2 mm 2.Callate the drift velo-
city of electrons. Assuming the temperature of the
wire to be 27°C, also calculate thermsvelocity at
this temperature. Which veloty is larger? Given
that Boltzman's constant = 1.38 x 10-23J K-1, den-
sityof copper 8.9g- 3,atomic mass of copper
=63. (Ans.1.1x1O-3ms-l, 1.17x105ms-1)
10.What is thedriftveloty of electrons in silver wire of
length 1 rn, having cross-sectional area 3.14 x10-6m 2
and carrying a rrentof10 A?Given atomic mass
of silver =108,densityofsilver =10.5x103kg m-3,
charge on electron = 1.6 x10-19C and Avogadro's
number = 6.023x1026per kg-atom.
(Ans. 3.399x1O-4ms-1)
11.When a potential difference of 1.5 V is applied
across awire of length 0.2 m and area of cross-
section 0.3 mm2,a rrentof 2.4 A flows through
the wire. If the number density of free electrons
in thewire is8.4x1028m -3,callate theaverage
relaxation time. Given that mass ofelectron
= 9.1 x10-31kg and charge on electron
= 1.6 x10-19e. (Ans.4.51x10-16s)
HINTS
1.I=enAvd =1.6x10-19 x8.5x1028x10-6x10-4
=1.36A.
I 2.7 -1
2.vd=enA= 1.6 x10-19 x9x1028 x2.5x10-7ms
= 0.75 x10-3ms-1 = 0.75 mms-1.
I 1.8A
3.Current density, j= -= 6 2
A0.5x10 m
= 3.6x106Am-2.
Drift speed,
v=1.=. 3.6x106
den1.6xIQ-19 x8.8 x1028
= 2.56 x10-4 ms ",
3.21
4.Relaxation time, r-..!!!L
-e2np
9.11x10-31
-(1.6x10 19)2x8.4x1028 x1.7x10 8
=2.49x10-14 s.
5.Diameter ofwire, D= 1.0 mm = 10-3 m
Area of cross-section,
A=nd=nx(10-3)2 = 7.854 x10-7m2
4 4
1 0.2
vd=-en-A=-1-.6-x-1-0......,,19O-x-8-.4-x-1-0""'i2Q8 -x-7-.8-5-4-x-1-0-"7
= 1.895x10-sms-t.
6.Driftvelocity, vd=_l_
enA
2
7.
1.6x10-19x1029x1x10-6
=1.25 x10-4ms-l.
R .d. I 4 4
eqUlre time, t=-= 4=3.2x10 s.
vd1.25x10-
I=!1.=~ = 3.33x10-2A
t5x60s
1 1
vd=enA=en(nr2)
3.33x10-2
-1.6x10 19 x5x1022x106x3.14x(103)2
= 1.326x10-6 ms-l.
8.Asthe two wires are connected in series, sorrent
through eachwire, I= 10A.
(i)Current density incopperwire,
. I 10x4
]=nd/4= 3.14x(0.16xlO 2)2
=4.976x106Am-2.
(ii) Area of cross-section of aluminium wire,
nd3.14x(0.2S x10-2)2
A=-=---'-------'-
4 4
=4.9x10-6m2
1 10
vd=-en-A=-1-.6-x-10-;Ot9"-x -10....,2;n9-x-4-.9-x-1-0-,6
=1.28x10-4 ms-t.
9.No.of atoms in 63 gram of copper = 6.023 x1023
No.of atoms in 8.9 gram or 1em!of copper
6.023x1023x8.9
63
No.of atoms per m 3of copper
6.023x1023x8.9x106
63···2xy§n~m}r±n2lyv

207. 3.22
Electron densi
ty,
6.023 x1023x8.9x106 88 028 -3
n= =.4xIm
63
AlsoI=30A, A=2mm2 =2x10-6m2,
e=1.6x1O-19C
:.Drift veloty,
I 30
v=-=--~;n----",-----,
denA 1.6 x10-19x8.48x1028x2x10-6
=1.1x10-3ms-1.
Thermsvelocity of electrons at 27°C (= 300K)is
given by
v= ~3kBT=3x1.38x10-23
rms m 9x10-31
=1.17x105ms-1
The rms veloty is about 108 times the driftveloty.
10.Mass of silver wire,
m=AlP=3.14x10-6x1x10.5x103
No.of electrons per unitvolume of silver,
6.023x1023 3.14x10.5xio-3
n= x-----,---
108 3.14 x10-6xl
= 5.8557 x1028
I
v-
d-enA
10
= 1.6 x10-19x5.8557x1028x3.14x10-6
=3.399x10-4ms-1
11.E=V= 1.5 V = 7.5 Vm-1.
I0.2m
Current density,
j=~= 2.46=8x106Am-2.
A0.3x10
ne2 't
Asj=crE=-- E
m
m.j 9.1x10- 31 x8x106
't--- -------".,,-------,-,,-,,--
- ne2 E - 8.4x1028x(1.6x10 19)2x 7.5
=4.51x10-16s.
3.14MOBILITY OF CHARGE CARRIERS
24.Define mobility of charge carrier. Write relations
between electric current and mobility for (i) a conductor
and(ii)a semiconductor. Hence write an expression for
the conductivity of a semiconductor.
Mobility. The conductivity of any material is due to
its mobile charge carriers. These may be electrons in
metals, positive and negative ions in electrolytes; and
electrons and holes in semiconductors.
PHYSICS-XII
Themobilityof a charge carrier isthe drift velocity
acquired byit ina unit electric field. Itis given by
v
I--l=-.fL
E
qE't
Asdrift veloty, vd=--
m
vd 't
I--l=-=q-
Em
er
I--l=_e
em
e
Foran electron,
For ahole,
Themobilities of bothelectronsand holes are
positive; although their drift veloties are opposite to
each other.
51unit of mobility = m2V-1s-1
Practical unitof mobility = 2V-Is-I.
1m2V-1s-1 =104 cm2V-1s-1
Relation between electric rrent and mobility for a
conductor
Inametallic conductor, the electric rrentisdue to
itsfree electrons and is given by
I=enAvd
But vd=I--leE I=enAl--le E
This is the relation between electric rrent and
electron mobility.
Relation between electric rrent and mobility for a
semiconductor
The conductivity of a semiconductor is both due to
electrons and holes. So electric current ina semi-
conductor is given by
I=Ie+Ih=enAve +epAVh
.=enAl--leE+ epAl--lhE
=eAE(nl--le+Pl--lh) ...(i)
wherenandParetheelectron and hole densities of the
semiconductor.
Conductivity of a semiconductor.Acording to
Ohm's law,
I=V=~=EA
Rpl/ A P
From equations (i) and(ii), weget
EA
-=eAE(nl--le+Pl--lh)
p
1
or -=e(nl--le +PI--lh)
p
But 1/ p is the electrical conductivity cr.Therefore,
cr=e(nl--le+Pl--lh)
...(ii)···2xy§n~m}r±n2lyv

208. CURRENT ELECTR
ICITY
Table 3.3 Mobilities in some materials at room
temperature, in cm2v-1s-1
Materials Electrons Holes
Diamond 1800 1200
Silicon 1350 480
Germanium 3600 1800
InSb 800 450
GaAs 8000 300
Formulae Used
1. Mobility, 11=vd=q't
Em
2.Electric rrent, 1=enAvd=enAIlE
3.Conductivity of metallic conductor, (J=nelle
4. Conductivity of a semiconductor, (J=nelle+pellh
Units Used
Conductivity (JisinSm-1 and mobility 11·in
m2V-1s-1.
Example 35. A potential diference of6Visapplied across
a conductor of length 0.12m Calculate thedriftveloci~ of
electrons, ifthe electron mobility is5.6x10-6~V-1S- .
Solution. Here V=6 V, I=0.12 m,
11=5.6x10-6~V-1s-1
Drift veloty,
_E- V_5.6x10-6x6 -1
vd-Il- Il .- - ms
I 0.12
=2.8x10-4 ms-1•
Example36.The number density ofelectrons in copper is
8.5x1028m-3. Determine the current owing through a
copper wire of length 0.2m, area of cross-section 1m~,
when connected to a batteryof3V.Given the electron
mobility=4.5x1O-6~V-lS-l and charge on electron
=1.6x10-19C.
Solution. Here n=8.5x1028m-3, I=0.2 m,
A=lmm2=10-6m2, V=3V,
1l=4.5xlO-6 m2V-1s-1, e=1.6x10-19c.
Electric field set up in the copper wire,
E=V=2..=15Vm-1
I0.2
Current,
1=enAIlE
= 1.6x10-19x8.5x1028x10-6x4.5x10-6x15
=0.918 A
3.23
Example 37. A semiconductor hasthe electron concen-
tration0.45x1012m-3and hole concentration 5x1020m-3.
Find its conductivity. Given : electron mobility
=0.135 ~V-ls-l andhole mobility =0.048 ~V-ls-l ;
e=1.6x10-19 coulomb.
Solution. Here n= 0.45x1012m -3,P= 5x1020m -3,
Ile=0.135 m2V-1s-1,llh =0.048 m2V-1s-1
Conductivity of the semiconductor is
(J=e(nlle +PilI!)
=1.6x10-19 (0.45x1012x0.135
+5x1020x0.048) Sm-1
= 1.6x10-7(0.06075+0.24x108) Sm-1
= 1.6x10-7x0.24x108Sm -1 =3.84 Sm-1.
flroblems For Practice
1.A potential difference of 4.5 V is applied across a
conductoroflength 0.1 m. Ifthe drift veloty of
electrons is1.5x10-4ms-1,find the electron
mobility. (Ans.3.33xlO-6m2V-1s-1)
2.Thenumber density of electrons incopper is
8.5x1028m -3.A rrentof 1 A flows through a
copperwire of length 0.24mandarea of
cross-section1.2 mm2, when connected to a battery
of3 V. Find the electron mobility.
(Ans.4.9x1O-6m2V-1s-1)
3.Mobilities of electrons and holes in a sample of
intrinsic germanium at room temperature are
0.54m2V-1s-1 and 0.18m2V-1s -1 respectively. If the
electron and hole densities are equal to
3.6x1019m-3,calculate the germanium conductivity.
[BITRanchi1997j(Ans.4.147Sm-1)
HINTS
1.E=V= 4.5 V =45 Vm-1.
IO.lm
v1.5x10-4ms-1 6 1
11=...!l. = =3.33x10- m2V-1S- •
E 45 Vm-1
V
2.I=enAIlE=enAll. -
I
II 1x0.24
:.11=enAV=1.6x10-19x8.5x1028x1.2x10-{ix3
=4.9x10-6m2V-1S-1.
3H 0542V-1 -1 018 2V-1-1
.ere11e=. m s, 11 h=. m s,
n=p=3.6x1019m-3
Conductivity,
(J=e(nile+P11h)=en(11e+11h)
=1.6x10-19x3.6x1019(0.54+0.18)
=4.147 Sm-1.···2xy§n~m}r±n2lyv

209. 3.24
3.
15TEMPERATURE DEPENDENCE OF
RESISTIVITY
25. Explain the variation of resistivity of metals,
semiconductors, insulators and electrolytes with the
change in temperature. Define temperature coeficient of
resis tivi ty.
Temperature dependence of resistivity.The resisti-
vity of any material depends on the number densityn
of free electrons and the mean collision time1:.
m
P=--
ne21:
1.Metals. For metals, the number densitynof free
electrons is almost independent of temperature. As
temperature increases, the thermal speed of free elec
trons increases and also the amplitude of vibration of
the metal ions increases. Consequently, the free elec
trons collide more frequently with the metal ions. The
mean collision time1:decreases. Hencethe resistivity of a
metal(poc1/1:)increases and the conductivity decreases
with the increase in temperature.
For most of the metals, resistivity increases linearly
with the increase in temperature, around and above
the room temperature. In such cases, resistivityPat
any temperatureTis given by
P=Po [1+a.(T - To)] ...(1)
wherePois the resistivity at a lower reference
.temperatureto(usually 20° qanda.is the coeffient
of resistivity. Obviously,
P-P
a.= 0
Po(T - To)
1dp
Po'dT
Thus, thetemperature coeficient of resistivitya.
may be defined as the increase in resistivity per unit
resistivity per degree rise in temperature.
The unit of a. is°C-1. For metals a. is positive. For
many metallic elements, a. is nearly 4x10-3°C-1. For
such conductors, the temperature dependence of p at
low temperatures is non-linear. At low temperatures,
the resistivity of a pure metal increases as a higher power
of temperature, as shown for copper in Fig. 3.20(a).
E
~
a:1.20
eo
I;::
a: .c
:~ 1.
.;;;
~
1.00
~
0.4
(a)
o 50 100 150
TemperatureT(K)....•
200 400 600 800
TemperatureT (K)....•
Fig. 3.20(a)Variation of resistivity p of copper with
temperature.(b)Variation of resistivity p of
nichrome with temperature.
PHYSICS-XII
Alloys have high resistivity. The resistivity of nich-
rome has weak temperature dependence [Fig.3.20(b)]
while that of manganin is almost independent of
temperature. At absolute zero, a pure metal has negli-
gibly small resistivity while an alloy (like nichrome)
has some residual resistivity. This fact can be used to
distinguish a pure metal from an alloy.
I
As R =p- i.e.,Rocp
A
Thus equation (1) can be written in terms of resis-
tances as
(b)
R,=Ro (1+a.t)
whereRt= the resistance attOC
Ra= the resistance at O°c, and
t= the rise in temperature.
2. Semiconductors and insulators. In case of insu-
lators and semiconductors, the relaxation time1:does
not change with temperature but the number density
of free electrons increases exponentially with the
increase in temperature. Consequently, the conductivity
increases or resistivity decreases exponentially with
the increase in temperature.
The number density of electrons at temperatureTis
given by
(T)_ -Eg/kBT
n - no e
wherekBis the Boltzmann constant andEgis the
energy gap (positive energy) between conduction and
valence bands of the substance.
1 .
Aspoc-,so we can wnte
n
_1_ = ~e-Eg /kBT
p(T)Po
E/k T
or p (T)=PoegB
This equation implies that the resistivity of semicon-
ductors and insulators rapidly increases with the decrease
in temperature, becoming infinitely large asT ~O.
At room temperature,kBT=0.03 eV. Whether the
non-conducting substance is an insulator or a semi-
conductor, depends on the size of the energy gap,Eg:
(i)If E::;1 eV, the resistivity at room temperature
is ~ot very high and the substance is a
semiconductor.
(ii)If E>1 eV, the resistivity at room temperature
is v~ry high (-103 nm) and the substance is an
insulator.
The coeffient of resistivity (u) is negative for
carbon and semiconductorsi.e.,their resistivity
decreases with temperature, as shown in
Fig.3.21.···2xy§n~m}r±n2lyv

210. CURRENT ELECTRIC
ITY
f
p
Fig. 3.21Resistivity of a semiconductor decreases
rapidly with temperature.
3.Electrolytes. Asthe temperature increases, the
interionic attractions (solute-solute, solvent- solute and
solvent-solvent types) decrease and also the viscous
forces decrease, theions movemore freely. Hence
conductivity increases or the resistivity decreases as
the temperature of an electrolytic solution increases.
26.Why alloys like constantan or manganin are used
for making standard resistors?
Use of alloys in making standard resistors. Alloys
like constantan ormanganin areused formaking stan-
dard resistance coils because of the following reasons:
(i)These alloys havehighvalue of resistivity.
(ii)Theyhavevery small temperature coeffient.
Sotheir resistance does not change appreciably
even for several degrees rise of temperature.
(iii)They are least affected by atmospheric
conditions like air, moisture, etc.
(iv)Their contact potential with copper is small.
Examples based on
Tern erarure Variarion of Resisrance
Formulae Used
Temperature coeffient of resistance
a= ~-~
~ (t2-t1)
Ht1 =O°Candt2 =tOC, then
R,-R
a= 0orR,=~(1+at)
Roxt
Units Used
Resistancesare in 0, temperatures in°Cor K.
Example38.(i) Atwhat temperature would the resistance
ofacopper conductor be double its resistance at O°C?
tii)Doesthistemperature hold for all copper conductors
regardless ofshape and size?
Given afor Cu =3.9·x 10-3°C-1.
R-R 2R-R 1
Solution. (i)a=''2 1= 0"0 =-
R1(t2-t1) Ro(t-O) t
3.25
t=2.= 1 =2560C
a3.9x10-3
Thus the resistance of copper conductor becomes
double at256°C
(ii)Sinceadoes notdepend onsize andshape of
the conductor, so the above result holds for all copper
conductors.
Example 39. The resistance of the platinum wire of a
platinum resistance thermometer at the icepointis50and
atsteampoint is5.39O.When the thermometer isinserted
in ahotbath,the resistance of theplatinum wireis5.975O.
Calculate thetemperature of the bath. [TERT]
Solution. HereRo=50,RlOO=5.230,Rt= 5.7950
As R,=Ro(1+at)
Rt-Ro=Roat
and RlOO- Ro=Roax100
On dividing (i)by(ii),we get
Rt-Ro=.L.
RlOO -Ro100
R-R
t= t"0x100
R100 -Ro
=5.795-5x100 = 0.795 x100 =345.650C
5.23 -5 0.23
... (i)
...(ii)
or
Example40.Anichrome heating element connected toa
220Vsupply draws an initial current of2.2Awhich settles
down after afew seconds to asteady value of 2.0 A. Find the
steady temperature of theheating element. Theroom
temperature is30°Cand the average temperature coeficient
of resistance of nichrome is1.7x10-4per°C.
Solution. HereV=220V,II=2.2A,I2=2.0A,
a=1.7x1O-4°C-1
Resistance at room temperature of30°C,
R=V=220 =1000
1II2.2
Resistance at steady temperature,
~=V=220=1100
122.0
R -R
a=''2 1
Rl(t2-t1)
__~-Rl _110-100 -5880C
t2tl- - -
Rla 100x1.7x10-4
As
Steady temperature,
t2=588+tl=588+30 =618°C.
Example 41. Anelectric toaster uses nichrome (an alloy of
nickel and chromium) for its heatingelement. When a
negligibly smallcurrent passes through it, itsresistance at···2xy§n~m}r±n2lyv

211. 3.26
room temperature (27.0°
C)isfound to be75.3O.When the
toaster isconnected to a 230 Vsupply, thecurrent settles
after afew seconds to a steady value of2.68A.Whatisthe
steady temperature of the nichrome element?The t'em-
perature coeficient of resistance of nichrome averaged over
the temperature range involved is1.70x10-4°C-1.
[NCERT]
Solution.HereR1=75.3 0, t1=27°C
R=230 = 85.8 0t-?
"22.68 '2 -.
__ ~-R1_ 85.8 -75.3 -8200C
t2 t1 - - -
R1a 75.3 x1.70x10- 4
Steady temperature,
t2'"820+t1=820+27 =847°C.
Atthe steady temperature, the heating effect due to
the rrent equals heat loss to the surroundings.
Example 42.Theresistance of a tungsten filament at
150°C is133ohm.What will be its resistance at500°C?
Thetemperature coeficient of resistance of tungsten is
0.0045periC.
Solution. HereR1S0=1330, a=0.0045°C, Rsoo=?
or
R,=Ra(1+at)
R1S0= Ra(1+ax150)
133 = Ro (1+0.0045x150)
Rsoo =Ra(1+ax500)
Rsoo= Ra(1+0.0045x500)
Dividing(2)by(1),we get
Rsoo= 1+0.0045x500 = 3.25
133 1+0.0045x150 1.675
Rsoo =3.25x133 =258O.
1.675
Now
and
or
or
Example 43. The resistance of a conductor at20°C is
3.150 and at100°C is3.75 O.Determine the temperature
coeficient of resistance of the conductor. What will be the
resistance of the conductor atO°C?
Solution.Rl=Ra(1+at1)and ~=Ra(1+at2)
Ondividing,
..&=1+atl
~ 1+at2
orR1(1 +at2)= ~(1 +at1)
a= ~-R1
Rlt2 --:~ t1
or
Heretl=20°C, Rl =3.150,
t2= 100°C, ~ = 3.750
PHYSICS-XII
3.75 -3.15
..a=---------
(3.15x100)-(3.75x20)
0.60 = 0.60 = 0.00250C-1•
315-75 240
R =~ 3.15 =3.00.
o1+atl1+0.0025x20
Example 44. Astandard coil marked 20isfound to have
a resistance of 2.1180at30°e.Calculate thetemperature at
whichthe marking iscorrect. The temperature coefficient of
the resistance of the material ofthe coil is0.0042°C-1.
Solution. Rl=Ra(1+at1)and ~=Ra(1+at2)
Rl=1+atl
~ 1+at2
Here,Rl =2 0, ~ =2.1180, t2=30°C, tl=?
2 = 1+0.0042xtl= 1+0.0042xtl
2.118 1+0.0042x30 1.126
or1+0.0042t= 2x1.126 = 2.252
1 2.118 2.118
••t=_1_ [2.252 -1]= 0.104 ::.150C
10.0042 2.118 0.0042 x2.118
...(1)
i.e.,the marking will be correct at15°C
Example 45. Apotential diference of200Visapplied to a
coilatatemperature of 15°Cand the current is10A.What
willbethemeantemperature of the coil whenthe current has
fallen to 5A,the applied voltagebeing same as before ?
Given a=_1_oC-1 atO°e.
234
Solution. In the second case, the rrent decreases
dueto the increase in resistanceon heating.
V200
Now R1S=I=10 =200
Lettbe the temperature at which current falls to
5A.Then
...(2)
or
R= 200 =400
t5
As Rf=Ra(1 +at)
R1S=Ra(1+~) or
234
Rt=Ra(1+2~4) or
Dividing(2)by(1),
2=234+t
249
t= 498-234 = 264°C.
...(1)
Rx249
20=_"-"----0 __
234
40 = Ro(234+t) ...(2)
234···2xy§n~m}r±n2lyv

212. CURRENT ELECTRICITY
Example46.Theres
istances of iron and copper wiresat
20°Care3.9 0and4.1 0respectively. Atwhat temperature
will the resistances be equal?Temperature coeficient of
resistivity for ironis5.0x10-3K-1and for copper itis
4.0x10-3K-1.Neglect any thermal expansion.
Solution. Let resistance of iron wire attOe
= Resistance of copper wire attOe
~o [1+a(t-20)]=R;o [1+a'(t-20)]
3.9[1 + 5.0 x10-\t-20)] = 4.1 [1 + 4.0x10-3(t -20)]
[3.9 x 5 - 4.1x 4]x10-3x(t-20) = 4.1-3.9
t-20 = 0.2 =64.5
3.1x10-3
t= 64.5 + 20 = 84.5°C.
Example47.A metal wire of diameter2mmand length
100mhas a resistance of0.54750at20°Cand0.805 0at
150°C. Find (i) the temperature coeficient of resistance
tii)resistance atO°C(iii)resistivities at0°and20°C.
Solution. Here r= 1 mm = 10-3m, 1=100m,
t1=20°C, R1=0.54750, t2=150°(, ~ =0.8050
(i)Temperature coeffient of resistance is
~ - R1 0.805 - 0.5475
a= =------
R1(t2 -t1)0.5475(150-20)
=3.6x10-30(-1.
(ii)Resistance at OO( is
R= ~ = 0.5475 = 0.5475
ol+at1 1+3.6xl0-3x20 1.072
= 0.51070.
(iii)Resistivity at O°c,
_ RaA _ Ra x11:? _0.5107x3.14x (10-3)2
Po --1- - I - 100
= 1.60x10-8Om.
Resistivity at 200( is
P20=Po(1+at)
= 1.60x10-8(1 + 3.6 x10-3 x20)
= 1.60x10-8x1.072 = 1.72x10-8Om.
j2)roblems ForPractice
1.A platinum wire has a resistance of 100 at Oo( and
of 20 0 at273°C. Find its temperature coeffient of
resistance. (Ans. _1_ 0C-1)
273
2.A standard coil marked 30 is found to have a true
resistance of 3.115 0 at 300 K.Callate the tempe-
rature at which marking is correct. Temperature
coeffient of resistance of the material of the coil is
4.2x10-3°e1. (Ans.290.2 K)
3.27
3.The resistance of a silver wire at O°Cis 1.25 O. Up to
what temperature it must be heated so that its
resistance is doubled? The temperature coeffient
of resistance of silver is 0.00375 °C-1. Will the
temperature be same for all silver conductors of all
shapes? (Ans. 26~ C, Yes)
4.The resistance of a coil used in a platinum-resis-
tance thermometer at O°C is 3.000 and at 100°C is
3.75 O. Its resistance at an unknown temperature is
measured as 3.15 O. Callate the unknown
temperature. (Ans.200q
5.The temperature coeffient of a resistance wire is
0.0012.soC-1. At 300K,its resistance is 10. At what
temperature the resistance of the wire will be 20 ?
[lIT 80]
(Ans.1127K)
6.The temperature coeffient of resistivity of copper
is 0.004°C-1. Find the resistance of a 5 m long
copper wire of diameter 0.2 mm at 100°C, if the
resistivity of copper at O°Cis 1.7 x10-8Om.
(Ans.3.80)
HINTS
1.a=Rf -Ra=20 - 10= ~0C1.
Raxt10x273 273
2.Heret=300-273=27°C
Rz7=Ra(1+ax27)
3.115= Ra(1+4.2x10-3 x27)
and 3 =Ra(I+4.2x10-3xt)
Dividing (2) by (1), we get
3 1+4.2xl0-3xt
3.115=1 + 4.2x10-3x27
...(1)
...(2)
5.
This gives,
t=17.2°C =17.2 + 273=290.2K.
3,Proceed as in Example 38, page 3.25.
4. Rf=Ra(1+at)
~oo=Ra(1+ ax100)
3.75=3.00 (1+ax100)
3.75_1=100a
3
a= ~ = 0.0025bC1
3x100
and
R -R
t=_f_"_O
Raxa
300K=300-273=27°C
Rz7=Ra(1 + ax27)=10
Rf=Ra(1+axt)=20···2xy§n~m}r±n2lyv

213. +1 +1
Pure Cu
S04solution
V
metal with Cuelectrodes
Vo
Fig.3.24V-Igraphfor a watervoltameter.
-V +V
(iii)p-njunction diode. It consists of a junction of
p-type and n-type semiconductors (For details, refer to
-I -I
+1
3.28
1+at2
..
1+27a 1
or 1+at=2+54a
t =1+54a=1+54x0.00125=8540C.
a 0.00125
=854+273=1127 K.
6.PlOO=Po(1+at)=1.7x10-8(1+0.004 x 100)
=2.38x1O-8Qm
I 238x10-8x5
R=p it?=3.14x(0.lx10 3)2':::3.BQ.
or
3.16LIMITATIONS OF OHM'S LAW :OHMIC
AND NON-OHMIC CONDUCTORS
27. State the conditions under which Ohm's law is
not obeyed in a conductor. What are ohmic and
non-ohmic conductors ?Give examples of each type.
Limitations of Ohm's law. Ohm's law is obeyed by
many substances under certain conditions but it is not
a fundamental law of nature.
Ohmic conductors. The conductors which obey Ohm's
law are called Ohmic conductors. For these conductors,
thelinear relationship between voltage and rrent
(Vex: I)holds good. The resistance (R=V /I) is
independent ofthe current Ithroughthe conductor. In
these conductors, the rrent Igetsreversed in direc
tion when the p.d.Visreversed, but themagnitude of
current changes linearly with voltage. Thus the V-I
graph for ohmic conductorsisastraight line passing through
the origin. A metallic conductor for small rrents and
the electrolyte like copper sulphate solution with
copper electrodes are ohmic conductors, as shown in
Figs.3.22(a)and(b)respectively.
Fig.3.22Ohmicconductors.
Non-ohmic conductors. The conductors which do not
obey Ohm's law are called non-ohmic conductors. The
resistance of such conductors is not constant even at a
given temperature, rather it is current dependent.
Non-ohmic situations may be of the following types:
(i)The straight line V-Igraph does not pass
through the origin.
PHYSICS-XII
(ii)V-Irelationship is non-linear.
(iii)V-Irelationship depends on the signofVfor
the same absolute value ofV,and
(iv)V-Irelationship is non-unique.
Examples of non-ohmic conductors.(i)Metallic
conductor. For small rrents, a metallic conductor
obeys Ohm's law and its V-Igraph is a straight line.
But when large rrents are passed through the same
conductor, it gets heatedup and its resistance increases.
V-Igraph no longer remains linear, i.e.,conductor
becomes non-ohmic at higher rrents, as shown in
Fig.3.23.
Fig.3.23V-Igraphfor a metallicconductor.
(ii)Water voltameter. Here a back e.m.f. is set up
due to the liberation of hydrogen at the cathode and
oxygen at the anode. No rrent flows through the
voltameter untilthe applied p.d. exceeds the back
e.m.f.Vo(1.67V for water voltameter). SoV-Igraph is
a straight line but not passing through the origin, as
shown in Fig. 3.24. Hence the electrolyte (water
adified with dil. ~S04) is a non-ohmic conductor.
+V
-1
Fig.3.25V-Igraphfor a junction diode.wwwbnotesdrivebcom

214. CURRENT ELECTRICITY
chapter14Vo
l.II).A voltageVis applied across the
junction. The resulting rrentIis shown in Fig. 3.25.
Obviously,Iis not proportional toV.Further, very
little rrent flows for fairly high negative voltage
(called negative bias) and a rrent begins to flow for
much smaller positive (forward) bias. Thus the junction
diode allows rrent to flow only in one directioni.e.,it
acts as a rectifier (converts a.c.into d.c.).
(iv) Thyristor. It consists of four alternate layers ofp
and n-type semiconductors. Its V-Irelationship is as
shown in Fig. 3.26. It can be easilyseen that(a)theV-I
relation is non-linear, (b)V-Irelationship is different
for positive and negative values ofV,and(c)in certain
portions, there are two or more values of rrent for
thesamevalueof voltage, i.e.,theV-Irelationship is
not unique. The regionABisinteresting because the
rrent carried by the device increases as thevoltage
decreases, i.e.,aisnegative in this region.
+v-v
-1
Fig. 3.26V-Irve for a thyristor.
(v) Gallium arsenide.Fig.3.27shows theV-Igraph
for thesemiconductor GaAs. Itexhibits non-linear
behaviour. Moreover, after a certain voltage, the
rrent decreases as the voltage increases. That is, if!1V
is positive thenMis negative and hence the effective
resistance(=!1V / M)isnegative.
I
(rrtA)"]
~
•...
;:J
u
•..
'"
~.§
,01)
c'"o •.•
Z
'"'"
:E ~s:
I r\l..a.J "b.ol
I~.~ Q)I
IZ ~ ~:
I
I
I
I
I
I
I
I
Voltage V(V) ~
Fig. 3.27 V-Igraph for GaAs.
3.29
3.17SUPERCONDUCTIVITY
28.What issuperconductivity ?Whatisitscause?
Superconductivity. In 1911,ProfKamerlingh Onnes at
the University of Leiden (Holland), observed that the
resistivity of merry suddenly drops to zero at a tempe-
rature of about4.2 Kand it becomes a superconductor.
0.16
t
a
~0.08
o 2 4 6
T(K)~
Fig.3.28Merry loses complete resistance at 4.2 K.
Thephenomenon of complete loss of resistivity by
certain metals and alloys when they are cooled below
acertain temperature iscalledsuperconductivity.
The temperature atwhich asubstance undergoes a
transition from normalconductor to superconductor in
azeromagnetic fieldiscalled transitionorcritical
temperature(Te).
A rrent once set up in asuperconductor persists
for a very long time without anyapparentchange inits
magnitude.
Cause of superconductivity. It isbelieved that near
the transition temperature, aweak attractive force acts
on the electrons which brings themcloser to form
coupled pairs. Such coupled pairs are not deflected by
ionic vibrations andso move without collisions.
29.WhatisMeissener effect insuperconductors ?
Meissener effect. In 1933, Meissner andOchsenfeld
observed that if a conductor is cooled in a magnetic
field to a temperature below the transition tempe-
rature, then at this temperature, the lines of magnetic
induction Bare pushed out of the spemen, asshown
in Fig. 3.29. Thus B becomes zero inside a supercon-
ductingspemen.
...•
B
Fig. 3.29 Meissner effect in a superconductor.···2xy§n~m}r±n2lyv

215. 3.30
The expu
lsion of the magnetic flux from asupercon-
ducting material when itiscooledtoatemperature
below the critical temperatureinamagnetic fieldis
called Meissner effect.
Meissner effect indicates that asthe supercon-
ductivity appears ina material, it becomes perfectly
diamagnetic.
30.Whatishigh Tc superconductivity ?Mention
important applications of superconductors.
HighTcsuperconductivity.A rrent onceset up in
a superconducting loop can persist foryearswithout
any applied emf. This important property of super-
conductors can have important practical applications.
A serious difficulty in their use is thevery lowtem-
perature at which they must be kept. Sentists all over
the world arebusyto construct alloys which would be
superconducting at room temperature. Superconduc-
tivity at around 125Khas already been achieved and
efforts arebeing made toimprove uponthis temperature.
Table 3.4 Critical temperatures of some
superconducting materials
Material Tc (K)
Hg 4.2
Pb2Au 7.0
YBa2Cu307 90
T12Ca 2Ba 2Cu3010 120
Applications of superconductors. The possible
applications of superconductors are
1.For produng high magnetic fields required for
research work in high energy physics.
2.For storage of memory in high speed computers.
3.In the construction of very sensitive galvano-
meters.
4.In levitation transportation (trains which move
without rails).
5.In long distance power transmissionwithout
any wastage of power.
3.18RESISTANCES IN SERIES AND PARALLEL
31.What do you mean by equivalent resistance of a
combination of resistances ?
Equivalent resistance of a combination of resis-
tances. Sometimes, a numberof resistances are
connected in arit inorder to get adesired value of
rrent in the circuit. Resistances can be connected in
series, in parallel ortheirmixed combinationcan be used.
If acombination of two or more resistances inany electric
circuitcan be replaced by a single resistance such that there
PHYSICS-XII
isno changeinthe current in the circuit and in the potential
diference between theterminals of the combination, then the
single resistance iscalled theequivalent resistance of the
combination.
32.When aretheresistances said to be connected in
series?Findanexpression forthe equivalent resistance of
a number of resistances connected in series.
Resistances in series.Ifa number of resistances are
connected end to endsothat the same current ows through
each one of them in succession, then they are said to be
connected in series. Fig.3.30shows threeresistances R1,
~ and ~connected inseries. Whenapotential dif-
ference Visapplied across the combination, the same
rrentIflowsthrough each resistance.
IE;:.:JI
v
Fig. 3.30Resistances inseries.
ByOhm's law,thepotential drops across the three
resistances are
VI=IR1, V2=I~, V3=l~
IfRsis the equivalent resistance of theseries
combination, then we must have
V=IRs
ButV=Sum of the potential drops across the
individual resistance
or V=VI +V2+V3
IRs=IRI+I~+l~
Rs=RI+ ~ + ~
or
or
The equivalent resistance ofnresistances connected
inseriewillbe
Rs=RI+ ~ + ~ + .....+ RII
Thuswhena number of resistances are connected in
series, their equivalent resistance isequal to the sum of the
individual resistances.
Laws of resistances in series
(i)Current through each resistance is same.
(ii)Total potential drop=Sum of the potential drops
across the individual resistances.
(iii)Individual potential drops aredirectly proportional
to individual resistances.
(iv) Equivalentresistance =Slimoftheindividual resistances.
(v)Equivalent resistance islarger than the largest
individual resistance.www3xy°o«n¥s•o3myw

216. CURRENT ELEC
TRICITY
33. When are the resistances said to be connectedin
parallel ?Find the equivalent resistance of a number of
resistances connectedinparallel.
Resistances in parallel. If a number of resistances are
connected in between two common points so that each of
them provides a separate path for current, then they are said
to be connected in parallel.Fig. 3.31 shows three resis-
tancesR1, ~ andR3connected in parallel between
points Aand B.LetVbe the potential difference
applied across the combination.
R1
B
v
Fig. 3.31 Resistance in parallel.
Let11,12and13be the rrents throughtheresis-
tances R1,R2andR3respectively. Then thecurrent in
themain rcuit must be I=II+12+13
Since all the resistances have been connected
between the same two pointsAandB,therefore, poten-
tial dropVis same across each of them. By Ohm's law,
the rrents through the individual resistances will be
V V V
11=- ,12=-,13=-
R1 ~ ~
IfRisthe equivalent resistance of the parallel
combin~tion, then wemust have
But
I=~
Rp
1=11 +12+13
V V V V
-=-+-+-
RpR1 ~ ~
1 1 1 1
-=-+-+-
Rp Rl ~ R3
or
or
The equivalent resistance Rpofnresistances
connected in parallel isgiven by
1 1 1 1 1
-=-+-+-+ .....+-.
RpRlR2 ~ Rn
Thus when a number of resistances are connected in
parallel, the reciprocal of the equivalent resistance of the
parallel combination isequal tothe sum of the reciprocals of
the individual resistances.
3.31
Laws of resistances in parallel
(i)Potential drop across each resistance issame.
(ii)Total current=Sum of the currents through
individual resistances.
(iii)Individual currents are inversely proportional to the
individual resistances.
(iv)Reciprocal of equivalent resistance =Sum of the
reciprocals of the individual resistances.
(v) Equivalent resistance isless than thesmallest
individual resistance.
E I b d•
." Combfnation of Resistances in
,.. . Series and Parallel
Formulae Used
1.The equivalent resistanceRsof a number of resis-
tances connected in series is given by
Rs=R,.+~+~+·..
2.The equivalent resistanceRpof a number of resis-
tances connected in parallel is given by
1 1 1 1
-=-+-+-+ ...
Rp R,.~ ~
3.For two resistances in parallel,
Currents through the two resistors will be
I= ~ IandI=R,.I
1R,.+~ 2R,.+~
Units Used
All resistances are in ohm(n).
Example 48.Awire of resistance 4Risbentintheformof
a circle (Fig. 3.32).What isthe efective resistance between
the ends of the diameter? [CBSE D 10)
2R
A~B
Fig. 3.32 Fig. 3.33
Solution.As shown inFig.3.33,the two resistances
of value 2 Reachare in parallel witheach other. So the
resistance between the ends Aand Bof a diameter is
R'=2Rx2R
2R+2R A
=R.
Example 49. Findthe value
of currentIinthe circuit
shown in Fig. 3.34.
[CBSE F 03, lIT 83)
2V
B 300 c
Fig. 3.34···2xy§n~m}r±n2lyv

217. 3.32
Solution. In
the given rit,the resistance of arm
ACB(30 + 30 = 600) is in parallel with the resistance of
armAB(=30 0).
Hence the effective resistance of the rit is
R = 30x60=20 0
30 +60
V 2
Current, I=-=-=0.1A
R20
PHYSICS-XII
c
A E
Fig. 3.36
the voltage drop acrossthe
given below with e=60V,
Example 50. Determine
resistor R}inthe circuit
RI=180,Rz=100.
Solution.Asthe resistancesR3andR4are inseries,
their equivalent resistance
= 5+10=150.
e=60V
Fig. 3.35
The series combination of R3andR4is in parallel
withRz.Their equivalent resistance is
R,=10x15 =150=60
10 + 15 25
The combination R'is inseries with RI.
.'. Total resistance of the rcuit,
R=6 + 18=240
I=~ = 60=2.5A
R24
Current,
•. Voltage drop acrossR}
=IRI=2.5x18V=45V.
Example 51.A letter A consists of a uniform wire of
resistance 1ohm per em.The sides of theletter are each 20em
long and the cross-piece inthemiddle is10em long while the
apex angle is60°.Find the resistance of the letterbetween the
two ends of the legs.
Solution.Clearly,
AB=BC=CD= DE= BD=10 em
R}=Rz= ~ =R4=Rs= 10 0
AsRzand ~ are in series, their combined resis-
tance = 10+10 = 20O.This combination is in parallel
withRs(=100).
Hence resistance 'between points Band D is
given by
1 1 1 3
-=-+-=- or
R 20 10 20
Now resistances RI,RandR4form aseries com-
bination. So resistance between the ends Aand E is
R'=10+20+10 =26.67 O.
3
Example 52.Aset of n identical resistors, eachofresis-
tanceR0,whenconnected inserieshave an effective resis-
tanceX 0andwhen the resistors are connected inparallel,
theireffective resistance isYO.Find the relation between
R,XandY.
Solution. The effective resistance of the nresistors
connected in series is
X=R+R+R+.....nterms =nR
The effective resistance Yof thenresistors
connected in parallel is given by
1 1 1 1 n
-=-+- +-+ ....nterms =-
YR R R R
Y=R
n
XY=nR.R=R2.
n
or
Example53.Aparallel combination of three resistors takes
acurrent of7.5A from a30V supply. Ifthetworesistors
are100and120,find the third one.
[Punjab 91; Haryana 94]
Solution. HereRp=V=~ =4 0
I7.5
1 1 1 1
But -=-+-+-
Rp R}».~
1 1 1 1
-=-+-+-
4 10 12 ~
1 1 11 1
---=-
~ 4 60 15
or
or ..~ =150.
Example 54.When a current of 0.5Aispassed through
tworesistances inseries,thepotential diference betweenthe
endsof the seriesarrangement is12.5V.On'connecting
them in parallel and passing a current of 1.5A, the potential
diference between their endsis6V.Calculate thetwo
resistances.°°°2uv{lzkyp£l2jvt

218. CURRENT ELECTRI
CITY
Solution.Forseries combination, V= 12.5 V, 1=0.5 A
Rl+ R2 = 12.5 =25.0 D
0.5
For parallel combination, V =6.0V,I=1.5A
Rp=Vor Rl Rz = 6.0 = 4.0
I Rl + Rz 1.5
or RlRz=4(Rl+Rz)=4x25=100
(Rl- Rz)2 = (Rl+Rzl-4 RlRz
=(25)2-4 x100 =225
Rl- Rz= 15 ...(2)
Solving (1) and (2), Rl = 20 D, Rz = 5 D.
Example55.Two square metal plates A andBare ofsame
thickness and material. ThesideofBistwice that ofA.These
are connected in series, asshown in Fig. 3.37.Find the ratio
RA /RBof the resistance of the two plates.
Fig. 3.37
Solution. LetIbe the side of thesquare plateAand
21that of square plateB.Letdbe the thickness of each
plate.
R_pl_ pI_p
A-A-lxd-d'
RA=p /d=1: 1.
RBpid
Example56.Three conductors of conductances Gl,G2and
G3are connected in series. Find their equivalent conductance.
Solution. As conductance is reprocal of resis-
tance,therefore
R=px21=~
B21xd d
1 1 1
Rl =-,Rz=-, R3 =-
Gl G2 G3
For the series combination, R = Rl+Rz+ R3
1_ 1 1 1 _G2G3+GlG3+GlG2
__ -+-+-_--=--"c-------"---"------"-----=-
G Gl G2G3 GlG2G3
or equivalent conductance,
G= GlG2G3
G2G3 + GlG3 + GlG2
Example57.A copper rod of length 20emand cross-
sectional area 2m~isjoined with a similar aluminium rod
as shown in Fig. 3.38.Find the resistance of the combination
between theends. Resistivity of copper =1.7x1O-8Dm and
resistivity of aluminium =2.6x10-8Dm
3.33
Copper
...(1)
Aluminium
Fig.3.38
Solution.Forcopper rod, p= 1.7x10-8Dm,
1=20 =20x10-2m,A=2 mm2 =2x1O-6m2
:. Resistance,
R = ~= 1.7x10-8x20x10-2=1.7x 10-3D
1A 2x10-6
For aluminium rod,
p=2.6x10-8Dm,I=20x10-2m,A=.2x1O-6m2
:. Resistance,
Rz= 2.6x10-8x20x10-2=2.6x10-3D
2x10-6
As the two rods are joined in parallel, their
equivalent resistance is
R = Rl Rz= 1.7x 10-3x2.6x 10-3
Rl +Rz1.7x10-3+2.6x 10-3
1.7x2.6x10-3
4.3
=1.028x10-3D = 1.028mD.
Example58.A wireof uniform cross-section and length I
has a resistance of16 D.Itiscut into four equal parts. Each
partisstretched uniformly to length I and all the four
stretched parts are connected in parallel. Calculate the total
resistance of the combinationsoformed. Assume that
stretching of wire does notcauseany change in the density of
its material.
Solution. Resistance of each of the four parts of
length1/4= 4 D. When each part is stretched to lengthI,
itsvolume remains same.
or
V= A'l' =Al
A' I l/4 1
-=-=-
A I' I 4
RlA'111
-=-x-=-x-=-
R'I' A 4 4 16
R'=16x R=16x 4=64Dor
i.e.,resistance of each stretched part is 64 D. When
these four parts are connected in parallel, the total
resistance- R of the combination is given by
1111141
-=-+-+-+-=-=-
R 64 64 64 64 64 16
R=16D.or···2xy§n~m}r±n2lyv

219. 3.34
Example 59.Find
,inthe given network of resistors, the
equivalent resistance between the pointsAandB,between A
andD,and between AandC. [lIT]
Solution. The resistors
0
70
C
AD(= 3 0) andDC(=70)
are in series to give a total
resistanceR'=100. The
30 50
resistanceR'(= 10 0) and
the resistorAC(=100)
A B
are in parallel. Their equi-
valent resistance is
Fig. 3.39
R"=10x10=5 0
10+10
NowR"(= 50) andCB(= 50) are inseries,their
total resistanceR'"=10o.Finally, R'"(= 100) and
AB(= 100) are in parallel betweenAandB.Hence the
equivalent resistance between points AandBis
R= 10x10=5 O.
AB10 + 10
Similarly,
39 15
RAD=-0 andRAe=-0.
16 4
Example60.Find the efective resistance between points A
andBfor the network shown in Fig. 3.40.
o30 E
c
F
A 30 B
Fig.3.40
Solution. At pointsAand D, aseries combination
of 30, 30 resistances (alongACandCD)is in parallel
with 60 resistance (alongAD), therefore, resistance
betweenAand D
1
1 1 0 =30
--+-
3 +3 6
Similarly, resistance betweenAand E
1
1 1 =30
--+-
3 +3 6
Resistance betweenAand F
1
1 1 =30
--+-
3 +3 6
PHYSICS-XII
Finally, resistance betweenAand B
1
1 1 =20
--+-
3 +3 3
Thus theeffective resistance between AandBis
20.
Example 61. Findthe effectiveresistance ofthe network
shownin Fig. 3.41between the points AandBwhen (i)the
switch Sis open (ii)switch Sis closed.
60 120
A B
5
120 60
Fig. 3.41
Solution.(i)When the switchSis open, theresis-
tances of 60 and 12 0 in upper portion are inseries,
the equivalent resistance is18o.Similarly, resistances
in the lower portion haveequivalent resistance of 18 o.
Now thetworesistances of18 0 are in parallel between
points Aand B.
.'.Effective resistance between points AandB
= 18x18 = 90.
18 + 18
(ii)When theswitchSis closed, the resistances of
6 0 and 12 0on the leftare in parallel. Their equivalent
resistance is
Similarly, the resistances on the right have
equivalent resistance of 4o.Nowthetwo resistances
of 40are in series.
..Effectiveresistance between points Aand B
=4+4=SO.
Example62.Calculate the current shown by the ammeter
A inthe circuit shown in Fig. 3.42. [CBSE002000]
50
10V
Fig.3.42www9notesdrive9com

220. Now we have resistances of 50,100 and 50 Fig. 3.45
connected in parallel, so Solution
. In the steady state (when the capator is
1 1 1 1 1 fully charged), no rrent flows through the branch
R =5"+10+5"=2" CEF. The given rit then reduces to the equivalent
circuit shown in Fig. 3.46.
R=2 0 3Q
Br-_---.J'AVA\iA"v--.[_1 -,C
y y y
>
30:
3Q ~ 3Q
Af-[-J_~[-IJV\iYY"v---D-'---~--"VYV\ YV'---' F
CURRENT ELECTRICITY
Solution.The equivalent rit is shown in Fig. 3.43.
10Q
5Q IOQ
10V
Fig. 3.43
For the two 100 resistances connected in parallel,
. . 10 x10
equivalent resistance=--- =5 0
10 + 10
For two such combinations connected in series,
equivalent resistance = 5 + 5 = 10 0
or
Also V=10V
V 10
.'.Current, I=-=-= 5 A.
R 2
Example63.Calculate the value of the resistance Rin the
circuitshowninFig.3.44sothat the current inthercuit is
0.2 A. Whatwould be thepotential diference between points
AandB? [CBSE00 12]
c
5Q
o
Fig.3.44
Solution.
1 1 1 1
--=-+-+--
RBA 15 30 5+ 5
6 1
-=-
30 5
RBA=50
3.35
ByOhm's law,
0.2A= 6-2 =_4_A
R+I0+5 R+15
4
or R + 15 = -= 20 or R = 50
0.2
VAB= 0.2xRAB=0.2x5 =1.0 V.
Example64.In the rcuitshown inFig.3.45,find the
potential diference acrossthecapacitor.
B
3Q
Ec
3Q
AI----'\IV\.,----'---'\/V\r----l F
o
15 V
15V
I
I
Fig. 3.46
The equivalent resistance of the rit is
6x3
R=--+3=50
6+3
Current drawn from the battery,
I=15V=3A
50
Current through the branch BCD,
I=_3_. xI=~x3=lA
16+3 9
Current through the armDF=I=3 A
P.D.across the capator
=P.D.between points CandF
= P.D. acrossCD+ P.D. across DF
=3x1+3x3=12V.°°°2uv{lzkyp£l2jvt

221. 3.36
Example65.A batt
ery ofemf10Visconnected to
resistances asshown in Fig. 3.47.Find the potential
diference between the points A andB.
H1 Asn
BIn
c o
3n
lOV
Fig. 3.47
4x4
Solution.Total resistance, R=--=2 0
4+4
VlOV
Current I=-=-- = 5 A
, R 20
As each ofthe two parallel branches has same
resistance (40), so the rrent of 5 A is divided
equally through them.
Current through each branch =5/2 =2.5 A
Now Vc -VA=2.5xl=2.5V
and Vc - VB= 2.5x3 = 7.5 V
VA-VB=(Vc - VB)-(Vc - VA)
= 7.5 -2.5 = 5.0V.
Example66.Whatisthe equivalent resistance between
points A andBof thecircuit shown in Fig.3.48? [lIT97]
A~B
Fig. 3.48
Solution.Obviously, the points Aand Dare
equipotential points. Also, the pointsBandCare equal
potential points. Sothe given network of resistances
reducestothe equivalent rit shown in Fig. 3.49.
2R
o>----+--c::::-\.~o
A ~c B
Fig. 3.49
Thethree resistances form a parallel combination.
Their equivalent resistance Reqis given by
_1_=_1_+ ~ +!= 1 + 1 + 2 ~ or R = R/2.
R 2R 2R R 2R R eq
eq
PHYSICS-XII
Example67.In the circuit shown inFig.3.50,Rl=4 0,
Rz= ~ = 150, R4 =300ande= 10V. Work outthe equi-
valent resistance of the circuit and the current in eachresistor.
II RI A [eBSE D2011]
B
Fig. 3.50
Solution. The resistances Rz, ~ and R4 are in parallel.
Theirequivalent resistance R' is given by
111111151
-=-+-+-=-+-+-=-=-
R'Rz ~ R415 15 30 30 6
or R' =60
Theresistance Rlisin series with R'.Hence total
resistance of the ritis
R=R1+R'=4+6=100
The rrentIIisthe rrent sent by the cellein the
whole rit.
e10
I=-=-=lA
1R10
Potential drop betweenAandB,
V=IIR' = 1x6 = 6 V
This is the potential drop across each of the resis-
tances Rz, ~ and R4 in parallel. Therefore, rrents
through these resistances are
V6 V6
12=-=-=0.4A;I3= -=-=0.4A
Rz15 . ~ 15
V 6
I4=-=-=0.2A.
R4 30
and
Example68.Findthe equivalent resistance between the
points A and Bofthe network of resistors shown in Fig. 3.51.
Solution. The resistors R2~ 30
Rland Rz are in series.
Theirequivalent resistance
=3+3=60
A~----~~r------PB
The6 0resistance is in
parallel with ~, so that
their equivalent resistance
=6x3=20
6+3
30
Fig. 3.51
Now the 20 resistance is inseries with.R4'So the
total resistance of the upper portion=2+3=5 O.°°°2uv{lzkyp£l2jvt

222. CURRENT EL
ECTRICITY
Similarly, total resistance of the lower portion
=5Q
Nowwehave three 5Qresistors connected in
parallel between the points AandB.Hence the equi-
valent resistance Rofthe entire network isgiven by
1 1 1 1 3 5
-=-+ - + -=-or R =-Q.
R5 555 3
Example 69. Findthe effective resistance between points A
andBof thenetwork of resistors shown inFig.3.52.
cSolution. By
symmetry, the potential
drops across GCandGD
are equal, so no rrent
flows in the arm CD.
Similarly, no rrent A
flows in the armDE.
Hence the resistances in
the armsCDandDEare
ineffective. The given
rit reduces to the
equivalent rit shown
in Fig. 3.53.
H
K
E
Fig. 3.52
G~~VVV- __C~~VVV---oH
H~-oBAO---+'"
E
Fig. 3.53
Resistance of armGH=r+r=2r
rxr rxr
Resistance of arm 1/=--+--=r
r+r r+r
Resistance of armFK=r+r=2r
Theabovethree resistances are in parallel between
pointsAand Band their equivalent resistance R is
given by
1 1 1 1 2
-=-+ - + -=- R =0.5r.
R2r r 2r r
Example70.Aregular hexagon with diagonals ismade of
identical wires, each having same resistance r,as shown in
Fig.3.54.Find the equivalent resistance between the points
A andB.
Fig. 3.54
3.37
Solution. As shown in Fig. 3.55, thegiven hexagon
hasa line of symmetry C1CC2•Soall points on this
linehave the same potential i.e.,potential atC1=
potential atC=potential atC2.Hence the points Cl,C
andC2can be made to coinde with each other.
A
B
Fig.3.55
r/2 : r/2
B
r/2:r/2
After this isdone, the ritsplits into identical parts,
joined in series between the points AandB.One such
part between AandCis shown in [Fig. 3.56] which, in
turn,is equivalent to the ritshown in Fig. 3.57.
0 r/2 4r/3
Cj Cj
r
A
C C
A
4r/3
G
C2 C2
r/2
Fig.3.56 Fig.3.57
From Fig. 3.57,theequivalent resistance R'between
the points AandCis given by
~ = ~+ ~ +~= 10 or R' =4r= 0.4r
R'4rr 4r 4r 10
Astwo identical partsACand CB are joined in
series, hence the equivalent resistance of theentire
rit between points AandB is
R = R' +R'=0.4 r+0.4r= 0.8r.
Example 71. Find the C
equivalent resistance of
the circuit shown in Fig.
3.58between the points
A andB.Each resistor
has a resistance r.
A
Fig. 3.58
B···2xy§n~m}r±n2lyv

223. 3.38
Solution. B
y symmetry, potential drops acrossAC
andADare equal. Soresistance in arm CDisineffec-
tive.The given rit reduces to the equivalent rit
shown in Fig. 3.59.Clearly the equivalent resistance R
between points AandBis given by
1 1 1 1 4 2
-=-+-+-=-=-
R2r 2r r 2r r
or R =~ =0.5r.
2
C
r
r 0
A B
Fig. 3.59
Example72.Find the equivalent resistance of the circuit
shown inFig. 3.60 betweenthe pointsPandQ.Each resistor
has a resistance r.
A
p
B CQ
Fig. 3.60
Solution. Two resistances along each side of
triangle are in parallel.
The equivalent resistance of eachside
rxr r
r+ r 2
The given network of resistances reduces to the
equivalent ritshown in Fig. 3.6l.
A
PB CQ
Fig. 3.61
The resistances inarms BAandACare inseries.
PHYSICS-XII
Their equivalent resistance =r /2+r /2=r.This
resistance is in parallel with the resistance r /2 along
BC
:.Effective resistance between points PandQ
rx(r/2) r
r+(r/2)3
~roblems For Practice
1.Giventhe resistances of 1n, 2 n and3n.How will
youcombine them to get an equivalent resistance of
(i)11nand(ii)11n?
3 5 [CBSE F2015]
[Ans.(i)parallel combination of In and 2n in
series with 3 n(ii)parallel combination of
2 nand 3 n in series with 1n
2.Giventhree resistances of 30 n each. How can they
be connected to give a total resistance of (i)90 n
(ii)10o(iii)45o?
[Ans.(i)inseries (ii)in parallel(iii)two
resistances in parallel and one in series]
3.A 5 n resistor is connected in series with a parallel
combination of nresistors of 6 n each. The equi-
valent resistance is 7n. Findn. (Ans.3)
4.Auniform wire of resistance 2.20n has a length of
2 m. Find the length ofthe similar wire which
connected inparallel with the 2 m long wire, will
give aresistance of 2.0n. (Ans. 20 m)
5.A wire of 15n resistance is gradually stretched to
do ble its original length. It is then t into two
equal parts. These parts are then connected in
parallel across a3.0 volt battery, Find the rrent
drawn from ~e battery. [CBSEOD09]
(Ans.0.2 A)
6.Thetotal resistance of two resistors when connec-
tedinseries is 9nand when connected in parallel,
their total resistance becomes 2 n. Callatethe
value of each resistance. [Punjab 2000]
(Ans.6n,3 n)
7,Two wiresa,andb,each oflength 40 m and area of
cross-section 10-7 m2;are connected in series and a
potential difference of 60 V is applied between the
ends ofthis combined wire. Theirresistances are
respectively 40 nand 20 n. Determine foreachwire
(i)specificresistance, (ii)electric-field, and
(iii)rrent-density.
[Ans.(i)1.0x10-7nm, 5.0x10-8nm.
(ii)1.0 Vm -1,0.5 Vm-1
(iii)1.0x107Am-2,1.0x107Am-2]°°°2uv{lzkyp£l2jvt

224. CURRENT ELECTRICITY
8.T
hree resistances, each of 40, are connected in the
form of an equilateral triangle. Find the effective
resistance between its corners. (Ans.2.670)
9.Two resistors are inthe ratio 1 : 4. Iftheseare
connected in parallel, theirtotal resistance becomes
20O.Find the value of each resistance.
[Punjab 2000]
(Ans. 250,1000)
10.Five resistors are connected as shown in Fig. 3.62.
Findthe equivalentresistancebetween the points B
and C. [Punjab 011
(Ans.70/190)
A
9n
B
3n 5n
o zo
Fig.3.62
11.Four resistors of 120 each are connected in parallel.
Three such combinations are then connected in
series. What is the total resistance? If a battery of
9 V emf and negligible internal resistance is
connected across the network of resistors, find the
rrent flowing through each resistor. [Haryana 02]
(Ans.90, 0.25 A)
12.If the reading of the ammeter ~ in Fig. 3.63is 2.4 A,
what will the ammeters Azand ~ read? Neglect the
resistances of the ammeters. (Ans. 1.6 A, 4.0 A)
II20n
~"""-'V\./\r---{Al
IOn
Fig.3.63
13.The resistance of the rheostat shown in Fig. 3.64 is...(l
30o.Neglecting the meter resistance, find the
minimum andmaximum rrent through the
ammeter astheresistance ofthe rheostat is varied.
(Ans.0.18 A, 1.5 A)
6V
20n
Fig.3.64
3.39
14.Find the rrent through the 50 resistor in the
rit shown in Fig. 3.65,when the switch 5 is
(i)open and(ii)closed. [Ans.(i)0.2 A, (ii)0.6 A]
5n Ion
5
3V
Fig.3.65
15.The letter Aconsists of a uniform wire of resistance
10 -1.The sides of the letter are40 ern long and
the crosspiece 10 em long divides the sides in the
ratio 1 : 3 from the apex.Find the resistance of the
letter between the two ends of the legs.
[Punjab 9SC]
(Ans. 66.670)
16.Callate the equivalent resistance between points
Aand Bin each of the following networks of resistors:
[Ans. (a)120(b)40/30 (c)20
(d)10/30 (e)160if)50]
sn
sn
A 5n
5n 5n
AB
(a) (b)
C
A
IOn 10n
B
B
(c) (d)
Ion
2Qt87Q
Ao-"".;v~~N\r-"""'V'II'v"""---OB A 10n B
(j)
IOn Ion Ion
Ion
(e)
Fig.3.66···2xy§n~m}r±n2lyv

225. 3.40
17.Calc
ulate the resistance between points Aand Bfor
the following networks:
2 4 R
[Ans.(a)"30(b)"30(c)"30(d) 60]
20
A
(a)
B
(b)
(c)
Fig. 3.67
18.Findthe equivalent resistance of the networks
shown in Fig. 3.68 between the points AandB.
4 r
[Ans. (a)"3r (b)4(c) r]
(a) (b)
Fig. 3.68
(c)
PHYSICS-XII
19.Find the potential difference between the points A
and B for the network shown in Fig. 3.69.
(Ans.8.0 V)
60
2A 2.50
A
30
B
Fig. 3.69
20.In thecirit diagram shown in Fig. 3.70,a volt-
meter reads 30Vwhen connected across 4000 resis-
tance. Callatewhat thesamevoltmeter reads when
it is connected across 300 0 resistance. [lIT90]
(Ans. 22.5 V)
A
2A
4000
60V
B
Fig.3.70 Fig. 3.71
21.Find the potential difference between points Aand
Bi.e.,(VA-VB)in the network shown in Fig. 3.71.
[Punjab 93](Ans.1 V)
22.Inthe circuit shown inFig.3.72, ~=40,~= ~=50,
R4=100 andE.=6 V.Work out the equivalent resis-
tance ofthecirit and the rrent in each resistor.
[CBSE D11] (Ans. 60,1 A,0.4 A, 0.2 A)
Fig.3.72 Fig. 3.73
23.Findthe equivalent resistance between points A
and BinFig.3.73.
(Ans.7.S0)
c
24.Letter Aas shown in
Fig.3.74has resistances on
eachside of arm. Callate the
total resistance between two
ends ofthelegs. .
[Himachal 93] A
(Ans.28.750) F'
19.3.74
Ewww4not~s{riv~4zom

226. CURRENT ELECTR
ICITY
25.Find the resistance between the points (i)AandB
and(ii)AandCof the network shown in Fig. 3.75.
[Ans. (i)27.50 (ii)30OJ
Ion Ion 10n
IOn Ion
Ion Ion Ion
Fig.3.75
26.A combination of four resistancesis shown in Fig.3.76.
Callate the potential difference between the
points PandQ,and thevalues ofcurrents flowing
in the different resistances.
(Ans.14.4V,0.8 A, 1.6 A)
2.4A
Ion
p 4n
Q
Fig.3.76
27.In Fig. 3.77, X, YandZare ammeters andYreads
0.5A.(i)What are the readings in ammeters Xand
Z? (ii)What isthe total resistance of the rit ?
[Ans. (i)1.5 A, 1.0 A(il)4OJ
x
L-----'-t+ '1'11-------'
Fig.3.77
28.In the rit shown in Fig. 3.78, the terminal
voltage ofthe battery is 6.0V.Find thecurrentI
through the 180 resistor. (Ans.0.25 A)
6n
sn
iso I2n
L----'--_~'II---------'
6.0V
Fig.3.7S
3.41
29.In thecirit shown in Fig. 3.79, the battery has an
emf of 12.0V andan internal resistance of 5 R/ 11. If
the ammeter reads 2.0 A, what is the value of R ?
(Ans.60)
112 V
1
1
1 5Rill 1
-- 1
Fig.3.79
30.Find the ammeter reading in the rit shown in
Fig. 3.80. (Ans. 3 A)
llV
p
sn
Fig.3.80
HINTS
1.
(1)When parallel combination of 10 and 20
resistors is connected in series with 30 resistor,
the equivalent resistance is
R=R+~= ~Rz +~
p ~+Rz
=1x2+3=~+3=110.
1+2 3 3
(ii)When parallel combination of 20 and 30
resistors is connected in series with 10 resistor,
the equivalent resistance is
R=Rz~ +~=2x3+1=~+1=110.
Rz+~ 2+3 5 5
3.Total resistance=5+~ =70,son=3.
n
4.Resistance per unit length of the wire
_2.2-llA -1
--- •• m
2
Let R' be the resistance that should be connected in
parallel to resistance, R = 2.200, so that effective
resistance,Rp= 2.0O.Then
1 1 1 1 1 0.1 1
-=---=--- -=- .. R'=220
R'Rp R 2 2.2 2.2 22
Length of the wire needed=22 =20m.
1.1···2xy§n~m}r±n2lyv

227. 3.42
5.W
hen the wire of 15 0 resistance is stretched to
double its original length, its resistance becomes
R' =n2R = (2)2xl,S = 60 0
Resistance of each half part = 60/2 = 300
Whenthe two parts are connected in parallel, their
. 30x30
equivalent resistance =--- = 15 0
30+30
Current drawn from 3.0 V battery,
1 =V= 3.0=0.2 A.
R15
6.~+ ~ =90
~~ =2
~+~
or ~ ~ = 2 (~ + ~) = 2x9 = 18
~_~ =~(~+~)2_4 ~ ~
=~81-72 = 30
Onsolving (1) and (2),
...(1)
...(2)
~=60,~=30
RA 10-7
7.(i)p=-= 40x--=1.0x10-7Om
a 1 40
10-7
andPb= 20x-- =5.0x10-8 Om
40
(ii)Total resistance, R =Ra+ ~ = 40 + 20 = 600
The rrent in the wires, 1 =V=60 = 1.0 A.
R60
:. Potential differences between the ends of wiresa
andbare
Va= 1xRa=1.0x40 = 40V
and Vb= 1x~ = 1.0 x20 = 20V
Electric fields in the two wires are
=Va= 40 =1.0Vm-1
Ea
la40
and E"=Vb=20 =0.5 Vm-1
lb40
(iii)The rrent in each wire is the same. Also, the
area of cross-section of each wire is same. Hence the
rrent-density in each wire is
1 1.0 1 0 107 A-2
JA=JB= A=10-7=' x m.
R=(4+ 4) x 4 = 32 = 2.67 O.
8.
(4 + 4) + 4 12
9.Let the two resistances be Rand 4 R. Then
Rx4R . 4
--- = 200 or-R = 200
R+ 4R 5
R =250 and 4 R = 100O.
PHYSICS-XII
10.Resistance in branchADe
~ = 3 + 7= 100
This resistance is in parallel with the 100 resistance
in branchAe.Theirtotal resistance is
10x10
~ = 10 + 10 = 50
This 50 resistance is in series with the 90 resis-
tance in branchAB.Their equivalent resistance is
R.,=5+9=140
This 140 resistance is in parallel with the 50
resistance in branch Be.Hence the equivalent
resistance between BandCis
R=14x5=70O.
14+519
11.The rit diagram is shown in Fig.3.81.
120 120 120
120 120 120
120 120 120
120120 120
I
Fig. 3.81
Effective resistanceR'of four resistances of 120
each connected in parallel is given by
1 1 1 1 1 4
-=-+-+-+-=-
R' 12 12 12 12 12
or R'=30
Total resistance of the network,
R = R' + R'+R'=3+3 + 3 =9n
9V
Current inthe rcuit, I=-= 1A
90
Current through each resistor
= ~I=~x1=0.25A
4 4
12.P.O.across 200 =p.o.across 300
orIIx20 = 12 x30
20 20
or 12= 30II=30x2.4 =1.6 A
and I=II+ 12= 2.4 + 1.6 = 4.0 A
13.Equivalent resistance ofthe 50 and 200 resis-
5x20
tances connected in parallel =-- =4 O.This
5+20
resistance is connected in series with the rheostat
whose minimum and maximum resistances are 00
and 300.···2xy§n~m}r±n2lyv

228. CURRENT ELECTRICITY
When the rheostat is austed at the rrurumum
resistance of00,current will be maximum.
6V
I---15A
max-40 -
.
When the rheostat is austed at the maximum
resistance of300,current will be minimum.
6V
I.= =0.18A
rrun(4+ 30)0
14.(i)When switch5is open, resistances of50and
100are in series.
3V
Current,I= = 0.2A
(5+ 10)0
(ii)When switch5is closed, no current flows
through100resistance.
3V
:.Current,I=-=0.6A
50
15.Refer to Fig. 3.82.Clearly
BC=CD= BD=10em
AB= DE=30cm
and
~=~=Rs=100
1)=R4=300
C
Fig. 3.82
Series combination of ~ and ~ is in parallel with
Rs. Their equivalent resistance
= (10 + 10)x 10 = 200 = 200 = 6.670
(10 + 10)+ 10 30 3
This resistance is in series with 1) andR4.So the net
resistance is
R= 30 + 6.67+ 30 = 66.670.
8x8
16.(a)R=8+ -- =12O.
8+8
(b)R= 5 + 10x5 + 5 = 40 O.
10+5' 3
(c)R=(3+3) 3=20.
(3 + 3) + 3
3.43
(d)All the three resistances are connected in
parallel between pointsAand B.
1 1 1 1 3 10
-=-+ - + - =-orR =-O.
R10 10 10 10 3
(e)The given network is equivalent to the net-
work shown in Fig.3.83.
10x15
R= 10+ -- =160.
10+ 15
IOQ
IOQ 5Q
A B
Fig. 3.83
if)Resistance in branchADC=2+4=6 O.This
resistance is in parallel with60resistance in
armAC
Their equivalent resistance
=6x6=30
6+ 6
The series combination of this30resistance
and70resistance in arm BCis in parallel with
100resistance in armAB.
R= 10x10 =5 O.
10 + 10
17.The corresponding equivalent circuit diagrams are
given below:
A
2Q
2Q
2Q 2Q
2Q
Fig.3.84
B
(a)www4§±ttssrxvt4r±£

229. 3.44
18.(
a)The equivalent network for 3.68(a) is shown in
Fig.3.85(a).
r4
R=r+-=-r.
3 3
r
r
r
A
r r
r
B A
r
(a) (b)
Fig.3.85
(b)The equivalent network for3.68(b)is shown in
Fig.3.85(b).
~=~+~+~+~=~ or R=~.
Rrrrrr 4
(c)The current divides symmetrically in the two
upper and the two lower resistances. Sothe
resistances in thevertical armare ineffective. The
given network reduces ~
to the equivalent r r
network shown in A B
r r
Fig.3.86.
R=2rx2r=r. Fig.3.86
2r +2r
3x3
19.R =-- +2.5 = 1.5 +2.5 = 4.00
3+3
V= R1= 4.0x2 =8.0V.
20.p.o.across 4000 resistance=30V
PD.across 3000 resistance = 60- 30 = 30V
This shows that potential drop is same acrossboth
resistances.
Let R be the resistance of thevoltmeter. Then
equivalent resistance of Rand 4000 connected in
parallel should also be 300O.
Rx400
--- = 300 or R = 12000
R+400
When thevoltmeter is connected across the 3000
resistance, their equivalent resistance isgiven by
R'= 1200x300 = 2400
1200+ 300
Total resistance in thecircuit = 240+400 = 6400
C inthe circuit.T 60 3A. .urrentillt e circuit, = -= -
640 32
Reading of thevoltmeter
=JR'=~ x240 = 22.5V.
32
PHYSICS-XII
21.Current through each branch =2/2 = 1A
Vc-VA=lx2=2V
Vc-VB=lx3=3V
VA-VB=(VC-VB)-(VC-VA)=3-2=1V.
111111151
--=-+-+-=-+-+-=-=-
RAB ~ R:,R45510 10 2
22.
B
RAB=20
R=1\+RAB =4+2=60
e6V
11=-=-=lA
R60
2
14=-=0.2A.
10
23.The equivalent circuit is shown inFig.3.87. The
effective resistance between points CandD
3x6
=--+ 8=100
3+6
3n
A
c
8n
o
B
30n
Fig.3.87
Nowthe100and300resistances are in parallel.
The equivalent resistance between points AandB
10x30
=--=7.50.
10 +30
24.Proceed as inExample 51onpage 3.32.
25.(i)The equivalent circuit is shown in Fig. 3.88.
10n E io n H 10n
10n io o
F10n G
Fig.3.88
Resistance of the arm EFGH = 10 + 10+10 = 300
Thisresistance is parallel to the 100resistance of
armEH.
Equivalent resistance between points EandH
=10x30 = 7.50
10+30
Hence total resistance between points AandB
= 10+7.5 + 10=27.5nwww4§±ttssrxvt4r±£

230. CURRENT ELECTRICITY
(ii)Theequi
valent circuit is shown in Fig. 3.89.
IOn EIon H
A
IOn Ion
FIOn G Ion B
Fig. 3.89
Resistance of armEHG = 10 +10= 200
Resistance of arm EFG=10+10 = 200
These two 200resistances arein parallel.
:.Effective resistance between points EandG
=20x20 = 100
20+20
Hence total resistance between points AandC
=10+10+10=30O.
26.The resistances of40,100and40are in series.
Their equivalent resistance =18o.Thisisin parallel
with90resistance.
Equivalent resistance between PandQ,
18x9
R=--=60
18+9
P.D.between PandQ=IR= 2.4x6 =14.4 V
12=V= 14.4 = 1.6A
R 9
and II=2.4 - 1.6 =0.8 A.
27.P.D.across 60 =P.D.across30
6x0.5=3x12
Current throughZ,
12=1.0 A
Current throughX= 0.5+1.0 = 1.5 A
6x3
Total resistance =--+2=4o.
6+ 3
28.Total resistance in the upper branch
6x12
=8+--=120
6+12
Total resistance in thecircuit,
12x12
R = 18+~~ = 18+6 = 240
12+12
Current through 180resistor = ~ =0.25 A.
24
29.The resistancesR,2 Rand3 Rare in parallel
between the points PandQ.Their equivalent
resistanceR'isgiven,by
~=~+_1_·+~=~ orR'=6R
R'R 2 R 3 R 6R 11
Now6R/11and5 R/11areinseries.
3.45
6R 5R
:.Total resistance of thecircuit=-+-=R
11 11
Resistance, R=§.= 12 = 6O.
I 2
30.Theresistances of (5+ 7)= 120, 60and80arein
parallel between points PandQ.Their equivalent
resistance R'isgiven by
~ = ~ + ~ + ~=~orR' = ~0
R'126 8 8 3
R'is inseries with 10resistance.
:.Totalresistance =~ +1=110
3 3
e11
Current, I= -=-- =3 A.
R 11/3
3.19INTERNAL RESISTANCE OF ACELL
34. What isinternal resistance ofa cell?On what
factors does it depend?
Internal resistance. When the terminals of a cell are
connected by a wire, an electriccurrent flows in the
wire from positive terminalof the cell towards the
negative terminal. But inside the electrolyte of the cell,
the positive ions flow from the lower to thehigher
potential (or negative ions from the higher to thelower
potential) against the background of other ions and
neutral atoms of the electrolyte. Sotheelectrolyte offers
some resistance to the flow of current inside the cell.
Theresistance offered bythe electrolyte of a celltothe
flow of current between its electrodes iscalledinternal
resistance of the cell.
The internal resistance of a cell depends on following
factors:
1.Nature of the electrolyte.
2.It isdirectly proportional to the concentration of
the electrolyte.
3.It is directly proportional to the distance
between the two electrodes.
4.It varies inversely as the common area of the
electrodes immersed in the electrolyte.
5.It increases with the decrease in temperature of
theelectrolyte.
The internal resistance ofafreshly prepared cell is
usually lowbut its value increases aswedraw more
and more current from it.
3.20RELATION BETWEEN INTERNAL
RESISTANCE, EMF AND TERMINAL
POTENTIAL DIFFERENCE OF A CELL
35. Defineterminal potential difference of a cell. Derive
arelation between the internal resistance, emf andterminal
potential difference of a cell. Draw (i)evs.R(ii)V vs.R
(iii)V vs.Igraphs for a cell and explain their significance.www4§±ttssrxvt4r±£

231. 3.46
Terminal potent
ial difference. Thepotential drop
across the terminals of a cellwhen acurrent isbeing drawn
from itiscalled its terminal potential difference (V).
Relation betweenr,e andV.Consider a cell of emf e
and internal resistance r connected to an external
resistance R, as shown in Fig. 3.90.Suppose a constant
currentIflows through this circuit.
......
R
:IeYY
----------
Cell
Fig.3.90Cellof emf e and internal resistancer.
By definition of emf,
e=Work done by the cell in carrying a unit charge
along the closed circuit
=Work done in carrying a unit charge fromAto B
against external resistance R
+Work done in carrying a unitcharge fromBto
Aagainst internal resistancer
or e=V+V'
By Ohm's law,
V=IRandV'=Ir
e=IR+Ir=I(R+r)
Hence the current in the circuit is
I=_e_
R+r
Thus to determine the current in the circuit, the
internal resistancercombines in series with external
resistance R.
The terminal p.d. of the cell that sends currentI
through the external resistance R is given by
V=IR=~
R+r
or terminal p.d. =emf - potential drop across the
internal resistance
Again, from the above equation, we get
r=e~v =~~~=(e~v)R.
Special Cases
(i) When cell is on open circuit, i.e.,I=0, we have
~pen-e
PHYSICS-XII
Thus the potential difference across the terminals of the
cellisequal toitsemf when no current isbeing drawn from
the cell.
(ii) A real cell has always some internal resistancer,
so when current is being drawn from cell, we have
v<e
Thusthepotential difference across theterminals ofthe
cellina closed circuit isalways lessthanits emf
Characteristic curves for a cell. When a cell of emfe
and internal resistance risconnected across avariable
load resistance R,itsfunctioning can be represented by
the following three graphs :
(i) e versusRgraph.The emf of a cell is equal to the
terminal p.d. of the cell when no current is drawn from
it.Hence emf e is independent ofRand e-R graph is a
straight line, as shown in Fig. 3.91(a)
jl------
w
Fig.3.91(a)e vs.Rgraph for a cell.(b) Vvs.Rgraph for a cell.
(ii)Vversus R graph. Ina closed circuit, the
terminal p.d of the cell is
V=IR=(_e_)R=_e_
R+r 1+l:..
R
As R increases,Valsoincreases.
When R ~ 0, V=0
When R=r, V=e/2
When R ~00, V=e
HenceV-Rgraph is as shown in Fig.3.91(b).
(iii)VversusIgraph.AsV=e-Ir
~ V=-rI+e <=:>y=mx+c
Hence, the graph between VandIis a straight line
with a -veslope, as shown in Fig. 3.91(c)
For point A,1=0
Hence,
A
VA=e
=intercept on the y-axis j
>
For pointB,V=a
e=IBr
e
Hence, r =_ Fig.3.91(c)Vvs.Igraph
IB for a cell.
=negative of the slope ofV-Igraph.
o B
l~www4§±ttssrxvt4r±£

232. CURRENT ELECTRICITY
Examples
based on
Grouping of Cells
Formulae Used
1.EMF of a cell, e =w
q
2.For a cell ofinternal resistance r,the emf is
e=V+Ir=I(R+r)
eR
3. Terminal p.d. of a cell,V=IR=--
R+r
4.Terminal p.d. when a current is being drawn from
the cell,
v=e- Ir
5.Terminal p.d. when the cell is beingcharged,
V=e+ Ir
[e -V]
6.Internal resistance of a cell, r= R---v-
Units Used
EMF e and terminal p.d. Vare in volt(V),internal
resistancerand external resistance R in nand
currentIin ampere (A).
Example 73. For driving acurrent of3Afor5minutes in
an electric circuit, 900Jofwork is to bedone.Find the emf of
the source inthe circuit.
Solution. The amount of charge that flows 'through
the circuit in 5 minutes is
q=Ixt= 3x5x60 = 900 C
Asemf is the work done in flowing a unit charge in
the closed circuit, therefore
e =w= 900J=1.0 V.
q900 C
Example 74.A voltmeter of resistance 998 Qisconnected
across a cell of emf2Vand internal resistance 2Q.Find the
p.d. acrossthe voltmeter, that across the terminals of thecell
and percentage error in the reading of thevoltmeter.
Solution. Here e = 2 V,r= 2 Q
Resistance ofvoltmeter,
R =998Q
Current in the circuit is
..».
R+r
2V
Cell
:2V
I
2Q :
Voltmeter
R=998Q
(998+2)Q
=2x10-3A
Fig. 3.92
3.47
The p.d. across the voltmeter is
V= IR
= 2x10-3x998 =1.996 V
The same will be the p.d. across the terminals of the
cell.The voltmeter used to measure the emf ofthe cell
will read 1.996 volt. Hence the percentage error is
e ~Vx100=2-~.996x100=0.2%.
Example 75. Inthe
circuitshown in Fig. 3.93,
the voltmeter reads 1.5V,
when the keyisopen.When
thekeyisdosed, the
voltmeter reads1.35V and R
ammeter reads 1.5AFind
the emf and the internal Fig.3.93
resistance of thecell.
Solution. When the key is open, thevoltmeter
reads almostthe emf of the cell.
e=1.5 V
Whenthe key is closed, voltmeter reads the
terminal potential difference V.
V= 1.35 V, I= 1.5 A, r=?
r= e-V= 1.5 -1.35 = 0.1Q.
I 1.5
Example 76.Acell ofemf2V and internal resistance
0.1Qisconnected to a3.9Qexternal resistance. What will
be thep.d.across the terminals of thecell? [CBSE D OlC]
Solution. Heree=2 V,r=O.lQ, R=3.9Q
e 2
Current, 1=-- = = 0.5 A
R+r3.9+0.1
P.D.across the terminals of the cell,
V=IR=0.5x3.9=1.95V.
Example 77.The reading on a
highresistance voltmeter when a
cell is connected across it is
2.2 V. When the terminals of the
cell are also connected to a
resistance of5Q,thevoltmeter
reading drops to1.8V.Find the
internal resistance of the cell.
[CBSEOD 10]
Solution. Here e=2.2 V,
Fig. 3.94
R=5Q,V=1.8V
+
r----{ V}-----,
+-
R=5Q K
Internal resistance,
r= R(e;VJ= 5 C.21~81.8)Q =1.1 Q.www4§±ttssrxvt4r±£

233. 3.48
Example78.A
drycell ofemf1.6V and internal resis-
tance0.10 n is connected toa resistance ofRohm.The
current drawn from the cell is2.0AFind the voltage drop
acrossR.
Solution. Here e= 1.6 V, r= 0.10 n,1=2.0 A
Voltage drop across Rwill be
V=e -Ir
= 1.6 - 2.0x0.10 =1.4V.
Example 79. A battery of e.m.f't',and internal resistance
'r',givesacurrent of0.5A with an external resistor of
12ohm and acurrent of0.25A with an external resistor of
25ohm. Calculate (i) internal resistance ofthe cell and
(ii)emf of the cell. [CBSE D 02; OD 13C]
Solution. EMF of the cell, e=I(R+r)
Infirstcase, e=0.5 (12+r)
insecond case, e= 0.25 (25+r)
0.5 (12+r)= 0.25 (25+r)
On solving, we getr=1n
Hence e= 0.5 (12+1) =6.5 V.
Example80.A battery of emf3voltand internal resis-
tancerisconnected in series with a resistor of 55nthrough
an ammeter of resistance 1n.The ammeter reads 50mA.
Draw thecircuit diagram andcalculate the value of r.
[CBSE D 95;Haryana 02]
Solution. Total resistance
= 55+1+rn = 56+rn
Current 55Q
=50mA
=50x10-3 A
emfr=S V
R. emf
esistance=---
Current
R
, ,
, ,
, 1
56+r= 3 =60Fig.3.95
50x10-3
r= 60-56 =4n.
Example 81.(a) A car has a fresh storage battery of emf
12Vand internal resistance 5.0x10-2n.If the starter
motor draws acurrent of90A,what istheterminal voltage
of thebattery when the starter ison?
(b) After long use, the internal resistance of thestorage
battery increases to 500 n. What maximum current canbe
drawn from the battery ?Assume the emf of thebatteryto
remain unchanged.
(c)If the discharged battery ischarged by an externalemf
source, istheterminalvoltageof thebattery during charging
greater or less than its emf12 V? [NCERT]
PHYSICS-XII
Solution. (a)Heree= 12 V, I= 90 A,
r=5.0x1O-2n
.'.Terminalvoltage,
V =e -Ir=12 - 4.5 =7.5 V.
(b)The maximum current can be drawn from a
battery byshorting it.
Then V=0
e12
I=-=-A =24 mA
maxr500
and
Clearly, the battery is useless for starting the car
and must be charged again.
(c)During discharge of the accumulator, the current
inside the cells (of the accumulator) is opposite to what
it is when the accumulator discharges. Thatis,during
charging, current flows from the+ve to-ve terminal
inside the cells. Consequently, during charging
V=e+Ir
Hence Vmustbe greater than 12 Vduring charging.
Example 82. A battery of emf 12.0 Vandinternal resis-
tance0.5nisto be charged by a battery charger which
supplies110Vd.c.Howmuch resistance must be connected
in series with the battery tolimit the charging current to
5.0 A?What will be thep.d.across theterminals of the
battery during charging ?
Solution. For charging, the positive terminal of the
charger is connected to the positive terminal of the battery.
Netemfe'= 110 -12.0 = 98 V
Battery charger
.------f 110 Vt-----,
5.0A
r------------ ..
:12.0 V 0.5 Q: R
'-----',--'-1+IIII - ,
, :
' J
Fig.3.96
IfRistheseries resistor, then the charging current
will be
I=~=~A
R+rR+0.5
Given 1=5.0 A, therefore
~ = 5.0 or R=19.1n
R+0.5
The terminal p.d. of the battery during charging is
V=e+Ir= 12.0 +5.0x0.5=14.5 V
If the series resistor R were not included in the
charging circuit, thecharging current would be 98/0.5
= 196 A, which is dangerously high.www4§±ttssrxvt4r±£

234. CURRENT ELECTR
ICITY 3.49
Example 83. A cellofemf1.5V andinternal resistance
0.5Qisconnected to a (non-linear) conductor whose V-I
graphisshowninFig.3.97(a). Obtain graphically the
current drawn from the cell and its terminal voltage.
I I
Fig. 3.97
Solution. HereE.=1.5V,r=0.5Q
Terminal voltage of the cell, V=E.-Ir
For different currents, terminalvoltages can be
determined as follows:
I=O, V=1.5-0=1.5V
I=1A, V=1.5-1x0.5=1.0V
I= 2 A, V=1.5 -2x0.5=0.5V
I= 3 A, V=1.5-3x0.5=0
TheV-Igraph for the cell is astraight lineAB,as
shown in Fig. 3.97(b). Thisstraight line graph intersects
the given non-linearV-Igraph at current=1 Aand at
voltage=1 V.
:.Current drawn from the cell=1 A
Terminalvoltage of the cell=1 V.
Example 84. Potential differences across terminals of a cell
weremeasured (involt) against different currents (inampere)
flowing through the cell.Agraphwasdrawn whichwasa
straight line ABC, as shown inFig.3.98.Determine from
the graph
(i)emf of the cell
(ii)maximum current obtained from the cell, and
(iii)internal resistance ofthe cell. [CBSE D 1IC]
2.0
t
%1.2
~
:::.0.8
Fig. 3.98
1.6
I
I
I
---+-------
I
I
I I
---T-------r-------
I I I
I I I
I I I
<.A
0.4
o
0.04 0.12 0.20
I(ampere) ~
0.28
Solution. (I)The potential difference corresponding
to zero current equals the emf of the cell.
:.EMF of the cell,E.=1.4V.
(ii)Maximum current is drawn from the cell when
the terminal potential difference is zero.
Imax=0.28 A.
(iii)Internal resistance,
E.1.4 V
r=--=--=SQ.
Imax 0.28 A
Example 85. Find the current drawn from a cell of emf1V
and internal resistance 2 /3Qconnected to the network
given below. [CBSE D OlC)
D c
Fig. 3.99
Solution. The equivalent circuit is shownbelow.
Hl
A B
Hl In
p
In
D c
IV
Fig.3.100
Resistance in armAB=1Q
lxl lxl 1 1
Resistance in armPQ=--+-- =-+-=1Q.
1+1 1+1 2 2
Resistance in armDC=1Q
These three resistances are connected inparallel.
Their equivalent resistanceRis given by
1 1 1 1 3 1
-=-+- +-=-or R=-Q
R1 1 1 1 3
Currentdrawn from the cell,
E. 1 V
I=-- = (1 2) = 1 A.
Rwr_+_ Q
3 3www4§±ttssrxvt4r±£

235. 3.50
Example 86. A uniform wire
of resistance120 iscut into
three piecesinthe ratio1: 2:3and the three pieces are
connected to form a triangle. A cell of emf8V and internal
resistance10 isconnected across the highest of the three
resistors. Calculate the current through each part of the
circuit. [CBSE OD 13C]
Solution. In Fig. 3.101, RAB= 2 0,RBC= 4 0 and
RAC=60.
B
BV r=10
Fig.3.101
The series combination of 20 and 40 (of equi-
valent resistance 60) is in parallel with the 60
resistance. The equivalent resistance is
R=6x6=30
6+6
Current,I=_E_ =8 V=2 A
R+r(3+1)0
The resistancesRBAC andRBCof the parallel
branches are equal.
.. I ABC=lAC=1 A.
~roblems For Practice
1.The emf of a cell is 1.5V. On connecting a 140
resistance across the cell, the terminal p.d. falls to
1.4V. Calculate the internal resistance of the cell.
[Haryana 01] (Ans.10)
2.The potential difference across a cell is 1.8V when a
current of 0.5A is drawn from it. The p.d. falls to 1.6V
when a current of 1.0A is drawn. Find the emf and the
internal resistance of the cell. (Ans.2.0 V, 0.40)
3.The potential difference of a cell in an open circuit
is 6 V, which falls to 4 V when a current of 2 A is
drawn from the cell. Calculate the emf and the
internal resistance of the cell. (Ans. 6 V, 10)
4.In the circuit shown in Fig. 3.102, the resistance of
the ammeterAis negligible and that of the
voltmeterVis very high. When the switch 5 is open,
the reading of voltmeter is 1.53 V. On closing the
switch 5, the reading of ammeter is 1.00A and that
of the voltmeter drops to 1.03 V. Calculate: (i)emf
PHYSICS-XII
of the cell(ii)value ofR(iii)internal resistance of
the cell. [Ans. (i)1.53V(ii)1.030(iii)0.50OJ
5
Fig.3.102
5.The potential difference between the terminals of a
battery of emf 6.0 Vand internal resistance 10
drops to 5.8Vwhen connected across an external
resistor. Find the resistance of the external resistor.
(Ans.290)
6.The potential difference between the terminals of a
6.0Vbattery is 7.2Vwhen it is being charged by a
current of 2.0 A. What is the internal resistance of
the battery? (Ans. 0.60)
7.A battery of emf 2Vand internal resistance 0.50 is
connected across a resistance of 9.50. How many
electrons pass through a cross-section of the
resistance in 1 second? (Ans. 1.25x1018)
8.A cell of emf E and internal resistanceris connected
across a variable load resistorRIt is found that
whenR=40, the current is 1A and whenRis
increased to 90, the current reduces to 0.5 A. Find
the values of the emf E and internal resistancer.
[CBSE D 15]
(Ans.5 V, 10)
9.The emf of a battery is 4.0 V and its internal resis-
tance is 1.5 O. Its potential difference is measured
by a voltmeter of resistance 1000 O. Calculate the
percentage error in the reading of emf shown by
voltmeter. (Ans. 0.15%)
10.The emf of a battery is 6 V and its internal resistance
is 0.6 O. A wire of resistance 2.40 is connected to
the two ends of the battery, calculate(a)current in
the circuit and(b)the potential difference between
the two terminals of the battery in closed circuit.
(Ans.2 A,4.8 V)
11.The two poles of a cell of emf 1.5 V are connected to
the two ends of a 100 coil. If the current in the
circuit is 0.1A, calculate the internal resistance of
the cell. (Ans. 50)
12.The potential difference across the terminals of a
battery is 8.5 V, when a current of 3 A flows
through it from its negative terminal to the positive
terminal. When a current of 2 A flows through it in
the opposite direction, the terminal potential
difference is 11V. Find the internal resistance of the
battery and its emf. (Ans. 0.50, 10 V)www3°otssrrwvs3qo~

236. CURRENT ELECTRIC
ITY
13.In the circuit shown in Fig. 3.103, a potential
difference of 3 V is required between the points A
andB.Find the value ofresistance R,.. (Ans.30)
I
12 V,1Q
-'-
Fig.3.103
HINTS
(e -VJ(1.5-1.4)
1.r=R----v-= 14 1.4 =10.
2.EMF of acell,e=V+Ir
When1=0.5 A,V= 1.8 V..e= 18 +0.5r (1)
WhenI= 1.0 A, V= 1.6 V..e= 1.6+1.0r(2)
Solving (1)and(2),weget
e=2.0Vandr=0.4O.
3.e=P.D.measured inopen circuit =6 V
e-V6-4
r=--=--=ln.
I 2
4.(i)e= P.D. measured inopen circuit = 1.53V.
(..)R-V -1.03-1 03 0
II --- -. •
I 1.00
(iii)r=R(e-VJ= 1.03(1.53 - 1.03) =0.50O.
V 1.03
5.R=/V= lx5.8 = 5.8 =290.
c-V6.0-5.8 0.2
6.During charging, V=e+Ir
:.7.2=6.0+ 2xror r=0.60.
e 2
7.1=--= =0.2A
R + r 9.5 + 0.5
It 0.2xl 18
n=-= 19= 1.25x10
e1.6 x10
8.Proceed as in Example 79 on page3.48.
9.Proceed as in Example 74 on page 3.47.
e 6
10.Current, I=--= =2 A.
R +r2.4+0.6
P.D.between the two terminals of the battery is
V=IR"=2x2.4 V=4.8 V.
11.AsI=_e_
R+r
0.1=~
10+ r
orr=50.
3.51
12.When current flows through the cell from its
negative to positive terminal,
V=e-Ir
or 8.5=e-3r ...(i)
When current flows through the cellfrom its positive
tonegative terminal, p.d. across radsto itsemf.So
V=e+Ir
or 11=e+2r ...(ii)
Onsolving equations (i)and(ii),we get
r=0.50 ande=10V.
13.Currentin the main circuit,
1=e
R,.+Rz+r
Since apotential difference of3 Visrequired across
R,.,therefore
or
IR,.=3volt
eR,. =3or
R,.+Rz+r
R,.=30.
_12_R,.-'--. =3
R,.+8+1
or
3.21COMBINATIONS OF CELLS IN SERIES
AND PARALLEL
36. Why do we often use a combination of cells ?
Combination of cells. Asingle cell provides a feeble
current. Inorderto get ahigher current in a circuit, we
often use a combination ofcells, two or more cells.A
combination of cells iscalled abattery.Thebattery cells
areusedin torches, transistor sets,automobiles, etc.
Cells can be joinedinseries, parallel or inamixed way.
37.Whatdoyou mean by a series combination of
cells?Twocellsof different emfs and internal resis-
tancesare connected in series.Findexpressions for the
equivalent emf andequivalent internal resistance of the
combination.
Cells in series. When the negative terminal ofone cell is
connected to the positive terminal of the othercell and soon,
thecellsaresaid to be connected in series.
As shown in Fig. 3.104, suppose two cells of emfs S,
ande2and internal resistances r1andr2are connected
inseries between points AandC.LetIbe the current
flowing through theseries combination.
Fig.3.104 A series combination of two cells is equivalent to a
single cell of emfeeqandinternal resistance'eqwww4§±ttssrxvt4r±£

237. 3.52
LetVA' VBandVcbe the potentials at pointsA,B
andCrespectively. The potential
differences across the
terminals of the two cells will be
VAB=VA- VB=e1-Irl A
and VBC=VB - Vc=e2 -Ir2
Thus the potential di.fference between the terminals
AandCof the series combination is
VAC=VA - Vc=(VA -VB)+(VB-VC)
=(e1-Ir1)+ (e2 -Ir2)
VAC=(e1 +e2)-I(r1 +r2)
If we wish to replace the series combination by a
single cell of emfs;and internal resistance~q'then
VAC=s: -I~q
Comparing the last two equations, we get
eeq= e1+ e2 and~q=r1+r2
or
We can extend the above rule to a series combi-
nation of any number of cells:
1.The equivalent emf of a series combination ofn
cells is equal to the sum of their individual emfs.
eeq= e1+ e2+ e3+ + ell
2.The equivalent internal resistance of a series
combination ofncells is equal to the sum of
their individual internal resistances.
req=r+r2+r3+ +r"
3.The above expression foreeqis valid when then
cells assist each otheri.e.,the current leaves each
cell from the positive terminal. However, if one
cell of emfe2,say, is turned around 'in
opposition' to other cells, then
eeq=e1-e2+e3+ +en·
or
38. What do you mean by a parallel combination of
cells? Two cells of different emfs and internal resistances
are connectedinparallel with one another. Find the
expressions for the equivalent emf and equivalent internal
resistance of the combination.
Cells in parallel. When the positive terminals of all cells
are connected to one point and all their negative terminals to
another point, the cells are said to be connected in parallel.
As shown in Fig. 3.105, suppose two cells of emfs
e1ande2and internal resistancesr1andr2are connec-
ted in parallel between two points. Suppose the
currentsIlandI2from the positive terminals of the
two cells flow towards the junctionB1,and currentI
flows out. Since as much charge flows in as flows out,
we have
PHYSICS-XII
Fig. 3.105A parallel combination of two cells is equivalent to
a single cell of emfeeqand internal resistancereq'
As the two cells are connected in parallel between
the same two points1\and ~, the potential difference
Vacross both cells must be same.
The potential difference between the terminals of
first cell is
V=V~ - VB2 =e1-I1r1
e-V
I__1__
1-
r1
The potential difference between the terminals of
e2is
or
Hence
If we wish to replace the parallel combination by a
single cell of emfeeqand internal resistance~q,then
V=eeq-I~q
Comparing the last two equations, we get
e=e1r2+e2r1 and r=r1r2
eq r+1: eq r+r.
1 2 1 2
We can express the above results in a simpler way
as follows:
and
eeq_e1e2
---+-
~q r1 r2
1 1 1
-=-+-
~qr1r2www4§±ttssrxvt4r±£

238. CURRENT ELECTRICITY
F
or a parallel combination ofncells, we can write
eeq _e1e2 en
---+-+ +-
'eq r1r2 rn
and
1 1 1 1
-=-+-+ +-.
'eqr1r2 ~J
39.Derive the condition for obtaining maximum
current through an external resistance connected across a
series combination of cells.
Condition for maximum current from a series
combination of cells. As shown in Fig. 3.106, supposen
similar cells each of emfeand internal resistance rbe
connected in series. LetRbe the external resistance.
Fig. 3.106 A series combination ofncells.
Total emf ofncells in series
=Sum of emfs of all cells =ne
Total internal resistance ofncellsinseries
=r+r+r+nterms =nr
Total resistance in the circuit =R+nr
The current in the circuit is
1=Total emf
Total resistance
ne
R+nr
R
Fig. 3.107Equivalent
circuit.
Special Cases
(i)If R»nr,then
1=ne
R
=ntimes the current(eIR) that can
be drawn from one cell.
(ii)If R«nr,then
1=ne=~
nrr
=the current given by a single cell
Thus,when external resistance ismuch higher thanthe
totalinternalresistance, thecells should be connected in
seriestoget maximum current.
3.53
40. Derive the condition for obtaining maximum
current through an external resistance connectedtoa
parallel combination of cells.
Condition for maximum current from a parallel
combination of cells. As shown inFig.3.108,suppose m
cells each of emfeandinternal resistance rbeconnec-
tedin parallel between points AandB.LetRbethe
external resistance.
A
R
Fig.3.108A parallel combination of m cells.
Sinceall the minternal resistances are connected in
parallel, their equivalent resistance R'is given by
1 1 11m
- =-+-+-+ mterms =-
R'rrr r
R'=!-.-
m
or
Total resistance in the
circuit
~ -"t=:- - --r---:
IC. m I
I
I I
I I
L _
=R+R'=R+!-.-
m
R
As the only effect of joining
mcellsin parallel is to get a
single cell of larger sizewithFig.3.109Equivalent circuit.
the samechemical materials, so
total emf of parallel combination
=emf due to single cell =e
The current in the circuit is
e
1=---
R+rim
me
mR+r
Special Cases
(i)If R«!-.-,then
m
1=me=mtimes the current due to a single cell.
r
(ii)If R»!-.- ,then
m
I=~ =the current given by a single cell.
R
Thus,whenexternal resistance ismuchsmaller than the
netinternal resistance, thecells should be connected in
paralleltoget maximum current.www4§±ttssrxvt4r±£

239. 3.54
41.Derive the condition for obtaining maximum
current through an external
resistance connected across a
mixed grouping of cells.
Mixed grouping of cells. In this combination, a certain
number of identical cells are joined inseries, and all such
rows arethen connected inparallel with each other.
As shown in Fig.3.110, suppose ncells, each of emf
eandinternal resistance r,are connected in series in
each row and msuch rows are connected inparallel
acrossthe external resistance R.
A,:-.q:-- mrows ~...;;,
I~I------------~
I~I------------~
R
Fig.3.110Mixed grouping of ceUs.
Total number of cells
=mn
Netemf of each row ofn
cells in series=ne
Asmsuch rows are
connected in parallel, so net
emfofthe combination =ne
R
Fig. 3.111Equivalent circuit.
Net internal resistance of
each row of ncells =nr
Asmsuch rows are connected in parallel, so the
total internal resistance t'ofthe combination is given
by
1 1 11m
-=-+-+-+ mterms =-
r nr nr nr nr
or r=nr
m
Total resistance of the circuit
=R+r=R+nr
m
The current through the external resistance R,
1= Totalemf ne
Total resistance R +nr /m
mne
mR+ nr
Clearly, thecurrentIwill be maximum if the
denominator i.e., (mR +nr)is minimum.
PHYSICS-XII
Now
mR+nr= (.Jri1R)2+(.Jnr)2
= (.Jri1R)2+(.Jnr)2-2.Jri1R.Jnr+2.jmR.Jnr
=(.Jri1R -.Jnr)2+z.JmnRr
Asthe perfect square cannot be negative, so
mR+nrwill be minimum if
i.e., .jmR-.Jnr=0
or mR=nr
R=nr
m
or
or External resistance
=Total internal resistance of the cells.
Thus, ina mixedgrouping of cells,the current through
theexternal resistance will be maximumiftheexternal
resistanceisequal to the total internal resistance of the cells.
Examples based on
0-- -_ •.. Grouping ofCells
Formulae Used
ne
1=--
R+nr
ne
2.Forncells in parallel, I=--
nR+r
1.Forncells in series,
mne
3.For mixed grouping, I=---
mR+ nr
where n=no. ofcellsinseries in one row,
m=no. ofrows of cells in parallel.
4.Formaximum current, the external resistance
mustbe equal to the total internal resistance.
i.e., nT=R
m
or nr=mR
Units Used
EMFand terminal p.d. are in volt(V),internal
resistance (r)and external resistance R inQ,
current inampere (A).
Example 87. (a) Three cells of emf2.0 V,1.8Vand1.5V
are connected inseries. Their internal resistances are
0.0512,0.712 andInrespectively. If the batteryisconnec-
ted to an external resistor of 412via avery low resistance
ammeter, what would be thereading inthe ammeter?
(b) If the three cells abovewerejoinedinparallel, would
they be characterised by a definite emf and internal
resistance (independent of external circuit) ?If not, how will
you obtain currents in different branches? [NCERT]www4§±ttssrxvt4r±£

240. CURRENT ELECTR
ICITY
Solution. (a)The circuit diagram is shown in
Fig.3.112.
I
r----------, ~---------I r----------,
0.05QI: 0.7QI: 1Q
I~~I-'V\J'\r-"""""
~~-~------: ~!.~ y- - - ---:~!.§-~--- --
Fig. 3.112
3.55
Resistance of external circuit =Total resistance of
two resistances of 17 n connected in parallel
or R = Rl~ = 17x 17 n =8.5n
Rl+ ~17+17
17Q
As the three cells have been connected inseries to
an external resistor of 4 n,therefore
Total emf = (2.0 +1.8+1.5)V= 5.3V Fig. 3.113
Total resistance
Current,
=(0.05+0.7+1+4)n = 5.75 n
1= emf
resistance
= 5.3 A=O.92A.
5.75
(b)No,there is no formula for emf and internal
resistance of non-similar cells, joined in parallel. For
this situation, we must use Kirchhoff's laws.
Example 88. A cellof emf 1.1V andinternal resistance
or
0.5 nisconnected to a wire of resistance 0.5n.Another cell
of the same emf isconnected in series but the current inthe
wireremains the same. Findthe internal resistance ofthe
secondcell.
Solution.In first case:
Total emf, Eo=1.1 V
Total resistance, R = 0.5 +0.5 = 1 n
:.Current, I=!= 1.1 = 1.1 A
R 1
Insecond case :
Total emf,
Eo==1.1+1.1=2.2V
Total resistance,
R=0.5+0.5+r= (1+r)n
whereristhe internal resistance ofthe second cell.
C 2.2
urrent, 1=--=1.1 or r=ln.
l+r
Example 89.Two identical cells ofemf1.5Veach joined
in parallel providesupply to anexternal circuit consisting of
two resistances of17neachjoinedinparallel. A very high
resistancevoltmeter reads theterminalvoltage ofcellsto be
1.4 V. Calculate theinternal resistance of each cell.
[eBSE 0 9SC]
Solution. Here Eo=1.5 V, V=1.4 V
Letl'be the total internal resistance of the two cells.
Then
As the two cells of internal resistance rneach have
been connected in parallel, therefore.
1 1 1 1 2
-=-+- or-=-
l'r r 0.6r
r=0.6x2=1.2n.
Example 90. Four identicalcells, each of emf2V,are
joined in parallel providing supply of current toexternal
circuit consisting of two 15nresistors joined in parallel.
Theterminalvoltage of the cells,asread by an ideal
voltmeter is1.6volt. Calculate theinternal resistance of each
cell. [CBSE 002]
Solution. As shown in Fig. 3.114, four cell are
connected inparallel to the parallel combination of two
15nresistors.
lSQ
Fig. 3.114
HereEo=2 V, V=1.6 V
The external resistance provided. bytwo 15n
resistors connected in parallel is
15xIS
R=--=7.5n
15+15www4§±ttssrxvt4r±£

241. 3.56
Ifr'is
thetotal internal resistance of the four cells
connected in parallel, then
l'=R(e -VJ=7.5(2 -1.6) = 15n
V 1.6 8
Ifristheinternal resistance of each cell, then
1 1 1 1 1 4
-=-+-+-+-=-
l'r r r r r
15
or r=41'= 4x-=7.5 n.
8
Example 91.Inthecircuit
diagram given in Fig. 3.115,
the cells £1and£2have emfs
4 V and8V and internal
resistances 0.5 nand 1.0 n
respectively. Calculate' the
current ineachresistance.
4.50
60
[CBSE DISC]
Fig. 3.115
Solution. Effective emf of the circuit
=e2-e1=8- 4=4V
Total resistance of the circuit
3x6
= 1+ 0.5+ 4.5 n +--= 8 n
3+6
:.Current in the circuit, I=~=0.5 A
Current through 4.5 n resistance = I=0.5A
p.o.across the parallel combination of 3 nand 6 n
resistances is
3x6
V=R'I=-- x0.5 = 1 V
3+6
Current through 3n resistance =~=!A
3n 3
=~=!A.
so6
Current through 6n resistance
Example 92. InFig.3.116, e1and e2 arerespectively
2.0 V and 4.0 V and the resistances r1,r2and Rare
respectively 1.0 n, 2.0nand5.0 n. Calculate the current in
the circuit. Also calculate (i) potential difference betweenthe
points b anda,(ii)potential difference between a andc.
:------e---: :-e-------:
c I lL~ 2 Ib
I ~ I
: Y1 :: Y2:
---------- ----------
R
Fig.3.116
Solution.Asemfs e1 and e2 are opposing each
otherand e2 >e1,so
Netemf=e2 -e1=4-2 =2 V.
PHYSICS-XII
This emfsends circuit Iin the anticlockwise
direction.
Total resistance =R+r1+r2= 5+1+2 =8n
Current in the circuit
Net emf 2
----- =-=0.25 A.
Total resistance 8
(i)Currentinside the cell e2 flows from-ve to +ve
terminal, so the terminal p.d. of this cell is
Va-Vb=e2-Ir2
=4.0-0.25x2.0 =3.5V.
(ii)Current inside the cell e1 flows from +veto-ve
terminal. Hence the terminal p.d. ofthis cell is
Va- Vc=e1+Ir1
=2.0+0.25x1.0 =2.25V.
Example 93.In the twoelectric circuits shown in
Fig.3.117,determine thereadingsof ideal ammeter (A) and
the ideal voltmeter (V). [CBSE DISC]
Fig.3.117 (a) (b)
Solution. In the circuit (a)
Total emf = 15 V, Total resistance =2n
15V
Current, 1=-- = 7.5 A
2n
Asthe current Iflows from-veto+ve terminal
inside the cellof6V,the terminal p.d. of the cells is
V=e-Ir=6-7.5x 1 =-1.5 V
:.Reading of ammeter =7.5 A,
Reading of voltmeter = -1.5V.
In the circuit(b)
Netemf=9-6=3V,Total resistance=2n
3V
Current, I=-= 1.5 A
2n
As the currentIflows from +ve to -ve terminal
inside the 6 V cell, so the terminal p.d. of the cell is
V= e +Ir= 6 + 1.5 x1 = 7.5 V
Reading of ammeter=1.5A,
Reading ofvoltmeter =7.5 V.«««2vw~m{lzq§m2kwu

242. CURRENT E
LECTRICITY
Example 94. Anetwork of resistances isconnected to a
16V battery with internal resistance of10,as shown in
Fig.3.118.
(a) Compute equivalent resistance of the network,
(b) Obtain thecurrent ineachresistor, and
(c) Obtain the voltage dropsVAB,VBCandVco'
[NCERT;CBSE F 10]
40 120
I10
B
16 V
r---------.
, 10', ,
,
,
1 --_ ••
Fig. 3.118
Solution. (a)Asthe two 4 0 resistances are in
parallel, their equivalent resistance is
R=4x4=20
14+4
..
Also, the 120and60resistances are in parallel,
their equivalent resistance is
~=12x6 =40
12+6
Now the resistances R1, ~and 10 are inseries.
Hence the equivalent resistance ofthenetwork is
R = Rl +~ +1 =2+4+1=70.
(b)The total currentin the circuit is
I=_e_=~=2A
R+r 7+1
or
The potential difference betweenAandBis
VAB=4II=412
II=12
ButII+12=1=2A
II=12=1A
The potential difference between C and D is
Vco= 1213= 614i.e.,14=213
But 13+14=I=2 A
13+213=2 A
2
13="3A
and
4
14=-A
3
(c) VAB= 4xII=4x1 =4 V,
VBC=1x'I = 1x2 =2 V,
2
Vco= 12x13= 12x3"=8 V.
3.57
Example 95. A20V battery of internal resistance 10is
connected to three coils of 12 0,6 0 and4 0in parallel, a
resistor of50and a reversed battery (emf =8V and internal
resistance =2 0),as shown in Fig. 3.119.Calculate the
current in each resistor and the terminal potential difference
across each battery. [CBSE00OlC]
10 20V
~---------I
r--l __ -!:--'\.fV'v~111-+:-"---'
50 c:
N
,....;
8V, 20
.---------~
:1II---JV'If\r-L' ----..------'
Fig.3.119
Solution. Equivalent resistance R' of 12 0,60,40
resistances connected in parallel is given by
1 1 1 1 6 1
-=-+-+-=-=-
R' 12 6 4 12 2
R'=20or
Total resistance = 1 + 5 + 2 + 2 = 10 0
Net emf =20-8=12
Current in the circuit, 1=12=1.2A
10
So the current through each battery and 5 0resistor
is1.2A.
P.D. across the parallel combination of three
resistors is
V'=IR'=1.2x2 =2.4 V
Current in 2 0 coil=2.4 =0.2A
12
Current in 6 0 coil = 2.4 = 0.4 A
6
Current in 4 0 coil = 2.4 =0.6 A.
4
Terminal p.d. across 20Vbattery,
V=e-Ir= 20 -1.2 x1 =18.8 V
Terminal p.d. across8 Vbattery,
V'=e'+lr=8 + 1.2 x2 = 10.4 V.
Example 96.36 cellseach of internal resistance 0.50and
emfl.5 Veach are used to sendcurrent through an external
circuit of2 0resistance. Find the best mode of grouping
them and the current through the externalcircuit.
Solution. Here e= 1.5 V, r= 0.5 0, R=2 0
Total number of cells, mn=36 ...(1)www4§±ttssrxvt4r±£

243. 3.58
For maximu
m current in the mixed grouping,
nr=R or nx0.5=2 .... (2)
m m
Multiplying equations (1) and (2), we get
0.5n2=72 or n2=i44
n= 12 and m= 36 = 3
12
Thus for maximum current there should be three
rows in parallel, each containing 12 cells in series.
:.Maximum current
mne 36x1.5 =4.0 A.
mR+nr3 x 2+12x0.5
Example97.12cells, each of emf1.5V and internal
resistance of0.50,are arranged in m rows each containing
n cells connected in series,asshown. Calculate the values of
nand mfor which this combination would send maximum
current through an external resistance of1.5O.
[CBSESamplePaper08]
~R=1.5n~
~~'----~~i
i 1
1 1
1--_- - - - - - --- - - - ~ mrows
1 I'
1 1 \
~ - - - -1111- ---11- - - -_I
(ncellsin eachrow)
Fig.3.120
Solution.For maximum current through the
external resistance,
External resistance
= Total internal resistance of the cells
R=nr
m
1.5=n~~.5 [mn=12]
or
n
or 36 =n2 orn=6andm= 2.
Example98.In the given circuit in the steady state, obtain
the expressions for (i) the potential drop (ii) the charge and
(iii)the energy stored in the capacitor, C [CBSEF15]
V R
F
PHYSICS-XII
Solution.In the steady state (when the capacitor is
fully charged), no current flows through the branchBE.
Net emf =2V-V=V
Net resistance = 2 R+R = 3 R
C . h . .I V
:. urrent In t e CIrCUIt, =-
3R
Potential difference acrossBE
V 4
=2V-Ix2R =2V --x2R =-V
3R 3
(i)Potential difference across C = ±V-V=V .
3 3
(ii)Charge on the capacitor, Q= CxV=CV.
3 3
(iii)Energy stored in the capacitor
= ~C(V)2 CV2
2 3 18
~roblem5 For Practice
1.Three identical cells, eachof emf 2V and internal
resistance 0.2 nareconnected inseries to an
external resistor of 7.4 n. Calculate the current in
thecircuit. (Ans. 0.75 A)
2.Three identical cells each of emf 2 V and unknown
internal resistance are connected in parallel. This
combination is connected to a 5 n resistor. If the
terminal voltage across the cells is 1.5V, what is the
internal resistance of each cell? [CBSE OD99]
(Ans.5n)
3.Two cells connected in series have electromotive
force of 1.5 Veach. Their internal resistances are
0.5nand 0.25n respectively. This combinationis
connected to a resistance of 2.25n. Calculate the
current flowing inthe circuit and the potential
difference across the terminals of each cell.
(Ans.1.0 A, 1.0 V, 1.25 V)
4.When 10 cells inseries are connected to the ends of
a resistance of 59n, the current is found to be
0.25 A, but when the same cells after being
connected in parallel are joined to the ends of a
0.05n, the current is 25 A. Calculate the internal
resistance and emf of each cell. (Ans. O.ln, 1.5 V)
5.Find the minimum number of cells required to
produce an electric current of 1.5 A through a
resistance of 30n. Given that the emf of each cell is
1.5 V and internal resistance 10 o.
(Ans.120 cells, 60 cells in one row
and two rows in parallel)
Two identical cells, whether joined together in series
or in parallel give the same current, when connec-
ted to an external resistance of 1 n. Find the inter-
nal resistance of each cell. [ISCE95](Ans. 1n)
A
V
C
B
~
E
6.
2V
2R
C 0
Fig.3.121«««2vw~m{lzq§m2kwu

244. CURRENT ELECTRIC
ITY
7.Asetof 4 cells, each ofemf2V and internal resis-
tance1.5 nareconnected across an external load of
10 n with 2 rows, 2 cells in each branch. Calculate
thecurrent in each branch and potential difference
across 10 n. [Karnataka91C]
(Ans.0.175 A,3.5 V)
HINTS
nE. 3x2 6
1.1=--= =-=0.75A.
R+nr7.4+3x0.2 8
2.HereE.= 2V, V=1.5 V, R= 5n
Ifr'isthe totalinternal resistance of thethree cells
connected in parallel, then
r'=R[E.-V]= 5[2- 1.5] = ~n
V 1.5 3
111 1 33
But-=-+-+-or-=-..r=5 n.
r' r r r 5r
3.Total resistance, R= 0.5+0.25+2.25= 3.0n
Total emf, E.=1.5+1.5 = 3.0 V
C . h..1E3.0lOA
urrent m t e circuit, =-=- =.
R3.0
P.D. across first cell,
VI=E -bi=1.5- 1.0 x0.5=1.0 V
P.D.across second cell,
V2=E -1r2= 1.5 - 1.0 x0.25 =1.25 V.
4.For series combination. The current is
lOE
---=0.25A
59+10r
Forparallel combination. The current is
lOE
----=25A
10x0.05+r
Onsolving the above two equations,
r= 0.1 nand E=1.5 V.
nr nx1
5.As-= R:.--=30 or 1I=30m
m m
nE. nx1.5
I=- or 1.5 = -- orn=60
2R 2x30
. .m= 60/ 30 =2andmn=120.
6.When the cells are connected in series, (Fig. 3.122),
the current in thecircuit is
I=~
51+2r
e
,
e
I
IsIs
R=lQ
-JVVv-
r.
Fig. 3.122
When the cells are
connected in
parallel (Fig.3.123),
thecurrent in the
circuitis
3.59
s
-JVVv-
re
Ip i,
.J\.N\r
r
R=lQ
I=_--::e_
P rxr
1+--
r+r
2E.
2+r
Fig. 3.123
Given'.>', ..
1+2r 2+r
or1+2r=2+rorr=1n.
7.Thecircuit diagram is shown below.
-JVVv-
r
e,
R=lOQ
Fig. 3.124
Heree=2 V,r= 15 n, R= 10 n,11= 2, m=2
ne 2x2 4
1=--= =-=0.35A.
Rnr 2x1.5 11.5
+- 10+--
m 2
The two branches areidentical.
:.Current ineach branch = 0.35 = 0.175 A.
2
Potential difference across R
=IR=0.35x10= 3.5 V.
3.22HEATING EFFECT OF CURRENT
42.Whatisheating effect of current?Explain the
causeof heatingeffectof current.
Heating effect of current. Consider apurely
resistive circuiti.e.,a circuit which consists of only
some resistors and a source of emf. The energy of the
source gets dissipated entirely in the form ofheat
produced in the resistors. Thephenomenon of theproduc-
tion of heat inaresistor bytheflow of anelectric current
through it iscalledheating effect of current or Joule heating.
Cause of heating effect of current. When a potential
difference isapplied across the ends of a conductor, its
free electrons get accelerated in the opposite direction
of the applied field. But the speed of the electrons does
not increase beyond a constant drift speed. This iswww4§±ttssrxvt4r±£

245. 3.60
beca
useduring the course of their motion, the
electrons collide frequently with the positive metal
ions. The kinetic energy gained by the electrons during
theintervals of free acceleration between collisions is
transferred to themetal ions atthe time of collision.
The metal ionsbegin to vibrate about their mean
positions more and moreviolently. The average kinetic
energy of theions increases. This increases the
temperature ofthe conductor. Thus the conductor gets
heated due to the flow of current. Obviously, the
electrical energy supplied by the source of emf is
converted into heat.
3.23HEAT PRODUCED BY ELECTRIC
CURRENT: JOULE'S LAW
43. Obtain an expression for the heat developed ina
resistor bythe passage of an electric current through it.
Hence stateJoule'slaw ofheating.
Heat produced inaresistor. Consider a conductor
ABof resistance R,shown inFig.3.125. A source of emf
maintains a potential difference Vbetween its ends A
and Band sends asteady currentIfrom AtoB.Clearly,
VA>VBand the potential difference across ABis
V=VA-VB>0
Fig.3.125Heat produced in a resistor.
Theamount ofcharge that flows fromAto B in
timetis
q=It
As the chargeqmoves through a decrease of
potential of magnitude V,itspotentialenergy
decreases bythe amount,
U=Final P. Eat B- InitialP.E.atA
=qVB-qVA =-q(VA -VB)=-qV<O
If thecharges move through the conductor without
suffering collisions, their kinetic energy would change so
thatthe total energy is unchanged. By conservation of
energy, thechange in kinetic energy must be
K=-U=qV=Itx V=VIt>O
Thus, in case, charges were moving freely through
the conductor under the action of theelectric field,
their kinetic energy would increase as they move.
However, we know that on the average, the charge
carriers orelectrons do not move with any acceleration
PHYSICS-XII
but with a steady driftvelocity. This is because of the
collisions of electrons with ions and atoms during the
course of theirmotion. Thekinetic energy gained by
theelectrons is shared with the metal ions. These ions
vibrate more vigorously and the conductor gets heated
up. The amount of energy dissipated as heat in
conductor intimetis
or
2
H=VItjoule=12Rtjoule=V tjoule
R
VI 2 2
H=_tcal=IRtcal= ~ cal
4.18 4.18 4.18 R
Theabove equations are known as Joule's law of
heating. According tothislaw,theheat produced in a resistor
is
1.directly proportional to thesquare of current fora
givenR,
2.directly proportional to theresistance Rfor a givenI,
3.inversely proportional. to theresistance Rfor a given
V,and
4.directly proportional to the time for whichthe
current flows through the resistor.
ForYour Knowledge
>The equation: W =VItisapplicable to the conversion of
electrical energy into any other form, butthe equation:
H=[2Rtisapplicable onlyto the conversion of electrical
energy into heat energy in anohmic resistor.
>Joule's law of heating holds good even for a.c.circuits.
Only current andvoltage haveto be replaced by their
rms values.
>Ifthe circuit is purely resistive, the energy expended
bythesource entirely appears as heat. But if the
circuithas an active element like a motor, then a part
of the energy supplied by the source goes to do useful
workand the rest appears asheat.
---------/
3.24ELECTRIC POWER
44.Define theterm electric powerand state its SI
unit.
Electric power. The rate at which work isdone by a
source ofemfinmaintaining an electric current through a
circuit iscalledelectric power of the circuit. Or,the rate at
which an appliance converts electric energy into other forms
of energy iscalled itselectric power.
If a current Iflows through acircuit for time tat a
constant potential difference V,then the work doneor
energy consumed is given by
W=VItjoulewww4not•s»riv•4~om

246. CURRENT ELECT
RICITY
:.Electric power,
W 2 V2
P=-=VI=I R=-
t R
or Electric power=currentx voltage.
SI unit of electric power.The SI unitof electric
power iswatt(W).Thepower of an applianceisone wattif
it consumes energy at the rate of1joule per second.Or,the
powerofa circuit is onewatt if 1ampere of current
flows through it on applying a potential difference of 1
volt across it.
1 1joule 1joule1coulomb
watt= = x ----
Isecond1coulomb 1second
or 1watt=1voltx1ampere
Thebigger units of electric power are kilowatt
(kW) andmegawatt(MW).
1kW=1000Wand 1MW =106W
The commercial unit of powerishorse power(hp)
Ihp=746w.
3.25ELECTRIC ENERGY
45. Define the term electric energy. State its 51 and
commercial units.
Electric energy. Thetotal work done (or the energy
supplied) bythe source ofemf in maintaining an electric
current in a circuit for a given timeiscalledelectric energy
consumed inthe circuit. It depends upon the power of
the appliance used in thecircuit and the time for which
this power is maintained.
Electric energy,
W=P.t=VItjoule=12Rtjoule
TheSI unit of electric energy isjoule(J).
1joule=1voltx1amperex1second
=1wattx1second
Commercial unit of electric energy. The commercial
unit of electric. energy iskilowatt hourorBoard of
Trade(B.O.T.) unit.One kilowatt hour is defined asthe
electric energy consumed byan appliance ofl kilowatt in one
hour.
or
1kilowatt hour=1kilowattx1hour
=1000wattx3600s
=3,600,000 joules
1kWh=3.6x106J
The electric metres installed in our houses mea-
sure the electrical energy consumed in kilowatt hours.
Another common Unit of electric energy iswatt
hour. Itis theelectric energy consumed byan appliance of
one watt in one hour.
1watt hour=1wattx1hour=3.6x103J
3.61
3.26POWER RATING
46.Whatismeant by the power rating of a circuit
element? Briefly explain how can we measure theelectric
power of an electric lamp ?
Power rating.The power rating of an electricalappliance
isthe electrical energy consumed per second by the appliance
when connected across the marked voltage of themains. Ifa
voltageVapplied across a circuit element of resistance
Rsends current Ithrough it,then power rating of the
element will be
V2
P=- =I2R=VIwatt
R
Measurement of electric power. To measure the
electric power of an appliance, sayan electric lamp, we
connect a batteryandanammeter inseries with the
electric lamp and a voltmeter in parallel with it, as
shown in Fig. 3.126.Suppose thevoltmeter readsV
volts and the ammeter reads Iamperes, then power
rating of the electric lamp will be
P=VIwatt
Fig.3.126Tomeasure electric power of
an electric lamp.
3.27POWER CONSUMPTION IN A
COMBINATION OFAPPLIANCES
47. Prove that the reciprocal of the total power con-
sumed by a series combination of appliances isequal to the
sum of the reciprocals of the individual powers of the
appliances.
Power consumed by a series combination of
appliances. Asshown inFig.3.127, consider a series
combination of three bulbs of powers PI'P2and P3 ;
which have been manufactured forworking on the
samevoltageV.
Fig. 3.127Seriescombination of bulbs.www4§±ttssrxvt4r±£

247. 3.62
The res
istances of the three bulbs will be
V2 V2 V2
RI=p:'~=P:' ~=P:
1 2 3
Asthe bulbs are connected in series, so their
equivalent resistance is
R=RI+~+~
If P is the effective power of the combination, then
V2 V2 V2 V2
-=-+-+-
P PI P2 P3
1 1 1 1
or -=-+-+-
PPI P2P3
Thusfor aseries combination ofappliances, the
reciprocal of the effective power isequal to the sum of the
reciprocals of the individual powers of the appliances.
Clearly, whenNbulbs ofsame power Pare
connected in series,
P
Peff=N
Asthebulbs are connected in series, the current I
through eachbulb will be same.
1= V
Rl+~+~
Thebrightness of the three bulbs willbe
P"-I2R p'=PR P'=I2R
1 - I' 2 "2' 3 "3
AsRex ~ ,the bulbof lowest wattage (power) will
P
havemaximum resistance and itwill glow with maxi-
mum brightness. When the current in the circuit exceeds
the safety limit, the bulbof lowest wattage will be
fused first.
48. Prove that when electrical appliances are
connected in parallel, thetotal power consumed is equal
tothe sum of the powers of the individual appliances.
Power consumed by a parallel combination of
appliances. As shown in Fig.3.128,consideraparallel
combination of three bulbs of powersPI'P2andP3,
which have been manufactured for working on the
same voltage V.
~-----oVo-----~
Fig. 3.128Parallel combination ofbulbs.
PHYSICS-XII
The resistances of thethree bulbs will be
V2 V2 V2
RI=p:' ~=P:' ~=p:
123
As the bulbs are connected inparallel, their
effective resistance R is given by
1 1 1 1
-=-+-+-
R RI ~ ~
Multiplying both sides by V2,we get
V2 V2 V2 V2
-=-+-+-
R RI ~ ~
or P=PI+P2 +P3
Thusfora parallel combination of appliances, the effective
power isequal to the sum of the powers of the individual
appliances.
IfNbulbs, each of powerP,are connected in
parallel, then
Peff=NP
The brightness of the three bulbs will be
V2 V2 V2
PI=~,P2= ~,P3= ~
As the resistance of the highest wattage (power)
bulb isminimum, it will glow with maximum bright-
ness. If the current in the circuit exceeds the safety
limit, thebulbwith maximum wattage will be fused
first.Forthis reason, the appliances in houses are
connected in parallel.
3.28EFFICIENCY OF A SOURCE OF EMF
49. Define efficiency of a source ofemf Write an
expression for it.
Efficiency of a source of emf.The efficiency of a source
of emfis defined astheratio of the output power tothe input
power. Suppose a source of emf eandinternal
resistance risconnected to an externalresistance R.
Then its efficiency will be
Output powerVI V
11= -
Input powerel -E
R
11=--
R+r
IR
I(R+r)
or
50. (a) A battery of emfe and internal resistance ris
connected across a pure resistive device (e.g.,an electric
heater or an electric bulb) of resistance R.Show that the
power output of the device is maximum when there is a
perfect 'matching' between the external resistance and the
source resistance (i.e., whereR=r).Determine the
maximum power output.www4§±ttssrxvt4r±£

248. CURRENT ELECTRICITY
(
b)Whatispower output of the source aboveifthe
battery isshorted ?Whatisthe power dissipation inside
thebattery inthatcase? [NCERT)
Maximum power theorem.itstatesthatthe output
power of asource of emfismaximum when the external
resistance in the circuit isequal to the internal resistance of
thesource.
Let emf of the battery
Internal resistance
Resistance of the device
=r
=R
.'.Current through the
device,
Total emf
1=-----
Total resistance
R+r Fig. 3.129
.'.Power output of the resistive device will be
P=I2R =(_e_J2 R
R+r
e2R e2R
(R+r)2 (R-r)2+4Rr
Obviously, the power outputwill be maximum
when
R -r=0orR=r
Thus, the power output of the device is maximum
when there is a perfect matching between the external
resistance and the resistance of the source, i.e.,when
R =r.This proves maximum power theorem.
Maximum power output ofthesource is
Pmax=~= e2 [Putting R=rin Eq. (i)]
(r+r)4r
(b)When the battery is shorted, R becomes zero,
therefore, power output =O.In this case, entire power
of the battery is dissipated as heat inside the battery
due to its internal resistance.
Power dissipation inside the battery
=I2r=(~rr=er2.
51.Show that the efficiency of a battery when
delivering maximum power is only50%.
Maximum efficiency of a source of emf. For a
source of emf,
Input power = eI
Output power=VI
3.63
.'. Efficiency
VI V IR R
11=-=-= =--
eIeI(R+r)R+r
When the source delivers maximum power, R=r
r1
.. 11=--=-=50%
r;+-r2
Thus theefficiency of a source of emf isjust 50%
when itisdelivering maximum power.
3.29EFFICIENCY OF AN ELECTRIC DEVICE
52. Define efficiency of anelectric device. Write an
expression for the efficiency of an electric motor.
Efficiency of an electric device. The efficienClJof an
electric deviceisdefined as the ratio of the output powerto
the input power
Output power
11=--"---"---
Input power
Foran electric motor,we can write
···(0
Output mechanical power
11=--"--------=---
Input electric power
Here,input electrical power
=Output mechanical power+Powerlostasheat
53.(a) An electric motor runs on a d.c. source of emfe
and internal resistance r. Show thatthepoweroutput of
the source is maximum when the current drawn by the
motor ise/2r.
(b) Show that power output of electricmotor ismaximum
when the back emfisone-half the source emf provided the
resistance of the windings of the motor is negligible.
(c)Compare and contrast carefully the situation in this
exercise with that in Q.SO(a) above. [NCERT)
(a)Output power from a source connected to an
electric motor. Let the current drawn by themotor be I.
Then
Power output ofthesource, P=eI-I2r
P· . hdP0
ISmaxImum w en-=
dI
e-2Ir=0 or I= ~
2r
or
Hencethe power output of the sourceismaximum when
the current drawn by the motorise/2r.
(b)Here,emf of source = e
Internal resistance of source =r
Back emf of motor =e'
Resistance of motor =R"'O
Astheexternal resistance R is negligible, therefore
.th..e-e'
currentille circuit =-- .
rwww4§±ttssrxvt4r±£

249. 3.64
And power outp
ut ofthe motor
=Power output ofthe source
=eI-Pr
From part(a),this is maximum when
e-e'es
1=-
2r
or ore'=~
2r 2r
Hencethepower output of electric motor ismaximum
when the backemf is. one-half the source emf
(c) The condition inQ.50(a) is forapassive resistor
in which the entire electric energy is converted into
heat while the condition inQ.53(a)is for a non-passive
resistor(e.g., electric motor) inwhich the supplied
electric energy changes partly into heatand partly into
mechanical work. So the former is a special case ofthe
latter.
3.30APPLICATIONS OF HEATING
EFFECT OF CURRENT
54.Discusssome practical applications of the heating
effect of current.
Applications of heating effect of current. Some of
the important applications of Joule heating are as follows:
l.Household heating appliances. Many electrical
appliances usedin daily lifeare based on the heating
effect of current such as room heater, electric toaster,
electric iron, electricoven, electric kettle, geyser, etc.
The designing of these devices requires the selection of
a proper resistor. The resistor should have high
resistance so that most ofthe electric power is
converted into heat. In most of the household heating
appliances, nichrome element is usedbecause of the
following reasons :
(i)Its melting point is high
(ii) Its resistivity is large
(iii)It is tensile, i.e.,it can be easily drawn into wires.
(iv)It is not easily oxidised by theoxygen of the air
when heated.
2.Incandescent electric bulb.It is animportant
application of Joule heating in producing light. It
consists of a filament offine metallic wire enclosed in a
glass bulb filled withchemically inactive gases like
nitrogen and argon. Thefilament material should
have high resistivity and high melting point.
Therefore, tungsten (melting point 33800q is used for
bulb filament. When current is passed through the
filament, it gets heated. toa high temperature and
emits light. Most of the power consumed by the
filament is converted into heat and only asmall part
PHYSICS-XII
of it appears as light. A bulbgivesnearly 1 candela of
lightenergy for the consumption of every wattof
electric power.
3.Electric fuse.It is a safety device used to protect
electrical appliances from strong currents. A fuse wire
must havehighresistivity and low melting point. It is
usually made from an alloy of tin (63%) and lead
(37%). It is put inseries with the live wire of the circuit.
When thecurrent exceeds the safetylimit, thefuse
wire melts and breaks the circuit. The electric
installations are thus saved from getting damaged.
The fuse wire of suitable current rating (1 A,2 A,
3 A, 5 A, 10 A etc.) should be usedin the circuit
depending on the load in the circuit. Forexample,
whenweuse an electric iron of1 kW electric power
withelectric mains of 220 V, a current of (1000/220) A
i.e.,4.54 A flows in the circuit. This requires a fuse of
5 A rating.
4.Electric arc. Itconsists of two carbon rods with a
small gapbetween their pointed ends. Whenahigh
potential difference of 40 - 60V is applied between the
two rods, veryintense light is emitted by the gap. We
know that E=-dV / dr. Clearly, Ewill be large if the
gapissmall. When the electric field exceeds the
dielectric strength of air, ionisation of air occurs. This
causes a big spark to pass across thegap.
5. Other devices. Many other devices are based on
theheating effect ofcurrent such as electric welding,
thermionic valves, hotwire ammeters and voltmeters.
55.Explainwhyiselectric powertransmitted at high
voltagesand low currents to distant places.
High -oltage po 'er transmission.Electric power is
transmitted from power stations to homes and
factories through transmission cables. These cables
have resistance. Power is wasted in them as heat. Let
us see how can weminimise this power loss.
Suppose power P isdelivered to a load R via
transmission cables of resistance Rt.IfVis thevoltage
across loadR andIthe current through it, then
P=VI
The power wasted in transmission cables is
p2R
P.=12R=__1
1 1 V2
Thus the power wasted in the transmission cables
isinversely proportional to the squareofvoltage.
Hence to minimise the power loss, electric power is
transmitted to distant places at high voltages and low
currents. These voltages are stepped down by
transformers before supplying to homes and factories.www4§±ttssrxvt4r±£

250. CURRENT ELE
CTRICITY
For Your Knowledge
~Theemission oflight by a substance when heated to a
high temperature is called incandescence.
~A heater wire is made from a material of large resis-
tivity and high melting point while a fuse wire is made
from a material of large resistivity and low melting point
~The loadinanelectric circuit refers to the current
drawn by the circuit from the supply line. Ifthe
current inacircuitexceedsthe safe value, we say that
the circuit is overloaded.
~The temperature upto which a wire gets heated (i.e.,
steady state temperature 9) is directly proportional to
the square ofthecurrent andis inversely proportional
tothe cube of itsradius but is independent of its length.
]2
9ex:-
r3
~When the resistances are connected in series,theI
current] througheach resistance is same. Consequently,
Pex:R (':P=12R)
and Vex:R (.: ~=]R)
Hence in aseriescombination of resistances, the
potential difference, power consumed and hence heat
produced will belarger in the higher resistance.
~When the resistances are connected inparallel, the
potential difference Vis same across each resistance.
Consequently,
1
Pex: -
R
and
1
Iex: -
R
Hence in aparallel combinationof resistances, the
current, power consumedand hence heat produced
will belarger in the smaller resistance.
~,--------------~
Formulae Used
1.Heat produced by electric current,
12Rt
H=fRtjoule=-- cal
4.18
orH=VItjoule=VItcal
4.18
W V2
2.Electric power, P=- =VI=12R= -
t R
3.Electric energy, W=Pt=VIt=/2Rt
Units Used
CurrentIis in ampere, resistance Rinohm, time t
in second, powerPin watt, electric energy in joule
or in kWh.
3.65
Example 99. Anelectric current of 4.0Aflows through a
120resistor. Whatisthe rate at which heat energyis
produced in the resistor? [NCERT]
Solution. Here I=4 A, R =120
Rate of production of heat energy,
P =I2R=42 x12 =192 W.
Example 100. How many electrons flow through the
filament of a120V and60Welectric lamp per second?
Givene=1.6x10-19C.
Solution.HereP =60W,V=120V,t=1s
1=P=~=O.5A
V 120
But I=!1.=ne
t t
No.of electrons flowing per second is
It0.5x1 18
n=-= 19=3.125x10 .
e1.6xlO-
Example 101.A heating elementismarked210V,630W.
Whatisthe current drawn by the element when connected to
a210Vd.c.mains? Whatisthe resistance of the element?
[NCERT]
Solution.HereP =630 W,V=210 V
P630
Current drawn 1=-=-=3 A.
, V 210
. V210
Resistance of the element,R=-= - =70 O.
I 3
Example 102.A10Vstorage battery of negligible internal
resistanceisconnected across a500resistor made of alloy
manganin. How much heat energyisproducedinthe resistor
in1h?What'isthe source of this energy? [NCERT]
Solution. Here V= 10V, R= 500,t=1h=3600s
Heatenergyproduced in1his
H=V2t= 10x10x3600=7200J.
R 50
The source of this energy is the chemical energy
stored in the battery.
Example 103.An electric motor operates on a50Vsupply
and draws a current of12AIf the motor yields a mecha-
nical power of150 W, whatisthe percentage efficiency of the
motor? [NCERT]
Solution.Input power =VI= 50 x12 =600 W
Output power =150 W
Efficiency of motor
=Outputpower x100=150x100
Input power 600
=25%.www4§±ttssrxvt4r±£

251. 3.66
Example104.An elec
tricmotor operating on a50V d.c.
supply draws a current of12 A. Ifthe efficiency of the motor
is 30%, estimate the resistance of the windings of the motor.
[NCERT]
Solution. Here V= 50V,I= 12 A, Yl=30%
Asthe efficiency of electric motor is 30%,therefore,
power dissipated as heat is
70
P= 70% ofVI=-x50x12 W = 420 W
100
But power dissipated as heat, P=[2R
PR= 420
R= 420 = 420= 2.9O.
12144
or
Example105.(a) A nichromeheating element across
230Vsupply consumes 1.5 kWof power and heats up to a
temperature of 7500C.A tungsten bulb across the same supply
operates at a much higher temperature of1600°C in order to
be able toemit light. Does itmean that the tungsten bulb
necessarily consumes greater power?(b)Whichof the two
has greater resi~J!mce_: a 1kWheater or aiootvtungste..n
.bulb,bothmarked for230V? [NCERT]
Solution. (a)No,the steady temperature acquired
by aresistor depends not only on the power consumed
butalso its characteristics such as surface area, emissi-
vity, etc., which determine its power loss due to radiation.
(b)HereV=230 V, PI=1 kW =1000 W,P2=100 W
R=V2=230x230 0 = 52.9 0
1PI 1000
~ =V2=230x230 0 =5290
P2 100
Thus the 100 W bulb has agreater resistance.
Example106.Anelectric power station (100MW)
transmits power to adistant load through long and thin
cables. Which of the two modes of transmission would result
inlesserpower wastage: power transmission of:(i)20AJOOV
or(ii)200V? [CERT]
Solution. Let R be the resistance of transmission cables.
Here P=100 MW = 100x106W
(i)VI= 20AJOOV
P100x106
:.Current, II=-= = 5000 A
VI20AJOO
Rate of heat dissipation at 20AJOOV is
PI=IiR=(5000)2 R =25x106R watt.
(ii)V2=200 V
. 100x106 5
..Current, 12= 200 = 5 x10 A
PHYSICS-XII
Rate of heat dissipation at200Vis
P2=I~R=(5x105)2R=25x1010Rwatt
Clearly, PI<P2 •
Hence there will be lesser power wastage when the
power is transmitted at 20,000 V.
Example107.Two ribbons are given with the following
particulars:
Ribbon
I
A B
Alloy IConstantan Nichrome
Length (m) 8.456 4.235
Width(mm) 1.0 2.0
Thickness (mm) 0.03 0.06
Temp. coefficient of egligible egligible
resisti vity(OC-1 )
Resistivity (Om) 4.9x10-7
I
1.1x10-6
For afixed voltage supply, which of thetworibbons corres-
ponds toa greater rate of heat production? [CERT]
Solution.Since R = pi
A
Resistance of constantan ribbon,
R=4.9x10-7x8.456 0 = 138.1 0
11.0x10-3x0.03 x10-3
LetVbe the fixedsupply voltage. Then the rate of
production of heat in constantan ribbon,
V2 V2
P,= - =-- watt
1Rl138.1
Resistance ofnichrome ribbon,
~= 1.1 x10-6x4.235 0 =38.80
2.0x10-3x0.06x10-3
Rate ofproduction of heat in nichrome ribbon,
V2 V2
P2=-=--watt
».38.8
Clearlynichrome ribbon has greater rate of produc-
tionofheatbecause of its lesserresistance.
Example108.A heater coil israted100W,200V.Itiscut
into two identical parts. Both parts are connectedtogether in
parallel,tothe same source of200 V. Calculate the energy
liberated per second in the new combination.
[CBSE OD 2000]
Solution. Resistance of heater coil,
R= ~ =200x200=400 0
P 100
Resistance of either half part =200 0www4§±ttssrxvt4r±£

252. CURRENT ELECTRICITY
Equivalent res
istance when both parts are connected
in parallel,
R'= 200 x200= 100 Q
200+200
Energy liberated per second when combination is
connected to a source of200V
V2 200x200
=- = =400J.
R' 100·
Example109.Anelectric bulbismarked 100 W,230 V.If
the supply voltage drops to 115V,what is theheat and light
energy produced by the bulb in 20min? Calculate the
current flowing through it. [NCERT; CBSE F 94]
Solution. If the resistance of thebulb be R, then
Rate of production of heat and light energy,
V2
P=-
R
R=V2= 230x230 = 529 Q
P 100
When the voltage drops to V'=115V,thetotal
heat and light energy produced by the bulbin20min
willbe
V,2
H=Pxt=--xt
R
115x115
= x20x60 =30,000 J= 30kJ.
529
I=V'= 115 = ~A.
R52923
Example110.An electric bulb rated for500 Wat100V is
used in circuit having a200V"supply. Calculate the
resistanceRthat must be putinserieswith the bulb,sothat
the bulb delivers 500 W. [IIT 87]
Solution. Resistance ofthebulb,
R=V2= 100x100 = 20 Q
P 500
Current,
Current through the bulb,
1= V= 100 =5A
R20
Forthe same power dissipation, the current
through bulb must be5 A.
When the bulb is connected to 200 Vsupply, the
saferesistance of the circuit should be
R' =V'= 200=40 Q
I 5
Resistance required to be put in series with the
bulbis
R' -R= 40 - 20 = 20 Q.
3.67
Example 111.Themaximum power rating of a20Q
resistoris2.0kW.(That is,thisisthemaximum power the
resistor can dissipate (as heat) without melting or changing
in some other undesirable way). Would you connect this
resistordirectlyacrossa300Vd.c.source of negligible
internal resistance? Explain your answer.
[Haryana 97C ; NCERT]
Solution. Maximum power rating of the given20Q
resistor,
P'=2.0kW
When connected to 300Vd.c. supply, the power
consumption or rate of production of heat Wouldbe
P= ~ = 300x300W= 4500W= 4.5kW
R 20
This power consumption exceeds the maximum
power rating of the resistor. Hencethe20Qresistor
must not be connected directly across the300Vd.c.
source. For doing so, a small resistance of 10Qshould
be connected in series with it.
Example112.An electric heater and an .electric bulb are
rated500 W,220 V and100 W,220 V respectively. Both are
connected in seriesto a220V d.c.mains. Calculate the
power consumed by (i)the heater and (ii)electric bulb.
[CBSE D 97]
Solution. Resistances of heater and bulb are
Rl=V2= 220x220 = 484 =96.8 Q
PI 500 5
~ =V2= 220x220 =484Q
P2 100
Total resistance of series combination is
Rl+ ~=96.8+484 = 580.8 Q
Current,I=V= 220:::00.38 A
R580.8
(i)Power consumed by heater is
PI =PRl=0.382 x96.8 = 13.8 W.
(ii)Power consumed by bulb,
P2=P~=0.382 x484 =69.89w.
Example113.Two heaters are marked 200V,300Wand
200 V, 600 W. If the heaters are combined in series and the
combination connected to a200V d.c. supply, which heater
will produce more heat ? [NCERT]
Solution. Resistances of the two heaters are
R =V2= 200x200 = 400 Q
IP13003
~ =V2= 200x200 = 200 Q
P2 600 3www4§±ttssrxvt4r±£

253. 3.68
Fors
eries combination,
600
R1+ ~ =3=2000
V 200
:.Current 1=-=- =1A
, R 200
Power dissipations in the two heaters are
P{= 12R1=12x400 = 400 W
3 3
P;=12 ~ = 12x200 = 200 W
3 3
.. P{ =2P;
Thefirst heater (of 300 W) produces more heat than
the second heater.
Example 114. In a part of thecircuitshown in the
Fig. 3.130,the rate of heat dissipation in 40resistoris
100J/s.Calculate theheatdissipated in the30resistor in
10seconds. [CBSE Sample Paper 03]
R, R2
40 20
30
Fig.3.130
Solution.Let11 bethe current through the series
combination of ~ and~ and 12bethe currentthrough
~.
r.o.across(R1+ ~)=r.o.across ~
.. (4 +2) 11=3 12 or 12 =211
Rate ofheat dissipation in 4 0 resistor
=I~R1=I~x4 =100 Js-1
~100
.. 11=4=.J25 =5 A
and 12=211 =10 A
Heatdissipated in 30resistor in 10 s
= Ii~t= (10)2x3x10 =3000 J.
Example 114. The resistance of each of the three wires,
shown in Fig.3.131, is 4O.This combination of resistors is
connected to a source of 4 0
emfe. Theammeter shows 4 0
a reading of1ACalculatep Q
the power dissipated inthe
circuit. e
[CBSE F 03]
Fig.3.131
PHYSICS-XII
Solution. Total resistance between the points P
and Q,
4x4
R=--+4=2+4=60
4+4
Current in the circuit, I = 1 A
Power dissipated in the circuit,
P =r2R=12 x6 =6W.
Example 116. A houseisfitted with20lamps of60 W
each,10fans consuming0.5A each and an electric kettle of
resistance 110O.If the energyissupplied at220V and costs
75 paise per unit, calculate themonthly bill for running
appliances for6hours a day. Take1month=30days.
Solution. Power of 20 lamps of 60 W each
= 20x60=1200 W
Power consumed by 10 fans at 0.5Acurrent
=10xVI=10x220x0.5=1100 W
Power consumed by electric kettle of110 0 resistance
V2=220x220 =440W
R 110
Total power of theappliances
=1200+ 1100 + 440 =2740 W =2.74 kW
Total time forwhich appliances are used
=6x 30=180h
Total energy consumed
=P.t=2.74 kWx180 h
= 493.2 kWh or units
.'.Monthly bill=493.2 x0.75 =~ 369.90.
Example 117. Therearetwo electric bulbs rated60 W,
110 Vand100W,110V.They are connected in series with a
220 Vd.c.supply. Will any bulb fuse? What will happenif
they are connected in-parallel with the samesupply?
Solution. Currents required by the two bulbs for
the normal glowness are
I=P1=~ =0.55 A
1V110
and 1 =P2=100 =0.91 A
2V110
Theresistances of the two bulbs are
R =V= 110 = 202 0
1110.55
and ~ =V= 110 =1210
120.91
When the bulbs are connected in series across the
220 V supply, the current through each bulbwill be
I =V 220 = 0.68 A
R1 +~ 202 + 121«««2vw~m{lzq§m2kwu

254. CURRENT ELECTRICITY
AsII<Iand1
2>I,so that 60 W bulb will fuse
while the 100 W bulb will light up dim.
When the bulbs are joined in parallel, their
equivalent resistance is
R'=RlRz= 202x121 =76Q
Rl+Rz202+121
Current drawn from the 220 Vsupply will be
I'= ~=220 :::. 3 A
R'76
In the two bulbs of resistances Rl (:::. 202 Q) and
Rz(=120 Q), thecurrent of 3 A will split up into
roughly 1 A and2 A respectively. Hence both the bulbs
willfuse.
Example 118.The resistance of a240Vand200 W
electric bulb when hot is10times the resistance when cold.
Finditsresistance at room temperature. If the working
temperature of the filament is2000°C, find the temperature
coefficient ofthe filament.
Solution.Resistance of the hotbulb is given by
R'=~ = 240x240 = 288 Q
P 200
Resistance of bulb at room temperature,
R=!3'..= ~~ = 28.8 Q
10 10
R'= R(l+at)
288 =28.8(1 +ax2000)
-90C-1_4510-3°C-1
a--- -.x .
2000
Since
or
Example 119.Athin metallic wire of resistance100 Q is
immersed in a calorimeter containing 250g of water at10°C or
and a current of0.5ampere is passed through it for half an
hour.If thewater equivalent of the calorimeteris 10g, find
the rise of temperature.
Solution. Here m= 250 g,l= 0.5 A,
t=30min=1800s, w=10g
.'. Heat produced
=[2Rt=(O.5l x100x1800J= 45000J
Heatgained by water and calorimeter
=(m+w)c9=(250+10)x1x9cal
=260 x4.29joule
260x4.2x9 = 45000
Risein temperature; 9 = 45000 = 41.2°C.
260x4.2
Example 120. A copper electric kettle weighing 1000g
contains 900g ofwater qt20° e. It takes12minutes to raise
the temperature to100°e. If electric energy is supplied at
210V, calculate the strength ofthe current, assuming that or
10%heat is wasted. Specific heat of copper is 0.1.
3.69
Solution. Water equivalent of copper kettle is
w= MassxSpecific heat = 1000x0.1 = 100 g
Alsom=900 g,
9=02-91=100 -20 =80°C
Heat required,
H =(m+w)c9 = (900+100)x1x80 =80,000 cal
Heat produced
=V I t= 210xIx12x60 cal =36000Ical
4.2 4.2
Usefulheat
= 90% of 36000I
= 90x36000 [ = 32400Ical
100
..324001=80,000
Current,I= 80000 =2.469A.
32400
Example 121. A coil of enamelled copper wire of resis-
tance50Qisembeddedina block of ice and a potential
difference of210Vapplied across it.Calculate the rate at
which ice melts. Latent heat of iceis 80calpergram.
Solution. Here R= 50 Q,V= 210 V,t= 1s,
L=80calg-1
Heat produced,
H=~ = 210x210x1 = 210 cal
4.2R 4.2x50
Supposemgram of ice melts per second. Then
mL=H
m=H=210 =2.62 gs ".
L80
Example 122.An electric kettle has two heatingcoils,
when one of the coilsisswitched on, the kettle begins to boil
in6minutes and when the otherisswitched on, the boiling
begins in8minutes. Inwhattime will the boiling beginif
both the coils are switched on simultaneously (i) inseries
and(ii)in parallel ? lIlT]
Solution.LetRl and Rzbe the resistances of the
two coils,Vthe supply voltage and H,the heatrequired
to boil the water.
H =V\=V2x6x60 cal
JR1 4.2Rl
V\ V2 x8x60
For the second coil, H=-- = cal
JRz 4.2».
V2x6x60 V2x8x60
4.2 Rl 4.2 Rz
Rz=~=i
Rl63
For the first coil,www4§±ttssrxvt4r±£

255. 3.70
(i)When t
he coils are connected in series,
effective resistance = R1+ ~.
Let the boiling occur in timet1min.
Then
V \x60 V2x 6x60
--~-- =H=-----
4.2(R1 + ~) 4.2R1
or t1=6 (R1;1~)=6(1+ ~J
=6(1+ ~)min = 14min.
(ii)Whenthetwocoilsare connected in parallel,
effective resistance = R1~
R1+~
Letthe boiling occur in timet2min.Then
V2t2x60 V2x6x60
----,------'''-------, -H-----
(R1».J- -4.2R1
4.2-~~
Rl+~
RR 1
or t2= 6x 1''2 =6x-,---------,-
(R1+~)R1(1+~J
6x (\)min = 3.43 min.
1+-
4
Example123.The heater coilof an electric kettle israted
at2000W,200V.Howmuchtime will ittake in raising the
temperature of1litre of water from 20°Cto100°C, assuming
thatonly80%of the total heat energy produced by the heater
coilisused in raising the temperature of water. Density of
water=19em-3and specific heat ofwater=1calg-l°C-1.
Solution. Here P =2000 W,
Volume of water = 1 litre = 1000 cm3
Mass of water,
m=Volumexdensity
= 1000em3x1g em -3=1000 g
Rise in temperature,
8=82 -81=100-20=80°C
Heat gained by water
=me8=1000x1x 80 =80,000 cal
Lettbe the time taken to increase the temperature
from 20° to 100°C
Then total heat produced by heating coil
=Pt=2000tjoule
PHYSICS-XII
Useful heat produced
=80%2000t= 80x2000tJ
100
= 80x2000tcal
100x4.2
or
Useful heat produced = Heat gainedby water
80x2000t= 80000
100x4.2
t= 80000x100x4.2 =210s.
80x2000
Example 124.One kilowatt electric heateristo be used
with 220 V d.c.supply.(i)What is the current in the heater?
(ii)Whatisitsresistance? (iii)What is thepower dissipated
in the heater? (iv) Howmuch heat in calories isproduced
persecond? (v) How many grams of water at 100°C will be
converted perminute into steam at 100° C,with the heater?
Assume that the heat losses due to radiation arenegligible.
.Latent heat ofsteam =540 calper gram [liT)
Solution. HereP= 1 kW = 1000 W,V= 220 V
(i) Current,I=£= 1000 =4.55A.
V220
(ii)Resistance, R=V2=220x220 =48.4Q.
P 1000
(iii)Power dissipated in heater = 1000 W.
(iv)Heat produced per second,
H=VIt=~ = 1000xl =240 cals-l.
J J 4.2
(v)Heat produced per minute,
H=240x60 = 14400 cal
Weknow that 540 cal of heat convert 1 g water at
100°C into steam at100°C
:.Mass of water converted into steam
•= 14400 =26.67 g.
540
Example125.The walls of a closed cubical box of edge.
50cm aremade of a materialofthickness 1mm and thermal
amductiuitv 4x10-4 cal s-lem-lOCI. Theinterior ofthe
box maintained at 100°C abovetheoutside temperature by a
heater placed insidethebox andconnected across a 400 V
d.c. source. Calculate theresistance of the heater.[lIT)
Solution. Here, K= 4xlO-4cal s-lcm -1oc-I,
82-81= 100°C, d=1 mm =0.1em
Surface area of the six faces of the cubical box,
A= 6x(50x50) = 15000 cm2www3°otssrrwvs3qo~

256. CURRENT EL
ECTRICITY
The amount of heat conducted out per second
through the walls of the cubical box is
H _KA(92-91)_ 4x 10-4 x 15000 x 100
I- d - 0.1
= 6000 cal = 6000x4.2J
IfRis the resistance oftheheater, then heat produced
per second
. H2 =I2Rt=V2= (400)2 [t=ls]
R R
Temperature inside thebox will be maintained by
the heater if
HI = H2 or (40~)2 = 6000x4.2
R= 400 x400=6.35Q
6000x4.2
Example 126.A10Vbattery of negligible internal
resistanceischarged by a 200Vd.c.supply. If theresistance
inthecharging circuit is 38Q,whatisthe value of charging
current ? [NCERT]
Solution. As the battery emf opposes the charging
emf, therefore,
or
net emf = 200 -10 = 190 V
Charging current,
I= Net emf =200 -10 = 5 A.
Resistance 38
Example 127. Adry cell ofemf1.6Vandinternal
resistance 0.10ohmisconnected to aresistor of resistance R
ohm.If the current drawn from the cellis2A, then
(i)whatisthe voltage drop across R?
(ii)whatisthe energy dissipationinthe resistor?
Solution. Here e=1.6 V,r=0.10Q, I=2.0A
R+r=§.= 1.6= 0.8Q
I2.0
R=0.8-0.10 =0.70Q
(i)Voltage drop across R,
V=IR=2x0.70 =1.4 V.
(ii) Rateofenergy dissipation inside the resistor
=VI=1.4x2.0 =2.8W.
Example 128.A dry cell of emf 1.5V and internal
resistance 0.10Qisconnected across a resistor inseries with
a very lowresistance ammeter. When the circuit isswitched
on,the ammeter reading settles toasteady value of2.0 A
Whatisthesteady
(a) rate of chemical.energy consumption of the cell,
(b) rate ofenergy dissipation insidethe cell,
(c)rate of energy dissipation inside theresistor,
(d) power output of the source? [NCERT]
3.71
Solution. Here e = 1.5 V, r= 0.10 Q, 1=2.0A
(a)Rate of chemical energy consumption of the cell
=eI=1.5 Vx2.0 A =3.0 W.
(b)Rate of energy dissipation inside the cell
=12r=(2)2 x 0.10 W =0.40 W.
(c)Rate of energy dissipation inside the resistor
=eI-I2r=3.0 -0.40 = 2.6W.
(d)Power output of the source
= Power input to the external circuit
=eI - I2r=2.6 W.
Example129.A seriesbattery of10lead accumulators,
each of emf2V and internal resistance 0.25ohm,ischarged
bya220Vd.c.mains.To limit the charging current, a
resistance of 47.5ohmisusedinseries inthecharging
circuit. Whatis(a) the power supplied by the mains and
(b)power dissipated as heat? Account for the difference of
powerin(a) and(b). [CBSE Sample Paper 98]
Solution. emf of the battery = 10x2 = 20V
Internal resistance of the battery
=10x0.25 =2.5 Q
Total resistance =r+R= 2.5+47.5 = 50.0 Q
Asthe battery emf opposes the charging emf,
:.Effective emf =e -V=220 -20 =200 V
Ch. Effective emf 200 4 A
argmg current = =-=
Total resistance 50
(a)Power supplied by the mains
=VI=220x4=880W.
(b)Power dissipated as heat
=p(R+r)= 42x50 =800 W.
The difference of power =880 -800 =80 W, is
stored in the battery in the form of chemical energy.
Example 130. A series battery of6lead accumulators each
ofemf2.0 V and internal resistance 0.50 Q ischarged by a
100 V d.c. supply. Whatseries resistance should beused in
the charging circuit inordertolimit the current to 8.0A?
Usingtherequired resistor, obtain (a)thepower supplied by
thed.c. source (b) the power supplied by the d.c. energy
stored inthe battery in 15min. [NCERT]
Solution. Here e = 2.0 V,r= 0.50 Q, V= 100 V,
I=8.0 A
As the batteryemf opposes the charging emf,
:.Effective emf =100 -2.0x6 =88 V
Let the required series resistance be ofRQ.
Then
total resistance = (0.50 x6+R) Q = (3+R)Qwww4§±ttssrxvt4r±£

257. 3.72
Now
1
=Totalemf
Total resistance
8=~
3+R
or
64
R=-O =80.
8
(a)Power supplied by d.c.source
=VI=100Vx8 A =800W.
24+8R=88 or
(b)Power dissipated as heat
=P(R+r)=82(8+0.50x6) W
=64x11W=704W.
(c)Powersupplied bythe d.c. energy stored in the
battery in 15 min
=(800-704)W x15 min
=96 Wx900s=86400J.
Example131.Power from a64Vd.c. supply goesto
charge a battery of 8lead accumulators each ofemf2.0V
andinternal resistance 1/8o.Thecharging current also
runs anelectric motorplaced inseries with the battery. Ifthe
resistance of the windings of the motoris7.00andthe
steadysupplycurrent is3.5A,obtain
(a)themechanical energy yielded by the motor,
(b) the chemical energy, stored inthe battery during
charging in 1h. [CERT]
Solution. emfofthebattery,
Eb=2.0x8 V=16V
d.c.supply voltage, Es=64V
Internal resistance of the battery,
1
r=-x80=10
8
Resistance of motor, R= 7.00
Let backemfof motor = Em
Both the backemf Em ofthemotorand the emf Ebof
thebattery act intheopposite direction ofthesupply emf
E Therefore, netcurrent inthecircuit must be
s.
I=Netemf =Es-Eb-Em
Netresistance r+R
64-16-E
or 3.5= m
8
or Em = 48-28=20V.
(a)Mechanical energy yielded bymotor in 1h
=Em.It=20x3.5x3600J=252000J.
(b)Chemical energy stored in the battery in 1 h
=Eb.It= 16x3.5x3600J=201600 J.
PHYSICS-XII
Example132.A24Vbattery ofinternal resistance 4.00
isconnected to avariable resistor. Atwhatvalue of the
current draum from the batteryistherateof heat produced
in the resistor maximum ? [CERT]
Solution. HereE =24 V,r=4.0 0
Let the variable resistor be R.Therate of heat
produced in the resistor will be maximum when
External resistance = internal resistance
or R=40
Required current,
emf 24
I= =--A=3.0A.
resistance 4+4
Example133. 4cells ofidentical emfE,internal
resistance r,are connected in series to a variable resistor. The
following graph showsthe variation of terminal voltage of
the combination with the current output.
(i)What istheemf of
eachcellused?
(ii)Forwhatcurrent from
thecells,does maxi-
mum power
dissipation occur in
the circuit?
(iii) Calculate theinternal
resistance of eachcell.
[CBSE 0006C]
Solution. WhenI=0,
5.6
t
~ 4.2
2
(3
..::.2.8
:::.
1.4
o'----'----'_-'--->L_
0.51.0 1.5 2.0
I(ampere) -t
Fig. 3.132
totalemf = terminal voltage
4E=5.6V
or E=1.4
WhenI=1.0A,V= 2.8 =0.7 V
4
Internal re istance
The output power is maximum, when
externaJ resistance = internal resistance = 4 r
Total emf 4 E
Total resistance 4r+4r
=~=~=lA.
2r2xO.7
Example134.Two batteries, each of emfEand internal
resistance r,are connected in parallel. Ifwe take current
from this combination inan external resistanceR,then for
what value of Rmaximum power will be obtained? What
will be thispower?www4§±ttssrxvt4r±£

258. CURRENT E
LECTRICITY
Solution.The situation is shown in Fig. 3.133.
e
R
Fig. 3.133
Net emf of the parallel combination of two cells =E,
Total resistance in the circuit
=rxr+R=!.+R
r+ r 2
Hence current in the circuit is
I=_E,_=~
!.+R r+2R
2
Power dissipated in the resistance Ris
P=PR=(2E,)2R= 4 E,2R
(r+2R)2 (r-2R)2+8rR
PowerPwill be maximum when the denominator or
has a minimum value. Thishappens when
(r -2R)2=0orR=!.
2
E,2
P=(2E,l r/2
max (r+rl 2r
Example 135. Two wires made of tinned copper having
identical cross-section ( =1O-6m2) andlengths 10emand
15 em are to beused as fuses.Showthatthefuses will melt at
the same value of current in eachcase. [NCERT]
Solution.The temperature of the wire increases up
to a certain temperature ewhere the heat produced per
second by the current equals heat lost (by radiation)
persecond.
But heat produced by the current
=I2R=l2P.!..-=I2p/
A 1t?
Ifhis heat lost per second per unit surface area of
the wire and if we ignore the heat loss from the end
faces of the wire, then heat loss per second bythe wire
=hxcurved surface area of the wire
=hx21trl
When the steady state temperature is attained,
[2pI
hx21tr/=-.-2
1tr
or
Pp
h=-23 ...(i)
21tr
3.73
Nowhisindependent of Iand the values of rand p
are same for both wires,hence steady state tempe-
ratureewill depend only onI i.e., the two fuses will
melt at the same values of current.
Example136.A fuse with acircular cross-sectional radius
of0.15mm blows at15 AWhat should bethe radius of
cross-section of afuse made of the same material which will
blowat30A? [NCERT]
Solution.Herer1=0.15mm,II=15A,r2=?
12=30 A
From Eq.(i),theheat lostper second per unit
surface area of the wire is
Pp
h=--
21t2r3
For a fuse wire of the given material andthe
given value ofh,
r3ocI2
3 2
1'2_12
3--2
r1 II
"I2 3(30)2
r2"=~xr1= - x(0.15)3
II- 15
r2=(4)1/3 x0.15mm
= 1.5874 x0.15mm=0.24 mm.
'Problems For Practice
or
1.Calculate the current flowingthrough aheater
rated at2kW when connected to a 300Vd.c.
supply. [CBSEF94Cl
(Ans. 6.67A)
2.Calculate the amount ofheatproduced per second
(incalories), when a bulb of100W-220 Vglows
assuming that only 20% of electric energy is
converted into light. J=4.2Jcal-1. [Haryana 011
(Ans.19.05call
3.An electric heating element to dissipate 480watts
on240 Vmains is to be made from nichrome ribbon
1mm wide and thickness 0.05mm.Calculate the
length of the ribbon required ifthe resistivity of
nichrome is 1.1x10-6Om. (Ans. 5.45m)
4.100W,220 Vbulb is connected to 110 Vsource.
Calculate the power consumed by thebulb.
[Roorkee861
(Ans.25W)
5.How many electrons flow per second through an
electric bulb rated 220 V, 100W?[BITRanchi98]
(Ans.2.84x1018)
6.Anammeter reads a current of 30A when it i.s
connected across theterminals of a cell of emf 1.5V.www4§±ttssrxvt4r±£

259. 3.74
Neglect
ing the meter resistance, find the amountof
heat produced in the battery in 10seconds?
(Ans.107.14cal)
7.A coil of resistance 1000 is connected acrossa
battery of emf 6.0 V. Assume that.the heat deve-
loped in the coil is used to raise its temperature. If
the thermal capacity of coilis4.0JK-1, howlong
would ittake to raise the temperature of the coil by
15°C? (Ans.2.8 min)
8.A generator is supplying power to a factory by
cables of resistance 20O. If the generator is
generating 50 kW power at 5000 V, what is the
power received by the factory? [Punjab 96C]
(Ans.48 kW)
9.Two bulbs are marked 220 V, 100 W and 220 V,
50 W respectively. They are connected in series to
220 Vmains. Find the ratio of heats generated in
them. (Ans.1 : 2)
10.Ina house having 220 V line, the following applia-
nces are working: (i)a60W bulb(ii)a1000W heater
(iii)a 40 W radio. Calculate(a)thecurrent drawn by
heater and(b)the current passing through the fuse
line. [MNREC 86]
(Ans.(a)~~A(b)5 A)
11.Three equal resistances connected in seriesacrossa
source of emf consume 20 W. If the same resistances
are connectedin parallelacross the same sourceofemf,
what will be the power dissipated? [Punjab 99]
(Ans.180 W)
12.An'electricheater consistsof 20m length of manganin
wire of 0.23 m2cross-sectional area. Calculate the
wattage of the heater when a potential difference of
200 V is appliedacross it.Resistivity of manganin
=4.6x10-7Om. (Ans.109W)
13.Aline having a total resistance of 0.20 delivers
10 kW at 220 V to a small factory. Calculate the
efficiency of the transmission. (Ans.96%)
14.A motor operating on 120V draws a current of2 A.
Ifthe heat is developed in the motor at the rate of
9 cals-1,what is its efficiency? (Ans. 84.425%)
15.A 500 W electric heater is designed to work with a
200 V line. If the voltage of the line drops to160V,
then what will be the percentage loss of the heat
developed? (Ans.36%)
16.A 50 Wbulb is connected in a 200 V line. Determine
the current flowing in it and its resistance. If 10% of
the total power is converted into light, then what will
bethe rate ofproduction of heat?
TakeJ=4.2Jcal-1
(Ans.0.25 A, 8000, 10.7 cals-l)
PHYSICS-XII
17.Two bulbs rated 25 W, 220 Vand 100 W, 220 V are
connected in seriesto a440 Vsupply. (i)Show with
necessary calculations which bulb if any will fuse.
(ii)What will happen if the two bulbs are connected
in parallel tothe same supply?
[Ans.(i)25 W bulb will fuse
(ii)Both the bulbs will fuse]
18.A servo voltage stabiliser restricts thevoltage
output to 220V±1%.If an electric bulb rated at
220 V, 100W is connected to it, what will be the
minimum and maximum power consumed by it ?
(Ans.98.01W, 102.01W)
19.Aroom is lighted by 200 W, 124 V incandescent
lamps fed by a generator whose output voltage is
130 V.The connecting wires from the generator to
the user are made of aluminium wire oftotal length
150 m and cross-sectional area15 mm2. How many
such lamps can be installed ?Whatis the total
power consumed by theuser? Specific resistance of
aluminium=2.9x10-8Om. (Ans.12,2.4 kW)
20.Twowires AandBof same material and mass, have
their lengths in the ratio 1:2.On connecting them,
one atatime to thesame source ofemf,the rate of
heat dissipation in Bis found tobe 5W.Whatisthe
rate of heat dissipation inA? (Ans. 20W)
21.Two electricbulbs rated as 100W, 220 V and 25 W,
220 V are connected in seriesacross 220 Vline.
Calculate (i)current through (ii)potential difference
across and (iii)actual powers consumed infilament
ofeach bulb.
(Ans.(i)1\ A(ii)44 V, 176V, (iii)4 W, 16W)
22.The heater coil of an electric kettle isratedas2000W
at 200 V. How much time will it take to heatone
litre of water from 20°C to 100°C, assuming that
entire electric energy liberated from the heater coil
is utilised for heating water ? Also calculate the
resistance of the coil. Density of water is 1g ern-3.
(Ans.168 s,200)
23.Anelectric kettle was marked 500W, 230V and was
foundtoraise 1 kg of water at 15° C to the boiling
point in 15 minutes. Calculate the heat efficiency of
thekettle. (Ans.79.3%)
24.A copper kettle weighing 1000 gholds 1900 g of
waterat 19°C It takes 12 minutes to raise the
temperature to100°C. Ifenergy is supplied at 210V,
calculate the strength of current, assuming that 10%
of heatis wasted. Specific heat of copper
=O.lcals'eel. (Ans.5.0 A)
25.A 30 V storage battery is being charged by 120V d.c.
supply. A resistor hasbeen connected in series with
the battery tolimit the charging current to 15 A.www4§±ttssrxvt4r±£

260. CURRENT ELECTRIC
ITY
Findthe rate at whichenergy isdissipated in the
resistor. If the total heat produced could be made
available for heating water, howlong would it take
to bring 1 kg of water from 15°C to the boiling
point? Specific heatof water =lcalg-IoC-1 and
1cal=4.2J. [MNREC 84]
(Ans.1350Js-1,264.4s)
26.In the circuitshown in Fig. 3.134, eachofthethree
resistors of 4 n can have a maximum power of 20 W
(otherwise it will melt). What maximum power can
the whole circuit take? (Ans. 30 W)
H2
4Q
4Q
Fig.3.134
27.Find theheat produced per minute in each of the resis-
tors shown in Fig. 3.135. (Ans. 360J,720J,540J)
II6Q
9V in
Fig. 3.135
28.Calculate the current drawn from the battery of emf
15 V andinternal resistance 0.5 ninthe circuit
shown in Fig. 3.136. Also find the power dissipated
inthe 6 n resistor. [lIT]
(Ans.1.0 A, 3.375 W)
2Q I-II 7Q
15V,-=-
0.5Q-=-
lQ
8Q
Fig.3.136
29.In the circuit shown in Fig. 3.137, theheat produced
by4 n resistance due to current owing through it
is 40cal s-1.Find the rate at which heat is produced
in2 n resistance. (Ans.80 cal s-1)
2Q 3Q
Fig. 3.137
4Q 6Q
3.75
30.The 2.0n resistor shown inFig.3.138is dipped into
acalorimeter containing water. The heat capacity of
the calorimeter together withwater is 2000JK-1.
(a)Ifthe circuit is activefor 30 minutes, whatwould
betherise in the temperature of the water ?
(b)Suppose the 6.0n resistor getsburnt. What
would betherise in the temperature of the water in
thenext30minutes? (Ans. 5.8°C, 7.2°q
6V lQ
2Q
Fig.3.138
31.Three resistors ~,Rz and Rs each of 240 oare
connected across a 120 Vsupply, as shown in
Fig. 3.139. Find (i)thepotential diference across
eachresistorand(ii)the total heat developed across
thethree resistors in 1 minute.
[Ans.(i)VI=80 V, (ii)V2=V3= 40 V(iii)2400JJ
RI
120V
Fig. 3.139
32Aheating coil is connected in series with a
resistance R The coil is dippedin a liquidof mass
2 kg and specific heat0.5calg-1°C-I. A potential
diference of 200 Visapplied and the temperature
of the liquid is found to increase by 60°C in 20
minutes. If R is removed, the same rise in,
temperature is reachedin 15 minutes. Find the
value of R (Ans.22.14n)
33.A house is fitted with two electric lamps, each of
100W; one heater of resistance 110n and two fans,
each consuming 0.25 A. Ifelectric energy is supplied
at 200 V and each appliance works for 5 hours a day,
find the monthly bill at the rate of Rs. 3.0 per kWh.
[Punjab 98C](Ans. ~298.65)
34.An electric kettle has two coils. When one coilis
switched on, ittakes5 minutes to boil water and
when second coil is switched on, it takes 10
minutes. How long will it take to boilwater, when
both the coils are used inseries? [Punjab 01]
(Ans.15 minutes)www4not}szr±v}4yom

261. 3.76
35.A series b
attery of 6 lead accumulators, eachof emf
2.0 V and internal resistance 0.25 n is charged by a
230 V d.c. mains. To limit the charging current, a
series resistance of 53 n isused inthe charging
circuit. What is(i)power supplied by the mains
(ii)power dissipated asheat ? Account for the
diference in the two cases. [NCERT]
(Ans. 920 W, 872W)
36.A storage battery of emf8 V, internal resistance 1n, is
being charged by a 120 V d.c. source, using a 15n
resistor in series in the circuit. Calculate (i)the
current in the circuit, (ii)terminal voltage across the
battery duringcharging, and (iii)chemical energy
stored in the battery in 5 minutes. [CBSE 01, 08]
[Ans. (i)7 A,(ii)15 V, (iii)16800JJ
37.The following graph shows thevariation ofterminal
potential diference V,across a combination ofthree
cellsin series to a resistor,
versus the current, i: 6.0
(i)Calculate theemfof
each cell
t
:::.3.0
(ii)For whatcurrenti,will
the power dissipation
ofthecircuit be maxi-
.0
mum? [CBSE OD 08]
l.02.0i---t
(Ans.2.0 V, 1.0 A) Fig.3.140
PHYSICS-XII
5,n ~!!..=£!..= 100 x1 = 2.84x1018
eVe220x1.6x10-19
6.Ifristhe internal resistance of the cell, then
1=~ or r=§.=1.5=0.05n
r 130
H12rt(30)2 x0.05x10
=--= =107.14 cal.
J 4.2
7.Heat required bythe coil = Thermal capacity
xrise in temperature
= 4.0x15 = 60 J
Rate of production of heat,
V2 6x6
p=-=-- = 0.36 Js-1
R100
R.d. 60J 60 .
..eqUiretime = 0.36 Js-l 0.36x60rrun
'"2.8 min.
8.HereP=SO:kW :050x103W, V= 5000 V
Current supplied by generator,
P50x103
l=-= =lOA
V 5000
Power wasted as heat during transmission by
cables of 20.n resistance,
P'= [2R= (10)2x20 = 2000W= 2 kW
Power received bythe factory
= P'- P=50-2=48 kW.
220x220 220x220
9.1),= 100 =484n, ~ = 50 = 968n
Ratio ofheats produced whenconnected in series,
Ii=[21),= ~ = 484 =1: 2
~ [2~ ~ 968
R100050
10.(a)Current drawn by heater =-1.=-- = -A
V22011
Current drawn bybulb = ~ = ~ = ~A
V220 11
Cu d b -_!i~_,_40-_--12A
rrent rawn y radio-
V220 11
(b)Current passingthrough fuse for the line
50 3 2
=-+-+-=5A.
11 11 11
11.LetRbe the resistance ofeach resistor andetheemf
of the source.
Forseries combillatll: Rs=R + R+R=3 R
V2 V2 V2
P=- or20=- •.-=60W.
Rs 3 R R
Forparallel combinatn: Rp=R/3
V2 V2 3V2
p'=-=-- =--=3 x60=180W.
Rp R/3 R
HINTS
P 2kW 2000W
1.[=-=-- = =6.67 A.
V300V 300 V
2.Power of bulb, P=100 W
•.Electric energy consumed per second =100 J
Amount of heat produced per second
80
=80"/0of 100 J = 80 J=-cal=19.05 cal.
4.2
V2 V2 240x240
3.Power, P=-..R=-= = 120 n
R p 480
IArea of cross-section of the ribbon,
A= 0.05mm2 = 0.05 xJQ-om 2
Required length,
RA 120x0.05x10-6
1=-= 6 m=5.45 m.
p 11 x10-
Here P=100 W, V = 220 V
V2 220x220
:.Resistance of bulb, R= - = = 484 n
P 100
When the bulb is connected to 110 V source, the
power consumed by the bulbis
V,2 110x110
P'=-= =25W.
R 484www4not}szr±v}4yom

262. CURRENTELECTRICITY
12.F
irst find R = P ~ and then P = ~ .
A R
13.Let P' be the power loss in the transmission line in
the form of heat. Then
P'= [2R=(~rR=c~~~or x0.2
= 413.2 W = 0.4132 kW
Eficiency of transmission,
Power delivered by line
11= Power supplied to line
Power delivered
Power delivered+Power loss
10
---- = 0.96=96%.
]0+0.4132 .
14.Power supplied to line =VI= 120x2 = 240 W
Powerloss in the form of heat
=9cals-1 =9x4.2}s-1 =37.8 W
Power delivered byline = 240 - 37.8 = 202.2 W
Efi. Power delivered by line 202.2
ICIency,11=-------"---
Power supplied to line 240
= 0.8425 = 84.25%.
15.Here P = 500 W, V= 200 V
R =V2= 200x200 =80 n
P 500
When the voltage drops to 160 V, rate of heat
production is
P' =V,2= 160x160 = 320 W
R 80
% Drop in heat production
P-pi 180x100
=-- x100= =36%.
P 500
17.Proceed asinExample 117 on page 3.68.
(220)2
18.Resistance of thebulb, R =-- = 484 n
100
Variation in voltage =±1% of 220 V =±2.2 V
Minimum voltage = 220 -2.2 = 217.8 V
Mi. -_(217.8)2-_98.01W.
mmum power
484
Maximum voltage = 220 +2.2 = 222.2 V
Maximum power = (222.2)2 = 102.01 W.
484
19.Resistance of aluminium wire,
R _pi_2.9x10-8x150
- A-IS x10-6= 0.29 n
.. 130 -124
Current from the mam Ime = = 20.69 A
0.29
3.77
200
Currentthrough eachlamp =-=1.613 A
124
:.No. of bulbs which canbe used = 20.69 = 12.83.
1.613
No.of bulbs that should be installed =12.
Power consumed = 12x200 = 2400 W =2.4 kW.
20.As the two wires are of samematerial and mass,
their volumes must be equal.
:.~~ =a212or ~x1= a2 x21 or ~ =2a2
IfE,is the emf of the source, then rate ofheat
dissipation in wireBis
E,2
-=5 or
~
E,2
---=5 or
p.21/a2
or
E,2
--=5
P12/a2
E,2a
__2=10
pi
Rate of heat dissipation in wire Ais
E,2 €,2 E,2
- = -- =-.2a2 =2 x10=20W.
~ p1/ ~pI
21.Proceed as in Example 123 on page 3.70.
22.[=i= 2000 = 10 A
V 200
.. VIt200x10 xt
Heat produced m timet=-= cal
J 4.2
Heat gained by water = mdJ= 1000x1x80 cal
2000t
.. -- = 1000x80
4.2
1000x80x4.2
or t= = 168 s.
2000
R=V2= 200x200 =20n.
P 2000
23.Heat absorbed by water
= 1 x 4200 x (100 - 15) = 4200 x 85 J
Heat produced by electric kettle
=Pt= 500 x15x 60J
.. 4200 x 85
Heat eficiency = x 100 =79.3%.
500 x15 x 60
24.Proceed as in Example 120 on page 3.69.
. 120-30
25.Chargmg current,l= = 15
R
S· . R90
:.enes resistor, = - = 6 n
15
Rate of energy dissipation in the resistor,
P=[2R=(15)2x6 =1350}S-l.
Heat produced in resistor in time t=Heat absorbed
by water
1350xt= 1x4200x(100-15)
4200x85
t= = 264.4 s.
1350
orwww4not}szr±v}4yom

263. 3.78
26.Let I
be the current through a resistance of maximum
power 20W.Then
fR=W mfx4=W mf=5
Efective resistance betweenAandC,
4x4
R'=--+ 4=2 + 4=60
4+ 4
Themaximum power that canbe dissipated by the
circuit,
p=fR' =5x6 =30W.
27.Theequivalent resistance of the circuit is
6x3
R=--+ 1=2+ 1=30
6+ 3
Current drawn from the battery is
1=9V =3 A
30
As the current through 10 resistor is 3 A, so heat
produced in this resistorin1 minute (or 60 s) is
R=fRt=32x1x60 =540J
Current through 60resistor,
3
II=--x3=IA
6+3
..Heat produced in60 resistor
=12x6x60 =360J.
Current through 30 resistor,
12=I - II = 3 - 1 = 2 A
:.Heat produced in 30resistor
= 22x3x60 = 720J.
28.The distribution of current is shown in Fig. 3.136.
Applying Kirchhof's second law tothe loops 1 and
2,we get
(I-II)x(7 + 1+ 10) -IIx6 = 0
and IIx6+Ix(8 + 0.5 +2) =15
Onsolving theabovetwo equations, we get
II= 0.75 A and I=1.0 A
Power dissipated in the 60resistor which carries
current~is
P=IfR= (0.75)2 x 6 =3.375 W.
29.Resistance of the upper arm=2+3 = 50
Resistance of the lower arm=4+ 6 = 100
LetIbe the total current in the circuit. Then current
owing through the upper arm willbe
Ix10 21
II=5+10="3
Current owing through the lower arm.
Ix5 I
I=--=-
25+103
Heat produced per second in 20resistor,
11ex:112x2
PHYSICS-XII
Heat produced per second in 40 resistor,
~ex:l~x4
11112x2(21/3)2x2
.. -=--= =2
~ I~x4 (I/ 3) x4
or11=2 ~=2x40=SOcals-1.
30()Tal resi .th.. 6x2 1 5 rv
.aot resistance In eCIrCUIt=--+=-><
6+2 2
6 V 12
Total current, I= = -A
(5/2)0 5
Current through 20 resistance
=12x_6_ =1.8A
56+2
Heat producedin 20 resistance in30minutes
=(1.8)2 x2x30x60 = 11664 J
Rise in temperature
11664 J
=---=5.S Kor5.S°C.
2000 JK
(b)When the 60 resistor gets burnt,
6V
Current = = 2 A
(2+1)0
Heat produced in 20 resistor in 30minutes
=(2)2x2x30x60 = 14400J
Rise in temperature
14400 J
= I=7.2 Kor 7.2°C.
2000JK-
31.(i)Total resistance of the circuit,
~x~ 240x240
R=~+<'2 "3=240+ =3600
~+~ 240+240
Current drawn from the battery,
1=V= 120 =..!:A
R 360 3
1
P.D. across R,VI=~I=240x"3 =soV.
As ~=~,
so current through each of these resistors
111
=-x-=-A
2 3 6
1
P.D.across ~ or ~,V2=V3=240x(;=40 V.
(ii)Total heat developed in three resistors in
1 minute,
R= 12Rt=urx360x60=2400J.
32Here m=2 kg = 2000 g,C=0.5 cal g-1-c'.e=60°C,
tl=20min, t2=15m, R=?•••3xy°o~n}s·o3myw

264. CURRENT ELECTRICI
TY
..Heatgained by liquid
H =mce=2000x0.5x60=6x104cal
= 6x104x4.2J=2.52x105J
Letrbethe resistance oftheheating coil. Inthe first
case,the resistance Risin the circuit.
C I=~:. urrent,
R+r
Heat dissipated in time tI,
In thesecond case,the resistance Risremoved.
C 1=V
:.urrent,
r
(V)2 V2 t
Heat dissipated in time t2,H2=-;:rt2=~
As the liquidisraised to same temperature in both
cases, so
or
H=HI=H2
(R~rfrtl=(~rrt2
r2 t2153
(R+r)2=~= 20 ="4
r.J3 R+r 2
R+r2or-r- =.J3
R 2
-+1=-
r .J3
~ = 1.155-1=0.155 or r=~
r 0.155
H= H2
5(200)2 x15x60x0.155
2.52x10 =-'--~------
R
4x104x15x60x0.155
R= 5 =22.140.
2.52x10
or
or
or
or
As
or
33.Proceed as in Example 116 on page 3.68.
34.Proceed as in Example 122 on page 3.69.
35.EMF of the battery = 6 x2.0=12 V
Internal resistance of the battery = 6x0.25=1.50
Total resistance = 1.5 +53= 54.5n
Charging current
Efective emf 230 - 12
---=4.0A
Total resistance 54.5
(i)Power supplied by themains
=VI=230x4.0 = 920 W.
(ii)Power dissipated as heat
=12(R +r)=(4)2x(53+ 1.5) = 872 W.
3.79
Thedifference : 920 - 872 = 48 W, is the power
stored in the accumulator in the form of chemical
energy of its contents.
36.Total emf = 120 - 8 =112V
Total resistance = 1+15=160
. Total emf 112
(I)Current, 1= = -= 7 A.
Total resistance 16
(ii)Terminal voltage during charging,
V=e+Ir=8+-7x1=15V.
(iii)Chemical energy stored in the battery in
5minutes
=eIt = 8x7x(5x60) =16800 J.
37.(i)Total emf the three cells in series
= P.O. corresponding to zero current = 6.0V
:. EMF of each cell = 6.0/3 = 2.0 V
(ii)When i=1.0 A, V= 3.0/ 3 = 1.0 V
e-V2.0-1.0
.. r=--= = 1.00
i 1.0
Theoutput power is maximum, when
externalresistance = internal resistance=3r
. Total emf 3E.e
1 = =---=-
maxTotal resistance3r+3r 2r
=~=1.0A.
2x1.0
3.31KIRCHHOFF'S LAWS
Introductory concepts.In 1942, a German physicist
Kirchhoffextended Ohm's law to complicated circuits
andgavetwolaws, which enable us to determine
current inanypart of such a circuit. Before under-
standing these laws, we first define a few terms.
1. Electric network. The term electric network isused
for acomplicated system of electrical conductors.
2. Junction.Any pointinan electric circuit where two
ormoreconductors are joined together isajunctn.
3. Loop or Mesh.Any closed conducting path in an
electric networkiscalled a loop or mesh.
4. Branch.Abranch isany part of the network that lies
betweentwo junctns.
56.Statethe twoKirchhoffe laws for electrical
circuits and explain them giving suitable illustratns.
Also state the sign conventns used.
Kirchhof's first law or junction rule. Inanelectric
circuit, the algebraic sum of currents at anyjunctn is zero.
Or,the sum of currents entering a junctn isequal to the
sum of currents leaving that junctn.
Mathematically, this law may be expressed as
L 1=0www4not}szr±v}4yom

265. 3.80
Signcon
vention forapplying junction rule:
1. The currents owing towards the junction are
taken as positive.
2.The currents owing away from the junction
are taken as negative.
Figure3.141represents a
junctionJin a circuit where
four currents meet. The
currentsIIand12owing
towards the junction are
positive, while the currents
13and14owing away from
thejunction are negative,
therefore, by junctn rule: Fig. 3.141Junction rule:
11+12=13+14•
or II+12 - 13 - 14=0
or II+12=13+14
i.e.,Incoming current=Outgoing current
First law is also called Kirchhofs current law (KeL).
Justification. This law is based on the law of
conservation of charge.When currents in a circuit are
steady, charges cannot accumulate ororiginate at any
point of thecircuit. So whatever charge ows towards
the junction in any time interval, an equal charge must
ow away from that junction in the same time interval.
Kirchhof's second law or loop rule. Around any
closed loop of a network, thealgebraic sum of changes in
potential must be zero. Or,thealgebraic sum of the emfs in
any loop of a circuit isequal tothe sum of the products of
currents and resistances in it.
Mathematically, the loop rule may beexpressed as
Lt.V=0orLe=L IR
Sign conventn for applying looprule:
1. We can take any direction (clockwise or anti-
clockwise) as the direction of traversal.
2. The emf of cell is taken as positiveifthe
direction of traversal is from its negative to the
positive terminal(through the electrolyte).
Fig.3.142Positive emf. Fig. 3.143Negative emf.
3. The emf of a cell is taken as negative if the
direction of traversal is from its positive to the
negative terminal,
4.The current-resistance(IR)product is taken as
positive if the resistor is traversed in the same
direction of assumed current.
PHYSICS-XII
1+
--+--'VV'v-
----.
V=+IR
Fig.3.144Positive potential drop across a resistor.
5.The IRproduct is taken asnegative if the
resistor is traversed in the opposite direction of
assumed current.
I+ - I
--+--'VV'v-
....--
V=-IR
Fig. 3.145 Negative potential drop across a resistor.
Illustration. Letusconsider the circuit shown in
Fig.3.146.
Fig. 3.146An electrical circuit.
R3
o Evv
R2J
e2
12
1
vv
I
RlJe1
II
I.
vvv
I'
c F
B A
InFig.3.146, traversing in the clockwise direction
around the loopABCFA, we find that:
Algebraic sum ofcurrent resistance products
=IIRl- I2~
Algebraic sum of emfs =eI-e2
Applying Kirchhofs loop ruleto closed path ABCF A,
we getel-e2=fIRl -I2~
Similarly, applying Kirchhof s second rule to mesh
CDEFC, we get
e2=I2~+(11+12)~
Second law is alsocalledKirchhofs voltage law (KVL).
Justification.This law isbased on the law rof
conservationof energy. As the electrostatic force is a
conservative force, so the work done by it along any
closed path must be zero.
Formulae Used
1.LI=a (Junctn rule)
or Totalincoming current =Totaloutgoing current
2. Le=L IR
(Loop rule)
Units Used
CurrentIis in ampere, resistance R in ohm and
emfeinvolt.www5}~trsqrvvr5p~z

266. CURRENT ELECTRI
CITY
Example137.Network PQRS (Fig. 3.147) made as
under: PQhas abattery of4Vand negligible restance
with positive terminal connectedtoP, QRhas a restance of
60n.PShas a battery of 5Vand negligible restance with
positive terminal connected toP,RShas a restance of
200n.Ifamilliammeter, of 20nresistanceisconnected
betweenPandR,calculate the reading of the milliammeter.
[NCERT]
200n
R
5V
60n
J
p
Q
4V
Fig.3.147
Solution. Applying Kirchhoff's second law to the
loopPRQP,we get
20[1+60[=4
Similarly, from the loopPSRP, weget
200 (l- II)-20[1 = -5
40[-4411 =-1
Multiplying (i)by2 and(ii)by3, we get
120[+4011=8
and 1201-132II=-3
Subtracting(iv)from(iii),we get
172[1=11
[=~ =0.064 A
1172
or
... (iii)
...(iv)
or
Thus the milliammeter of 20owill read 0.064 A.
Example138.UsingKirchhoff'slaws in theelectrical network
shown in Fig. 3.148,calculate the values ofII'12and13,
[CBSE D 2000q
ABC
F E o
Fig. 3.148
Solution. Applying Kirchhof's first law at junction B,
II+[2=13 ...(1)
Applying Kirchhof's second law to loops ABEFA
andBCDEB, we get
213+511=12 (2)
- 213 -3[2=-6 (3)
3.81
Solving equations (1), (2) and(3), we get
48 18 66
II=-A,12=-A,[3=-A
31 31 31
Example 139. Findthe potential difference acrosseachcell.
andthe rate of energy dsipation inR.[Fig.3.149(ti)}.
[CBSE Sample Paper 11]
e1=12V'1=2o
R=4n
...(i)
Fig.3.149(a)
Solution. Applying Kirchhof's laws,
For closed loopADCBA
12=4(Il+12)+2 II=6II+412
Forclosed loop ADEFA,
6=4(Il+12)+II=411+512 ...(il)
...(i)
...(ii)
R=4n
A•.......•.. -----'\N\r--___. 0
F'-- ...•.. ---l:1---..JVV'v----' E
12
Fig.3.149(b)
Solving(i)and(i),we get
[ =18A and I=-~A
17 2 7
P'D,across R=V
=(II+I2)R
(18-6) 48
=-7- x4 volt='7volt
P'D.across each cell =p.o.across R=48 V
7
Energy dissipated in R=4nresistor
2(12)2
=(II+[2) R='7 x4}
=576J=11.75J.
49www4not}szr±v}4yom

267. to writ
ethe
in the circuit
[CBSEOD 10]
Fig.3.153(a)
3.82
Example140. Two cells ofemfs1.5V and2.0V and
internal resistances 1nand 2nrespectively are connected
in parallel so as tosendcurrent in the same directn
through anexternal restance of 5n. [CBSE OD05]
(i)Draw thecircuit diagram.
(ii)Using Kirchhoff's laws, calculate
(a)current through eachbranch ofthe circuit.
(b) p.d. across the5nrestance.
Solution. (I)The circuit diagram is shown in Fig. 3.150.
J
B
r----------
:I1Q
FQ- .....• ~---l
:Iv v :
:__1~~": ~~ .:
c
J
Fig.3.150
(ii)(a)LetIIandIzbe the currents as shown in
Fig. 3.150. Using Kirchhof's second law for the loop
AFCBA, we get
2 Iz-1Il=~\ -E.l=2-1.5
or 2Iz-II=0.5 ...(1)
For loopCFEDC, wehave
1Il+5(11+Iz)=e,=1.5
or 5Iz+611=1.5 ...(2)
Solving equations (1) and (2), we get
1 9
II=34 A, Iz=34 A
.'. Current through branch BA,
1
II=-A
34
Current through branch CF,
9
Iz=-A
34
Current through branch DE,
10
11+Iz=- A
34
(b)P.D.acrossthe 5oresistance
10
=(11+Iz)x5 = -x5 V=1.47V.
34
Example 141. Use Kirchhoffs rules
expressns for thecurrents I1,Iz and13
diagram shown in Fig. 3.151.
PHYSICS-XII'
11
e1~2V
'1~4Q
12
e2~1V
'2~3Q
)
13
e3~4 V
r3~2Q)
Fig.3.151
Solution. ByKirchhof's junction rule,
13=II+Iz ...(i)
From upper loop,
3Iz-4I1 =2-1=1 ...(ii)
From lower loop,
3Iz+2I3=4-1=3 ...(iii)
Onsolvingequations (i), (ii) and(iii),we get
II=~A Iz =2.A I=~A
13 13 313
Example142.Apply Kirchhoffs rules to theloopsACBPA
andACBQA to write the expressn for the currents II'Iz
and13in the network shown inFig.3.152. [CBSE OD 10]
Solution. ByKirchhoff's e1 ~6V
junctionrule, P
13=II+12...(1)
From loopAQBP A,
0.5 II-12=6-10=-4
...(ii)
c
From loop ACBP A,
1213+0.5II=6
...(iii)Fig.3.152
Onsolving equations (i), (ii)and(iii),we get
I=-84 A I = 106 A I= _22 A
137 237 337
R~12Q
Example143. UseKirchhoffs rulestodetemine the
potential difference between the points A andDwhen no
current flows in the arm BEof the electric networkshown in
Fig. 3.153(a). [CBSE OD 15]
3Q E
F,-.JV\I\r-.----,D
::AL~
B
6V 4V
R
cwww5}~trsqrvvr5p~z

268. CURRENT E
LECTRICITY
Solution. No current flows through the armBE.
LetIbe the current along the outer loop as shown
inFig.3.153(b).
3Q E
F.-~~~ __ -.~~D
I
~:e R1
T3V,
A Ir---:----i
6V B 4 V
R
c
Fig. 3.153 (b)
Applying Kirchhof' loop rule to the loop AFEBA,
(2 +3)1+ RlxO=1+3+6
I=2A
FromAtoDalong AFD,
VAD=2x2-1+3x2 =9 V.
Example 144.In the circuit Fig. 154,assuming point A to
beat zero potential, use Kirchhoffs rules to determine the
potential at point B.
0
4V
lA 3A
II
B
R 2Q R1
2V
3A
II
A C
Fig.3.154
Solution.From the loop BDCR1 B,we get
2x2+3Rl=4orRl=0
3.83
Solution. LetI},12and13be the currents asshown
in Fig. 3.155.Kirchhoff's second rule for the closed
loopADCA gives
10-4( II-12)+2{12+13-II) -II= 0
or 711-612-213=10 ...(1)
Forthe closed loop ABCA, weget
10-4 [2 -2(12+13)-11=0
or 11+612+213=10 ...(2)
For the closed loop BCDEB, we get
5-2(I2+13)-2(I2+13 -II)=0
or 211 -412 -413=-5 ...(3)
On solving equations (1),(2)and(3),weget
5 7
I}=2.5A,12=8"A,13=18A
The currents in the various branches of the network
are:
5
IAB=-A;
8
7
lAD=1-A;
8
1
ICA=2- A;
2
7
IDEB=1-A
8
1
IBC=2-A.
2
ICD=0;
Example 146. In thecircuitshown inFig. 3.156(a), E,F,
eandHare cells of emf2V,1V,3Vand1V,andtheir
internal restancesare2Q,1Q,3Qand1Q,respectively.
Calculate (i)the potential difference between Band Dand
(ii)the potential difference across the terminals of each of the
cellseandH. [CBSE D 04C;CBSE Sample Paper 08]
D""-----~
Example 145. Determine the current in each branch of the
networkshown inFig.3.155. [NCERT] Fig.3.156(a)
B
D
Fig.3.155
Solution. InFig.3.156(b),thenetwork has been
redrawnshowing the ernfs and internal resistances of
the cells explicitly.
2Q 2 V I
1:~I---<II!-:-Il-H"""~2
1 VIF I,~:slll:
",,---•.._G-I C
D
123V 3Q
Fig.3.156(b)www4not}szr±v}4yom

269. 3.84
(i)Applying Kirchhof
'sfirst law atjunction D,
we get
I=II+12 ...(i)
Applying Kirchhof's second law to loop ADBA, we
get
21+I+2II= 2-1
or 3I+2I1=1 ...(i)
Applying Kirchhof's second law to loopDCBD
3 12+12-2II=3 - 1
or 412-2I1 =2 ...(ii)
On solving equations (i),(i)and(ii),we get
1 6 5
II= - 13A,12=13AandI=13A
P.D.between the points Band D,
2
VI=IIx2=-V.
13
(ii)PD. between the terminals of G (giving current),
6
V2=e-I?r= 3 --x3=1.615V
- 13
PD.between the terminals of H (taking current),
V3=e'+12r'= 1+~x1 = 1.46V.
13
Example147.In a Wheatstone bridge, p=ln,
Q=2 n, R=2n,5=3nand Rg = 4 n.Find the current
through the galvanometer in the unbalanced positn of the
bridge, when a battery of2 Vandinternal restance 2nis
used.
Solution. The circuit for the given Wheatstone
bridge is shown in Fig. 3.157. LetI,IIandIgbethe
currents as shown.
B
2V
E
2Q
Fig.3.157
Applying Kirchhof's second law to loopABDA,
we get,
IIx1+1gx4 - (I -II)x2 =a
or 311-2I+4Ig=0 ...(1)
PHYSICS-XII
Applying Kirchhof's second law toloopBCDB,
we get
(II-1 )x2-(I-l1+1 )x3-1 x4=0
g . g g
5I1-31-9Ig =0 ...(2)
Applying Kirchhof's second law toloopADCEA,
we get
or
2(1-II)+3(I-II+Ig)+2I=2
- 5II+71+3Ig=2 ...(3)
Adding (2) and(3),
41-6Ig=2 ...(4)
Multiplying (1) by 5 and (2) by 3 andsubtracting,
weget
-1+471g=aor
From (4),
4x47Ig-6Ig = 2 or 182 Ig=2
21
I=-=-A.
g18291
Example148.Thefourarms of a Whetstonebridge
(Fig.3.158)havethe following resistances:
AB=100n, BC=10n, CD=5nandDA=60n.
A galvanometer of15nrestance isconnected across
BD.Calculate the current through the galvanometer when a
potential difference ofl0Vismaintained across AC.
[NCERT]
B
o
lOV
Fig.3.158
Solution.Applying Kirchhof's second law to loop
BADB, we get
100I1 +15Ig-6012 =a
or 20I1+3Ig-1212=0 ...(1)
Considering the loopBCDB, we get
10(II-Ig)-15Ig -5 (I2+Ig) =0
1011-30Ig -512 =0
2 II-61g-12=a ...(2)www4not}szr±v}4yom

270. CURRENT ELECTR
ICITY
or
Considering the loop ADCEA, we get
6012+5(12 +Ig}=l0
6512+5Ig=10
1312+Ig=2
Multiplying Eq. (2}by10,we get
2011-60Ig -1012 =0
From equations(I)and(4),we get
631-212 =0
g 63
12=2Ig=31.5Ig
Substituting the value of12inEq.(3),weget
13 (31.51g) +Ig=2
410.5Ig=2
2
I=---A=4.87 mA.
g410.5
or
Example 149. Twocells ofemfs1.5Vand2Vand
internal resistances 2 0and10respectively have their
negative terminals joined by awire of 60andpositive
terminals bya wire of40resistance. Athird resistance wire
of80connects middle points of these wires.Drawthe
circuit diagram. Using Kirchhoff laws, find the potential
difference at theendofthis third wire. [CBSE D 2000c]
Solution. As shown in Fig.3.159, thepositive
terminals of cells e1ande2are connected to the wire
AEof resistance 40andnegative terminals tothe wire
BDof resistance 6 O. The80wire isconnected
between the middle pointsFandCof the wires AEand
BDrespectively.
4
Rl=~=2=20
6
R=R=-=30
342
and
The distribution of current in various branches is
shown in Fig. 3.159.
RJ=30
II
O- .•....---"vv'\r-- ...•..--{)C
[2
R4=30
Fig.3.159
Applying Kirchhof's second law totheloop
ABCF A,we get
3.85
... (3}
I]x'1+IIxR] + (11 +I2)R+I]xR3 =e1
IIx2 +IIx2+(11 +12)x8 +IIx3=1.5
1511+812 =1.5
Applying Kirchhoff's second lawtothe
CDEFC,we get
12xr2+ 12x~+(II+12)xR+12xR4=e2
12x1 +12x2+(11 + 12)x 8 +12x3=2
811+1412=2
or 4 II+712=1
On solving equations (i)and(ii),we get
5 18
1[ =146A and 12= 146A
Current through the 80resistance wire is
I+I =2+~=E...A
12146146 146
PD.across theends of 8 0resistance wire
=E...x8=1.26V.
146
Example 150. AB,BC,CDandDA are restorsof L, 1,2
and20respectively connected inseries.Between Aand Cis
a1volt cell of resistance 2 0,A beingpositive. Between B
andDisa2Vcell af1 0restance,Bbeing positive. Find
the current ill each branch of the circuit.
Solution. The circuit arrangement and current
distribution is shown in Fig.3.160.
B
...(i}
loop
...(4}
...(ii}
A C
20 I3 -II20
[3-I2
IVU
F E
20
Fig.3.160
Applying Kirchhoff's second law toloops BADB,
BCDB andADCEFA, we get
1 .12+2.13+1.II=2
or 11+12+213=2 ...(1}
or 1(/1-12}-2(13-Il}+11=2
or 4II -12-213= 2 ...(2}
and213+2(13-11}+2(13-12}=1
or - 2II -212+613=1 ...(3}
Solving equations (I),(2)and(3),we get
II=0.8A,12=0.2 A and13=0.5Awww4not}szr±v}4yom

271. 3.86
Currents in
diferent branches are
lAB=IZ=0.2 A;
IBC=II - IZ =0.6 A;
'co=II-13=0.3 A ;
lAD=13=0.5 A;
IEF=13 -1Z=0.3A.
Example 151 .Find the equivalent restance between the
terminals A andBinthenetworkshown in Fig. 3.161.
Giveneach restor R isofl0n.
1 K 11 L 12
A~~~~~~~~~~~r-QM
1-11
R
11-12
R
12
R
J
RJ
R
1-11 1-12
P B
o
U
-- - -- ---- - -11- ---- - - --- - - - --- - --- - ~
e
R R N
Fig.3.161
Solution.Imagine a battery of emft,having no
internal resistance, connectedbetween the points A
andB.The distribution of current through various or
branches isas shown in Fig. 3.161.
Applying Kirchhoff's second law toloopKLOPK,
we get
II R+(I1-Iz)R-2(I-II)R=O
4II - Iz =2I
Similarly, from the loopLMNOL, we have
2IzR-(I-Iz)R-(II-Iz)R=0
-II+4Iz=I
From the loop AKPONBEA, we have
2(I - II) R+(I- Iz)R=t
Solving equations (1)and(2),we get
3 2
II="5IandIz="5I
or
or
Substituting these values in equation (3),we get
2(I-~I) R+(I-~I) R=t
~IR=t ...(4)
5
IfR'isthe equivalent resistance between AandB,
then
or
From(4)and(5),
IR'=t ...(5)
ti:= ~IR
5
R'= ~R= ~x10=14o,
5 5
or
PHYSICS-XII
Example 152.Two squares ABCD and BEFC have the
sideBCincommon. The sides are of conducting wires with
restances as follows: AB, BE,FCandCDeach2o;AD,
Be,EFeach1 n.A cell of emf2V and internal resistance
2nisjoinedacross AD. Find the currents invarious
branches ofthe circuit. .
Solution. The current distribution in various branches
of the circuit is shown in Fig. 3.162.
1 A11 B 12
r-~~-.~~~~-;~~vv~E
1 -11
2Q
11 - 12
2Q
12
2V 2Q
IQ IQ IQ
2Q
11
2Q
12
F
0 C
Fig. 3.162
...(1)
Applying Kirchhof's second law to the loop
containing the cell andAD,weget
2xI+1x(I - II)=2
3I - II=2
From the loopABCDA,we get
2xII +lx(II-Iz)+2x II-1x(I-II)=0
-I+6II -Iz=0
Similarly, from the loopBEFCB, weget
2xIz+1xIz+2xIz-1x(II -Iz)=0
- II+6Iz=0
Solving equations (1),(2)and(3),we get
I=70 AI=12 AI=~A
99'199'z99
Currents in diferent branches are
12 2
lAB=Ieo=II=99 A, IBE=IEF=ICF=Iz=99A
58. 10
lAD=I-II=99A, IBC=II -Iz=99A
70
Current through the cell =I= -A
99
or ...(1)
...(2)
or ...(3)
...(2)
...(3)
Example 153.Twopoints A andBaremaintained at a
constant potential difference of 110V.Athird point is
connected to Abytwo resistances of100and200nin
parallel, and toBby a single restance of 300n.Find the
current in each restance and the potential difference
betweenAandCandbetweenCandB.
Solution.The circuit arrangement and the current
distribution is shown in Fig. 3.163.
Applying Kirchhof's second law to the loop
DEFGHID, we get
IIx100-(I - II)x200=0
or 300II-2001=0 ...(1)•••3xy°o~n}s·o3myw

272. CURRENT E
LECTRICITY
c
I--+---~B
Fig. 3.163
Similarly, from loop ADIHGCBA, we get
(I -11)200+Ix300=110
or 5001-200II=110 ...(2)
Solving equations (1)and(2),we get
3 1
I=10AandII=5"A
.'.Current through 1000resistance
1
=II=-A
5
Current through 2000resistance
1
=I-Il=-A
10
Current through 3000resistance
=I=-2.A
10
P.D. between Aand C=PD. across 100nresistor
1
=IIx100=-x100=20 V
5
PD.between C and B=PD. across 300 0 resistor
=Ix300=2.x300=90V.
10
Example 154.Abattery of 10Vandnegligible internal
resistance connected across thediagoYfally opposite corners
ofa cubical network consisting of12 restors eachof restance
10.Determine the equivalent resistance of the network and
thecurrent along each edgeof the cube. [NCERT]
Solution. Let61bethe current through the cell.
Since the paths AA', AD andABaresymmetrically
placed, current through each of them issame, i.e.,21.At
the junctions A',BandDthe incoming current 21splits
equally into the two outgoing branches, the current
through each branch isI,asshown in Fig 3.164. At the
junctionsB',CandD',these currents reunite andthe
currents along B' C',D' C'andCC'are21each. The total
current at junction C'is61again.
Applying Kirchhoff's second law to the loop
ABCC' EA, we get
- 2IR-1R-2 1R.+.e=aore=5IR
where R is the resistanceof each edge andeis the emf
of the battery.
3.87
61
E
Fig. 3.164
..The equivalent resistance of the network is
R'=Total emf= ~=5 IR= ~R
Totalcurrent 61 6I 6
ButR=10
R'=~O
6
Total current in the network is
e10
6 I=-=-=12AorI=2 A
R'5
6
Thecurrent owing in each branch can be read of
easily.
Example 155. Twelve wires each havinga resistance of r0
are connected toform a skeleton cube; find the resistance of
the cube between thetwo corners of the same edge.
Solution. Let a current x+2Yenter the junctionA
of the cubeABCDEFGH. From thesymmetry of the
parallel paths, current distribution will be as shown in
Fig.3.165.
H G
2(y-z)
y-z
y-z
Ef------...•..--(F
y-z. z
Y D~ ~Y
Y " z
,~
c
x
x+2y A B x+2y
Fig. 3.165
Applying Kirchhof's second law to theloop
DHGCD,we get
(y -z) r+2(y -z) r+(y-z) r-zr=a
4
or4yr-5zr=0 or5z=4y orz=-y
5www4not}szr±v}4yom

273. 3.88
Applying Kirchh
off's second law to the loopABCDA,
we get
or
xr-yr -zr-yr=0
x-2y-z=0
4
x-2y--y=0
- 5
or
or
14 5
x='5y;y=14x
LetRbe the resistance across AB.Then
PD.across AB=xr
or ( x+ ~~x)R=xror
7
R=-rn.
12
12 R=r
7
Hence
Example 156. Eleven equalwires eachof resistance rform
theedgesof an incomplete cube.Find the totalresistance
from one end of the vacant edgeof the cube to the other.
Solution. LetAandBbe thevacant edges ofthe
cube. Let an emfE.applied acrossABsenda current 2x
in the circuit. Since the pathsADandAEaresymme-
trical, the current2xatAisdivided into two equal
partsxandx.At other points, the current is divided as
shown in Fig.3.166,sothatagain the currents combine
atBto give current2x.LetRbe the total resistance of
the cube between AandB.
D~ ____ ~
x
x
C
X
/ Y
/~
2x
/
/
A E.
2x
Fig. 3.166
Applying Kirchhof's second lawto the loop,
ABCDA,we get
xr+yr+xr=E.
From Ohm's law,
E.=2x.R
2xr+ yr=2xR
Applying Kirchhof's second law to the loop
EFGHE, we get
yr -(x-y)r-2(x - y)r-(x -y)r=0
or y-x+y-2x+2y-x+y=0
PHYSICS-XII
or -4x+5y=0
4
y=-x
5
4
2xr+-xr=2xR
5
or
14r =2R
5
or
Hence R=1.4rn.
Example 157.Twelvewires each havingarestance of1n
are connected to form acube. Find the resistance of the cube
between two cornersof a diagonal of one face of acube.
Solution. Imagine a battery connectedbetween
points AandCso that a current of1A enters junction
A.This current is divided equally along ABandAD.
The distribution of current in various branches is
shown in Fig. 3.167. These currents finally addsothat a
current of1A ows out of junction C.
E
1-2x-z
H
I
,=
y
1-2x-z+yB
x
X I
I
I
I
Iy+z
F)..__.-_
Y /
/~
x-y
x-y
D
Fig. 3.167
Applying Kirchhof's second law to the loop
AEFDA, we get
- (1-2x)-z+y+x=0 ...(1)
Similarly, from the loop BHGCB, we have
-y-(1-2x - z+y) -(1-2x+2y)+(x - y)=0
...(2)
Again, from the loop FGCDF, we have
-(y+z)-(1-2x+2y)+(x-y)-y=0
Onsolving equations (1), (2)and(3),we get
31·
x=3"A,y=0,Z=8'A
Now VAC=VAB+VBC
3 3 6 3
=lx-+1x-=-=-V
8 8 8 4
Equivalent resistance between AandC,
R-VAC .;3 / 4_ 3o
--[---1--4 .
...(3)•••3xy°o~n}s·o3myw

274. CURREN
T ELECTRICITY
Example 158. Inthenetwork as shown in Fig. 3.168,each
resistance r isof2n.Find the effective resistance between
points A and B.
Fig.3.168
Solution.Thedistribution ofcurrent is shown in
Fig.3.169.By symmetry, current inarmAE=current in
armEB.Asthe current in arm CEis equal tothe current
C12 -13 0
A E B
Fig.3.169
in armED,so the resistance ofthe networkwill not be
..afected if the wire CEDis disconnected from the wire or
AEBat the point E,as shown in Fig. 3.170.
C 0
A B
Fig. 3.170
Resistance of wire ABD=r+r=2r
. 2rx r 8r
Resistance of WIre ACDEB =r+-- +r=-
2r+r 3
As these two resistances arein parallel, so the
equivalent resistance Rbetween points AandBis or
given by
1 1 3 7 8r
-=-+-=- orR=-
R 2r8r8r 7
8x216
Givenr=2n,therefore, R=-- =-n.
7 7
Example 159. Calculate the equivalent restance between or
the points AandBin the network shown inFig.3.171.
R
A
Fig. 3.171
R r
3.89
Solution.Suppose a cell of emfeis connected
betweenAand B. Then the given circuit can be
represented by an unbalanced Wheatstone bridge as
shown in Fig. 3.172. The distribution of current is also
shown.
A B
Fig. 3.172
Applying Kirchhoff's second law tothe loop 1, we
get
II r+12r-(I-II)R=0
II ( r+R);t-12 r - IR=0
From the loop2,wehave
... (1)
(II-12)R-(I-II+12)r-'I2 r=O
or Il(R+r)-I2(R+2r)-Ir=0
Solving equations (1)and(2),we get
I=R+r I
1R+3r
... (2)
I=R-r 1
2R+3r
Similarly, from the loop3,we have
and ...(3)
(I-I})R+(I-I}+12)r=e
- II(R+r)+12r+1 (R+r)=e
B
Substituting thevaluesofI}and12from equation
(3),we get
2
_ (R+r) 1+ (R-r) I+(R+r)l=e
R+3r R+3r
3rR+?I=e
R+3r
Equivalent resistance betweenAand B,
R'=~=3rR+?
I R+3r
r(3R+r)
(R+3r)www4not}szr±v}4yom

275. 3.90
j2)roblems
ForPractice
1.Apply Kirchhoff's rulesto the loops PRSPand
PRQPto write the expressions for the currents II'12
and13in the circuit shown inFig. 3.173.
[CBSE OD 10]
(Ans.~ A---±-AIIAJ
860 '215 '172
200n
Fig.3.173
60n
5V
LL- ...---H--.........J Q
4V
2.Use Kirchhof's rules to determine the value of the
current 11owing in the circuit shown in Fig.3.174.
[CBSE D13C]
(Ans.I}":0.75 A)
30n h
11
13
20n
13
20V
a d
12
bc
gf
20n
e
80V
12
Fig.3.174
24V
11
3.Using Kirchhof's
laws, determine the
currents II'12and13
for the network shown 13
in Fig. 3.175.
[CBSE D99C]
(Ans.3 A, -1.5 A, 4.5A)Fig.3.175
4.The circuit diagram shown in Fig.3.176 hastwo
cellse}ande2withemfs4 Vand2 V respectively,
eachone having an internal resistance of 2 Q.The
external resistance R is of 8Q.Find the magnitude
anddirection of currents flowing through the two
cells. [ISCE 98]
(Ans. 11= ~A,12= - ~AJ
27V
6n
4n
R
Fig.3.176
PHYSICS-XII
5.Fig. 3.177 shows ncellsconnected to form a series
circuit. Their internal resistances are related to their
emfsasIi=aei,whereais a constant. Find(i)the
current through the circuit and (ii) the potential
diference between the terminals of ith battery.
[Ans.(i)J:.(ii)0]
a
el '1e2 '2 e3'3 en'n
L~~~--- ~:=J
Fig.3.177
6.Two cells of emfs 3 V and 4 V and internal resis-
tances 1Qand 2Qrespectively are connected in
parallel so as to send current in the same direction
through anexternal resistance of5Q.
(i)Draw thecircuit diagram.
(ii) Using Kirchhoff's laws, calculate (a)thecurrent
through each branch of the circuit. (b)p.d.
across the5Qresistance. [CBSE OD 95, 96 C]
(Ans. (a) ~A,~ A, ~ A(b)2.35 vJ
17 17 17
7.In theelectric network shown in Fig.3.178,use
Kirchhof's rules to calculate the power consumed
bytheresistance R =4Q. [CBSED14C]
el=12V'I=2n (Ans.9 W)
B I. C
II I'
II+12 R=4n
A 0
E
e2~6 V
F
12
I'
Fig.3.178
8.Anetwork of resistors is connected to a battery of
negligible internal resistance, as shown in Fig.3.179.
Calculate the equivalent resistance between the
pointsAand D, and the value of the current 13,
(Ans. 1.25Q,0.5 A)
2n11-13
B c
2n
2n
I=2A
~---+-I f-----J
Fig.3.179www5}~trsqrvvr5p~z

276. CURRENT ELECTRICI
TY
9.Using Kirchhoff's rules, determine the value of
unknown resistance R in the circuit shown in
Fig.3.180 sothat no current flows through 4Q
resistance. Also find the potential difference
betweenAand D. [CBSE D12]
(Ans.3 V)
F1Q
E o
R1Q 4Q
r---I-:
B
9V 3V
A c
Fi.3.180
10.Find the current flowing through each cell inthe
circuit shown in Fi. 3.181. Also calculate the
potential difference across theterminals of each
cell. (Ans.0,- 3 A, 3 A, 3 V)
13V 1Q
Fi.3.181
11.In the network shown in Fig. 3.182, (i)calculate the
current ofthe 6 Vbattery and(ii)determine the
potential difference between the points AandB.
[Ans.(i)2 A(ii)4 Vj
A B
C 0
14V
4Q 12
4V II
B E
10V
6Q
4Q 13
6V
A F
2Q
Fig.3.183
2V
Fi.3.182
12.In the network shown in Fig. 3.183, find(i)the
currentsII'12and13and(ii)the potential difference
between the points B and E.
[Ans.(i)II=2 A, 12= -3A,13=-1A(ii) -2 Vj
13.Calculate the potential difference between the
junctions Band D in the Wheatstone's bridge shown
in Fig. 3.184. [Roorkee 89](Ans.0.2 V)
3.91
B
o
e
2V
Fi.3.184
14.In the given electrical networks shown in
Figs. 3.185(a)and(b),identical cells each of emft,
are giving same current 1. Find the values of the
resistors ~ and ~ in thenetwork (b).
(Ans. 9.9Q, ~1Q)
Fig.3.185
15.What does the ammeter Aread in the circuit
shown in Fig. 3.186? Whatifthe positions of the
cell and the ammeter are interchanged?
(Ans. ~A ~A)
11'11
I
5V
40
2Q
Fi.3.186
E 50
B
16.In the circuit shown in
Fig.3.187, determine
5Q 50
thecurrent inthe
resistance CD and C 0
equivalent resistance
between the pointsA
50 5Q
andB.Theinternal
resistanceof cell is F
negligible.
(Ans. 7Q,0.4 A)
14V
Fi.3.187www4·•tssrrxvs4p•£

277. 3.92
17.A certain length
of a uniform wire of resistance 120
is bent into a circle and two points, a quarter of
circumference apart, are connected to a battery of
emf 4 V and internal resistance 10. Find the current
in the different parts of the circuit.
(Ans. 12 A ~ A)
13'13
18.In Fig. 3.188, ABCDAis a uniform circular wire of
resistance 2O.AOCandBODare two wires along
two perpendicular diameters
of the circle, each having
same resistance 10. A
battery of emf€and internal
resistanceris connected A1----+---1 C
between the pointsAandD.
Calculate the equivalent
resistance of the network.
15
(Ans. 140)Fi.3.188
B
D
19.In the circuit shown in Fig. 3.189, find the currents
1,II'12and 13. Given that emf of the battery=2V,
internal resistance of the battery=2Qand resis-
tance of the galvanometer=4O.
( I=47 AI=17 AI=30 A 1= _J..A)
Ans. 91' 1 91'2 91 '3 91
B
Fi.3.189
20.Determine the current flowing through the
galvanometerGof the Wheatstone bridge shown in
Fig.3.190. (Ans. 0.0454A)
B
1=1 A
A
lA
Fi.3.190
21.The terminals of a battery of emf 3 V and internal
resistance 2.50 are joined to the diagonally opposite
comers of a cubical skeleton frame of 12wires, each
of resistance 3o.Find the current in the battery.
(Ans.0.6 A)
PHYSICS-XII
22.Twelve identical wires each of resistance 60 are
arranged to form a skeleton cube. A current of 40mA
is led into cube at one comer and out at the diago-
nally opposite comer. Calculate the potential diffe-
rence developed across these comers and the effec-
tive resistance of the network. (Ans.0.2V,50)
23.Twelve identical wires each of resistance 60 are
joined to form a skeleton cube. Find the resistance
between the comers of the same edge of the cube.
(Ans.3.50)
24.Find the currentsII'12and13through the three
resistors of the circuit shown in Fi. 3.191.
(Ans.Zero in each resistor)
C D
Ion
A,----H--.----jll--,.--H-----,
3V
Ion Ion
3V 3V 3V
Fi.3.191
HINTS
1.By Kirchhoff's junction rule,
13=II+12 ...(i)
From loopPRSP, 2013+20012 =5 (ii)
From loopPRQP, 2013 +6011=4 (iii)
Onsolving equations(i),(ii)and(iii),we get
39 4 11
II=860A,12=215A,13=172A
2.Applying Kirchhoff's junction rule at 'a',we get
13=II+12
Applying Kirchhoff's loop rule to the loop ahdcba,
we get
3011+2013=20
or 3011+20(11+12)=20
or 5011+2012 =20
511+212=2 ...(i)
Again, from the loopagfedcba,we get
2012+2013 =80+20
or2012+20(11+12)=100
or 2011+4012=100
or II+212=5 ...(ii)
Subtracting(ii)from(i),we get
411=-3 orII=-0.75 A
The negative sign showsthat the actual direction of
currentIIis opposite to that shown in the given
circuit diagram.www4yztp€o·tvp4nzx

278. CURRENT ELECTRICITY
3,Traversing the upper
and lower loops anticlock-
wise, we get
211+ 612 = 24 - 27
or 211+ 612 = -3 ...(1)
and 413 - 612 = 27
or4(Il -12)- 612 = 27
or 411- 10 12= 27 ...(2)
On solving(1)and (2), we get
11= 3 A, 12=-1.5A
13= ~ - 12= 3 + 1.5 = 4.5 A
4.Applying Kirchhoff's second law to loop1,we get
111- 12'2=el-e2
or 211- 212 = 4 -2
or 11- 12= 1
Similarly, from loop 2, we get
12'2+(II+12)R=e2
or 212 + 8(II+ 12)~ 2
or 411 +512=1
On solving equations (1) and (2), we get
2
II=3A
(From -ve to +ve terminal insidee1)
1
12=-3A
(From +ve to -ve terminal insidee2)
5.Suppose a currentIflows in the circuit in the
indicated direction. Applying Kirchhoff's loop
law,
...(1)
...(2)
and
11 + 1'2 + 1'3 + ... +lr.,=e1+e2+e3+ ...+en
1=e1+e2+e2+ ... +en
1+'2+r3+"'+'n
_ e1 +e2+e3+ +en _1
- a(e1+e2+e3+ +en) - -;:.
(ii)P.D.between the terminals of ith battery
1
=ei-I,;=ei--.aei= O.
a
8.HereRBCD= 2 + 2 = 4 n.Itis in parallel with 2 n
resistance inBD.Their equivalent resistance
4x24AThis resi .. .ith2A
= -- = -s z.s resistanceIS illsenesWI ,.
4+2 3
or
resistance inAB.Their equivalent resistance
= 2 + 4/ 3 = 10/ 3 n. This resistance is in parallel
with 2 n resistance inAD.The equivalent resistance
betweenAand D,
3.93
10x2 5
RAD=fa---= - = 1.25 n
3+2 4
e= lR = 2 x 1.25 = 2.5 V
Applying Kirchhoff's second law to the lower
rectangular loop,
212=e=2.5V
or 12= 1.25 A
Now II+12=I
.. 11=I-12= 2 - 1.25 = 0.75 A
From loopBCDB,we get
2(II -13)+2(II -13) -213=0
or 411-613= 0
4 4
or 13= "611="6 x 0.75 = 0.50A
9.Applying Kirchhoff's loop rule to the loopAFEBA,
(1+1)1+4xO=-6+9
1= 1.5 A
H2 E I
F D
InJ 4nJ
R
A£ c
B
9V 3V
Fi. 3.192
From the loopBEDCB,we get
1.5R+4xO=-3+6
R=2n
VAD= (1+ 1)x1.5=3 V.
10.Applying Kirchhoff's first law at the junctionB,we
get
...(1)
Applying Kirchhoff's second law to the loop
AErB~A,we have
11xl- 12x2=(10-4)
II -212=6. ...(2)
Similarly, from the closed loopA~BE3A,we have
12x2 - 13 x 1 = 4-13 or 2 12- 13=-9
...(3)
Solving equations (1), (2) and (3), we get
II=0, 12=-3 A, 13=3 Awww4·•tssrrxvs4p•£

279. 3.94
Thus, the current in the 10Vcell is zero. The
current given by the 13Vcell to the circuit is 3 A,
and the current taken by the 4Vcell from the circuit
is 3 A.
As there is
no current in the 10Vcell,so the
potential difference across its ends is equal to its
e.m.f. .e.,10V.Since all the three cells are in
parallel, the potential difference across the
terminals of each is 10V.
11.(i)The distribution of current in various branches of
the circuit is shown in Fig. 3.193.
A B
2V 4V
6V
Fi. 3.193
Applying Kirchhoff's second law to loop 1,
312+(II-12)=2
or ~1+212=2 ...(i)
From loop (2), we get
(II-12)+2(II+13)=6
or 311-12+213=6 ...(ii)
From loop (3), we get
413+2(II+13)=4
or 211+613=4
or II+313=2 ...(iii)
On solving equations(i),(ii)and(iii),we get
II=2A
(ii) VA -VB=e,2+e,1- e,3= 2+6 -4 =4 V.
12.(i)Applying Kirchhoff's first law at thejunction E,
13=II+12 ...(i)
From the loopBCDEB,we get
-611+412=-14-10
or - 311+212= -12 ...(ii)
From the loopABEFA,we get
611+213=10
or 311+13=5 ...(iii)
On solving equations(i),(iz)and(iii),we get
I1=2A,I2=-3A,I3=-lA
(iii)P.O.between points BandE
=10-611 =10-6x2=-2V.
13.Applying Kirchhoff's second law to the loop ABCEA,
we get
IIx1+IIx1=2..II=1.0A
PHYSICS-XII
Similarly, from the loopADCEA,we have
12x1.5+12x1=2
2
12=-=0.8A
2.5
Potential difference between the points Aand B is
VA- VB=1.0 Axl 0 = 1.0 V
Potential difference between the pointsAandDis
VA- VD=0.8 Ax1.50 = 1.2 V
:.Potential difference between the points B and0is
VB-VD=(VA -VD)-(VA -VB)
= 1.2-1.0 =0.2 V.
14.From the network of Fig. 3.194(a),E.= 11I
Fi. 3.194
In the network Fig. 3.194(b), the main currentI
I
passes throughR,.,a part 10 through the 110
. dth .. I I 91
resistor an e remmnmg current, --=-
10 10
through the resistorRz.
Applying Kirchhoff's law to the loop 1, we get
I 9I 11
-x11--xRz=0orRz=-0
10 10 9
Similarly, from the loop 2, we get
91 9111
IR,.+-x Rz=e,orIR,.+-x-=11I
10 10 9
.. R,.=9.90.
15.From Kirchhoffs first law, I=II+12
1
5V
60
I,
A
20
Fi. 3.195
Applying Kirchhoff's second law for the loop 1 of
Fi. 3.195, we get
Ilx4+Ix2=5
or IIx4+(II+12)x2=5
or 611+212=5 ...(1)
Similarly, from the loop2,we get
12x6 -IIx4=0
or 4II=612, •..(2)www4·•tssrrxvs4p•£

280. CURRENT ELECTRICITY
Solving equations (1) and (2), 12= ~A
11
This will be the reading of the ammeter.
On interchanging the
cell and the ammeter, the
circuit takes the form as shown in Fi. 3.196. Again,
we can show that
5
12=11 A.
5V
4Q
6Q
2Q
Fi. 3.196
16.Proceeding as in Example 151, we obtain equivalent
resistance between pointsAandBas
R'='ZR ='Zx5 = 70 [.:R=50]
5 5
Main current,
I= ~ = 14 = 2A
R 7
Current through 50 resistance in armCD
321 1
= II -12="51 -"5I="5I="5x2 = 0.4A.
18.The current distribution is shown in Fi. 3.197.
B
r
D
Fi. 3.197
Applying Kirchhoff's law to different loops, we get
R(I-II)+ R(I-I2)+(R+r)I=e (1)
RII+R(II - 13) -R(I - II)=0 (2)
RI3 -R(I2 - 13) - R(II-13)=0 (3)
RI2 -R(I -12)+R(I2 - 13) =0 (4)
On simplifying and solving these equations,
·27
11=12,13='312, 1='3 12
7
and -12r+5 12R=e
3
3.95
IfR'is the equivalent resistance of the network, then
l(r+R')=e
7 7
:. -12r+512R=I(r+R')=-I2(r+R')
3 3
R'= 15 R = 15x0.5 = 15o.
7 7 14
or
19.Applying Kirchhoff's first law at the junctionA,
I=II+12 ...(i)
Applying Kirchhoff's second law to the loop ABDA,
we get
211+ 413 - 12= 0 ...(ii)
From the loopBCDB, we get
3(II-13)-2(I2 +13)-413 =0
or 311- 212- 913= 0 ...(iii)
From the loopABCEA, we get
211+ 3(II -13)+ 2(II+12)= 2
or 711+212-313=2 ...(iv)
On solving equations(i), (ii)and(iii),we get
17 30 1
II=91 A, 12=91A and13= -91A
20.From the loopABDA, we get
5~+101g-(I-II)15=0 [l=IA]
or 201I+lOIg=15
or 411+21g=3 ...(i)
From the loopBCDB,we get
1O(II-1g)-20(1-1I +Ig)-10Ig=0
3011 -401g =20
or 311-41g=2 ...(ii)
On solving equations(i)and(ii),we get
Ig=~A= 0.0454A
22
21.Proceeding as in Example 154, weobtain theeffective
resistance,
But R=30, therefore,
R'=~ R
6
5x3
R=--=2.50
6
Total resistance of the circuit = 2.5 +2.5 = 5.00
Current,
I= emf = 2= 0.6 A.
total resistance 5.0
22.Proceeding as in Example 154, we obtain the
effective resistance, R'= ~R
6
5x6
But R = 60, therefore, R' =--= 50
6
P.D. developed=ResistancexCurrent
=5x(40x10-3) =0.2 V.www4·•tssrrxvs4p•£

281. 3.96
23.Proceeding as in
Example155,we obtain effective
. R 7
resistance,=-r
12
7x6
Butr=60,therefore,R=--=3.5o.
12
24.From the loopABGHA, we get
lOll=3-3orII=o.
From theloopBCFG8,we get
1012 -lOll=3 - 3or12=O.
From the loopCDEFC, we get
1013-1012 =3 - 3or13=O.
3.32POTENTIOMETER
57.What a potentmeter?Give its constructn
and principle.
Potentiometer.Anideal voltmeter which does not
change the original potential difference, needs to have
infinite resistance. But a voltmeter cannot be designed
to have an infinite resistance. Potentiometer is one
such device which does not draw any current from the
circuit and still measures the potential difference. So it
acts as an ideal voltmeter.
A potentmeter isa device used to measure an unknown
emf or potential difference accurately.
Construction. As shown inFig.3.198, a potentio-
meter consists of along wire ABofuniform cross-
section, usually 4to 10 m long, of material having high
resistivity and low temperature coefficient such as
constantan or manganin. Usually, 1m long separate
pieces of wire are fixed on a wooden board parallel to
each other. The wires are joined in series by thick
copper strips. A metre scale is fixed parallel to the
wires. The ends Aand B are connected to astrong
battery, a plug key Kanda rheostatRh.This circuit,
called drivingorauxiliary circuit, sends a constant
current Ithrough the wire AB.Thus, the potential
gradually falls from Ato B. A jockey can slide along
thelength ofthe wire.
100
+
Battery ~
r-
K·
200
300
Rh
B400
1"'!!!!lI""I"III""llIrd""I""I",,lr,
Fi. 3.198Principle ofa potentiometer.
PHYSICS-XII
Principle.Thebasic principle of apotentmeter isthat
when aconstantcurrent flows through a wire of uniform
cross-sectnalarea and compositn, the potential drop across
any length of the wire directly proportnal to that length.
InFi.3.198, if we connect a voltmeter between the
endAand the jockey J,it reads the potential difference
Vacrossthe lengthIof the wireAJ.ByOhm's law,
V=IR=l.p~ [-:R=p~J
For a wire of uniform cross-section and uniform
composition, resistivitypand area of cross-section A
are constants. Therefore, when a steady currentIflows
through the wire,
Ip=a constant, k
A
Hence V=kIorVocI
This is the principle of a potentiometer. A graph
drawn betweenVandIwill be astraight line passing
through the origin0,as shown in Fi.3.199.
v
Fig. 3.199 Potential dropVoclengthI
Potential gradient. Thepotential drop per unit length
of the potentmeter wire isknownas potential gradient. It is
given by
k=V
I
51 unit of potential gradient =Vm-1
Practical unit of potential gradient =V em-1.
3.33APPLICATIONS OF A POTENTIOMETER
58. With the help of a circuit diagram, explain how
can a potentmeter be used to compare theemfs of two
primary cells.
Comparison of ernfs of two primary cells.Fig.3.200
shows the circuit diagram for comparing the ernfs of
two cells. A constant current is maintained in the
potentiometer wireABbymeans of a battery of emfE
through a keyKand rheostatRh.LetE1andE2be the
ernfs of the two primary cellswhich are to be
compared. The positive terminals of these cells are
connected to the endAof the potentiometer wire and
their negative terminals are connected to a highwww4·•tssrrxvs4p•£

282. CURRENT ELECTRICITY
resistance
box RB., a galvanometer G and a jockey I
through a two way key.A high resistance R is inserted
in the circuit from resistance box R.B. to prevent
excessive currents flowing through the galvanometer.
Fi. 3.200Comparing emfs of two cells by a
potentiometer.
As the plug is inserted between aandc ,the celli\
gets introduced in the circuit. The jockey Iismoved
along the wire ABtill the galvanometer shows no
deflection. Let the position of the jockey beI]and
length of wireAI1=11,Ifkisthe potential gradient
along the wireAB,then at null point,
E,1=kl]
By inserting the plugbetween bandc,thenull point
is again obtained for cellE,2'Let the balancing length be
AI2=12,Then
Hence,
E,2=kl2
E,2 12
E,]=T;
If one of the twocellsis a standard cellof known
emf, then emf of theother cellcan be determined.
~_12 ~
c.2--.c.1
11
Inorder toget the null point on the potentiometer or
wire, it isnecessary that the emf, E,of the auxiliary
battery must be greater than bothE,1andE,2'
59. With thehelp of acircuitdiagram, explain how
can a potentmeter be usedto measurethe internal
restance of a primarycell.
Internal resistance of a primary cell by a
potentiometer.As shown in the Fi.3.201, the +ve
terminalof the cellof emfE,whose internal resistance r
isto be measured is connected to the endAofthe
potentiometer wire and its negative terminal to a
galvanometer G and jockey J.A resistance box RB. is
connected across the cell through a key Kz.
3.97
+
Battery ~
1-
K1•
Rh B
Fi. 3.201To determine the internal resistance
of acell by a potentiometer.
Close the key K1.Aconstant currentflowsthrough
the potentiometer wire. With key Kzkeptopen, move
the jockey along ABtillit balances the emfE,of the cell.
Let11be the balancing length of the wire. If kis the
potentialgradient, then emfofthe cellwill be
E,=kl1
With the helpofresistance boxRB., introduce a
resistance R andclose key Kz.Find the balance point
for the terminal potential difference Vof the cell. If12is
the balancinglength, then
V=kl2
E,_11
V Zz
Letrbe the internal resistance of thecell.If current I
flows through cell when itis shuntedwith resistance
R,then from Ohm's law we get
E,=I(R+r)and V=IR
~=R+r=ll
V R 12
rI
1+-=~
R12
L=11 -12
R 12
.'.Internal resistance,
[1-I ]
re=Ry.
60. Whyisapotentiometer preferred overa voltmeter
for measuring the emf of a cell ?
Superiority of a potentiometer to a voltmeter.
Potentiometer isa null method device. At null point, it
doesnot draw any current from the cell and thus there
isno potential drop due to theinternal resistance of the
cell. It measures the p.d. in an open circuit which is
equal to the actual emf of the cell.www4·•tssrrxvs4p•£

283. 3.98
On the oth
er hand, avoltmeter draws asmall
current from the cell for its operation. So it measures
theterminal p.d. in aclosed circuit which is less than the
emf of a cell. That is why a potentiometer is preferred
over a voltmeter for measuring the emf of a cell.
3.34SENSITIVENESS OF A POTENTIOMETER
61. What do you mean by the sensitivity of a
potentmeter? How can we increase the sensitivity of a
potentmeter?
Sensitivity of a potentiometer.Apotentiometer is
sensitive if
(i)itiscapable of measuring very small potential
differences, and
(ii) itshows a significant change inbalancing length for
a small change in thepotentialdifference beingmeasured.
The sensitivity of a potentiometer depends on the
potential gradient along its wire. Smaller the potential
gradient, greater will bethesensitivity of thepotentiometer.
The sensitivity of a potentiometercan beincreased
byreducing the potential gradient. Thiscan be done in
two ways:
(i)For a given potential difference, thesensitivity
can be increased by increasing the length of the
potentiometer wire.
(i)For a potentiometer wire of fixed length, the
potential gradient can be decreased by reducing
the current in the circuit with the help of a rheostat.
For Your Knowledge
~A potentiometer can be regarded as anideal voltmeter
with infinite restancebecause it does not draw any
current from the source of emf at the null point.
~The principle of potentiometer requires that(i)the
potentiometer wire should be of uniform cross-
section and(ii)the current through the wire should
remain constant.
~The emf of the auxiliary battery must be greater than
the emf of the cell to be measured.
~The balance point cannot be obtained on the potentio-
meter if the fall of potential along the potentiometer
wire due to the auxiliary battery is less than the emf of
the cell to be measured.
~The positive terminals of the auxiliary battery and the
cell whose emf is to be determined must be connected
to the zero end of the potentiometer.
~Other uses ofapotentiometer. Any physical quantity
that can produce or control a potential difference can
be measured using a potentiometer. Thus, a potentio-
meter can be used to measure and control stress,
temperature, radiation, pH, frequency, etc.
PHYSICS-XII
Formulae Used
eI
1. For comparing emfs of two cells,e2=..1.
1 ~
2.For measuring internal resistance of a cell,
r= ~-12xR
Lz
3.Potential gradient of the potentiometer wire,
k= V
I
4.Unknown emf balanced against lengthI,e=k I
Units Used
The emfse1ande2are in volt, lengths ~ and12of
potentiometer wire in metre.
Example 160.Apotentmeter wireis10mlong and has a
restance of18 Q.Itisconnected to a battery of emf5 Vand
internal restance2Q.Calculate the potential gradient
along the wire.
Solution. Here1=10 m, R=18Q,e= 5 V,r=2Q
Current through the potentiometer wire,
1=_e_=_5_ =2.=.!.A
R+r18 +2 20 4
:.Potential gradient =IR=.!.x18=0.45 Vrn-1.
1410
Example161.Apotentmeter wire issupplied a constant
voltage of3V. A cell of emf108 V isbalanced by the voltage
drop across 216emof the wire. Find the totallength ofthe
potentiometer wire.
Solution. Here e=3v.s,=1.08 V,11=216ern,I=?
eIe 3x216
Ase1=z; .'.1=e1x11=108 =600 em.
Example 162. Two cells of emfse1ande2(e1>e2)are
connected as shown in Fig. 3.202.
~~~
e, e,
Fig.3.202
When a potentmeter isconnected between AandB,the
balancing lengthof the potentiometer wireis300em.On
connecting the same potentmeter between A and C,the
balancing lengthis100em.Calculate the ratofe1ande2.
[CBSE D94]www4yztp€o·tvp4nzx

284. The sensit
ivity of a potentiometer wire can be AO-----------.-----O
increased by decreasing potential gradient either
through increasing length of the potentiometer wire or
through increasing resistance put in series with the
main cell. Fi. 3.205
CURRENT ELECTRICITY
Solution. As emf a:balancing length of the
potentiometer wire
.'.When the potentiometer is connected between
Aand B, E.1 o;300
When potentiometer is connected betweenAand C,
E.1- E.2o:100
E.1- E.2_ 100
Hence
-E.-1-- 300
E.2 1 2
-=1--=-
E.1 3 3
or
E.3
or~---3' 2
E.2- 2 - ..
Example 163.In Fig.3.203,a long uniform potentmeter
wireABishaving a constant potential gradient along its
length. The null points for the two primary cells of emfsE.1
andE.2connected in the manner shown are obtained at a
dtance of120em and300cmfrom the endA.Find (i)e1/e2
and (i) positn of null point for the cellE.1.
Howisthe sensitivity of a potentmeter increased?
[CBSE D 12]
t---- 300ern
120 ern1
A B
Fig.3.203
Solution. (i)Letkbe the potential gradient in
volt/ern. Then
E.1+E.2=300k
e1=210k
e1_7
e2-3
(ii)As E.1=210k
.'. Balancing length for cell E.1is
E.
11=~=210 em
k
ande1-E.2=120k
and E.2 =90k
Hence,
3.99
Example 164. In a potentmeter, a standard cell of emf
5 V and of negligible restance maintains a steady current
through the potentmeter wire of length5m Two primary
cells ofemfsE.1andE.2are joined in series with(i)same
polarity, and(ii)opposite polarity. The combinatnis
connected through a galvanometer and a jockey to the
potentmeter. The balancing lengths in the two cases are
found to be350em and 50em respectively.
(i)Draw the necessary circuit diagram.
(ii)Find the value of the emjs of the two cells.
[CBSE D 04C)
Solution. (i)The circuit diagram is shown in
Fi. 3.204.
E.=5V
I----{.
AQ----,---~B
Fi. 3.204
(ii)Herek= 5V= 5V=_1_Vcm-1
5m500 em 100
Infirstcase,
1
E.1+E.2=kl1=- x350
100
or E.1+E.2=3.50V ...(i)
Insecond case,
1
E.1-E.2 =kI2=- x50 =0.50V ...(ii)
100
On solving(i)and(i),we get
E.1=2.0Vande2= 1.50V.
Example 165. A10metre long wire of uniform
cross-sectn of200restanceisused as a potentmeter
wire.Th wireisconnected in series with a battery of 5V,
along with an external restance of480o.If an unknown
emfE.isbalanced at600em of th wire, calculate (i) the
potential gradient of the potentmeter wire and(ii)the
value of the unknown emfe. [CBSE D 06]
5V
480Q
II
R
]
600 ern ~I
B
E.www4yztp€o·tvp4nzx

285. 3.100
Solution. Current
in
potentiometer wire is
1= V
RAB+R
the circuit or through the
5 V =0.01 A
(20+480)0
Resistance of potentiometer wire,
RAB=200
.'. PD.across the wire,
V=ll'AB =0.01x20 =O.~ V
Length of potentiometerwire,
1=10 m =1,000 em
:.Potential gradient,
k=V= 0.2 V=0.0002Vcm-1
I1,000 ern
Unknown emf balanced against 600 cm length is
E,=kl'= 0.0002 x600 =0.12V.
Example166.In the circuit diagram given below, AB a
uniform wire of restance15ohm and length one metre. It is or
connected to a series arrangement ofcellE,1of emf2.0 Vand
negligible internal restance and a restorR.Terminal A is
also connected to an electrochemical cellE,2of emf75mV
and a galvanometer G.In E,
th set-up, a balancing ..--~ rl--~'V\I\r----,
pointisobtained at30em
mark from A. Calculate theA~----r-----<l B
restance ofR.IfE,2were
to have an emf of300 m V,
where will you expect the
balancing point to be?
[CBSE D 99C]
Fi. 3.206
Solution. Current through the potentiometer wire,
1= E,1 _2_
R+RABR+15
Resistance of the 30 em length of wire, which
balances the emf E,2'is
R'= ~x30=4.50
100
Now, E,2=Potential drop acrossR'
75x10-3 =_2_ x4.5
R+15
R = 2x4.5_15 = 120-15 =105 O.
75x10-3
or
For E,2= 300 m V, the balancing length is given by
E,2 300
12=e .11=-x 30 = 120 em
(;1 75
As the length of the potentiometer wire is just
100 em, so this balance point cannot be obtained on the
wire.
PHYSICS-XII
Example 167.The length of a potentmeter wire 5mIt
isconnected to a battery of constant emf For a given
Leclanche cell, the positn of zero galvanometer deflectnis
obtained at100emIf the length of the potentmeter wire be
made8m instead of5m,calculate the length of wire for zero
deflectn in the galvanometer for thesamecell.
[CBSE F 97]
Solution. Here I=5 m,11= 100 cm = 1 m,l'= 8 m,
I~=?
LetE,be the emf of the Leclanche cell.
In first case, E,=IR 11
I
E,=IR/~
l'
...(1)
In second case, ... (2)
Comparing equations (1)and(2),
!l=ll
l' I
, 11 , 1
11=-xI=-x8 =1.6m.
I 5
Example168.A potentmeter wireof length 100em has
a restance of 10 O. It connectedinseries with a
restance and a battery of emf2V and of negligible internal
restance. Asource of emf10mVisbalanced against a
length of40em of the potentmeter wire. Whatisthe value
of the external restance? [lIT]
Solution. Fi. 3.207 shows a potentiometer wire of
length 100 em connected in series to a cell of emf 2V
andan unknown resistance R.The cell of emf 10 mV
balances length A]=40 cm of the wire.
R
2V
A~---7r-------6B
lOmV
Fig.3.207
Resistance of wireAl= ~x40 = 4 0
100
Current through wireAI,
1=10mV= 10x10-3V= 2.5x10-3A
40 40
The same current flows through the potentio-
meter wire and through the external resistance R.
Total resistance=(R+10) 0www4~»trsqrvvr4p»z

286. CURRENT ELECTRICITY
o
r
2.5x10-3A = 2 V
(R+10)0
2
R+I0= =BOO
2.5x10-3
R=BOO-10=790O.
Example 169. ABis1metrelong uniform wire of 100
resistance. Other dataareas shown inFig. 3.20B. Calculate
(i)potential-gradient along ABand(ii)length AO, when
galvanometer showsnodeflectn. [CBSE D 2000q
Fi. 3.208
Solution. (i)Total resistance of the primary circuit
= 15+ 10=250, emf=2V
:.Current in the wire AB,
I=3..-=0.08A
25
P.D.across the wireAB
=Currentxresistance of wireAB
= 0.08 x10=O.BV
Potential gradient
= P.D. =~=0.008Vem-l.
length 100
(ii) Resistance ofsecondary circuit
=12+0.3 =1.50
emf =1.5 V
Current in the secondary circuit = 1.5 = 1.0A
1.5
Thesame is the current in 0.3 0 resistor.
P'D.between pointsAand 0
= P.D. across 0.30 resistor in the
zero-deflection condition
=Currentxresistance
= 1.0 x0.3 =0.3 V
LengthAO
Potential difference
Potential gradient
0.3V
-----0-1 =37.5 em.
O.OOBVem"
3.101
Example170.A cellgives a balance with 85emof a
potentmeter wire.Whentheterminals of the cell are
shorted through a restance of7.5 0,the balance isobtained
at75emFind the internal resistance of the cell. [ISCE 951
Solution. Here 11 = 85 em, 12= 75 em, R = 7.5 0
Internal resistance,
r=R(/1-12) =7.5(85-75) = 10.
12 75
Example171.Whenarestor of 50isconnected across
cell,its terminal p.d. isbalanced by 150emof potentmeter
wireandwhen a restor of100restance isconnected
across the cell, theterminal p.d. isbalanced by175emofthe
potentmeter wire. Find the internal resistance of the cell.
Solution. In the first case, r=Rl(I~11J
I
r_1 = 1-11 ...(1)
Rl
In the second case,
r=RzC~212J
I
r-.L=1-12
Rz
Subtracting(2)from(1),
r[~1- ~]=I-II-I+12
r=12-11 = 175 -150 = 25 =20
~_.!L 150_175 12.5 .
RlRz 5 10
~roblems For Practice
...(2)
1.A potentiometer wire is 10 m long and a potential
difference of6V is maintained between its ends.
Find the emf of a cellwhich balances against a
length of 180 em of the potentiometerwire.
(Ans.1.08V)
2.The resistance of a potentiometer wire of length
10 mis20 O. A resistance box and a 2 volt
accumulator are connected in series with it. What
resistance should be introduced in the boxto have a
potential drop of one microvolt per millimetre of
thepotentiometer wire? [Kerala 94)
(Ans. 39800)
3.Inapotentiometer arrangement, a cell of emf
1.20volt gives a balance point at 30 cm length of the
wire. This cell is now replaced by another cell of
unknown emf. If the ratioof the emfs of the twowww4·•tssrrxvs4p•£

287. 3.102
cells is 1.5, calculate
the difference in the balancing
length of the potentiometer wire in the two cases.
[CBSE D 06C](Ans.10 em)
4.Two cells of emfse1ande2are connected together
in two ways shown here. The 'balance points' in a
given potentiometer experiment for these two com-
binations of cells are found to be at 351.0 em and
70.2 em respectively. Calculate the ratio of the emfs
of the two cells.[CBSESample Paper 08](Ans.3 : 2)
5.A potentiometer has 400 em long wire which is
connected to an auxiliary of steady voltage 4 V. A
Leclanche cell gives null point at 140em and Daniel
cell at 100 cm. (i)Compare emfs of the two cells.
(ii)If the length of wire is increased by 100ern, find
the position of the null point with the first cell.
[Ans.(i)7 : 5,(ii)175 em]
6.With a certain cell,the balance point is obtained at
60 cm from the zero end of the potentiometer wire.
With another cell whose emf differs from that of the
first cell by 0.1 V, the balance point is obtained at
55 ern mark. Calculate the emf of the two cells.
(Ans.1.2 V, 1.1 V)
7.A potentiometer wire has a potential gradient of
0.0025volt/em along its length. Calculate the length
of the wire at which null-point is obtained for a
1.025 volt standard cell. Also, find the emf of
another cell for which the null-point is obtained at
860 cm length. (Ans. 410 em, 2.15 V)
8.ABis a potentiometer wire of length 100 em. When
a celle2is connected acrossAC,whereAC=75em,
no current flows frome2.Find(I)the potential
gradient alongABand(ii)emf of the celle2.The
internal resistance of the celle1is negligible.
[Ans. (i)0.02 volt/ern (ii)1.5 V]
+ -
ACf-------~:::..--4:l B
Fi. 3.209
9.A cell can be balanced against 110ern and 100em of
potentiometer wire respectively when in open circuit
and in circuit shorted through a resistance of 10O.
Find the internal resistance of the cell. (Ans. 10)
10.A potentiometer wire of length 1 m has a resistance
of 10O. It is connected to a 6 V battery in series with
a resistance of 5 O. Determine the emf of the
primary cell which gives a balance point at 40 em.
[CBSE D 14]
(Ans. 1.6 V)
PHYSICS-XII
11.A standard cell of emf 1.08 V is balanced by the
potential difference across 91 cm of a metre long
wire supplied by a cell of emf 2 V through a series
resistor of resistance 2 O. The internal resistance of
the cell is zero. Find the resistance per unit length of
the potentiometer wire. (Ans. 0.030 em-I)
12.Potentiometer wire PQof 1 m length is connected to
a standard celle1.Another cell,e2,of emf 1.02V is
connected as shown in the circuit diagram with a
resistance'r'and a switch,S. With switch 5 open,
null position is obtained at a distance of 51 em from
P. Calculate(i)potential gradient of the potentio-
meter wire and(ii)emf of the celle1.(iii)When
switch 5is closed, will null point move towards P or
towards Q? Give reason. [CBSEOD04]
(Ans.0.02V em-\ 2 V, no effect)
P~------------.---~Q
Fig.3.210
13.A batterye1of 4 V and a variable resistanceRhare
connected in series with the wireABof the
potentiometer. The length of the wire of the
potentiometer is 1 metre. When a celle2of emf
1.5 volt is connected between pointsAand C, no
current flows throughe2.Length ofAC=60em.
(i)Find the potential difference between the ends
Aand Bof the potentiometer.
(ii)Would the method work, if the batterye1is
replaced by a cell of emf of 1 V ?
[CBSE D 03]
[Ans.(i)2.5 V,(ii)No]
Rh
4V
~-----100cm-----~
C
A~------------~------~B
Fi. 3.211
14.The potentiometer wire of length 200em has a
resistance of 20O. It is connected in series with a
resistance 100 and an accumulator of emf 6 Vwww4~»trsqrvvr4p»z

288. CURRENT ELECTRICITY
having negligible resistance. A source of
2.4 V is
balanced against a lengthtr;of the potentiometer
wire. Find the value ofL [CBSE F 03)
(Ans.120 cm)
R'=10Q K
~--J""f\r---i r----{ •
6V
t.-----L~
A~----~rC~--~B
2.4 V
Fig.3.212
15.A potentiometer wire carries a steady current. The
potential difference across 70 em length of it balances
the potential difference across a 2 0 coil supplied by
a cell of emf 2.0 V and an unknown internal
resistancer.When a 10 coil is placed in parallel
with the 20 coil, a length equal to 50 cm of the
potentiometer wire is required to balance the
potential difference across the parallel combi-
nation. Find the value ofr. (Ans. 0.5 0)
HINTS
2.Resistance of the potentiometer wire,R= 200
Length of the potentiometer wire = 10 m = 104 mm
Required potential gradient,k= 1flV mm-1
Potential drop along the potentiometer wire,
V=kl= 10 flV mm-I x104mm = 104flV =1O-2V
Current through the potentiometer wire,
I=V= 10-2 = 5 x 10-4A
R20
IfR'is the required resistance to be introduced in
the resistance box, then
.:»:
R+ R'
or
5xl0-4 = __ 2_
20+ R'
R'=39800.
ei= 120 V, ~ = 30 em
.s.=i=15
e212
I=l.= 30 = 20 em
215 15
Difference in the balancing lengths,
~ - 12=30 - 20=10 em.
4.Proceed as in Example 162 on page 3.98.
5.(i)Here ~ = 140cm,12= 100 em
.. ei =i= 140 =Z= 7: 5.
e,12100 5
or
3.Here
Also
3.103
(ii)Letebe the emf of the auxiliary battery andIbe
the length of potentiometer wire. Thene= 4 V and
[= 400 em.
e, ~
.. ""[=[or
..ei=1.4 V
When length is increased by 100 em, new length,
l'= 400+100 = 500 em
ei_ ~ 1.4_~
Now""[-i' or4-500
1.4x500
:. New balancing length, ~ = = 175 em.
4
6.Let the emf of the two cells beeande-0.1 Then
e60
e -0.1=55
ei= 140 =2
4 400 20
e=1.2 V.
emf of the other cell = 12 - 0.1 = 1.1 V.
7.(i)I= ~ = 1025 = 410 em.
k 0.0025
(ii)e'=kl'= 0.0025x860 = 2.15V.
. ~ 2V 4
8.(1)k=-=--=0.02Vem .
~ 100cm
(ii)e2=kl2= 0.02 x 75 = 1.5V.
9.r=R(~-/2J = 10(110 -100) =10.
12 100
10.I= V 6 V =0.4A
RAE+ R (10+ 5)0
V=IRAE= 0.4x10 = 4.0 V
k=V= 4.0 V = 4.0 V = 0.04 V em-I
I1m 100cm
Unknown emf balanced against 40 em of the wire,
e=kl'= 0.04 V em-I x40em = 1.6 V.
11.Letrohm be the resistance per ern of the
potentiometer wire. Then
k=IRAB=eRAB 2 x 100rV cm-I
I 1(R+RAB) 100(2+100r)
As the emf of 1.08 V balances against a length of
91 em, so
k= 1.08 V em-I:. 2 x100r 108
91 100 (2+100r)91
On solving, r= 0.029::::' 0.03 0 em-1.
e102 V 1
12.(i)k=J=-- = 0.02 V em- .
1251em
(ii)ei=klPQ= 0.02 V cm-1x100 em = 2 V.
(iii)With switch S closed, the null point is not
affected because no current flows through the
celle2at the null point.www4·•tssrrxvs4p•£

289. 3.104
13.(I)As~AB=i
c,,
212
..P.D.betweenAand B,
~e100CII\
VAB=- .c.2=--- x 1.5 =2.5 V.
12 60cm
(ii)No,this methodwould not. work when e1=1V,
because then e1<e2and null point cannot be
obtained through the potentiometer wire.
6 6
14.lAB= 10+20 = 30 = 0.2 A
VAB=lABRAB= 0.2x20 =4 V
Potential gradient,
k= VAB'=~ = 0.02 Vern-I
1200crn
B 1.1 h Potential difference
a ancmg engt,L=-------
Potential gradient
2.4V
----'1 =120ern.
0.02Vern
15.Infirstcase.Current sent by the2.0Vcell through
2n coil,
I= e 2
1Total resistance 2 +r
Potential drop across 2 n coil,
2, 4
VI =Rll =2x--=--
2+r 2+r
But VIocZOern
4
--oc70
2+r
In second case.The combined resistance of the
parallel combination of 2n and In coil,
2xl 2
~=2+1=3n
Currentsent by the cell through the parallel
combination,
I= e
gTotal resistance
2 6
2+3r(2/3)+r
Potential drop acrossRp ,
2 6 4
V=RI=-x--=--
2 P 23 2+3r 2+3r
ButV2oc50 ern
4
:. -- oc50
2+3r
Dividing (i)by(ii),'we get
4 2+3r 70
-- x--=- or r=0.5n.
2+ r 4 50
PHYSICS-XII
3.35WHEATSTONE BRIDGE
...(i)
62. Whatisa Wheatstone bridge? When isthebridge
said to be balanced? Apply Kirchhoff's lawsto derive the
balance conditn of the Wheatstone bridge.
Wheatstone bridge.Itisan arrangement of four .
restances used to determine one of these resistances quickly
and accurately in terms of the remaining threeresistances.
Thismethod was firstsugested by a British physicist
SirCharles F.Wheatstone in 1843.
A Wheatstone bridge consists of four resistances P,
Q,Rand S ; connected to form thearmsof a
quadrilateralABCD. A battery of emfeisconnected
between points ,AandCand asensitive galvanometer
between Band D, as shown in Fig. 3.213.
Let S be the resistance to bemeasured. The
resistance R is soadjusted that there is no deflection in
the galvanometer. Thebridgeissaidtobalanced when
the potential difference across thegalvanometer is zero
so that there is no current through the galvanometer.
In the balanced condition of the bridge,
P R
-=-
QS
Unknown resistance,S =Q .R
P
Knowing the ratio of resistances PandQ,and the
resistance R, we can determine the unknown
resistance S. That is why the arms containing the
resistances P and Q are calledratio arms, the armAD
containing Rstandard arm and the arm CD containing 5
theunknown arm.
B
D
Fi. 3.213Wheatstone bridge.
Derivation of balance condition from Kirchhoff's
laws.In accordance with Kirchhoff's first law, the
...(ii) currents through various branches are as shownin
Fig.3.213.
Applying Kirchhoff's second law to the loop ABDA,
we getwww4~»trsqrvvr4p»z

290. CURRENT ELECTRICITY
where G i
sthe resistance of the galvanometer. Again
applying Kirchhoff's second law to the loop BCDB, we
get
(II -Ig)Q-(I2+Ig)5-GIg=0
In the balanced conditionofthebridge, Ig=O.The
aboveequations become
IIP-I2R=0 or IIP=I2R (i}
and IIQ-I2S=0 orIIQ=125 (il)
On dividing equation (i)byequation (ii),we get
P R
Q5
This proves the condition for the balanced
Wheatstone bridge.
63. What do you mean bysensitivity of a Wheatstone
bridge? On what factors doesit depend ?
Sensitivity of a Wheatstone bridge. AWheatstone
bridgeissaid to be sensitive ifitshowsa largedeflectnin
the galvanometer for a small change of resistance inthe
restance arm.
Thesensitivity of the Wheatstone bridge depends
on two factors :
(i)Relative magnitudes oftheresistances in the
four arms of the bridge. The bridge is most
sensitive when all the four resistances areof the
same order.
(ii)Relative positions of battery and galvanometer.
According to Callender for the greatersensitivity of
the Wheatstone bridge,thebattery should besoconnec-
ted that the resistance inseries with the resistance tobe
measured isgreater than the resistance in parallel with it.
According to Maxwell for the greater sensitivity of
the Wheatstone bridge, outof thebattery and the galvano-
meter, the onehaving thehigher resistance should be
connected between the junctnof thetwohighest restances
and the junctn of the two lowest resistances.
64.Whataretheadvantages of measuring resistance
by Wheatstone bridge method over other methods ?
Advantages of Wheatstone bridge method. The
bridge method has following advantages over other
methods for measuring resistance :
(i)It isa null method. Hence the internal resistance
of thecell and the resistance of the galvano-
meter do not affect the null point.
(ii)Asthe method does not involve any measure-
ment of current and potential difference, so the
resistances of ammeters andvoltmeters do not
affect the measurements.
3.105
(iii)The unknown resistance can be measured to a
veryhigh degree of accuracy byincreasing the
ratio of the resistances in arms PandQ.
For Your Knowledge
~When the Wheatstone bridge is balanced, the po-
tential difference between the points B and D is zero.
~The Wheatstone bridge is most sensitive when the
resistances in the four arms are of thesame order.
~Wheatstone bridge method is not suitable for the
measurement of very low and very high resistances.
~In the balanced Wheatstone bridge, the resistance in
arm BD is ineffective. The equivalent resistance of the
balanced Wheatstone bridge between the points A
and C will be
R=(P+Q)(R+5)
eqP+Q+R+S
~If the bridge is balanced, then on interchanging the
positions ofthe galvanometer and the battery there is
no effect on the balance ofthe bridge. That is why the
arms BD and AC are calledconjugate arms ofthe
bridge.
~The Wheatstone bridge is the simplest example of an
arrangement, thevariants of which are used for a
large number ofelectrical measurements. The
important applications of Wheatstone bridge are
metre bridge,Carey-Faster's bridge and post office box.
3.36METRE BRIDGE OR SLIDE WIRE BRIDGE
65.What is a metre bridge? Withthehelpof acircuit
diagram, explain how itcanbeused to find an unknown
resistance. Explain the principle of the experiment and
givetheformula used.
Metre bridge or slide wire bridge.Itisthe simplest
practical applicatn of the Wheatstone bridgethatisused to
measure an unknown resistance.
Principle.Its working is based on the principle of
Wheatstone bridge.
When the bridge is balanced,
P R
-=-
Q5
Construction.It consists of usually one metre long
magnaninwire of uniform cross-section, stretched
along a metrescale fixed over a wooden board and
with its two ends soldered to two L-shaped truck
copper stripsAandC.Between these two copper
strips, another copper strip is fixed so as to provide
two gaps abandaIbI.A resistance box R.B. is connected
in thegapaband the unknown resistance 5 iswww4·•tssrrxvs4p•£

291. 3.106
connected in thegapa1b1
.Asource of emfeis
connected across AC. A movable jockey and a
galvanometer are connected across BO,as shown in
Fig.3.214.
K
.-----~------+~II~--~(·r-~r-----~
s
C
Fi. 3.214Measurement of unknown resistance
by a metre bridge.
Working. After taking outasuitable resistance R
from the resistance box,the jockey is moved along the
wireACtill there is nodeflection in the galvanometer.
This is the balanced condition of theWheatstone
bridge. If P andQare the resistances of the parts AB
andBCof thewire,then for the balanced condition of
the bridge, we have
P R
-=-
Q5
Let total length ofwireAC=100em andAB=Iem,
thenBC=(100-I)cm. Since the bridge wireisof
uniform cross-section, therefore,
resistance of wireexlength of wire
P resistance of AB
Qresistance of BC
or
crI I
o(100-I) 100-I
whereois the resistance per unit length of the wire.
Hence
or
R I
5100-1
~ =R(1~0-I)
Knowing IandR,unknown resistance 5can be
determined.
Determination of resistivity. If risthe radius ofthe
wire andI'its length, then resistivity of its material
will be .
SA5x1t?
P=-1-'=--1'-.
PHYSICS-XII
E I B d•
(i) Wheatstone Bridge'
(ii) Slide Wire Bridge
Formulae Used
1.For a balanced Wheatstone bridge, !..=13.
Q5
IfXis the unknown resistance
!..=B.orX=RQ
QX p.
2.In a slide wire bridge, if balance point is obtained
at1emfrom the zero end, then
!..=B.=_1_ orX=(1001- I)R
QX 100-1
Units Used
All resistances are in ohm and distances in em.
Example 172.Findout the magnitude of restance Xin
the circuit shown inFig.3.215,when no current flows
through the5Qrestor. [ISCE 98]
6V
x 18n
Fi. 3.215
Solution. As no current flows through the middle
5Qresistor, the circuit represents a balanced Wheat-
stone bridge.
X2
186
2
orX= -x18=6Q.
6
Example 173.P,Q,Rand 5arefourrestance wires of
restances2,2,2and3ohms respectively. Find out the
resistance with which5must be shunted inorderthat bridge
may be balanced.
Solution.For abalanced Wheatstone bridge,
P R
-=-
Q5
ButP=2Q, Q=2Q, R=2Q .. ~=~
2 5
i.e.,resistance 5must have a total resistance of2Q.In
arm 5, theresistance of 3Q must be shuntedwitha
resistancersothat the combined resistance is of2Q.
1111111
i.e., -+-=-or-=-- -=-
r32 r236
:. Required shunt, r=6Q.www5notesdrive5com

292. CURRENT ELE
CTRICITY
Example174.In a Wheatstone bridge arrangement, the
rat armsPandQare nearlyequal. Thebridge isbalanced
whenR= 500O.On interchanging PandQ,thevalueofR
for balancingis505O.Find the value ofXand the rat
PIQ.
Solution.Forbalanced Wheatstone bridge,
P R
-=-
Q X
In the first case,R=500 0
P500
-=-
Q X
In the second case whenPand Q are interchanged,
R=5050
Q = 505
P X
Multiplying equations (1)and(2),
1 = 500x505
X2
or X=~500x505
=502.50
Substituting the value of Xin(1),we get
P 500
-=--
Q 502.5
=_1_=1: 1.005
1.005
Example175.The galvanometer, in each of the two given
circuits, does not show any deflectn. Find therat of the
restors Rland ~, used in these two circuits.
[CBSEOD 13]
3.0V
1.20n
~----~Gr-----~
Circuit 1 Circuit 2
Fig. 3.216
Solution. Incircuit1,the Wheatstone bridge is in
thebalanced condition, so
4 6
--= -:::>
Rl9
3.107
Incircuit 2, the interchange of the positions of the
battery and the galvanometer does not affect the
balance condition of the Wheatstone bridge, so
~=~
12 8
~ = 6x8 =40
12
...(1)
~=~=~
~ 4 2
=3: 2
Example 176.Calculate thecurrent drawn from the
battery by the network of restors shown in Fig. 3.217.
rCBSE OD 09, ISC]
...(2)
2n
2n
4V
Fi. 3.217
Solution. The given network is equivalent to the
circuit shown in Fig. 3.218.
•
B
D
4V
Fig.3.218
Now
10 20
---
20 40
i.e.,
P R
-=-
Q5
The given circuit is a balanced Wheatstone bridge.
The resistance of 50 in arm BD is ineffective. The
equivalent circuit reduces to the circuit shown in
Fig.3.219.
InBz n
c
z oD4n
4V
I
Fig. 3.219www4·•tssrrxvs4p•£

293. Resistances inABandBCare in series, their
equivalent resistance =
1+2 = 3 n.
Resistances in AOandOCare in series, their
equivalent resistance = 2 +4 = 6n
Theresistances of 3 nand 6n are in parallel.
The equivalent resistance R betweenAandCis
3x6
R=--=2n
.3+6
V 4
Current, I=-=-=2 A.
R 2
Example 177.Each of the restances in the network
shown in Fig. 3.220equals R.Find the resistance between
two terminals A and C. Fi. 3.222
3.108
D
Fig.3.220
Solution. The network shown in Fig. 3.221 is the
equivalent network of the given network.
D
A c
B
Fi. 3.221
Itis a balanced Wheatstone bridge because
R R
R R
PHYSICS-XII
Example 178. Apotential difference of 2Visapplied
between the points AandBshown in network drawn in
Fig.3.222. Calculate
c
D~----~~v-----~E
(i)the equivalent resistance of thenetwork between the
points A and B, and
(i) the magnitudes of currents flowinginthe arms
AFCEB and AFOEB. [eBSE OD 981
Solution. (i)Theequivalent network is shown in
Fi. 3.223. Itis a balanced Wheatstone bridge because
2n 2n
-=-
2n 2n
c
A B
D
Fig.3.223
Hence the pointsCand0are at the same potential.
The resistance in armCOis ineffective. The given
network reduces to the equivalent circuit shown in
Fi. 3.224.
HIC2Q
A B
2Q 2Q
D
Hence the points Band 0must beat the same
potential. The resistanceRinarmBOisineffective. Fig.3.224
Total resistance alongAOC=R+R = 2 R n
Total resistancealong ABC= R+R =2R n
These two resistances form a parallel combination.
Effective resistance between AandC
=2Rx2R =Rn.
2R+2R
Total resistance along FCE= 2+2 = 4 n
Total resistance alongFOE=2+2 = 4 n
These two resistances form aparallel combination.
:. Equivalent resistance between points AandB
=4x 4=2 n
4+4www4yztp€o·tvp4nzx

294. CURRENT ELECTRICITY
(..)
T I .h..V2V1 A
IIota current In t e CIrCUIt=-=-=
Rz n
Current through arm AFCEB
= Current through arm AFDEB
=..!.A=O.5 A.
2
Example 179. Find the value of the unknown resistance X,
inthefollowing circuit, ifno current flows through the
section AO. Also calculate the current drawn bythe circuit
from the battery of emf6Vandnegligible internal
restance. (Fig. 3.225) [CBSEOD 02]
A
A
o
B1t<---U-...J\fV\.,.---"" C
6V
2.4Q
6V
Fig.3.225 Fig.3.226
Solution.Theequivalent circuit for the given
network is shown in Fig. 3.226.
or
As no current flows through thesectionAO,so the
givencircuit is a balanced Wheatstone bridge.
Hence
2 3
4 X
3x4
or X=--.=6n
2
The resistance oflO n in sectionAOis not effective.
Total resistance alongBAC= 2+4 = 6 n
Total resistance alongBOC= 3 + 6 = 9 n
These two resistances form a parallel combi-
nation. The effective resistance between Band Cis
R = 6x9 = 18 = 3.6 n.
6+95
Total resistance in thecircuit = 3.6 + 2.4 = 6 n.
6V
Current,1=- =1 A
6n
Example 180. Sixequal resistors, each of value R,are
joined together as shown in Fig.3.227. Calculate the
equivalent restance across AB. If asupply of emfe
connected across AB, compute the current through the arms
DEandAB. [CBSESample Paper 03]
3.109
R
R
R R
C~J\/I.A,--.:::;.D-J\'I\"~-l E
R R
Bo--L---~F----l
Fi. 3.227
Solution. The equivalent circuits are shown below.
The resistance R in arm DE of the balanced Wheatstone
bridge is ineffective.
o
R R
R R
E
R
R
(a) (b)
A B A B
Fig.3.228
Theequivalent resistance R' across ABisgiven by
1 1 1 14 2
-=-+-+-=-=-
R'2R 2R R 2R R
R'=R/2
e.e 2e
Current through arm AB=-=-- = -.
R'R/2 R
Current through arm DE =o.
Example 181.Calculate the rat of the heat produced in
thefour arms of the Wheatstone bridge shown in Fig. 3.229.
. 40n 60n
Solution. As -- =-- B
IOn 15n
Thebridge is balanced.
.. p.o. acrossAB
=P.O.acrossAD
or o40 II=6012
l=60=1.5
1240
II=1.512
Heats produced in timetin different arms of
Wheatstone bridge are
HAB=I~ Rt =(1.512)2 x40xt=90Iit
HBC=I~xlOx t =(1.5I2)2xl0x t=22.5 Iit
HAD=Iix60xt= 60Iit
HDC=Iix15xt=15Iit
or
or
Fi. 3.229www4·•tssrrxvs4p•£

295. Whenthe resi
stance in the right gap is increased by
12.50,total resistance becomes 22.5 O. The balance
pointshifts towards zero end by 20 cm. Fi. 3.232
3.110
Hence the ratio of the heats produced in the four
arms is
HAB: HBC: HAD: HDC
=90Iit:22.5Iit:60Iit:15Iit
= 90:22.5 :60:15=6:1.5 : 4 : 1.
Example 182. Inthe following circuit, a metre bridgeis
showninitsbalanced state.The metre bridge wire has a
resistance of10emCalculate the value of the unknown
restance Xand the current drawn from the battery of
negligible internal resistance.
Solution. In balanced condition, no current flows
through the galvanometer.
Here P=Resistance ofwireAJ=40 0
Q=Resistanceof wire BJ=600
R=X,5=60
In the balanced con-
dition,
x
PR
A B---
Q5 /
40X
or -=- 6V
60 6
or X=40
Fi. 3.230
Total resistance of wire AB= 1000
Total resistance of resistances Xand6 0connec-
tedinseries =4+ 6 = 100
This series combination is in parallel withwireAB.
10x100 100
Equivalent resistance = =-0
10 + 10011
emf of the battery =6 V
.'.Current drawn from the battery,
I=emf=_6_=O.66A.
resistance 100/11
Example 183. With acertain resistance intheleftgap of a
slide wire metre bridge, thebalance pointisobtained when a
restance of100istakenoutfromthe restance box. On
increasing therestance from the restance box by 12.5 0, the
balancepoint shifts by20em.Find the unknownresistance.
Solution. With unknown resistance X in the left
gap and known resistance of 100 inthe right gap,
suppose the balance point is obtained at1em from the
zero end. Then
X 1
-=--
10100-1
PHYSICS-XII
X 1-20 1-20
22.5 100 - (l -20)120-1
Dividing (1) by (2),
22.5 I 120-1
--=--x---
10 100-1 I -20
Onsolving, we get 12-1201+ 3600=0
I=60em
...(2)
From(1),
X
10
60 60x10
orX=---=150.
40100 -60
Example 184.In metre bridge, thenullpointisfound at a
dtance of 60.0 em from A. If now a resistance of 50is
connected in series with 5,thenull point occurs at50em.
Determine the values of Rand5.[Fig.3.231]. [CBSE D 10]
A·~~------+- -r~B
Fi. 3.231
jlj'l'i1j'j,j,jij,j'jijijiiijljijljIPi'lIlijijijijljijijij'I'il
R60 3
Solution. Infirstcase, - =-=- ...(i)
5402
R 50
Insecond case, --=- ...(ii)
5+5 50
Ondi'din ("b (..)5+5_3
IV!gIJYus,-5- -2
or 25+10=35
m 5=WO
and R= ~5= ~x10=150
22
Example 185.In ametrebridge, the null pointisfound at
adtance of 40cmfrom A. If a restance of 120is
connected inparallel with5,the null point occurs at 50.0em
fromA.Determinethe values ofRand5.[Fig.3.232]
[CBSE D 10]
1Hl
.J\IV'y,
I
I
...(1)
A~~ 4- ~~B
jiiii"""""'!"""!"""j'I'I'I'j""""""'!'jilliljiilwww4yztp€o·tvp4nzx

296. CURRENT ELECTR
ICITY
Solution.
Infirst case,
R 40 2
-----
5603
R 50
lIS-50
5+12
Insecondcase,
R 12 2
5=6nor -=--=- or
5 5+12 3
and
2 2
R=-x5=-x6=4n
3 3
Example 186. A resistance R=2nisconnected to one of
thegaps in a metrebridge, which uses a wire of length 1m.
Anunknown resistance X >2nis connected in the other gap
as shown in the figure. The balancepoint is noticed at '1'from
the positive end of the battery. On interchangingRandX,it
isfound that the balance point further shifts by20em(away
fromendA).Neglecting the end correctn, calculate the
value of unknown resistance Xused. [CBSE OD OS]
X
A~~~~~----------~B
-- (100-/)em
Fig.3.233
Solution. Infirst case,
R I
---
X 100-1
Insecondcase,
X
R
1+20 1+20
80-I100-(I+ 20)
or
On multiplying the two equations,
l=_I_x 1+20
100-180-/
8000-1801 + 12= 12+ 201
2001 =8000
1= 40 em
Now X =I+20 R = 40 + 20x2 =3n.
80-1 80-40
or
or
3.111
null pointisfound toshift by10em towards theend A of the
wire.Findthe positn of null point if a restance of30 n
were connected in parallel with Y. [CBSE Sample Paper08]
X Y
AU-------~----------~c
111--1 ---~(.
Fi. 3.234
Solution. Infirstcase,
X60
Y40
X 3
or
Y2
Insecond case,
_X_=50=1
Y+ 15 50
X xY+ 15 = ~x 1
Y X 2
1+ 15 = ~
Y2
Y=30n
X=~Y=~x30=45n
2 2
When a resistance of 30 nis connected in parallel
withY,the resistance in the right gap becomes
Y'=30Y= 30x30 = 15 n
30+Y 30+30
or
or
Suppose the null point occurs atIem from endA
Then
X
15
I
100-I
or
45 I
-=--
15100-1
or 300-31=1
or 41 = 300 or 1=75em.
Example 188.When two known restances, Rand5,are
connected in the left and right gaps of a metre bridge, the
balance pointisfound at a dtance 11from the'zero end' of
x·
..J\N'v,
,
,
Example 187. Thegiven figure shows the experimental
setup of a metre bridge. The null pointisfound to be 60em
away from the end A withXand Y in positn asshown.
When a resistance of15nisconnected in series with Y, theFig.3.235
A B
IIj'jll'P,·I'lIl'lIl'lIiilijilijijljljljiltjljiiii'r'illliiiiljwww4·•tssrrxvs4p•£

297. 3.112
themetre
bridge wire. An unknown restance Xisnow
connected in parallel to the restance 5and the balance point
isnow found at a distance 12from the zero end of the metre
bridge wire. Obtain aformula forXin terms of11,12andS.
[CBSE004C, IOC ;0009]
Solution. Infirst case,
R
5
...(i)
Insecond case,
R 12
XS/(X+S) 100-12
Dividing (ii)by(i),we get
X+5= ~(100 -11Jor
X 11 100 -12
X= 5
~ (100 -11 J-1
11 100-12
problems ForPractice
1.Four resistances of 15Q, 12Q,4 Q and 10Q respec-
tively are connected in cyclic order to form a
Wheatstone bridge. Is the network balanced ?If
not,calculate the resistance to be connected in
parallel with the resistance of 10Q to balance the
network. (Ans.Bridge is not balanced, 10Q)
B
PHYSICS-XII
4.Calculate the equivalent resistance between points
Aand B of the network shown in Fig. 3.238.
[CBSE 0 99](Ans. 2 Q)
(..)Fig.3.238
...II
2.The Wheatstone's bridge
of Fig. 3.236 is showing
nodeflection in the
galvanometer joined
between the points B
andD.Compute the
value of R.
D
(Ans.25Q)
Fi. 3.236
3.(i)Calculate the equivalent resistance of the given
electrical network between pointsAandB.
(ii)Also calculate the current throughCDandACB,
if a 10Vd.c.source is connected betweenAandB,
and the value of R
is assumed as 2 Q.
[CBSE 0008]
[Ans.(i)RAB=RQ
(ii)lCD=0,
lACB =2.5 A]
c
Dt:F------J'IN\r-----O()E
Fi. 3.237
5.Calculate the equivalent resistance between the
points AandBof the network shown in Fig. 3.239.
(Ans.R)
Fi. 3.239
6.Calculate the resistance between the points AandB
of the network shown in Fig. 3.240. (Ans. 8Q)
Ion IOn
40n
Fi. 3.240
7.For the network shown in Fi. 3.241, determine the
valueof Rand the current through it, if the current
through the branchAOiszero. (Ans.6Q,0.5 A)
A
iznan
A D BBO'---oi I-~"".I\r--.::o C
lOV 2n
Fi. 3.241 Fi. 3.242
8.The potentiometer wireABshown in Fi. 3.242 is
40 ern long. Where should the free end of the
galvanometer be connected onABso that the
galvanometer may show zero deflection?
(Ans.16 em fromA)www4yztp€o·tvp4nzx

298. CURRENT ELECTRICITY
9.The potentiometer
wireABshown in Fi. 3.243 is
50 cm lon. WhenAD= 30em,no deflection occurs
in the galvanometer. FindR. (Ans.40)
x 15.
6Q R
B
J
100 em
A 0B 5V
16.
Fi. 3.243 Fig.3.244
10.Calculate the value of unknown resistance X and
the current drawn by the circuit, assuming that no
current flows through the galvanometer. Assume
the resistance per unit length of the wireABto be
0.010/ cm. (Fig.3.244) [CBSE D 01]
(Ans. 60,5.5 A)
11.In Fig 3.245, P = 30, Q= 20, R= 60, S=40 and
X=5nCalculate the currentl. [CBSE D921
(Ans.0.6 A)
c
I
r r
r
r r
o
6V
2.0 V
Fi. 3.245 Fi. 3.246
12.Each resistorrshownin Fig. 3.246 has a resistance
of 100 and the battery has an emf of 6V.Find the
current supplied by battery. (Ans. 0.6 A)
13.Find the equivalent resistance between the pointsX
and Y of the network shown in Fi. 3.247.
(Ans.100)
F'19·3.247
3.113
14.In a metre bridge, the length of the wire is 100em.
At what position will the balance point be obtained
if the two resistances are in the ratio 2 : 3?
(Ans. 40em)
In the metre bridge experimental set up, shown in
Fi. 3.248, the null point'D'is obtained at a distance
of 40 cm from endAof the metre bridge wire. If a
resistance of 100 is connected in series with X,null
point is obtained atAD= 60 cm. Calculate the
values ofXand Y. [CBSED 13]
(Ans. 80,120)
In a metre-bridge experiment, two resistances P
andQare connected in series in the left gap. When
the resistance in the right gap is 500, the balance
point is at the centre of the slide wire. If P and Q are
connected in parallel in the left gap, the resistance
in the right gap has to be changed to 120 so as to
obtain the balance point at the same position. Find
P andQ. (Ans. P= 300,Q=200)
17.In a metre bridge when the resistance in the left gap
is20 and an unknown resistance in the right gap,
the balance point is obtained at 40 cm from the zero
end.On shunting the unknown resistance with 20,
find the shift of the balance point on the bridge
wire. (Ans. 225em)
18.Fi. 3.248 shows experimental set up of a metre
bridge. When the twounknown resistancesXand Y
are inserted, the null pointDis obtained 40 cm from
the endA.When a resistance of 10 0 is connected in
series withX,the null point shifts by 10 cm. 'Find
the position of the null point when the 100
resistance is instead connected in series with
resistance'Y'.Determine the values of the resis-
tancesXand Y. [CBSE D 09]
X Y (Ans.X=200,
Y=300,
I'= 33.3 em]
o c
Fi. 3.248
HINTS
1.The four resistances' are
connected in acyclic order
, i I!t ,ij
as shown in Fi. 3.2491 i
,
15 10
~*-
124
I"Thp~ ':Yh~atstQne~ridgei.~; I'
not.balanced. To: balance
the network, suppose
B
As
c
'0
Fi. 3.249www4·•tssrrxvs4p•£

299. 3.114
resistance R is connected in parallel with 10n
resistance. Then
10R
15 10+ R
12 4
or
10R
--=5
lO+R
or R=100.
200
40
R = 250.
100
R = 20 ..
100+ R
2.
100
100 R
100+ R
or
3.Proceed as in Example178on page3.108.
1 2
4.As-=-
2 4
:. The given circuit is a balanced Wheatstone bridge
as shown in Fig. 3.250.The resistance of100is
ineffective.
B
A B
Fi. 3.250
:
We have(In + 20)and(2n + 40)combinations
in parallel.
3x6
R=--=20.
3+6
5.The given circuit is equivalent to the circuit shown
in Fi.3.251.
A B
Fi. 3.251
R R
-=-
R R
So it is a balanced Wheatstone bridge. We have
resistances(R + R)and(R + R)in parallel.
2R x2R
..Equivalent resistance= = R.
2R+2R
Here
6.Here10 = 10 :. Resistance of200is ineffective.
10 10
We have resistances of(IOn + 100), (IOn + 100)
and40nin parallel.
1 1 1 1 5
..-=-+-+-=- orR=SO.
R 20 20 40 40
PHYSICS-XII
7.As pointsAand0are at the same potential,
therefore
1 4
- =-orR = 4 x 1.5= 60
1.5 R
IfR'is the equivalent resistance of the network
betweenBandC,then
2.5x10
R'= +2=40
2.5+10
Current in the circuit,I=10=2.5A
4
Current throughR(= 60) = 2.5 x 2.5= 0.5 A.
2.5+10
8AD I
8.-=-=-- 1=16cm.
12DB40-/
96 _AD_30 R 4r.
.R -DB -50 - 30:.=s z.
10.Resistance of wireAJ= 60 x 0.01= 0.600
Resistance of wireBJ= 40 x 0.01= 0.400
When no current flows through the galvanometer,
P R 0.60 X
-=- or -=-
Q5 0.40 4
0.60x4
X=--=60
0.40
Total resistance ofXandRin series= 6 + 4 = 100
Total resistance of wireAB= 0.60+ 0.40= 1.00
The above two resistances are in parallel.
. . . 10 xI10
:. Total resistance of the circuit= -- = -0
10+ 1 11
EMF 5
Current,I= = -- = 5.5 A.
Resistance10/ 11
11.The circuit is a balanced Wheatstone bridge. Its
effective resistanceRis given by
1 1 1 3
-=--+ --=- or
R 3+2 6+410
V 2
:. Current, I=-= -- = 0.6A.
R10/3
12.As!.=!.,so the given circuit is a balanced
r r
Wheatstone bridge and the resistancerin the
vertical arm is ineffective. The circuit is then
equivalent to two resistances of2rand2rconnected
in parallel.
E. 1 . R 2rx2r
.'. qUlva ent resistance,=--- =r= 100
2r+2r
Current supplied by the battery of emf 6 V,
e6
I= - =-= 0.6A.
R10www4yztp€o·tvp4nzx

300. CURRENT ELECTRICITY
13.The equivalent circuit
is shown in Fi. 3.252.
A
Fi. 3.252
B
The resistances in armABare ineffective.
1 1 1 1
:.-=---+---=- or R=lOQ.
R 10+10 10+ 10 10
14.For a balanced metre bridge, X = _/-
R 100-/
2 /
-=--..
3 100-/
200
or 200- 2/ = 31 or / =-= 40 em.
5
But
X
R
2
3
X 40 2
15.Infirstcase:
Y 100-40 3
X+10 60
Insecondcase: -- = ---
Y 100-60
X103
or-+- =-
Y Y 2
103 X 3 2 5
or-=---= --- =-
Y 2 Y 2 3 6
10x6 2 2
.,Y=--=12Q and X=-Y=-x12=8Q.
533
16.When P and Q are connected in series in the left gap,
P+Q 50
50 100- 50
.. P+ Q= 50Q ...(1)
When P and Q are connected in parallel in the left
gap,
PQ
P+Q=50=1
12 50
PQ = 12(P + Q) = 12x50 = 600
(P -Q)2= (P +Q)2_4 PQ = 502 -4x600= 100
3.115
..P- Q= 10 ...(2)
Solving (1) and (2), P = 30Q and Q = 20Q.
17.If X is the unknown resistance, then
2
X
2x60
X=--=3Q
40
40
or
100-40
When resistance X is shunted with 2 Q resistor, the
effective resistance becomes
Xx2 3x2
X'=--=--=1.2Q
X+2 3+2
Now if the balance point is obtained at distancel '
from the left end, then
2 l' 2 I'
or-=---
X' 100-I' 1.2 100-I'
/'=62.5cm
Shift in the balance point
= /' -/ = 62.5 -40 = 22.5 em.
18.With the unknown resistances X and Y,the balance
point is 40 cm from the endA .
X
Y
40 2
100-40 3
or
With 10Q resistance in series with X, the balance.
point is at 40+ 10= 50emfrom the endA.
X+lO 50 =1
Y 100-50
Y=X+lO
~X=X+lO
2
X = 20Q. and Y = 20+ 10= 30Q.
or
or
or
When 10Q resistance is connected in series with Y,
let the balancing length beI'.
Then
X r
--
Y+ 10100-/'
20 v
or ---
30+ 10100-/'
or l'=33.3 em.www4·•tssrrxvs4p•£

301. GIDELINES To NCERT
EXERCISES
3.1.The storage battery of a car has anemfof12V. If the
internal resistance of the battery is0.40,whatisthemaximum
currentthatcan be dra'i/{,n from the battery?
Ans.Heree=12V>r=0.40
The current drawn £tom the battery will be maximum
when the external resistance in the circuit is zeroi.e.,
R=O.
I =§.J12
max r0.4
=30 A.
3.2.A battery of emf 10V and in ternal resistance 30is
connected to a resistor. If the current in thecircuitis05A,what
istheresistance of the resistor? Whatistheterminalvoltage of
thebattery when the circuit isclosed ?
Ans.As
J=_e_
R+r
s
or R +r=-
I
e 10
R=--r=--3=170
I 0.5
Terminal voltage,
V=IR=0.5x17=8.5V.www.notesdrive.com

302. CURRENT ELECTRICITY
3.3:(i) Three resistors
of1 0, 20and3 0are combined in
series. What is the total resistance of the combination?(ii)If the
combinationisconnected to a battery of emf12Vand negligible
internal resistance, obtain the potential drop across each
resistor.
Ans.(i)Rs= ~+ ~ + ~= 6O.
(") C . h . . It12 2 A
IIurrent In t e circuit, =-=-=
R 6
Potential drops across different resistors are
v;=I ~=2x1=2Y,
V2=I ~= 2 x 2 = 4V,
V3=I ~= 2x3 = 6 V.
3.4.(i) Three resistors2 0,40and50arecombined in
parallel. Whatisthe total resistance of the combination?(ii)If
the combinationisconnected to a battery of emf20V and
negligible internal resistance, determine the current through
each resistor, and the total current drawn from the battery.
. 1 1 1 1 1 1 1 19
Ans.(1)-=-+-+-=-+-+-=-
Rp ~ ~ ~ 2 4 5 20
R=20O.
p19
(ii)Currents drawn through different resistors are
t20 t20
II= ~ =2= 10A,12= ~ =4= 5 A,
t20
13=-=- =4A
~ 5
Total current drawn from the battery,
I=II+12+13= 10+5+4 = 19 A.
3.5.At room temperature(27°C), the resistance of a heating
element is100Q.Whatisthe temperature of the elementifthe
resistance is found to be117 0,given that temperature
coefficient of the resistor materialis1.70x10-40 C-1.
Ans. Here ~ = 1000, ~ = 1170, tl= 27°(,
a= 1.70 x 10-4 0(-1
a= ~ -~
~ «.-t1)
t _ t= ~-~ = 117 - 100 = 1000
21 ~ a 100 x 1.70 x 10-4
t2= 1000 +tl= 1000 + 27 = 1027°C.
3.6.A negligibly small currentispassed through a wire of
length15m and uniform cross-section 6.0x10-7m2and its
resistance is measured to be 5.0Q.What is the resistivity of the
material at the temperature of the experiment?
Ans.HereI= 15 m,A=6.0 x 10-7 m2,R = 5.00
RA' 5.0 x 6.0 x 10-7
Resistivity, p = - =------
I 15
As
= 2.0 x 10-70 m.
3.149
3.7.A silver wire has a resistance of2.10at27.5°C and a
resistance of2.7 0at100°CDetermine the temperature
coefficient of resistivity of silver.
Ans.Here ~ = 2.10,t1 = 27.5°(, R;=2.70,t2 = 1000(
Temperature coefficient of resistivity of silver,
a= ~ -~
~ (t2-t1)
2.7 - 2.1 0.6
2.1(100 - 27.5) 2.1 x 72.5
= 0.003940(-1.
3.8.A heating element using nichrome connected to a 230 V
supply draws an initial current of3.2 A which settles after afew
seconds to a steady value of2.8A.What is the steady
temperature of the heating element if the room temperatureis
27°C?Temperature coefficient of resistance of nichrome averaged
over the temperature range involvedis1.70 x 1O-4°C-1.
Ans. HereV= 230 V,II= 3.2 A,
12= 2.8 A,a= 1.70 x 10-4 0(-1
Resistance at room temperature,
~ =V= 230 = 71.8750
II3.2
Resistance at steady temperature,
~ =V= 230 = 82.1430
122.8
Now
a= ~ -~
~ (t2-t1)
t2-tl=~-~
~a
82.143-71.875
- 71.875 x 1.70 x 10-4
10.268x104= 840.350(
71.875x1.7
:.Steady temperature of element,
t2= 840.35+27 = 867.35°(.
3.9.Determine the current in each branch of the network
showninFig. 3.313.
Ion IOV
Fig. 3.313www5notesdrive5com

303. 3.150
Ans.LetI, II' 1
2,13be the currents as shown in
Fig.3.314.We apply Kirchhoff's second rule to different
loops.
B
LIon
Fig.3.314
For loopABDA,
lOll+513 - 512 =0
For loopBCDB,
5(11-13)-10(12+13) - 513= 0
For loopADCFGA,
512+10(12+13)+10(11+12) = 10 (":
1011-512+513 = 0
511- 1012 - 2013 = 0
1011+2512+1013= 10
Solving equations (1), (2)and(3),we get
4 6 2
11 = 17A,12 = 17A,13 = - 17A
or
11+12 =I)
...(1)
...(2)
...(3)
Currents in different branches are
4 6
lAB= 11 =17A, IBC=11-13=17A,
4
IDC= 12+13 =-A
17
6 2
IAD= 12 =-A, IBD= 13= --A
17 17
Total Current,
10
I=II+12=-A.
17
3.10.(i) In ametre bridge (Fig.3.315),the balance point is
found to be at39.5em from the end A, when the resistorYisof
12.5n.Determine the resistance ofX.Why are the connections
between resistorsina Wheatstone or metre bridge made of thick
copper strips?(jj)Determine the balance point of the bridge
x y
A~--~~~----------~C
Fig. 3.315
PHYSICS-XII
above ifXand Yare interchanged.(iii)What happens if the
galvanometer and cell are interchanged at the balance point of
the bridge? Would the galvanometer show any current?
[CBSE D05]
Ans.HereI = 35.9 ern,R = X = 7, 5 =Y= 12.5 n
As5- 100 -I R.2 _ 100 - 39.5 R
--- x .. 1.5- x
I 39.5
R= 12.5 x39.5 = 8.16 n
60.5
or
Connections are made by thick copper strips to
minimise the resistances of connections which are not
accounted for in the above formula.
or
(ii)WhenXand Yare interchanged,
R=Y= 12.5n, 5= X= 8.16n, I=?
100-I 100-1
5=-1- xR:.8.16=-1- x12.5
8.161= 1250 - 12.51
1250
1 =-- = 60.5 n ,from the endA.
20.66
As
or
(iii)When the galvanometer and cell are interchanged
at the balance point, the conditions of the balanced bridge
are still satisfied and so again the galvanometer will not
show any current.
3.11.A storage battery of emf B.O Vand internal resistance
0.5nisbeing charged by a 120 V de supply using a series
resistor of15.5 n. Whatisthe terminal voltage of the battery
during charging?Whatisthe purpose of having a series
resistorinthe charging circuit?
Ans.When the storage battery of8.0volt is charged
with adesupply of120V,the net emf in the circuit will be
E.' = 120 - 8.0 = 112 V
Current in the circuit during charging
1=~= 112 =7 A
R+r15.5+0.5
Theterminal voltage of the battery during charging,
V=E.+lr= 8.0+7x0.5 =11.5V
The series resistor limits the current drawn from the
external source. In its absence, the current will be dange-
rously high.
3.12.In a potentiometer arrangement, a cell of emf1.25V
gives a balance point at 35.0 em length of the wire. If the cellis
replaced by another cell and the balance point shifts to 63.0 em,
whatisthe emf ofthe second cell?
Ans.HereE.1= 1.25 V,4= 35.0 em,12 = 63.0em,
E.2=?
As
E.2 12
E.1=1;
e 12e 63x1.25
C,2 =-xC,1= =2.25V.
4 35www4not{syriv{4xom

304. CURRENT ELECTRICITY
3.13.Thenumb
er density of free electrons in acopper
conductor is 85x1028m"3.How long doesanelectron take to
driftfromone end of a wire 3.0 m long toits other end?Thearea
of cross-section of the wireis 2.0x10-6m2and itiscarrying a
current of3.0 A.
Ans.Heren=8.5x1028m-3, 1=3m,
A=2.0x10-6m2,e=1.6x10-19C,I=3.0 A
Drift speed,
I
v-
d-enA
3 -1
= ms
1.6x10-19x8.5x1028x2x10-6
3 ms-1 = 1.1x10-4ms-1
16x85x2x10
Required time,
I 3 4
t=-= 4 S= 2.73x10 s =7.57 h.
vd1.1x10-
3.14.Theearth's surface has a negative surface charge
densityof10-9 Cm-2. Thepotential difference of400 kV between
thetop of the atmosphere and the surface results (duetothe low
conductivity of the lower atmosphere) in a currentof only
1800 Aover the entire globe. If there were no mechanism of
sustaining atmospheric electric field, how muchtime (roughly)
wouldberequiredtoneutralise the earth's surface? (Radius of
the earth = 6.37x106m).
An.Surfacecharge density,
c=10-9Cm-2
Radius ofthe earth,
R=6.37x106m
Current, I=1800 A
Total charge of the globe,
q=surface areaxc= 41t R2cr
=4x3.14x(6.37x106)2x10-9
=509.65x103C
Required time,
q509.65x103
t= - = = 283.13 s= 283 s.
I 1800
3.15.(a)Six lead-acid type of secondarycells each of emf2.0 V
and internal resistance 0.0150are joined inseriestoprovide a
supplytoa resistance of8.5O.What are the current drawn from
the supply and its terminal voltage?
(b)A secondary cell after long usehas an emf of 1.9Vanda
large internal resistance of 380 0.What maximum current can
bedrawn from the cell? Could the cell drive the starting motor
ofacar?
Ans. (a)Here€= 2 V, r=0.0150, R = 8.50, n=6
When the cells are joined in series, the current is
I=~= 6x2 =~A=1.4A
R+nr8.5+6x0.015 8.59
3.151
Terminal voltage,
V=IR=1.4x8.5 = 11.9 V.
(b)Here€=1.9V,r= 3800
I=§.=~ A =0.005A
max r380
This secondary cell cannot drive the starting motor of
a car because that' requires a large current of about 100 A
for a fewseconds.
3.16.Twowiresofequal length, one of aluminium and the
otherofcopper have the same resistance. Whichof thetwo wires
islighter ?Hence explain why aluminiumwiresarepreferred
foroverhead power cables.
Given PAl= 2.63x1O-80m, PCu=1.72x10-8Om,
relative density of Al=2.7and that of Cu=8.9.
Ans.Mass=volumexdensity = Ald
=Pi.Id=pd 12
R R
I
[.,'R=P-]
A
As the two wires are of equal length and have the
same resistance, their mass ratio will be
ncu PCu dcu 1.72x10-8x8.9
-=---= 8 =2.1558=2.2
mAl PAldAl 2.63x10x2.7
i.e.,copper wire is 2.2 times heavier than aluminium wire.
Sincealuminium is lighter, it is preferred for long sus-
pension of cables otherwise heavy cable may sag down
due to itsown weight.
3.17. What conclusion can you draw from the following
observations on a resistor made of alloy manganin :
CurrentVoltageCurrentVoltage
I(A) V I(A) V
0.2 3.94 3.0 59.2
0.4 7.87 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.5
2.0 39.4 8.0 158.0
Ans.We plot a graph between current I (along y-axis)
and voltageV(alongx-axis)as shown in Fig. 3.316.
10 20 30 40 50 60 70 80 90 100110120130140150160170
V~
Fig. 3.316 V-Igraph for rnanganin.www5notesdrive5com

305. 3.152
Since the V-Igraph
is almost a straight line, therefore,
manganin resistor is an ohmic resistor for given ranges of
votlage and current. As the current increases from 0 to 8 A,
the temperature increases but the resistance of manganin
does not change. This indicates that the temperature
coefficient of resistivity of manganin alloy isnegligibly
small.
3.18.Answer the following questions:
(a) Asteadycurrent flows ina metallic conductor of
non-uniform cross-section. Say which ofthesequantities
isconstant along the conductor: current, current
density, electric field, drift speed? [CBSE DISC]
(b) Is Ohm's law universally applicablefor all conducting
elements ?If not, give examples of elements which
donotobeyOhm's law.
(c)Alow voltage supply from which oneneedshigh
current must have very low internal resistance.
Why?
(d) Whyahigh tension (H.TJ supply ofsay6kV must
haveaverylargeinternal resistance?
Ans.(a)Onlycurrent is constant because itis given to
besteady. Other quantities: current density, electric field
and drift speed vary inversely withareaofcross-section.
(b)No,Ohm's law is not universally applicable for all
conductingelements. Examples of non-ohmic elements
arevacuum diode, semiconductor diode, thyristor, gas
discharge tube, electrolytic solution, etc.
(c)The maximum current that can be drawn from a
voltage supply is given by
e
Imax =-
r
Clearly, Imaxwill be large ifris small.
(d)If the internal resistance is not verylarge,then the
current will exceed the safety limits in case the circuit is
short-circuited accidentally.
3.19.Choosethe correct alternative:
(a)Alloys of metals usually have (greater! lesser)
resistivity than thatof their constituent metals.
(b)Alloys usually have much(lower/higher) temperature
coefficients of resistance thanpure metals.
(c) The resistivity ofthe alloy manganin isnearly
independent of/increases rapidly with increase of
temperature.
(d) The resistivity of a typical insulator te.g., amber) is
greater than that of a metal by afactor of the order of
(1022/103 ).
Ans.(a)greater(b)lower(c)is nearly independent of
(d) 1022.
3.20.(a) Given n resistors each of resistance R,howwill you
combine them toget the ti)maximum, (ii)minimum effective
resistance?What isthe ratio of the maximum to minimum
resistance ?
PHYSICS-XII
(b) Given the resistance of Elf2Of30,how will you
combine them to get an equivalent resistance of:
(i)110 (ii)110 (iii)60 (iv)i..0?
3 5 11
[CBSE F 15]
(c)Determine the equivalent resistance of the following.
networks:
(a)
R
(b)
Fig. 3.317
Ans.(a)For maximum effective resistance, all then
resistors must be connected in series.
:.Maximum effective resistance,
Rs=nR
For minimum effective resistance, all thenresistors
must be connected in parallel. It is given by
1 1 1 1 n
-=- +-+ - + nterms=-
Rp R R R R
:.Minimum effective resistance,
R=~
pn
Ratio of the maximum to minimum resistance is
~=nR= ~=n2: 1.
Rp R/n1
(b)HereR,.= 10, ~ = 20, R:,=30
(i)When parallel combination of 10and 20resistors
isconnected in series with30resistor [Fig. 3.318(a)], the
equivalent resistance is
R=R+R:,= R,.~ +R:,
P R,.+~
1x2 2 11
=--+3=-+3=-0.
1+2 3 3
(ii)When parallel combination of 20and30resistors
is connected in serieswith 10resistor [Fig. 3.318(b)], the
equivalent resistance is
~R:, 2x3 6 11
R= +R,.=--+1=-+1=-0.
~+R:, 2+3 5 5www5«°ttssrxvt5r°£

306. (iii)When the
three resistances are connected in series
[Fig.3.318(c)],the equivalent resistance is
R=~ + ~ +R:J=(1 + 2+3) 0 =6o.
(iv)When all the resistances are connected in parallel
[Fig.3.318(d)],
111111111
R= ~ + ~+R:J=1+"2+"3=6 Fig.3.321
CURRENT ELECTRICITY
lQ
~3AQA_
~vvv--
(a)
2Q
~AQA_
~vvv--
(b)
lQ
3Q
2Q
(c) (d)
Fig.3.318
Equivalent resistance, R= ~n.
11
(c)The network shown in Fig.3.317(a)isa series
combination of four identical units. One such unit is
shown in Fig. 3.319(a) and itis equivalent to a parallel
combination of two resistances of 20 and40as shown in
Fig.3.319(b).
(a) (b)
Fig.3.319
Resistance Rof one such unit is given by
1 1 1 2+1 3
-=-+-=--=-
R2 4 4 4
R=io
3
..Resistance of the total network (4 such units)
=4xi=16o.
3 3
(ii)The network shown in Fig.3.319(b)is a series
combination of 5 resistors, each of resistance R.
:. Equivalent resistance =5 R.
3.21.Determine the current drawn from a 12V supply with
internal resistance 0.50by thefollowing infinite network. Each
resistor has10resistance.
or
3.153
lQ 10 10 10 10 A
10
10 B
Fig.3.320
Ans.Let the equivalent resistance of the infinite
network be X. This network consists of infinite units of
three resistors of 1 0, 1 0,1O.The addition of one more
such unit across ABwill not affect the total resistance. The
network obtained by adding one more unitwould appear
as shown in Fig. 3.321.
x
10
10
B
Resistance betweenAandB
=Resistanceequivalent to parallel
combination ofXand 10
X xl X
--=--
xitX+1
Resistance between PandQ
X X
=1+--+1=2+--
X+1 xri
or
This must be equal to the original resistance X.
X
X=2+--
1+X
X2-2X-2=0
X=l±.J3or
Asthe value of resistance cannot be negative, so
X= 1+.J3= 2.7320
Current,
I= emf
Total resistance
E. 12
X+r2.732+0.5
=3.713 A
3.22.Figure 3.322shows a potentiometer witha cell of
2.0Vand internal resistance 0.400maintaining a potential
drop across the resistor wire AB. A standard cellwhich
maintains a constant emfof 1.02 V(jorverymoderate currents
upto afewA)gives abalance pointat67.3cm length ofthe wire.
To ensure very low currents drawn from the standard cell, a
very high resistance of 600 k0is put inseries withit, which is
shorted closetothe balance point. The standard ceil isthen
replaced by a cellof unknown emfE.and the balallce point found
similarly turns out to beat82.3emlengthofthe wire.
(a)What isthe value ofE. ?www4not{syriv{4xom

307. 3.154
2 V,0.4n
A
..---------...::-------4B
1.02V
600kn
Fig. 3.322
(b) What purpose does thehigh resistance of
600kflhave?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal
resistance of the driver cell ?
(e) Would the method work in the above situationifthe
driver cell of the potentiometer had an emf of 1.0 V
instead of 2.0 V?
if>Would the circuit work well for determining
extremely small emf, say of the order of a few m V
(such as the typical emf of a thermocouple) ?If not,
how will you modify the circuit ?
Ans.(a)e1=1.02V,~=67.3 em,e2=e=?,12=82.3 cm
Formula for the comparison of emfs by poten-
tiometer is
e212 e82.3
e1=I; 1.02=67.3
e=82.3x1.02=1.25 V.
67.2
(b)High resistance of 600 kn protects the
galvanometer for positions far away from the balance
point, by decreasing current through it.
(c)No,balance point is not affected by high resistance
because no current flows through the standard cell at the
balance point.
(d)Yes, the balance point is affected by the internal
resistance of the driver cell. The internal resistance affects
the current through the potentiometer wire, so changes
the potential gradient and hence affects the balance point.
(e)No, the arrangement will not work. Ifeis greater
than the emf of the driver cell of the potentiometer, there
will be no balance point on the wireAB.
if>The circuit as it is would be unsuitable, because the
balance point (foreof the order of a few mV)will be very
close to the endAand the percentage error in measure-
ment will bevery large. The circuit is modified by putting
a suitable resistor R in serieswith thewireABso that
potential drop across ABis only slightly greater than the
emf to be measured. Then the balance point will be at
larger length of the wire and the percentage error will be
muchsmaller.
or
PHYSICS-XII
3.23.Figure 3.323shows a potentiometer circuitforcomparison
oftworesistances. The balance pointwith astandard resistor
R=10.00 is found to be583em,whilethatwith the unknown
resistance Xis685em Determine the value ofX.What might
youdoifyoufailed tofind a balance point with the given celle?
At------~::---~B
R
x
Fig. 3.323
Ans.HereR=10.00, ~=58.3 em, X=?,12=68.5em
Lete1ande2be the potential drops across RandX
respectively andIbe the current in potentiometer wire.
e2 IXX
e1=IR=R
e2 12 X=~
~=I;..R ~
X=~.R=68.5 x10=11.75 0
~ 58.3
Then
But
or
If there is no balance point, it means potential drops
acrossRorXare greater than the potential drop across the
potentiometer wireAB.We should reduce current in the
outside circuit (and hence potential drops across RandX)
suitably byputting a series resistor.
3.24.Figure3.324showsa2.0 V potentiometer used for the
determination of internal resistance of a 1.5V cell. Thebalance
pointof the cell in open circuit is76.3em.When aresistor of9.5 0
isused inthe external circuit of the cell, thebalance point shifts
to64.8emlengthofthepotentiometer wire. Determine the
internal resistance of the cell.
2.0V
At------.,...--""""T""--4 B
,
@
,
,
,
,
Fig. 3.324
Ans.Here~=76.3 em, Iz=64.8 em, R=9.50
The formulafor the.internal resistance of a cell by
potentiometer methodis
r=R(~-12)=9.5(76.3 -64.8)=9.5x11.5~1.70.
12 64.8 64.8www5«°ttssrxvt5r°£

308. CURRENT ELECTRICITY 3
.155
Text Based Exercises
"YPE A : VERY SHORT ANSWER QUESTIONS (1 mark each)
1.Define electric current. What isthe51unit of electric
current?
2.Write the relation between a coulomb and an
ampere. [ISCE 96]
3.What does thedirection of electric current signify in
anelectric circuit?
4.What is electromotive force? State its 51 Unit.
[Punjab 2000]
5.State the condition in which terminal voltage across
a secondary cellis equal to its emf?
[CBSE D2000]
6.Define an emf of one volt.
7.StateOhm's law. [ISCE95]
8.Name the colours corresponding to the digits 4 and
7 in thecolour codescheme for carbon resistors.
[CBSE SP15]
9.Define resistance andstateits 51unit. [CBSED 92C]
10.Define Ohm.
11.Define conductance of a material. Give its 51unit.
[CBSE D02]
12.Define electrical conductivity of amaterial. Giveits
51unit. [CBSED 03,14]
13.Howmuch istheresistance of an air-gap?
14.Howmuch istheresistance of a closed plug-key?
15.Which metal has the lowest resistivity ?
16.Define resistivity of a material. Stateits 51unit.
[ISCE 93]
17.What is the order ofresistivity of aninsulator?
[Punjab97C]
18.What is theratioof the resistivity of a typical
insulator tothatofametal?
19.What isthe average velocity of free electrons ina
metalat roomtemperature?
20.Givethe order of magnitude of the number density
of free electrons in ametal.
21.Give the order ofmagnitude of thermal velocity
and driftvelocity of free electrons in a conductor
carrying current at room temperature.
22.What is the order of resistivity of conductor?
23.Define temperature coefficient of resistivity.
24.How does the random motion of free electrons in a
conductor get affected when apotential difference
is applied across itsends? [CBSED 14C]
25.Define theterm' drift velocity' of charge carriers in
a conductor and write its relationship with the
current flowing through it. [CBSED 14]
26.Write the expression for the driftvelocity of charge
carriers in a conductor of length'I'across which a
potentialdifference 'V'is applied. [CBSEOD 14C]
27.Howdoes one explain increase in resistivity ofa
metal with increase of temperature? [CBSEOD 14C]
28.Define the term mobility of charge carriers in a
conductor. Write its 51unit. [CBSED 14; OD 15]
29.Plota graph showing variation of resistivity of a
conductor (copper) with temperature.
[CBSE D14;F15]
30.Plota graph showing variation of current versus
voltage for the material GaAs. [CBSED 14]
31.Sketch a graph showing variation of resistivity of
carbon withtemperature. [CBSED06]
Or
Show on a graph, the variation of resistivity with
temperature for a typical semiconductor Si.
[CBSE DOS, 12,14]
32.Name two materials whose resistivity decreases
withthe rise of temperature.
33.How doesthe conductance of a semi-conducting
materialchange with rise in temperature?
34.Of copper and nichrome, which one has possibly
larger value of temperature coefficient of
resistance? [CBSED 95C]
35.How does resistivity of alloy manganin change
with temperature?
.36.Bow does the resistance" of an insulator change
with temperature? I
37.Name twoparameters which determine the
resistivity of amaterial.
38.How isthe conductivity ofan electrolyte affec-
ted by the increase oftemperature? [CBSED 95]
39.Ifpotential difference Vapplied across a conductor
isincreased to 2 V,how will the driftvelocity of the
electrons change? [CBSEOD2000C]
40.What is a non-ohmic device? State one example.
[Punjab02]
41.What is a linearresistor?
42.Give an example of non-ohmic devicewhich shows
upnegative resistance.www6notesdrive6com

309. 3.156
43.A cell of emf'e'and internal
resistance 't'draws a
current'1'.Write the relation between terminal
voltage 'V'in terms ofe,Iandr. [CBSE aD 13]
44.Twoidentical cells, each of emf e,having negligible
internal resistance r,are connected in parallel with
each other across an external resistance R.What is
the current through this resistance? [CBSE aD 13]
45.A4n non-insulated resistance wire isbent1800in
themiddle and the two halves aretwisted together.
What will be its new resistance? [CBSE D 10C]
46.Can Kirchhoff's laws be applied to both d.c. and a.c.
circuits?
47.Onwhat conservation principle is theKirchoff's
first law based ?
48.On what conservation principle is the Kirchhoff's
second law based ?
49.Name thedevice used for measuring the emf ofa
cell. [CBSE D 96]
50.Name the device used for measuring the internal
resistance ofa secondary cell. [CBSE D 96]
51.Define potential gradient. Give its 51unit.
52.Name the principle on which a metre bridge works.
53.What is a Wheatstone bridge? [CBSE D 03]
54.The given graph shows the variation of resistance
of mercury in the temperature' range 0<T<4 K
Name the phenomenon shown by thegraph.
[CBSEOD 03]
c
~0.16
u
§
'00.08
~
OL-~--~L-~--~~
246
Temperature (K)
Fig
55.If the resistances in the three successive arms of a
balanced Wheatstone bridge are1, 2and36n
respectively, what will be the resistance of the
galvanometer placed in the fourtharm?
56.CurrentIflows through a potential drop Vacrossa
conductor. What is the rate of productionofheat?
•[CBSE D 93C]
57.The rate of production of heat is given by P=VI.
Is this relation valid for a non-ohmic conductor?
58.How is electric ener related to electric power?
59.Of which physicalquantity is the unit kilowatt hour?
60.What do you mean by1unit of electric energy in
domestic-use ?
61.How many joulesofener are equivalentto 1kWh ?
PHYSICS-XII
62.How many kilowatt hours (kWh) are there in one
joule? [CBSEon99C]
63.Theapplied p.d. across a given resistance is altered
so that heat produced per second increases by a
factor of 9.By what factordoesthe applied p.d ..
change? [CBSEon99C]
64.Two electric bulbs are rated at 220 V-100W and
220V -60 W.Which oneof these has greater
resistance and why? [CBSE Sample Paper 03]
65.The maximum power dissipated in a 10,000n
resistor is1W. What is the maximum current?
[ISCE 93]
66.What is the safest voltage you cansafely put across
a 98n.0.5 Wresistor? [ISCE 97]
67.How much charge flows through a 250V,1,000W
heater in one minute? [ISCE 96]
68.Aheating element is marked 210V,630W.What is
the value of the current drawnbythe element when
connected to a 210Vdesource? [CBSE D 13]
69.Aheating element ismarked 210 V,630 W. Find the
resistance of the element whenconnected to a 210V
dc source. [CBSE D 13]
70.Two resistors of2nand 4nare connected in
parallel to a constant d.c. voltage. Inwhich case
more heat is produced? [CBSE D 98C ]
71.Two bulbs whose resistances arein the ratio 1 :2 are
connected in parallel to asource ofconstant
voltage. What will betheratioofpower dissipation
in these bulbs? [CBSE D 2000C]
72.Distinguish between kilowatt and kilowatt hour.
73.Which has a greater resistance-kW electric heater
or a 100 W filament bulb both marked for 220 V?
[CBSE D OlC]
74.The coil of a heater is cut into two equal halvesand
only one of them isused into heater. Whatis the
ratio of the heat produced bythis half coil to that by
original coil ?
75.What do youmean by themaximum power rating
of a resistor?
76.Express power transferred per unit volume into
joule heatin aresistor in terms of current density.
77.Write two special characteristics of the wire of an
electric heater. [CBSE D 94]
78.What are thecharacteristics ofafusewire?
79.What isthedifference between aheater wire and a
fuse wire?
80.Twoidentical heaters rated 220V,1000Ware
placedinseries with eachother across a 220Vline.
What istheircombined power?
81.Write an expression for the resistivity of a metallic
conductor showing its variation over a limited
rangeof temperature. [CBSE D 08C]www3yz•p·o§tvp3nzx

310. CURRENT E
LECTRICITY
82. v
83.
The plot of the
variation of potential
difference across a
combination of three
identical cellsiriseries,
versus current isas
shown in Fi. 3.326.
What is theemf of each
cell? [CBSE008]
A(i)series(ii)parallel combination of two given
resistors is connected one-by-one, across a cell. In
which case will the terminal potential difference,
acrossthe cell, have a higher value?
[CBSE0008C]
6V
o 1A
Fig
84.The I-V charac-
teristics of a resistor
are observed to
deviate from a
straight line for higher
values of current as
shown in Fig. 3.327.
Why?
V~
Fig. 3.327
85.
[CBSESP08]
Two identical slabs ofgivenmetal are joined
together, in two different ways, asshown in
Figs.3.328(i) and (ii).What is the ratio of the
resistances ofthesetwo combinations?
[CBSE010C]
(i) (ii)
Fig.328
Answers
3.157
86.A resistance R is connected across a cell, of emf e
and internal resistancer.Apotentiometer now
measures thep.d.,between the terminals of the cell,
asV.Statethe expressionfor'r'in terms ofe,Vand
R [CBSE011]
87.:Aparallel combination of two cells of emf's eland
e2,and internal resistances, 1.andr2,is used to
supply current to a load of resistanceRWrite the
expression for the current through the load in terms
of el, e2,1.andr2· [CBSESamplePaper2011]
88.Underwhatcondition can we draw maximum
current from a secondary cell? [CBSEF 10]
89.Write any two factors on which the' internal
resistance of acell depends. [CBSE0010]
90.Write two factors on which the sensitivity of a
potentiometer depends. [CBSE013C]
91.Graph showing the variation of current versus
voltage for amaterial GaAs is shown in the figure.
Identify the region of
(i)negative resistance
(ii) where Ohm's law is obeyed. [CBSE0 15]
t
c
~
::l
U
A~--~~--~------
Voltage V~
Fig
•
1.The electric current isdefined astherate of flow of
electric chargethrough any section of a conductor.
Totalcharge flowing
Electric current =-----'~---=
Time taken
I=i
t
The SI unitofelectric current isampere(A).
1coulomb
2.1ampere =------
Isecond
or
3.The direction of conventional current in an electric
circuit tells the directionof flow of positive charges
in that circuit.
4.Thework done per unit charge by a source in
taking thecharge once round the complete circuit is
called electromotive force or emf of the source. SI
unit of emf isvolt.
5.When no current is drawn from the cell, its terminal
voltage isequal to itsemf.
6.If an electric cell supplies an ener of 1 joule for the
flow of 1coulomb of charge through the whole
circuit (including the cell), then its emf is said to be
1volt.
7.Ohm's law states that the electric currentIpassing
through a conductor is proportional to the potentialwww6notesdrive6com

311. 3.158
differenceVapplied across its
ends, provided, the
temperature and other physical conditions remain
unchanged, i.e.,VocIorV=RI
whereRis called resistance of the conductor.
8.Yellow and violet respectively.
9.Resistance of a conductor is the property by virtue
of which it opposes the flow of current through it. It
is equal to the ratio of the potential difference
applied across the conductor to the current flowing
through it. SI unit of resistance isohm(n).
R=V
I
10.The resistance of a conductor is said to be 1 ohm if
1 ampere of current flows through it on applying a
potential difference of 1 volt across its ends
1 volt 1V
1 ohm= or 1n= - .
1ampere 1A
11.The ease with which a conductor allows a current to
flow through it is called its conductance. It is equal
to the reciprocal of resistance.
1
Conductance(G)= -----
Resistance(R)
SI unit of conductance is ohm-lor mho.
12.The conductivity of a material is equal to the
reciprocal of its resistivity.
Conductivity (0')=, 1
Resistivity (p)
SI unit of conductivity is ohm-1m-lor mho m-1.
13.Infinity.
14.Negligibly small.
15.Silver.
16.The resistivity of a material is the resistance offered
by a unit cube of that material. Its SI unit is n m.
17.The resistivity of an insulator like glass or rubber is
of the order of 108 -1015nm.
18.The ratio of 1022.
19.Zero. '
20.Number density of free electrons in a metal
=1029m-3.
21.Thermal velocity of free electrons=lOSms-l.
Drift velocity of free electrons=1 mm s-1,
22.Conductors have resistivities less thanlO-6nm.
23.The temperature coefficient of resistivity is defined
as the change in resistivity per unit resistivity per
degree rise in temperature. Mathematically,
P - Po 1 dp
a= -.-
Po(T-IQ)PodT
The unit ofais Co-lorK-l.
PHYSICS-XII
24.Random motion gets partially directed towards the
higher potential side.
Refer to point 17 of Glimpses on page 3.168.
eV,
vd=-;;;z
25.
26.
27.With the increase in temperature, the relaxation
timerdecreases and hence resistivity(p= ~J
ne,
increases.
28.
29.
30.
31.
32.
33.
Refer to point 21 of Glimpses on page 3.169.
See Fig3.20(a).
See Fig.27.
See Fig. 3.21 on page 3.25.
Germanium and silicon.
With the rise in temperature, the conductance of a
semi-conducting material increases exponentially.
Copper.
The resistivity of alloy manganin is nearly
independent of temperature.
The resistance of an insulator decreases with the
increase of temperature.
The resistivity of a material depends on(i)its
number density of free electrons,(ii)the relaxation
time.
The conductivity of an electrolyte increases with
the increase in its temperature.
Drift velocity,
eE, eV,
vd=----;;;=-;;;z
34.
35.
36.
37.
38.
39.
40.
Clearly, whenVis increased to 2V,drift velocity
also gets doubled.
A device which does not obey Ohm's law is called a
non-ohmic device. Semiconductor diodes, ther-
mistors, etc. are non-ohmic devices.
A linear resistor is one which obeys Ohm's law or
for which voltage-current graph is a straight line'
passing through origin. .
Thyristor.
V=e-IT
Effective emf of the parallel combination
=emf of anyone cell=e
Total emfe
1=-----
Total resistanceR
41.
42.
43.
44.
45.The length of the wire becomes half of the original
length while area of cross-section is doubled.
R'r /A'rA[/2A1
RI7A=T'A'=-[-.2A=4
R'=-.!R=-.!x4=1n.
4 4
orwww3yz•p·o§tvp3nzx

312. CURRENT ELECTRICITY
46
.Yes,Kirchhoff's laws can be appliedto both d.c.
and a.c. circuits.
47.Kirchhoff's first law is based on the law of conser-
vation ofcharge.
48.Kirchhoff's second law isbased on the law of
conservation of energy.
49.Potentiometer.
50.Potentiometer.
51.The potential drop per unit length of the poten-
tiometer wire is known as its potential gradient. Its
SI unit isvolt per metre(Vm-1).
52.The working of a metre-bridge is based on the
principle of Wheatstone bridge.
53.A Wheatstone bridge is an arrangement of four
resistances used to determine quickly and accu-
rately one of these resistances in terms of other
three resistances.'
54.Superconductivity.
55.Here P=H1,Q=20,R=360, 5 =?
For a balanced Wheatstone bridge,
P R
Q5
5=RxQ= 36x2 = 720.
P 1 -
56.Rate of production of heat is P=VI
57.Yes,it is valid.
58.Electric energy =Electric power xtime.
59.Kilowatt hour is theunit of electrical ener.
60.1unit of electric energy =1kWh. Thismeans that
when an appliance of power 1000 watt is operated
on mains for1hour, it consumes1unit of electric
energy.
61.1 kWh = 1 killowattx1 hour
= 1000 wattx3600 s =3.6x106J.
62.1J = 1 6kWh = 2.778x10-7kWh.
3.6x10
V2
63.Heat produced per second, P= -
R
or
64.
As the heat produced per secondincreases 9 times,
so the applied p.d. must increase 3 times the
original p.d.
V2 1
R=-.For a given voltage, Rex:-.
P . P
So 60 W bulb has greater resistance than 100 W
bulb.
3.159
65.I = ~Pmax=~1
max R 10,000
1
=- =0.01A.
100
66.Vmax= ~ PmaxR = ~0.5 x98=7V.
P 1000W
67.Current, l=-= =4 A
V250V
Charge that flows in 1minute,
q=It= 4x60=240 C.
68.[=!.. = 630=3 A.
V 210
69.R=V2= 210x210=700.
P 630
V2t
70.Heat produced, H=- i.e.,
R
1
n«-;
R
Thusheat produced in 2 0resistor is more than that
in40resistor.
71.Ii=V~/R,=Rz= ~= 2 :1.
PzV/Rz R, 1
72.Kilowatt is the unitofelectric power while kilowatt
hour is the unit of electric ener.
1 kilowatt = 1000 W =1000 Js-1
1 kilowatt hour = 3.6 x106J.
V2 220x220
73.Resistance of heater =-= = 48.40
Ii 1000
V2 220x220
---=4840
Pz 100
Thus the 100 W bulb has a greater resistance.
74.Letoriginal heat produced,
V2t
H1=R
Resistance of half coil=R / 2
Heat produced in half coil,
V2t 2V2t
H2=R/2=R..
Resistance ofbulb
H
_2=2:1
HI
75.The maximum power rating of a resistor is the
maximum power that it can dissipate in the form of
heatwithout undergoing melting.
76.P=[2R= [2.pi ;
A
Volume, V=Al
:. Power transferred per unit volume
=!.. =rZpl/A=(.!..)2P =J2(J
V Al A
whereJisthe current density.www6notesdrive6com

313. 3.160
77.A h
eater wire should have
(i)high melting point, and
(ii)high resistivity.
78.A fusewire must have high resistivity and low
melting point.
79.The melting point of a heater wire is very high
while that of a fuse wire isverylow.
80.Total powerPdissipated bythe series combination
isgiven by
1 1 1 1 1 1
-=-+-=--+--=-
P ~ ~1000 1000 500
orP=SOOW.
81.The resistivity pat any temperature Tis givenby
p=Po[1+a(T-1Q))
wherePois the resistivity at alower reference
temperature1Qandaisthetemperature coefficient
of resistivity.
82.Total emfof three cells inseries
=P.D. corresponding to zerocurrent =6 V
•.The emf of each cell = ~ =2 V.
3
83.In the case of series combination of thetwogiven
resistors, the the terminal p.d. will have a higher
value.
84.At higher values of current, theresistor getsheated
up and its resistance increases. The resistor
becomes non-ohmic and hence I-Vgraphdeviates
from the straight line.
PHYSICS-XII
I
85.For each slab, R=p-
A
21 1R
Rl=PA=2R ~=p2A=2
Rl=2R =4: 1.
».R/2
r -_(t-VV] R.86.Internal resistance,
t1'2+t21
R('t+ '2)+ 1'2
88.When the external resistance in the circuit is zero,
the current drawn from the secondary cell is max.
t
lmax=-
r
89.The internal resistance of a cell depends on
(i)the nature of the electrolyte and
(il)concentration of the electrolyte.
90.The sensitivity of a potentiometer depends on the
potential gradient along its wires. This, in turn,
depends on(I)length of the potentiometer wire and
(ii)thevalue of resistance put inseries with the
driver cell.
91.(i)Inthe regionDE.Idecreases with increasing V.
+ve~V
---= -veR
=veSl
(i!)AB/BCis the region where Ohm's lawis obeyed.
~YPE B:SHORT ANSWER QUESTIONS (2 or 3 marks each)
1.Distinguish between electromotive force and
terminal potential difference ofa cell. Whatare their
units? [CBSE0014C]
2.Explain how the averagevelocity of free electrons
in a metal atconstant temperature, in anelectric
field, remains constant even though the electrons
are being constantly accelerated by thiselectric
field?
3.Define the terms resistivity andconductivity and
state their 51 units. Draw a graph showing the
variation of resistivity with temperature for a
typical semiconductor. [CBSE005]
4.Define the electrical resistivity of a material. Howit
is related to the electrical conductivity ? Of the
factors, length, area of cross-section, nature of
material and temperature - which onescontrol the
resistivity value of conductor? [CBSEF98]
5.Explain the term'drift velocity' of electrons in a
conductor. Hence obtain the expression for the
current through a conductor in terms of'drift
velocity'. [CBSE0013, 13C, lSC]
6.Prove that the current density of a metallic
conductor is directlyproportional to thedriftspeed
of electrons. [CBSE008]
7.What is meant by drift velocity of free electrons?
Derive Ohm's law on the basis of thetheory of
electron drift. [CBSE003;Haryana94]
8.Arethepaths of electrons straight lines between
successive collisions (with positive ions of the
metal) in the (i)absence of electric field (ii)presence
of electric field ? Establish a relation between drift
velocity 'v/of an electron ina conductor of cross-
section' A',carrying current' l'and concentrationwww6notesdrive6com

314. CURRENT ELECTRICI
TY
'n'offreeelectrons per unit volume of conductor.
Hence obtain the relation between current density
and drift velocity. [CBSE aD 03]
9.Define relaxation time of electrons in a conductor.
Explain how itvaries with increase in temperature
of a conductor. Statetherelation between resistivity
and relaxation time. [CBSE D 2000]
10.A conductor of length 'I' is connected to ad.c.
source of potential 'V'.Ifthelength ofthe
conductor is tripled, by stretching it, keeping 'V'
constant, explain how do the following factors vary
in the conductor:
(i)Drift speed ofelectrons, (ii)Resistance and
(iii)Resistivity. [CBSE D 2000]
11.Write the mathematical relation between mobility
and drift velocity of charge carri~rs in a conductor.
Name the mobile charge carriers responsible for
conduction of electric current in
(i)an electrolyte (ii)anionised gas.[CBSE D 06]
12.Define the term current density of a metallic
conductor. Deduce the relation connecting current
density 0)and the conductivity ( o)of the
conductor, when an electric field E, is applied to it.
[CBSE D 06]
13.Define ionic mobility. Write its relationship with
relaxation time. Give its S1 unit. How does one
understand the temperature dependence of resis-
tivity of a semiconductor. [CBSEF10;oo13C]
14.Definethe terms(I)drift velocity, (ii)relaxationtime.
A conductor of length L is connected to adesource
of emfe.If this conductor is replaced by another
conductor of same material and same area of
cross-section but of length 3L,how will the drift
velocity change? [CBSE D 11]
15.Define relaxation time of the free electrons drifting
in a conductor. How is it related to the driftvelocity
of free electrons? Use this relation to deduce the
expression for the electrical resistivity of the
material. [CBSEon12]
16.Define the term resistivity ofaconductor. Give its
S1unit. Show that the resistance of aconductor is
given by
where the symbols have their usual meanings.
[CBSE co 02]
17.Define resistivity of a conductor. Plot agraph
showing thevariationofresistivity with tempe-
rature for a metallicconductor. How does one explain
such a behaviour, using the mathematical expression
of the resistivity of a material. [CBSE D 01, 08]
3.161
18.Draw aplot showing thevariation of resistivity of a
(i)conductor and (ii)semiconductor, with the
increase in temperature.
How does oneexplain this behaviour in terms of
number density of charge carriers and the
relaxation time ? [CBSE D 14C]
19.Define conductivity of a conductor and state its S1
unit.Stateand explain the variation of conductivityof
(a)good conductor (b)ionic conductor with
temperature. [CBSE D 01, 08]
20.Establish the relation between drift velocity of
electrons and the electric field applied to the
conductor. [Punjab02]
Or
Derive an expression for drift velocity of free
electrons in a conductor in terms of relaxation time.
[CBSE D 09, on 15]
21.Establish arelation between current and drift
velocity. [Himachal03;CBSEon15C]
22.Define the termresistance. Give physical expla-
nation oftheopposition offered by a conductor to
the flow of current through it. [Haryana94]
23.Explain the colour code for carbon resistors with
illustrations. [Haryana95, 98]
24.Three resistances ~,Rzand ~ are connected in
series. Find their equivalent resistance. [CBSE D 92]
25.Three resistances ~, Rzand~are connected in
parallel. Find the equivalent resistance of the
parallel combination.
[CBSE D 92 ;Himachal98C ;Punjab03]
26.What is superconductivity ? Explain. State two
applications ofsuperconductors. [Punjab03]
27.What are superconductors? Give two applications
of the phenomenon of superconductivity.
[Haryana94 ;Punjab95;CBSEF03]
28.What is Meissner effect ? What does it indicate
about the magnetic nature ofsuperconductors?
29.What are ohmic and non-ohmic resistors? Give one
example of each. [Haryana02]
30.Statethe conditions underwhich Ohm's law is not
obeyed in a conductor. [CBSE D 92]
31.What is internal resistance of a cell ? On what
factors doesit depend ?
32.Defineemf of a cell. Show that the voltage drop
across aresistor connected in parallel with a cell is
different from the emf of the cell. [CBSE OD'94C]
33.Whenabattery of emfeand internal resistance r is
connected to a resistance R, a current Iflows
through it. Derive the relation betweene,I,r andR
[CBSE D 92]www6notesdrive6com

315. 3.162
34.Figure
3.330shows acellof emfeand internal resis-
tancer,connected to avoltmeter Vand a variable
resistance RDeduce the relationship among V,e,R
andr.How willVvary when Risreduced.
[ISCE98]
R
,
,
,
,
,
,
Y',
,
,
v
Fig.330
35.Define internal resistance of a cell. Prove that
where R is the external resistance used.[Himachal99]
36.A cell of emf'e'andinternal resistance 'ris
connected across a variable loadresistor RDraw
the plots of the terminal voltage Vversus (i)Rand
(ii)the current I. [CBSE0 15]
37.Distinguish between emf (e)andterminalvoltage
(V)of a cell having internal resistance 'r.Draw a
plotshowing the variation of terminal voltage (V)
vs.thecurrent (I)drawn from the cell. Using this
plot, how doesone determine the emf and the
internal resistance of the cell? [CBSE00 14,14C]
38.A cell ofemfeand internalresistance r is connected
across avariable resistance RPlotgraphs showing
thevariation of(i)eand R, (ii)terminal p.d. Vwith
RPredictfrom thesecond graph under which V
becomes equal to e. [CBSE0 09]
39.Two identical cells, each of emfeand internal
resistancerare connected inparallel to an external
resistance RFind the expression for the total
current flowing in the circuit. [CBSEF 96]
40.Derive the formula for the equivalent EMF and
internal resistance for the parallel combination of
two cells with EMFse1ande2and internal resis-
tances1.and r2 respectively. What isthe corres-
ponding formulafor theseriescombination ?
[CBSESamplePaper 08]
41.Name anyone material having a small value of
temperature coefficient of resistance. Write one use
of this material. [CBSE0 97]
42.Define the terms electricener and electric power.
Give their units. [Haryana92; Punjab93]
43.Obtain the formula for the 'power loss' in a
conductor of resistance R,carrying a current I.
[CBSED 09C]
44.Two heating elements of resistances ~ and Rz
when operated at a.constant supply of voltage, V,
PHYSICS-XII
consume powers ~ andPzrespectively. Deduce the
expressions for the power of their combination
whentheyare,in turn, connected in (i)seriesand
(ii)parallel across the same voltage supply.
[CBSE0011]
45.Give four reasons why nichrome element is
commonly used in household heating appliances.
46.What is a safety fuse? Explain its function.
[Punjab99]
47.State the two Kirchhoff's rules used in electric
networks. How are these rules justified ?
[CBSE0 14,00 15]
48.State theworking principle of a potentiometer.
Explain, with the help of a circuit diagram, howthe
emfs oftwoprimary cells are compared by using a
potentiometer. How can the sensitivity of a
potentiometer be increased ?
[CBSE0 05,06C; 00 15C]
49.Statethe principle ofa potentiometer. With the help of
a circuit diagram, describe a method to find the inter-
nal resistance of a primary cell.
[CBSE0 03; 00 13]
50.You are required to find the internal resistance of a
primary cell in thelaboratory. Draw a circuit diagram
of the apparatus you willusetodetermine it. Explain
theprinciple oftheexperiment. Givethe formula
used. [CBSE0 08C]
51.Why isthe use of a potentiometer preferred over
that of avoltmeter for the measurement of emf of a
cell? [Himachal01]
52.Use Kirchhoff's rules to obtain conditions for the
balance condition in Wheatstone bridge.
[CBSE015]
53.Forthe circuit diagram of a Wheatstone bridge
shownin the figure, use Kirchhoff's lawsto obtain
itsbalance condition. [CBSE0 09C]
+
Fig
54.State, with the help ofa suitable diagram, the
principle on which the working of a metre bridge is
based. Under whatcondition isthe error in
determining the unknown resistance minimized ?
[CBSEF 10,13,00 15C]www3yz•p·o§tvp3nzx

316. CURRENT ELECTRIC
ITY
55.Drawacircuit diagram which can be used to deter-
mine theresistance ofa given wire.Explain the
principle of theexperiment and give the formula
used. [eBSE 0 03C]
56.Draw acircuit diagram using a metre bridge and
write the necessary mathematical relation used to
determine thevalue of an unknown resistance.
Why cannot such an arrangement be used for
measuringverylow resistances? [eBSE 0 06]
Answers
3.163
57.Draw acircuit diagram of a metre bridge to
compare tworesistances. Write the formula used.
Why is this method suitable only for two resistances
ofthe same orderof magnitude? [eBSE F 99]
58.Derive anexpression for the heat produced in a
resistor R whenvoltage drop across it is V.
[eBSE F 93]
••
1.
EMF Terminal Voltage
(i)It isthe potential dif- Itisthepotential dif-
ference between two ference between two
terminals of thecells terminals when a
when no current iscurrent passes through
drawn from it. it.
(ii)Itis a cause. It isan effect.
(iii)TheSIunitis volt. TheSI unit is volt.
2.Refer to thesolution of Probleml(b)on page3.122.
3.Refer to points 10 and 13 of glimpses on page 168.
4.Resistivity of a material is the resistance ofa
conductor of that material having unit length and
unit area of cross-section.
C d .. 1
on uctivity =----
Resistivity
Resistivity of conductor depends on the nature of
itsmaterial and its temperature.
5.Refer answer toQ.19 on page 3.16
6.Refer answer toQ.19 on page 3.16.
7.Refer answer to Q.19 on page3.16.
8.Refer answer to Probleml(e)on page 3.122 and
Q.19 on page 3.16.
9.The average time that elapsesbetween two successive
collisions of an electron in a conductor is called
relaxation time(r),It is rE!latedto resistivity p as
m
P=-2-'
ne't
With the increase in temperature, theelectrons
collide morefrequently with positive metal ions. So
their relaxation time decreases.
Drift speed, vd=eVr:Resistance, R=p..£..
ml A
WhenIis tripled
(i)drift-speed becomes 1/3 times the original vd
(ii)resistancebecomes3 times the original resistance
(iii)resistivity is not affected.
10.
11.Mobility =Driftvelocity
Electric field
or
v
Jl=--.!L
E
(i)The charge carriers in an electrolyte are
positive and negative ions. ,
(ii)The charge carriers in an ionised gas are
electrons and positively charged.
12.Refer answer to Q.12on page 3.7
13.Refer to point 21of Glimpses.
Resistivity, p=~
ne't
Astemperatureincreases, average speed of electrons
increases. This increases collision frequency and
decreases relaxation time 'toButnincreases more
rapidly with temperature. The increase innmore
than compensates the decrease in'toSopof
semiconductors decreases with temperature.
14.Refer to point 17 of Glimpses.
ee
vd=mL't
When length is increased to3L,drift velocity
becomes 1/3times the originalvd.
15.Refer answer toQ.19 on page 3.16.
16.Refer answer to Q.19 on page 3.16.
17.Resistivity of a material is the resistance of a
conductor of that material having unit length and
unit area of cross-section. The SIunit of resistivity is
ohm metre (am).
Resistivity of a conductor=~
ne't
With the increase in temperature, the amplitude of
vibration of positive ions increases. The electrons
suffer collisions more frequently. The relaxation
time r decreases. Hence the resistivity of a
conductor increases with the increase in
temperature.
18.Refer answer toQ.25 on page 3.24.
19.The conductivity of a conductor is the reciprocal of
itsresistivity. Its SI uni] is a-1m -1.www6notesdrive6com

317. 3.164
The conductivit
yof an ionic conductor increases
with the increase of temperature. Asthe tempe-
rature Increases, the electrostatic attraction between
cations and anions decreases, theionsare more free
tomove andso the conductivity increases.
20.Refer answer toQ. 18 on page3.15.
21.Refer answer to Q. 19on page 3.16.
22.Refer answer toQ.21on page3.17.
23.Refer answer to Q.15on page3.9.
24.Refer answer to Q.32on page 3.30.
25.Referanswer toQ.33on page3.31.
26.Referanswer to Q.28on page 3.29.
27.Refer answer to Q.30on page 3.30.
28.Referanswer to Q.29on page 3.29.
29.Referanswer toQ. 27on page 3.28.
30.Refer answer toQ.27on·page3.28.
31.Refer answer toQ. 34on page 3.45.
32.Refer answer toQ.35on page3.46.
33.Refer answer to Q.35on page 3.46.
34.Refer answer to Q. 35 on page 3.46.
eR e
We get V=R+r=1+(r /R)
Clearly Vdecreases whenRisreduced.
35.Refer answer to Q. 35on page3.46.
36.(i)See Fig. 3.91(b)(ii)See Fig. 3.91(c)on page 3.46.
37.Refer answer toQ. 35on page3.46.
38.Refer answer to Q. 35on page3.46.
e e 2Re
39.1= =--=--
R+_r_x_r
r+ r
R+r
2
2R+r
40.Referto the answers ofQ. 37,38on pages 3.51&3.52.
PHYSICS-XII
41.Alloy like manganin has a small value of tempe-
rature coefficient ofresistivity. It isusedfor making
standard resistances.
42.
43.
44.
45.
46.
47.
48.
49.
•50.
51.
52.
53.
54.
Referanswer to Q.44andQ.45.on pages3.60&3.61.
Referanswer to Q.44on page 3.60.
Referto answers ofQ.47, 48on page3.61&3.62.
Refer answer to Q.54(Application 1)on page 3.64.
Refer answer to Q. !?4(Application3)on page3.64.
Refer to point47ofGlimpses on page 3.172.
Refer answers toQ.57andQ. 58on page 3.96.
Refer answer to Q.59on page 3.97.
Refer answer to Q.59onpage3.97.
Referanswer to Q.60on page3.97.
Refer answer to Q.62on page 3.104.
Refer answer to Q. 62on page 3.104.
...---.
SeeFig.3.213.The working of ametre bridge is
based on the principle of Wheatstone bridge. When
thebridge isbalanced i.e.,nocurrent flows through
P R
the galvanometer arm,-=-
Q5
55.
56.
Error indetermination of resistance can be
minimised by adjusting thebalance point near the
middle of themetre bridge wire.
Refer answer to Q.65on page 3.105.
Refer answer toQ. 65on page 3.105.
Metre bridge becomes insensitive for very low
resistance. Moreover, the end resistances become
comparable tothe unknown low resistance and
cannot be neglected.
ReferanswertoQ.65on page3.105and Problem 71
on page 3.120.
Refer answer to Q. 43on page3.60.
57.
58.
~YPE C:LONG ANSWER QUESTIONS (5 marks, each)
1.Define the term resistivity and write its SI unit. 5.A mixedgrouping of cells hasmrows of cells
Derive the expression for the resistivity of a connected in parallel across an external
conductor in terms of number density of free resistance R.Each row containsncells in series.
electrons and relaxation time. [CBSE D 05] Each cell has emf eandiriternal resistance r.
2.What do you understand by the resistivity of a Show that the current in the circuit will be
conductor?Discuss its temperature dependence maximum when R=nr / m.
for a(i)conductor(ii)semiconductor, and 6.State Kirchhoff's lawsforanelectrical network.
(iii)electrolyte. [CBSED92C]
3.A battery ofncells, each of emfeand internal
resistance r,is connected across an external
resistance R. Find the current in the circuit. Discuss
the special cases when(i)R»nrand(ii)R«nr.
4.ncells, each of emfeand internal resistance rare
connected in parallel across an external resistance
R.Determine the condition for maximum current in
the circuit.
Using Kirchhoff's laws,find the relation between
the resistances of four arms ofaWheatstone bridge
when the bridge isbalanced.
Draw a circuit diagram to determine the unknown
resistance of a metallic conductor using a metre
bridge. [CBSEOD03C ;D13]
7.Define the term potentialgradient. Using this
concept, explain the method for comparison ofwww3yz•p·o§tvp3nzx

318. CURRENT ELECTR
ICITY
emfs of two primary cells using a potentiometer.
Establish the relation used. Write two possible
causesof potentiometer giving only one-sided
deflection.· [CBSED13]
8.(a)State the working principleof apotentiometer.
Draw acircuit diagram to compare theemfs oftwo
primary cells.Derive the fonriula used. (b)Which.
material is used for potentiometer wire and why?
(c)How can the sensitivity of a potentiometer be
increased? [CBSED 11C]
9.Deduce theconditionfor balance inaWheatstone
bridge. Using the principle of Wheatstone.bridge,
describe the method to determine the specific·
Answers
3.165
resistance of a wire in the laboratory. Draw the
circuit diagram and write the formula used. Write
any two precautions you would observe while
performingthe experiment. [CBSED04]
10.(a)State, with the help of a circuitdiagram, the
working principle of a. metre bridge. Obtain
the expression used for determining the
unknown resistance.
(b)What happens ifthe galvanometer andcellare
interchanged at the balance pointof the bridge?
(c)Why itis considered important to obtain the
balance point near the midpoint of the wire?
[CBSED 11C]
•
1.Refer answer to Q.19onpage3.16.
2.Refer answer toQ.25onpage3.24.
3.Refer answer to Q.39on page3.53.
4.Refer answer to Q.40on page3.53.
5.Refer answer to Q.41on page 3.54.
6.Refer answer to Q.62onpage3.104 and see
Fi.3.214on page 3.106.
7.Refer answer toQ.58on page 3.96and Problem 42
on page3.142. (c)
~YPE D :VALUE BASED QUESTIONS (4 marks each)
1.Mrs. Sharma parked her car and forgot toswitch off
the car headlights. Whenshe returned, she could
not start the car. Rohit, a passerby, came to her for
help. After knowing about her problem, he went to
a nearby garage and called mechanic Ramu. Ramu
noticed that the car battery hasbeen discharged as
the headlights were left onfor a long time. He
brought another battery from his garage and
connected its terminals tothe terminals of the car
battery. He succeeded in starting theengine and
then disconnected his battery. This is called 'jump
starting', Mrs. Sharma felt happy andthanked both
Rohit and Ramu. Answer the following questions
based on the above information:
(a)What values were displayed by Rohit ?
(b)A storage battery of emf12V and internal
resistance0.5nisto be charged by a battery
charger which supplies 110Vde. How much
resistance must be connected inseries with the
battery to limit thecharging current to 5 A.
What will be the p.d. across the terminals of
the battery during charging ? What is the
purpose of having a series resistor in the
charging circuit?
8.(a)Refer answertoQ.57andQ.58on page 3.96.
(b)Referto the solution of Problem 49on page 3.119.
(c)Referto the solution ofProblem 56on page3.119.
9.Refer answers toQ.62on page3.104andQ.65on
page3.105.
10. (a)Refer answer toQ.65on page3.105.
(b)There is no change in the position ofthe
balance of the bridge.
Referto the solution of Problem 65onpage3.119.,
2.Manish and Rajnish lived in anunauthorised
colony.They found that most people of that colony
stole powerfromtransmission lines using hooks.
They had read in the newspapers about different
fire accidents caused dueto electric short circuits.
Alongwith some oftheir friends and some respon-
sible representatives of that area, they visited house
tohouse ofthatcolony and made people aware of
the risks involved inshort circuitin. They also
explained the people the importance of paying
electricity bills.They succeeded in changing the
mindset ofthe people. Answer the following
questions based on the above information:
(a)What according to you, arethe values of
displayed by Manish and Rajnish ?
(b)Ahousehold circuit has a fuse of5 A rating.
Findthe maximum numberof bulbs of rating
60W -220Veach whichcan be connected in
this circuit.
3.Abhishek went to meet his grandfather who lived
ina village.Bothwereresting andgossiping under
a fan to get relief from the scorching heat of
summer. The lights suddenly went off. Onseeing
that, that all their neighbourers hadelectricity,www6notesdrive6com

319. 3.166
grandfather told Abhishek the fuse might have
blown up. Abhishek immediately changed the fuse.
Grandfather blessed Abhishek and told him that if
had he not come to the village, he would have to
sleep the whole night without fan. Abhishek realised
his grandfather's problem and decided to replace
the fuse with a circuit breaker which uses a solenoid
with a core. When the current exceeds a safety limit,
the breaker is activated and thus breaks the circuit.
The circuit can
be closed by a manual switch.
Answer the following questions based on the above
information:
(a)What were thevalues displayed by Abhishek?
(b)A low voltage supply from which one needs
high current must have very high internal
resistance. Why ?
4.Ameen had been getting huge electricity' bill for the
.:past few months. He was upset about this. One day
Answers
PHYSICS-XII
his friend Rohit, an electrical engineer by
profession, visited his house. When he pointed out
his anxiety about this to Rohit, his friend found that
Ameen was using traditional incandescent lamps
and using old fashioned air conditioner. In addition
there was no proper earthing in the house. Rohit
advised him to use CFL bulbs of 28 W instead of
1000 W - 220 V and also advised him to get proper
earthing in the house. He made some useful
sugestion and asked him to spread this message to
his friends also. reBSE DISC]
(a)What qualities/values, in your opinion did
Rohit possess ?
(b)Why CFLs and LEDs are better than traditional
incandescent lamps?
(c)In what way earthing reduces electricity bill ?
1.(a)Helpful, aware of his limits, ability to take quick
decisions.
(b)Net emf,e=110-12 =98V
IfRis the series resistor, then the charging current
will be
I=_E_= ~ A=5A (given) :.R=19.1n
R+rR+ 0.5
Terminal p.d. of the battery during charging,
V=E+Ir=12+5x0.5=14.5 V
If the series resistorRwere not included in
charging circuit," the charging current would be
~ =196A, which is dangerously high.
0.5
2.(a)Critical thinking, decision making, team spirit
and assertive communication.
P 60 3
(b)Current drawn by one bulb=-= - = -A
V220 11
Number of bulbs that can be safely used with
5 A fuse=_5_ =55=18.33
3/11 3
Hence 18 bulbs can be safely used with 5 A fuse.
3.(a)Empathy, dutifulness, determination, responsi-
bility and compassion.
(b)The maximum current that can be drawn from
a voltage supply is given by
E
Imax
r
Clearly, Imaxwill be large ifris small.
4.(a)Helpfulness, co-operative attitude and scientific
temperament.
(b)CFLs and LEDs have following advantages:
(i)Low operational voltage and less power
consumption.
(ii)Fast action and require no warm up time.
(c)In the absence of proper earthing, the
consumer can get extra charges in his bill for
the electrical ener not actually consumed by
his/her devices.www3yz•p·o§tvp3nzx

320. Current Electricity
GLIMPSES
1.Curre electricity.Thestud
yof electric charges
in motion is called curreelectricity.
2.Electric curre. The flow of electriccharges
through a conductor constitutes electric curre.
Quaitatively, electric curre across anarea
held perpendicular to the direction of flowof
charge is defined as the amou of charge
flowing across that area per unit time.
For a steady flow of charge, I=Q
t
If the rateof flow of charge varies with time,
then
1=lim~Q=dQ
M~O M di .
SI unit of curre isampere(A).
1 1 coulomb
ampere= or
1 second
3.Conveional and electronic curres. The
direction of motion of positive charges is taken
as the direction of conveional curre. Electrons
being negatively charged, so the direction of
electronic curre is opposite to that of the'
conveional curre.
4.Electric curre is a scalar quaity. Although
we represe curre with an arrow, yet itisa
scalar quaity. Electric curres donotobeythe
laws of vectoraddition.
5.Electromotive force (emf).The emf of asource
may be defined as the work done by thesource
in taking a unit positive charge from its lower
poteial terminal to the higher poteial
terminal. Or, it is the energy supplied by the
source in taking a unit positive charge once
round the complete circuit. It is equal to the
terminal p.d. measured in open circuit.
EMF=Work done ore=W.
Charge q
6.SI unit of emf is volt. If an electrochemical cell
supplies energy of1joule for the flow of1coulomb
ofcharge through the wholecircuit (including
the cell), then its emf is said to be one volt.
7.Ohm's law. The curre flowing through a con-
ductor is directly proportional to the poteial
difference across its ends, provided the tem-
perature and other physical conditions remain
unchanged.
V=R
I
Here R is called the resistance of the conductor.
v«Iorv=RIor
8.Resistance.It is the property by virtue of which
a conductor opposes the flow of charges
through it. It is equal to the ratio of the poteial.
difference applied across the conductor to the
curre flowing through it. It depends on the
lengthIand area of cross-sectionAof the
conductor through the relation:
I
R=PA'P=resistivity of the material.
9.SI unit of resistance is ohm (0).The resistance of
a conductor is 1 ohm if a curre of 1 ampere
flo,":,s through it on applying a poteial
difference of 1 volt across its ends.
1 ohm=1 volt or 10 =1 VA-1
1 ampere
(3.167)wwwtnotesdrivetcom

321. 3.168
10.Resistivity or specific
resistance.It isthe
resistance offered by a unitcube of the material
of a conductor.
RA
p==-
I
SI unit ofp ==nm
It depends on the nature of the material of the
conductor and the physical conditions like
temperature, pressure, etc.
11.Curre density.It is the amouof charge
flowing per second per unit area normaltothe
flow ofcharge. It is a vector quaity havingthe
same direction as that of the motion of the
positive charge.
For normal flow of charge,
. q / tI
t=rr=>:
AA
->->
In general, I==jAcase==J.A
SIunitof curre density ==Am-2.
12.Conductance.It isthe reciprocal of resistance.
1
Conductance
Resistance
or G==~
R
SI unit of conductance ==ohm-1==mho
==siemen(S).
13.Conductivity or specific conductance. It is the
reciprocal of resistivity.
Conductivity
1
Resistivity
1
or cr ==-
p
SI unit ofconductivity
==ohm-I m-I ==mho m-I ==S m-I
14.Resistivities of differe substances. Metalshave
low resistivities in the range of 10-8 nm to
10-6nm.Insulatorshave resistivities morethan
104nm.Semiconductors have intermediate
resistivities lying between 10-6nm to 104 nm.
PHYSICS-XII
15.Colour code for carbon resistors.
Colour Number Multiplier
Black 0 10°
Brown 1 HY
Red 2 102
Orange 3 103
Yellow 4 104
Green 5 105
Blue 6 106
Violet 7 107
Grey 8 108
White 9 109
B
-I.
o
How to remember colour code:
BR0YofGreat Britain hadVery Good Wife
-I. -I. -I. -I. -I. -I. -I.' -I. -I.
1234 5 6 7 8 9
Tolerence:
Gold
±5%
Nocolour
±20%
Silver
±10%
A set of coloured co-axial bands is pried on
the resistor which reveals the following facts:
(1)The first band indicates thefirstsignificant
figure.
(2)The second band indicates thesecond
significant figure.
(3)The third band indicates thepower of ten
with whichthe above two significa figures
must be multiplied to get the resistance
value in ohms.
(4)The last band indicates the tolerence in per
ceoftheindicated value.
16.Carriers of curre. In metals, freeelectrons are
the charge carriers. Inionised gases, electrons
and positively charged ions are the charge
carriers. In anelectrolyte, both positive and
negative ions are the charge carriers. In
semiconductors like Ge and Si, conduction is
due to electrons and holes. Aholeis a vaca
state from which anelectron has been removed
andactsas apositive charge carrier.
17.Drift velocity and relaxation time.The average
velocity acquired by thefree electrons of a con-
ductor in the opposite direction of the externally
applied electric field is called driftvelocity(vd)'
The average time that elapses betweenthe twowww4notwsvr~vw4uom

322. CURRE
NT ELECTRICITY (Competition Section)
successive collisions
relaxation time[r),
eE,
v - .
d--;;;- ,
of an electron is called
I=en A vd;j=envd
Heren=no.of free electrons perunitvolume or
freeelectron density and m=mass of an electron.
18.Other forms of Ohm's law. Interms of vector
-->
quaities like curre density jand electric
-->
field E,Ohm's law may be expressed as
--> -->
j=(JEor
--> -->
E=pj
19.
--> -->
The equation E=Pjleads to another state-
me of Ohm's lawi.e.,a conducting material
obeys Ohm's law when the resistivity of the
material does not depend on the magnitude and
direction of the appliedelectric field.
Temperature coefficie of resistance(a).It is
defined as thechange in resistance per unit
original rcsi.tance per degree rise m
temperature. It is given by
a=IS.-Rl
s,(t2 - t1)
If tl=ooe and t2=t=C,then
Rt-Ro
a= or R/=Ra(1+at)
Roxt
The unit of a isoe-1 orrc-I.
20.Effect of temperature on resistance.For metals a
ispositive i.e.,resistance of metals increases
with the increase in temperature.
For semiconductors and insulators, a is negative
i.e.,their resistance decreases with the increase
in temperature.
For alloyslike constaan andmanganin, the
temperature coefficie of resistance a is very
small. Sothey are used for making standard
resistors.
21.Mobility of a charge carrier.Themobility of a
charge carrier is the drift velocity acquired by it
ina unit electric field. It is given by
vds=
Il=-=-
Em
3.169
Foranelectron,
e'h
Ilh=-
m"
SI unit ofmobility=m2V-Is-1
Practical unitof mobility =cm2 V-Is-l.
22.Relation between electric curre and mobility.
For a.conductor,
For a hole,
I=enAlleE
For asemiconductor,
I=eAE(nlle+ Pllh)
and (J=e (nIle+PIlh )
where nand Parethe electron and hole
densities.
23.Ohmic conductors.Theconductors which obey
Ohm's laware called Ohmic conductors. For
theseconductors, V-Igraph is a straight line
passingthrough theorigi. For example, a
metallic conductor for small curres isan
Ohmic conductor.
24.Non-ohmic conductors. The conductors which
do not obey Ohm's law are called non-ohmic
conductors. The non-ohmic situations maybe of
thefollowing types:
(i)The straight lineV-Igraph does notpass
through the origin.
(ii)V-Irelationship isnon-linear.
(iii)V-Irelationship depends on the sign ofV
forthe same absolute value of V.
(iv)V-Irelationship isnon-unique.
Examples of non-ohmic conductors are water
voltameter, thyristor, a p-njunction, etc.
25.Superconductivity.The phenomenon of
complete loss of resistivity by certain metals
and alloys whenthey arecooled below a certain
temperature is called superconductivity. The
temperature at which asubstance undergoes a
transition from normal conductor to super-
conductor in a zero magnetic field iscalled
transition orcritical temperature (Tc)'
26.Meissner effect. Theexpulsion of the magnetic
flux from asuperconducting material when it is
cooled to a temperature below the critical
temperature in amagnetic field is called
Meissner effect.wwwCnotesdriveCcom

323. 3.170
27 ...Resistances in series.When a number of
resistances are conected in series, their
equivale resistance(R
s)is equal to the sum of
the individual resistances.
Rs=Rl+~ +~ +...
28.Resistances in parallel. When a number of
resistances are conected in parallel, the
reciprocal of their equivale resistance(Rp)is
equal to the sum of the reciprocals of the
individual resistances.
111 1
-=-+-+-+ ...
RpRl ~ ~
For two resistances in parallel,
R=Rl~
PRl+ ~
29.Division of curre in resistors joined in
parallel. The curre is divided in resistors,
conected in parallel, in the inverse ratio of
their resistances.
I= ~ . I
1Rl+ ~
1= Rl.1
2Rl+ ~
30.Iernal resistance(r).The resistance offered by
the electrolyte of a cell to the flow of curre
between its electrodes is called iernal
resistance of the cell. It depends on(i)nature of
the electrolyte,(ii)conceration of the
electrolyte,(iii)distance between the electrodes,
(iv)common area of the electrodes dipped in the
electrolyte and(v)temperature of the
electrolyte.
Relations between emf(e) ,terminal poteial
difference(V)and iernal resistance(r).The
poteial drop across the terminals of a cell
when a curre is being drawn from it is called
its terminal poteial difference. It is less than
the emf of the cell in a closed circuit.
31.
V=IR=~
R+r
r=e-V=R[e-V]'I=_e_'I =~
I V' R+r'max r
e=V+Ir ; V=e-Ir ;
Terminal p.d. of a cell when it is being charged
is
PHYSICS-XII
32.Cells in series.If,ncells of emfe~d iernal
resistancereach are conected in series, then
curre drawn through external resistanceRis
I=~
R+nr
33.Cells in parallel. Ifmcells are conected in
parallel, then curre drawn through external
resistanceRis
I=~
mR+r
34.Cells in mixed grouping.Ifncells are conected
in series in each row andmsuch rows are
conected in parallel, then curre drawn
through an external resistanceRis
mne
1=---
mR+ nr
For maximum curre, the external resistance
must be equal to the total iernal resistance, i.e.,
R=I1r
m
or mR=nr.
35.Heating effect of curre. The phenomenon of
the production of heat in a resistor by the flow
of an electric curre through it is called heating·
effect of curre or Joule heating. It is an
irreversible process.
Joule's law of heating.It states that the amou
of heatHproduced in a resistor is
(i)directly proportional to the square of
curre for a given R,
(ii)directly proportional to the resistanceRfor
a givenI,
(iii)inversely proportional to the resistance R
for a givenV,and'
(iv)directly proportional to the timetforwhich
the curre flows through the resistor.
Mathematically,this law can be expressed as
H=VItjoule
,..V2t
=12Rtjoule=Rjoule.
H=VItcal
4.18
36.
or
12Rt V2t
= -- cal=---cal.
4.18 4.18Rwww:notesdrive:com

324. CURRENT E
LECTRICITY (Competition Section)
37.Electric power.It is the rate at which an electric
appliance converts electric energy io other
forms of energy. Or, it is the rate at which work is
done by a source of emf in maiaining an
electric curre through a circuit.
Electric power,
W V2
P=-=VI=[2R=_
t R
38.51 unit of power is watt.The power of an
appliance is one watt if one ampereof curre
flows through it on applying a poteial
difference of 1 volt across it.
1 1 joule
watt=~'---
1 second
= 1 voltx1 ampere
or 1 W =1 Js-1 =1 VA
1 kilowatt (kW) = 1000 W.
39.Electric energy.It is the total work done in
maiaining an electric curre in an electric
circuit for a given time.
Electric energy = Electric powerxtime
W=Pt
=VItjoule =[2Rtjoule
40.Units of electric energy.The commercial unit of
electric energy is kilowatt-hour (kWh) or Board
of Trade (RO.T.) unit. It isthe electric energy
consumed by an appliance of power 1000 watt
in one hour.
1 kWh =1000 Wh
=1000Wx3600 s =3.6 x106J
Another unit of electric energy is watt hour.
1 watt hour = 1 wattx1 hour
=3.6 x 103 J.
41.Power rating.The power rating of an electrical
appliance is the electrical energy consumed per
second by the appliance when conected across
the marked voltage of the mains.
V2 2
P=-=[ R=VI.
R
42.Power consumed by a series combiriation of
appliances.The reciprocal of the effective
3.171
power is equal to the sum of the reciprocals of
the individual powers of the appliances which
have been manufactured for working on the
same voltage.
1 1 1 1
-=-+-+-+ .
PP1 P2P3
43.Power consumed by a parallel combinat on of
appliances.The effective power is equal to the
sum of the powers of the individual appliances.
P=P1+P2+P3+······
Efficiency of a source of emf.It is the ratio of the
output power to the input power. If a source of
emfeand iernal resistance r is conected to an
external resistance, then its efficiency will be
Output power
11=
Input power
44.
VIV R
e[IR+r
45.Maximum Power Thea ·em. It states that the
output power of a source of emf is maximum
when the external resistance in the circuit is
equal to iernal resistance of the circuiti.e.,
whenR=r.
e2
4r
46.
The efficiency of a source of emf is 50% when it
delivers maximum power.
Efficiency of an electric device.It is the ratio of
the output power to the input power.
Output power
11=
Input power
(i)For an electric motor,
Output mechanical power
11=
Input electric power
(ii)Input electric power
= Output mechanical power
+Power lost as heat
The power output of an electric motor is
maximum when its back emf is one-half the
source emf, provided the resistance of the
windings of the motor is negligible.wwwCnotesdriveCcom

325. 3.172
47.Kirchhoff's laws. These
laws enable us to
determine thecurrents and voltages in differe
parts of the electrical circuits.
First law or junction rule. In an electric circuit,
the algebraic sum of curres at any junction is
zero. Or,the sum of currents eering a junction
isequal to the sum ofthe curres leaving that
junction.
Mathematically,
LI=O
Justification.Thislaw is based on the law of
conservation of charge. When the curres in a
circuitare steady, charges canot accumulate or
originate atanypoiof the circuit.
Second law or loop rule. Around any loop of a
network, the sum of changes in poteial must
bezero.Or,the algebraic of the emfs in any loop
ofacircuit is equalto thesum of the products of
curresandresistances in it.
or
Justification. Thislaw is based on the law of
conservation ofenergy. As the electrostatic
force is a conservative force, the total work done
byitalong any closed path must be zero.
48.Gilvanometer.Itis a sensitive device to detect
curreina circuit. Itproduces a deflection
proportional to the electric curre flowing
through it.
49.Poteiometer.It isa device used to compare
emfs of two sources. Its working is based on
the principle that when a consta curre
flows through a wire of uniform cross-
sectional area and composition, the p.d. across
anylengthofthewire is directly proportional
to that length.
Vex:I
or V=kl
wherekisthepoteial drop per unitlength
which is called poteial gradie. Poten-
tiometer has twomain uses.
PHYSICS-XII
(i)To compare theemfsof two cells.If11and12
arethebalancing lengths of the poteiometer
wirefor the cells of emfse1ande2respectively,
then
(ii)Tofind the internal resistance r of a cell. If11
isthe balancing lengthofthe poteiometer
wire without shuand12thebalancing length
withshuR across thecell, then iernal
resistance of the cell will be
e-v 1-/
r=--x R=_I __2xR
V 12
50.Wheatstone bridge.Itis an arrangement of four
resistances P,Q,RandSjoined toform a
quadrilateral ABCD withabattery betweenA
andCanda sensitive galvanometer between B
andD.The resistances aresoadjusted that no
curre flows through the galvanometer. The
bridge is thensaid to be balanced. In the
balanced condition,
P R
QS
Knowing any three resistances, the fourth
resistance can be computed. A wheatstone
bridge is most sensitive when the resistances in
its four arms are of the same order.
51.Slide wire bridge or metre bridge.It is an
applicationof wheatstone bridgein whichRis
fixed and a balance poi is obtained by varying
P and Qi.e.,byadjusting the position of a jockey
on a100cm long resistance wire stretched
between two terminals. If the balance point is
obtained at length I,then
or
P R I
-=-=---
QS 100-1
S=(1001-IJR
SA
p=-
I
Sxnr2
Resistivity,www4notwsvr~vw4uom

326. C H A PT E R
MAGNETIC EFFE
CT
OF CURRENT
4.1CONCEPT OF MAGNETIC FIELD
1.Brieflyexplain the concept of magnetic field.
Concept of magnetic field.A magnet attracts small
pieces of iron, cobalt, nickel etc.Thespace around a
magnet withinwhich its influence can be experienced is
called its magnetic fiel.However, it isnowknown that
all magnetic phenomena result from forces between
electric charges in motion.
In order to explain the interaction between two charges
inmotion,it is useful to introduce the concept ofmagnetic
field, and to describe the interaction intwostages:
1.Amoving charge or a current sets uporcreates a
magnetic fieldin the spacesurrounding it.
2.The magnetic field exerts a force on a moving
chargeor a current in the field.
Like electric field, magnetic field is a vector field,
that is, a vector associated with each point inspace. We
~
use the symbolBfor a magnetic field.
4.2OERSTED'S EXPERIMENT
2.Describe Oersted's experiment leading tothe dis-
covery of magnetic effect of current. State Ampere's
swimming rule.
Magnetic effect of current : Historical note.The
relation between electricity and magnetism was first
noticed by an Italian Jurist. Gian Demenico Romagnosi in
1802. He found that anelectric current flowing ina
wire affects a magnetic needle, andpuished his
observations in alocal newspaper, Gazetta diTrentino.
However, his observations were overlooked. Thefact
thata magnetic field is associated with an electric
current was rediscovered in 1820 by aDanish Physicist,
Hans Christian Oersted. His observations are explained
below.
Oersted's experiment. Consider a magnetic needle
SNpivoted over a stand. Hold awireABparallel to
the needle SNand connect it to a cell and a plug-key,
as shown in Fig. 4.1.
Itis observed that:
1.When the wire isheldabove the needle andthe
current flows from the south to thenorth, the
northpole ofthemagnetic needle gets deflected
towards the west, as shown in Fig.4.1(a).
2. When thedirection of the current isreversed, so
that itflows from the north to the south, the
northpole ofthemagnetic needle gets deflected
towards theeast, as shown in Fig. 4.1(b).
(4.1)···4wx£n~m}r°n4lxv

327. 3. When the wire is
placed below theneedle, the
direction of deflection oftheneedle is again
reversed.
4. When thecurrent in the wire isstopped
flowing, themagnetic needle comesback into
its initial position.
Since a magneticneedle can be deflected by a
magnetic field only,itfollows from theabove
experiment that a current carrying conductor produces a
magnetic fieldaround it.
Ampere's swimming rule.This rule predicts the
direction of deflection of the magnetic needle in the
Oersted's experiment, it can bestated as follows:
Imagine a manswimmingalong thewire inthe
directionof the flow of the current withhisface always
turned towards the magnetic needle, then the north
pole of the needle will get deflected towardshisleft
hand, as shown in Fig. 4.2.
The direction can also be remembered with' the
help of the word SNOW. It indicates thatif the
current flows from South to Northandthe wire is
held Overtheneedle, thenorth pole is deflected
towards theWest. Fig. 4.3Biot-Savart law.
4.2
r----+~r-------~(.r---,
N'
AL---~--~~~~~r;~B
N
5'
(a)
-I-+-----i(· }----,
AL-~~~~~=-----~--~ B
5
.\J'
(b)
Fig. 4.1Deflection of amagnetic needle under
the influence of electric current.
PHYSICS-XII
Fig.4.2Ampere's swimming rule.
4.3BIOT-SAVART LAW
3. State and explain Bioi-Saoart law for the magnetic
fieldproduced by acurrentelement. Define the Sfunitof
magnetic field fromthislaw.
Biot-Savart law.Oersted experiment showed that a
current carrying conductor produces a magnetic field
around it. It isconvenient to assume that this field is
made of contributions from differentsegments of the
conductor. called current elements. A current element is
denoted by di,which has thesame direction as that of
current 1. From a series of experiments oncurrent
carrying conductors of simple shapes, two French
physicistsJean-Baptiste Biot andFelix Savart, in1820,
deduced an expression for the magnetic field of a
current elementwhich is known as Biot-Savart law.
Statement. As shown in Fig. 4.3, consider a current
-4
elementdlof aconductor XY carrying currentl.LetP
-4
be the point where the magnetic field dBdue to the
-4
current elementdlis to be calculated. Let the position
vector of point Prelative to elementdfbe7.Let 8 be
-4 -4
the angle between dlandr .
y
®P~~~2tuzkyjxo}k2ius

328. MAGNETIC EFFECT OF CURRENT
Acco
rding toBiot-Savart law, the magnitude of
-->
the fielddBis
4.3
-->
advances gives the directiondB.Thus the direction of
~
dBisperpendicular to andintothe plane of paper, as has
1.directly proportional to the current Ithrough the beenshown by encircled cross ®at point Pin Fig. 4.3.
conductor,
dBoc[
2.directly proportional to thelengthdl ofthecurrent
element,
dBo:dl
3.directly proportional to sin 8,
dBo:sin 8
4. inversely proportional to the sqU:lre of the distance r
of the point Pfrom the current element,
1
dBoc2"
I'
Combining all these fourfactors, we get
dBx Idl~in 8
r:
dB=K.Idl~n 8
The proportionality constant Kdepends on the
mediumbetween the observation point Pandthe
current element»nd the system of units chosen. For
freespace and in S1 units,
or
K=1-10=10-7 TmA-J(orWbm-1A -1)
41t
Here1-10is a constant called permeability offree
space. So the Biut-Savart law in S1 units may be
expressedas
dB=~Q. Idl~in~
41t r:
We can write the above equation as
dB=l-Io Idlrsin8
4n 1'3
-->
Asthedirection of dBisperpendicular tothe plane
--> -->
ofdlandr ,so from theabove equation, weget the
vector form of the Biot-Saoart lawas
-->!-.toldfx-;
dB=----;:----
41t 1'3
Direction ofdB.Thedirection ofasis the direction
--> -->
ofthe vector dlxr .It is given by rigM hand screw rille.
Ifweplace aright handed screw at point Pper-
pendicular to theplane of paper and turn its handle
--> -->
from dltor,then the direction in which the screw
Special Cases
1. If 8=0°, sin 8 =0, so that dB=O
i.e.,the magnetic field is zero at points on the
axis of the current element.
2.If 8=90°, sin 8=1,so thatdBis maximum.e.,
tilemagnetic field duetoa current element is
maximum in tJplane passing throllghthe element
andperpendicular toitsaxis.
S1 unit of magnetic field from Biot-Savart law.The
S1 unitof magnetic field is iesla,named after the great
Yugoslav inventor and scientistNikola Tesla. According
to Biot-Savart law,
dB=~.Jdlsin8
41t r2
IfI=IA,di=lm,r=lmand8=90° sothat sin8=1,
then
dB=~
41t
4nx10-710-7 I
----= tesa
41t
Tl1!lS oneteslais107timesthemagnetic field pro-
ducedbya conducting wire of length one metre and carryillg
current of one Il1npert! at adistance of one metre from it and
perpendicular toit.
4.4BIOT-SAVART LAWVS.COULOMB'S LAW
4.Give some points of similarities and differences
between Biot-Savart law for the magnetic field and
Coulomb's law for the electrostatic fiel.
Comparison of Biot-Savart law with Coulomb's
law. According toCoulomb's law, theelectric field
produced by a chargedelement at a distance risgiven
by
dE=_l_ dq
41t EO1;2
According toBiot-Savart law,the magnetic field
-~
produced by acurrent elementIdlat a distance risgiven
by
dB=!lo Tdlsin8
41t ,,2
On comparing the above two equations, we note
the following points of similarities and differences
between thetwolaws.···4wx£n~m}r°n4lxv

329. 4.4
Points of
similarity :
1.Bothfields depend inversely on the square of
the distance from the source tothepoint of
observation.
2.Botharelong range fields.
3. The principle of superposition is applicae to
both fields. This isbecausethemagnetic field is
linearly related to its source, namely, the current
element [dland the electrostatic field isrelated
linearly to its source, namely, theelectric charge.
Points of difference :
1.The magnetic field is produced by a vector
....•
source : the current element ldl .The electro-
static fieldisproduced by a scalar source: the
electric charge dq.
2.The direction ofthe electrostatic field is along
the displacement vector joining the source and
the field point. Thedirection of the magnetic
field is perpendicular to the plane containing
....•
the displacement vector randthe current
element[dl.
3.In Bio-Savart law, the magnitude of the
magnetic field is proportional to the sineof the
angle between the current element [dland
....•
displacement vector rwhile there is no such
angledependence in the Coulomb's law for the
electrostatic field. Along the axial line of the
currentelement 8 =0°, sin 8 =0 and hence
dB=O.
5.Write a relation between ).L0'Eaandc.
Relation between ).Lo' EOandc.We know that
_1_ = 9x109Nm2C-2
4nEO
and ).Lo=10-7Tm A-I
4rc
).LE=().L0)(4~J
o 04rc 1
=10-7x_1_ = 1
9x109 (3x108/
But3x108ms-1=speed of lightin vacuum(c)
1
).Lo EO=2
c
or
1
c=--~.
~).Lo EO
PHYSICS-XII
Examples Based on
Formula Used
).Lldlsine
Biot-Savart law, dB=~ 2
4rcr
Units Used
Magnetic field B is in tesla, currentIinampere
and distancerinmetre.
Constant Used
Permeability constant, ).Lo=4rcx10-7Tm A-1
Example 1.A wireplaced along thenorth-south direction
carries acurrent of8A from south to north. Find the
magnetic field due to a 1cm piece of wire at a point 200 cm
north-east fromthe piece.
Solution.The proem is illustrated in Fig. 4.4.
5
Fig. 4.4
As the distanceOPis much larger than the length
of thewire, we can treat the wire as a small current
element.
Here I= 8 A, dl=1 cm = 1x10-2m,
r=200cm =2 m, 8 =45°
dB=).Lo.Idlsin8
4n r2
4rcx10-7 8x1x10-2xsin45°
4rc 22
=1.4x10-9 T.
The direction of the magnetic field at point Pis
normally into the plane of paper.
••••• A
Example 2.Anelement ~I=~xiisplacedat the origin
and carries a large current 1=10 AWhatisthe magnetic
field on the y-axis at a distance of 0.5m. ~x=1em
[NCERT]
Solution. Heredl= ~ = 1 em = 10-2m, 1=10A,
r=y=O.5m, 8=90°, ).Lo/4rc=10-7TmA-1···4wx£n~m}r°n4lxv

330. -> A
3.AnelementL'
11=L'1xiis placed atthe origin (asshown
in Fig. 4.6)and carries a current I=2 A.Find outthe
magnetic field at a point Ponthe y-axis at a distance
of1.0m due tothe elementL'1x=1cm.Give also the
direction ofthe field produced. [CBSE D 09C] F·47
19..
(Ans. 2x1O-9T, in +z-direction)
MAGNETIC EFFECT OF CURRENT
y
O.5m
~------------~~x
Fig. 4.5
According to Biot-Savart law,
dB=1-10Idlsin 8
41t r2
10-7x10x10-2xsin90°
(0.5)2
=4x10-8 T
The direction of the fielddBwillbethe direction of
vectordlx-:.But
-;t ~ 1\ 1\ 1\ 1\ 1\
dlxr=LllixY j=L'1xY(ixj )=Sx Y k
)
Hence field dBisinthe+z-direction.
rproblems for Practice
1.A wire placed along east-west direction carries a
current of10A from west to east direction.
Determine the magnetic field due to a 1.8 cm piece
of wire at a point 300ern north-east from the piece.
(Ans.1.4x10-9T,normally out
of the plane of paper)
-> -> A
2.A small current element I dl ,withdl=2kmm
andI=2 A is centred atthe origin. Find magnetic
->
fielddBatthe following points:
(i)Onthex-axisatx=3 m.
(Ans. 4.44x1O-11J1)
(ii)Onthex-axisatx= -6 m.
(Ans. -1.11x1O-11J1)
(iii)On the z-axis atz=3 m. (Ans.O)
4.5
y
p
I~
o
x'..••- - - - ---~-_------_ x
~
: !11
,
~
z
Fig.4.6
HINTS
1.Proceed asinExample 1.
2.Proceed as in Example 2.
3.Proceed as in Example 2.
We shall now apply Biot-Savart law to calculate the
magnetic field due to(i)a straight current carrying
conductor and (ii)a circular current loop.
4.5 MAGNETIC FIELD DUE TO A LONG
STRAIGHT CURRENT CARRYING
CONDUCTOR
6.Apply Biot-Savart law to derive an expression for
the magnetic field produced at a point due to the current
flowing through a straight wire of infinite length. Also
draw the sketch of the magnetic fielde rules used
for finding the directionof this magnetic field.
Magnetic field due to a long straight current
carrying conductor.As shown inFig.4.7,consider a
straight conductor XY carrying current I.We wish to
find itsmagnetic field at the point P whose
perpendicular distance from the wire is a i.e.,PQ=a.
y ,
TQ
a 'h",
p
4>1
,
'"
,
...• ,
I r ,
,
i,
o
,
,
,
,
0
,
,
,
,
,
,
,
,
,
,
x'
Magnetic field due to a straight current
carrying conductor.www6notesdrive6com

331. B=flOI[sin<l
1+sin<12]
4na .
This equation gives magnetic field due to a finite
wire in terms of the angles suended at the Fig. 4.8
observation point by the ends of the wire.
4.6
Consider a small current element dTof the
~
conductor at O. Its distance from QisIi.e.,OQ=I.Letr
be the position vector of point Prelative to the current
--> -->
element and ebethe angle between dlandr.
According to Biot-Savart law, the magnitude of the or
~ -->
fielddBdue to the current elementdlwill be
dB=~.Idlsine
4n:,z
or
From right 11OQP,
e+~ =90°
o=900-~
sine=sin(90°- ~)=cos~
Also
a
cos<1>=-
r
or
a
r=--=asec<l>
cos<I>
I
tan<1>=-
a
As
I=atan<I>
On differentiating, we get
dl=asec2<l>d<l>
Hence dB=~I(asec2<I>d<l»cos<I>
4rt a2sec2<I>
p1
dB=_0_cos<l>d<l>
4na
According toright hand rule, thedirection ofthe
magnetic field at the Pdue to all such current elements
will be in the same direction, namely, normally into the
~
plane of paper. Hence the totalfield Batthe pointP
due to the entire conductor is oained by integrating
the above equation within the limits -<l1and<12.
cjI:z flI~
B=fdB=_o- fcos<l>d<l>
-Ij; 4na-<h
or
IlI '"
=_0_[sin<I>]:2
4n:a <h
~l1
=_0_[sin<12-sin (- <l1) ]
4n:a
or
PHYSICS-XII
Special Cases
1.iftheconductorXYisinfinitely long and the point P
lies near the midle of the conductor, then<1\= <12=rt/2.
B = ~~[sin90°+sin90°]
4n:a
Ilo1
B=-
2n:a
2.If the conductor XYis infinitely long but the pointP
lies near the endY(orX),then<1\=90° and<12=0°.
IlI Il1
B=_0_[sin90°+sin001 =_0_.
4na 4na
Clearly, the magnetic field dueto an infinitely long
straight current carrying conductor atits one endis
just halfofthat at any point near its middle, provided
thetwopoints are at the same perpendicular distance
from the conductor.
3.If the conductor isoffinite length Land the point
Plies 011its perpendicular bisector, then<l1=<12=<I>and
sin<I>=_Ll2 = L
~a2+(L/2)2 ~4a2 +L2
B=IloI[sin<1>+sin o]
4na
_IloI 2L
- 4na .~4a2+i3
B= floIL .
2naJ4a2 +z3
or
Direction of magnetic field.For an infinitely long
conductor,
i.e.,
B=Ilol
2n:a
1
Boc-
a
Clearly, the magnitude ofthe magnetic fieldwill be
same at all points located at the same distance from the
conductor. Hencethe magnetic lines of force of a straight
(a) (b)
Magnetic lines of foce of a straight current
carrying conductor.···4wx£n~m}r°n4lxv

332. MAGNETIC EFFECT OF CURRENT
current carryingconductor areconcentric circles with the
wire at the centre and in a plane perpendicular to the wire.
[A line of
force is a curve, the tangent to which at any
point gives the direction of magnetic field at that
point]. If the current flows upwards, the lines of force
have anticlockwise sense [Fig.4.8(a)]and if the current
flows downwards, then the lines of force have
clockwise sense [Fig.4.8(b)].
Rules for finding the direction of magnetic field
due to straight current carrying conductor. Either of
the following two rules can be.used for this purpose:
1. Right hand thumb rule. If we hold the straight
conductor in the grip of our right hand in such a way that
the extended thumb points in the direction of current, then
the direction of the curl of thefingers will give the direction
of the magnetic field(Fig. 4.9).
Fig.4.9Right hand rule for field due to a straight conductor
2. Maxwell's cork screw
rule. If a right handed screw be
rotated' along the wire so thatit
aances in the direction of
current, then the direction in
which the thumb rotates gives the
direction of the magnetic field
(Fig. 4.10).
Variation of magnetic field
with distance from straight
current carrying conductor.
For a straight current carrying
conductor,
]t
~-
I \
I I
" I<, ----~
Fig.4.10Corkscrew rule
for field due to a
straight conductor.
1
Boc-
a
Thus the graph plotted between the magnetic field
B and the distanceafrom the straight conductor is a
hyperbola, as shown in Fig. 4.11.
4.7
B
Distance---+
Fig.4.11Variation ofBwith distance from a straight conductor.
Formulae Used
1. Magnetic field due to a straight conductor of finite
length,
B=~oI(sin<Il+sin<12)
41ta
2. Magnetic field due to an infinitely long straight
conductor,
B=~oI
21ta
Units Used
Magnetic field B is in tesla, currentIin ampere
and distanceain metre.
Example 3. A current of10Aisflowing east to westina
long wire kept horizontally in the east-west direction. Find
magnetic field in a horizontal plane at a distance of
(i)10cm north
(ii) 20cm south from the wire;
and in the vertical plane at a distance of
(iii)40em downward and
(iv)50em upward
Solution.(i)Magnetic field in a horizontal plane at
10 cm north of the wire is
~ =1-101=41t x 10-7 x 10=2 x 10-5 T
21tr 21tx 0.10
According to right hand thumb rule, the direction
of the magnetic field will be downward in the vertical
plane.
(ii) Magnetic field at 20 cm south of the wire is
B=41tx 10-7 x 10=1 x 10-5 T
S 21txO.20
The magnetic field will point upward in the vertical
plane.···4wx£n~m}r°n4lxv

333. 4.8
(iii)Magnetic field40em just
down thewire is
Bv=4nx10-7 x10 =5x10-6 T
2nx0.40
Themagnetic field will point south in a horizontal
plane.
(iv)Magnetic field50cmjust above the wire is
Bu=4nx10-7 x10 = 4x10-6T
2nx0.50
The magnetic field will point north in a horizontal
plane.
Example4.Along straight wire carrying a current of30 A
isplaced in an external uniform magnetic field of4.0x10--4T
parallel tothe current. Findthemagnitude ofthe resultant
magnetic field at a point 2.0 em away from the wire.
Solution. HereI=30A,r=2.0em=2.0x10-2m
Field due tostraight current carrying wire is
~ =110I=4nx10-7 x30=3.0x 10--4T
2nr2nx2.0x10-2
This field will act perpendicular to the external
fieldB2=4.0x1O-4T. Hence the magnitude ofthe
resultant field is
B= ~lit+Bi=~(3x10--4)2+(4.0x10--4)2
=5x10--4T.
Example5.Figure4.12showstwo current-carrying wires
1and2.Find the magnitudes and directions of the magnetic
field atpointsP,QandR.
+---- 20em---
20A
lOem-
R
30A
10 em --- 10em
Q
-lOem
p
1 2
Fig.4.12
Solution.(i)According to right hand grip rule, the
fieldBlof wire1at pointPwill point normally outward
while the field ~ of wire2will point normally inward, or
hence
Bp=~ _ ~=11011 _11012
2nYl2TCr2
=4TCx10-7[~_ 30]
2n 0.10 0.30
=2x10-5 T,pointing normally outward.
PHYSICS-XII
(ii)At pointQ,bothBrand~ willpoint normally
inward,
:.R_=~+~=4nx10-7[~+30]
"1.l 2n 0.10 0.10
=10--4T, pointing normally inward.
(iii)At pointR, ~points normally inward and ~
points normally outward,
..~ = ~ _ ~ =4nx10-7 [30 _ ~ ]
2TC 0.10 0.30
=4.5x10-5T,pointing normally outward.
Example6.Two parallel wiresPandQplaced at a sepa-
ration ofr=6emcarry electric currents II=5A and
12=2Ain opposite directions as shown inFig.4.13.Find
thepoint on thelinePQwherethe resultant magnetic fieldis
zero.
11 12
--®~--------~0r------
P Q R
I- -----+·11+-- -- x----+l
Fig.4.13
Solution. At the required. point, theresultant
magnetic field will be zero when thefields duetothe
twowires have equal magnitude and opposite direc-
tions.Such point should lie either to the leftof P or to
the right of Q. But thewire Q has asmaller current, the
pointshould lie closer to and to the right of Q.Let this
point beRat distancexfromQ,asshown in Fig. 4.13.
Field due to currentIIat point R,
_ Il0II
~-2n(r+x)'
normally into the planeof paper.
Field due to current12at pointR,
~ =Ilo12,
2nx
normally out of the plane ofplane
But ~ = ~
r+xx
x=~
II- 12
2Ax6em
= =4 em.
5A-2A
Example7.Use Biot-Savart law to obtain an expression
for the magnetic field at the centre of acoil bent in the formof
asquareofside 2acarrying current 1.···4wx£n~m}r°n4lxv

334. MAGNETIC EFFECT OF CURRENT
Solution.R
efer to Fig.4.14.Magnetic field at 0 due
tofinite length of wireABis
~=1101(sina+sinB)
41ta
III -Ii«I
=_0_(sin45°+sin450)=__0_
41ta 41ta
A.----~f__---, D
a45°',
--45~VO
,
B C
r----2a---+I
Fig.4.14
The magnetic field at 0due to conductorsBC,CD
andDAwill also be of same magnitude and direction.
Therefore, resultant field at0is
B= 4 ~=4x.Ji11°I=.J2110I,
41ta 1ta
directed normally outwards.
Example8.A current of1.0Aisflowing inthe sides of an
equilateral triangle of side 4.5x10-2m. Find the magnetic
field atthe centroid ofthetriangle. [Roorkee 91]
Solution.The situation is shown in Fig. 4.15. The
magnetic field at the centre0due to the current
through side PQ is given by
~ =110I[sin91+sin92]
41ta
whereaisthe distance of PQ from 0 and 91, 92are the
angles as shown. The magnetic field due to each of the
three sides is the same in magnitude and direction,
therefore, total magnetic field at0is
311I
B=3 ~=_0_ [sin91+sin92]
41ta
R
Fig. 4.15
H~re1=1.0 A,91= 92=60°, 110= 41tx 1O-7Tm A-1
PS=tan9or1/2=tan600
OS 1 a
4.9
4.5x10-2
a= m
2tan60° 2.J3
B=3X41tx10-7 x1.0x2.J3[sin600+sin600]
41tx4.5x10-2
=6.J3x10-5[.J3+.J3]= 4x10-5T
4.5 2 2 '
directed normally outwards.
Example 9. Figure4.16showsaright-angled isosceles
L'l.PQR having its base equal to a. A current ofIampere is
passing downwards along a thinstraight wire cutting the
planeof paper normally as shown atQ.Likewise asimilar
wirecarriesanequalcurrent passing normally upwards at
R.Find the magnitude and direction of the magnetic
inductionBatP.Assume the wiresto beinfinitely long.
[ISCE 97]
p
r
Q~--------~a--------~·R
Fig.4.16
Solution. Let PQ=QR=r.In rightL'l.PQR,
a2=,z+r2=2,z or
Magnetic induction at point P due to the conductor
passing through Q,
III-Ii«I III
~ =_0_=__ 0_=_0_ ,acting along PR
2ttr21ta.Jina
Magnetic induction at point P due to the conductor
passing through R,
11I
~ =In° , acting along PQ
,,2tta
Asthe two fields at point P are acting along
perpendicular directions, the resultant magnetic
induction at point Pis
or
B=lloI
1ta
This field acts towards the midpoint of QR.···4wx£n~m}r°n4lxv

335. 4.10
~rO
blems for Practice
1.A straight wire carries a current of 3 A. Calculate
the magnitude of the magnetic field at a point 10em
away from the wire. [CBSED96]
(Ans.6 x 10-6 T)
2.At what distance from a long straight wire carrying
a current of 12 A will the magnetic field be equal to
3x10-5Wb m-2. (Ans.8 x10-2m)
3.The magnetic induction at a point P which is at a
distance of 4 cm from a long current carrying wire
is 10-3 T.What is the magnetic induction at another
pointQwhich is at a distance of 12 em from this
current carrying wire? (Ans. 3.33x10-4T)
4.What current must flow in an infinitely long
straight wire to give a flux density of 3x10-5T at
6 em from the wire? (Ans.9 A)
5.A vertical wire in which a current is flowing
produces a neutral point with the earth's magnetic
field at a distance of 10 em from the wire. What is
the current if BH=1.8x10-4 T? (Ans. 90 A)
6.Fig. 4.17 shows two long,
straight wires carrying
electric currents of 10 A
each in opposite direc-
tions. The separation be-
tween the wires is 5.0 cm.
Find the magnetic field at
a point P midway bet-
ween the wires.
1+5.0em-1
lOA
p
•
lOA
Fig.4.17
(Ans.1.6x10-6T)
7.Two long parallel wires are placed at a distance of
16 em from each other in air. Each wire has a
current of 4 A. Calculate the magnetic field at
midpoint between them when the currents in them
are(i)inthe same direction and(ii)in opposite
directions. [Ans.(i)Zero(ii)2x10-5T]
8.Two infinitely long insulated wires are kept per-
pendicular to each other. They carry currentsII=2 A
and12=1.5A. (i)Find the magnitude and direction
of the magnetic field at P. (i)If the direction of
current is reversed in one of the wires, what would
be the magnitude of the field B ?
[Ans.(i)2x10-5T,normally into
the plane of paper(i)zero]
,
,3cm',
Fig.4.18
PHYSICS-XII
9.A long straight wire carrying a current of 200 A,
runs through a cubical box, entering and leaving
through holes in the centres of opposite faces, as
shown in Fig. 4.19.Each side of the box is of 20 cm.
PQ C
Fig.4.19
Consider an element PQ of the wire 1 cm long at the
centre of the box. Calculate the magnetic field
produced by this element at the pointsA,B,C and
D. The pointsA,Band C are the centres of the faces
of the cube and D is the midpoint of one edge.
(Ans.20x10-6T, 20xl 0-6 T, 0, 7.07 x10-6T)
10.A long straight telephone cae containssix wires,
each carrying a current of 0.5 A. The distance
between the wires is negligie. What is the
magnitude of magnetic field at a distance of 10 cm
from the cae(i)ifthe currents in all the six wires
are in same direction(i)if four wires carry current
in one direction and the other two in opposite
direction. [Ans. (i)6.0x10-6T, (i)2.0x 10-6 T]
11.Calculate the magnetic induction at the centre of a
coil bent in the form of a square of side 10 em
carrying a current of 10 A. [Punjab 01]
(Ans.1.13x10-4 T)
12.A closed circuit is in the form of a regular hexagon
of sidea.If the circuit carries currentI,what is
magnetic induction at the centre of the hexagon?
[IPUEE 13]
(Ans.B=.J3:aoIJ
13.Two straight long conductorsAOBandCODare
perpendicular to each other and carry currents II
and12respectively. Find the magnitude of the mag-
netic field at a point P at a distanceafrom the point
oin a direction perpendicular to the planeABCD.
(Ans.~(12+12)1/2)
27ta1 2
14.Two insulating infinitely long conductors carrying
currentsIIand12lie mutually perpendicular to
each other in thesame plane, as shown in Fig. 4.20.
Find the magnetic field at the pointpea,b).
(Ans.~~(;-;) ,directed inwardJ°°°3vw}m{lzq«m3kwu

336. MAGNETIC EFFECT.OF CURRENT
00
,
,
,
, aP(
a,b)
00--- ---~O+---l1"2--' ----00
,
,
,
00
Fig.4.20
HINTS
1.B =JloI=41tx10-7x3 = 6.0x10-6T.
21tr21tx0.10
Jl0I41tx10-7x12 2
2.r=-= =8x10-m.
21tB 21tx3 x 105
3.Magneticfield due to a straight current carrying
conductor,
B=JloI
21tr
i.e.,
1
Boc-
r
}!Q=rp
s, TQ
rp 4 -3 -4
or BQ =TQ•Bp= 12x10 =3.33x10T.
4.I=21trB= 21tx6 x 10-2 x3x10-5=9A.
Jlo 41txlO 7
5.If neutral point is oained at distancerfrom the
wire, then
JloI=B
21tT H
I21tTBH 21tx0.10x1.8x10-4
or = = 7 =90 A.
Jlo 41tx10
6.According to right hand thumb rule, the direction
of magnetic field due to current in each wire is
perpendicular to and pointing into the plane of
paper. Hence total field at point P is
B=2xJloI
21tT
2X41tx10-7x10 -6
-----'2 -= 1.6x10 T.
21tx2.5x10
[r = ~ em = 2.5 x 10-2 m ]
When the currents are in same direction,7.(I)
B=~ -~
(ii)When the currents are in opposite directions,
B= ~+~.
4.11
JloII41tx10-7x2 -5
1\=21tIi=21tX4x102= 10 T,
normally into the plane of paper.
. 7
Hz=JloI2= 41tx10- x1.5 = 10-5 T
21tr2 21tx3x10-2 '
normally into the plane of paper
B=1\+Hz=2x10-5 T,
normally into the plane of paper.
(ii)When current in anyone wire is reversed, the
two fields will be in opposite directions, so that
B=zero.
8.(I)
9.HereI=200 A, PQ =dl=1cm=0.01 m
For point A or B,r= 10em = 0.1 m, e= 90°,
therefore
B = B=Jl0Idlsine
A B41tr2
10-7x200x0.01xsin 90° -6
---(-0-.1.....,)2.--- =20x10 T
For point C,e= 0°,therefore
_ Jl0Idlsin 0° _
Be --. 2 -0.
41t T
For pointD,
r =~102+102=lOJ2em=0.1J2 m, e=45°
B =Jl0Idlsin 45° =10-7x200x0.01x1
D41t·? (0.1.J2ix.[i
= 7.07 x10-6 T.
10.(i)Net current,I= 0.5x6 = 3.0 A, r=10 em = 0.1 m
BJlol 41tx10-7x3.0 -6T
=--= =6.0x10 .
21tr 21tx0.1
(ii)Net current,1=0.5x4-0.5x2 = 1.0 A
BIloI41tx10-7x1.0 '-6
=-= =2.0x10 T.
21tr 21tx0.1
11.Refer to Fig. 4.21.Magnetic field at 0dueto finite
wireAB,
10 em c
,
,
,
10ern
A 10 em
Fig. 4.21···4wx£n~m}r°n4lxv

337. 4.12
~ =~oI(sina+sin (3)
4na
41tx10-7x10
-----
(sin45°+sin45°)
41tx0.05
=2.83 x 10-5 T
Total magnetic induction at0,
B=41\=4x2.83x10-5=1.13x10-4T,
directed normally outward.
12.ON=a/=~a2_a: =~a
E D
, ,
, ,
, ,
, r
, /
, /
,,
,,a,,
F-- - ---- - - -~- -- - -- - - - -C
/',
'~
aII~:30d,a
" I "
,.,
'a/2 . a/2'
A N B
Fig. 4.22
Magnetic field at0due to current inABis
~ =~oI[sin a+sin13]
41ta'
~ I ~ I
o [sin30°+sin300]=_0_
41tJ3a /2 21tJ3a
J3~1
Total field at0=6 ~=__ 0_•
1ta
13.Magnetic field at P due to current in wireAOB,
.~ =~oII
2na
Magnetic field at P due to current in wireCOD,
Bz=~oI2
21ta
As the two conductors are perpendicular to each
other, so ~ andBzwill also be perpendicular to each
other. Hence the resultant magnetic field at Pis
B~N+Bi=[(~:r+ (",::rr
~ J:.2...(12+12)112.
21ta1 2
14.Magnetic field at point P due to currentII'
~ =~OIlrdirected normally inward
2na
Magnetic field at point P due to current12,
~I
Bz= ~ ,directed normally outward
21tb
PHYSICS-XII
Asb<a,s°Bz>~
Hence the net magnetic field at the pointP,
B=Bz - ~= ~0(12_.i),
21tba
directed normally inward.
4.6MAGNETIC FIELD AT THE CENTRE OF
CIRCULAR CURRENT LOOP
7. Apply Biot-Savart law to derive an expression for
the magnetic field at the centre of a current carrying
circular loop.
Magnetic field at the centre of a circular current
loop. As shown in Fig.4.23,consider a circular loop of
wire of radius rcarrying currentI.We wish to calculate
its magnetic field at the
centre0.The entire loop
can be divided into a
large number of small
current elements.
Consider a current
elementdtof the loop.
According to Biot-Savart
law, the magnetic field at
the centre0due to this
element is
Fig.4.23Magnetic field at the
centre of a circular current loop.
-+lloIdtx:
dB=-.--
41t r3
The field at point0points normally into the plane
of paper, as shown by encircled [email protected] direction
ofdtis along the tangent, sodt1..tConsequently, the
magnetic field at the centre0due to this current
element is
dB= lloI dlsin90°_~oI dl
41t r2 -41t .,1
The magnetic field due to all such current elements
will point into the plane of paper at centreO.Hence
the total magnetic field at the centre0is
B=fdB=fIl 0I .dl= ~0Ifdl
41t,1 41t,z
=lloI.1=lloI.21tr
41t,1 41t,z
B=lloI
2r
or
If instead of a single loop, there is a coil of N turns,
all wound over one another, then
IloNI
B=--
2a~~~2tuzkyjxo}k2ius

338. MAGNETIC EFFECT OF CURRENT
4.7MAGNETIC FIELD ON
THE AXIS OF
A CIRCULAR CURRENT LOOP
8.Apply Biot-Savart law tofind the magnetic field
dueto a circular current carryingloopata point onthe
axis of the loop.Statetherules used tofind the direction
of this magnetic fiel.
Magnetic field along the axis of a circular current
loop. Consider acircular loop of wire of radius aand
carrying current I,as shown in Fig. 4.24.Let the plane
of the loop be perpendicular to the plane of paper. We
~
wish to find field Bat an axial pointPat a distancer
from the centreC.
-->
dl
-->
S
\
\ -~
\
I
I
I
~l
I
I
--> I
dB I
dBcos ~- - --- - ~Q'
,
,
-->
dl
Fig. 4.24 Magnetic field on the axis of a
circular current loop.
Consider a current element dtat the top of the loop.
Ithas an outward corning current.
If-;be the position vector of pointPrelative to the
elements.then from Biot-Savart law, the field at
point P due to the current element is
dB=!:Q..Idl~ine
41t 5
Sincedt1. -;,i.e.,e=90°, therefore
dB=!:Q..!!!!..
,41ts2
The fielddBlies in the plane of paper and is
~ ~
perpendicular to5 ,as shown by PQ . Let<I>be the angle
between OP andCPoThendBcan be resolved into two
rectangular components.
1.dBsin<I>along the axis,
2.dBcos<pperpendicular to the axis.
For any two diametrically opposite elements of the
loop, the components perpendicular to the axis of the
loop will be equal and opposite and will cancel out.
Their axial components will be in the same direction,
i.e.,alongCPand get added up.
4.13
.',Total magnetic field at the point P in the direction
CPis
But
B=fdBsin<p
sin<p=~anddB=!:Q..!!!!..
s 41t 52
B=f!:Q..!!!!... ~
41t52 S
Since IloandIareconstant, and5andaare same for
all points on the circular loop, we have
B=lloIafdl=lloIa .2na=lloIa2
41tS3 41tS3 253
[.: fdl=circumference=2 1ta]
IlIa2
B= 0
2(1+a2)3/2
Asthe direction of the field is along +ve
X-direction, so we can write
~ lloIa2 ~
B= I
2(1+a2)3/2
If the coil consists ofNturns, then
lloNIa2
B=--;;'-----;..---;;c,-;;-
2(1+a2)3/2
or
Special Cases
1.At the centre of the current loop, r=0,therefore
B=lloN Ia2 =lloNI
2a3 2a
B=lloNIA
21ta3
or
whereA=na2=area of the circular current loop. The
field is directed perpendicular to the plane of the
current loop.
2.At the axial points lying far away from the coil,
r» a,so that
B=lloN I a2 =IloN IA
2? 21t?
This field is directed along the axis of the loop and
falls off as the cube of the distance from the current loop.
3.At an axial point at a distance equal to the
radius of the coil i.e.,r=a,we have
2 .
IloNIa IloNI
B- ---
- 2(a2+a2)3/2 -25/2a .
Direction of the magnetic field.Fig.4.25shows the
magnetic lines of force of a circular wire carrying
current. The lines of force near the wire are almost
concentric circles. As we move radially towards the···4wx£n~m}r°n4lxv

339. 4.14
centre of the
loop, the concentric circles become larger
and larger i.e.,the lines of force become less and less
curved. If the plane of thecircular loop is held
perpendicular to the magnetic meridian, the lines at
thecentre are almost straight, parallel and perpen-
dicular to the plane of the loop. Thus the magnetic field
is uniform at the centre of the loop.
~\ffk
~!!!!~
5
Fig.4.25Magneticlines of force of a circular current loop.
Rules for finding the direction of a magnetic field
due to a circular current loop.Either of the following
~
two rules can be used for finding thedirection ofB .
1.Right hand thumbrule.If we curl the palm of our
right hand around the circular wire with the fingers
pointing in the direction of the current, then the extended
thumbgives the direction of the magnetic field.
2.Clockrule.Thisrule gives thepolarity of any face of
the circular current loop.If the current round any face of the
coil isin anticlockwise direction, it behaves like a north pole.
If the current flows inthe clockwise direction, it behaves like
a south pole (Fig.4.26).
Fig.4.26Clockrule.
Variation of the magnetic field along the axis of a
circularcurrent loop.Fig. 4.27shows thevariation of
B
oDistance -t
Fig.4.27Variation ofBalong the axis of a
circular current loop.
PHYSICS-XII
the magnetic field along theaxis of a circular loopwith
distance from its centre. ThevalueofBismaximum at
the centre, and it decreases as we go away from the
centre, on eitherside of the loop.
Examples based on
r---m~
Formulae Used
1.Magnetic field at the centre of a circular loop,
B=1-101
2r
2.Magnetic field at an axial point of a circular loop,
1-1Ia2
B= 0
2(r2+a2)3/2 .
Units Used
Magnetic fieldBis in tesla, current inampere,
distancesrandain metre.
Constant Used
1-10=41tx 10-7Tm A-1.
Example 10.The plane of a circular coilishorizontal. It
has10turns each of radius8em.Acurrent of 2Aftows
through it.The current appears to flow clockwise from a
point above the coil. Find the magnitude and direction of the
magnetic field at thecentre of the coil due tothe current.
Solution.Here N = 10, r= 8 em=0.08 m, I=2 A
B= ~0NI=4nx10-7 x10x2=1.57x10-4T
2r 2x0.08
As the current flows clockwise whenseen from
above the coil, the magnetic field at the centre of the
coil points vertically downwards.
Example 11. In the Bohr model ofhydrogen atom, an
electron revolves around the nucleus in acircular orbit of
radius5.11x10-11mat afrequency of 6.8x1015Hz. What
isthemagnetic field set up at the centre of the orbit?
[Haryana 97C]
Solution.Ifnisthe frequency of revolution of the
electron, then
I=ne=6.8 x1015x1.6x10-19
=6.8x1.6x10-4A
B=~oI
2r
4nx10-7 x6.8x1.6x10-4=13.4 T.
2x5.11x10-11
Example 12. Theradius ofthe first orbitof hydrogen atom
is0.5A.Theelectron moves inanorbitwitha uniform
speed of 2.2x106ms-1. What isthemagneticfield~~~2tuzkyjxo}k2ius

340. MAGNETIC EFFECT OF CURRENT
produce
d at the centre of the nucleus due tothemotion of
this electron? Use~o /41t = 10-7 NA-2 andelectronic charge
=1.6x10-19C. [ISCE 98]
Solution. Herer=0.5A=0.5x10-10m,
v=2.2x106ms-1
Period of revolution of electron,
T=21tr= 2x22xO.5x 10-10 =~x10-15 S
v 7x2.2x106 7
Equivalent current,
1=Charge =!.- = 1.6 x10-19x7 = 1.12 x10-3A
Time T 10-15
Magneticfield produced at the centre of the nucleus,
B=~oI= 41tx10-7x1.12x10-3= 14.07T.
2r 2x0.5x10-10
Example 13. Ahelium nucleus is completing one round of
acircle of radius 0.8min2seconds. Show that the magnetic
field at the centre of the circle is10-19~otesla. Take
e=1.6x10-19C.
Solution. Thecharge on helium nucleus is+2e.The
revolving nucleus is equivalent to a current-loop.
Current,1=Charge=2e
TimeT
Magnetic field at the centre of the circle is
B _~oI_~o2e_~oe
-2;-~·T- rT
~ox1.6x10-19 -19
= =10~otesla.
0.8x2
Example 14.Themagnetic field due to a current-carrying
circular loop of radius 12em at its centre is0.50x10-4T.
Find the magnetic field due to this loop at a point on the axis
at a distance of5.0em from the centre.
• ~ I ~ Ia2
Solution. B=_0_and B.=_--=-,o'---c=--==
centre 2a 'axial2(a2+,2)3/2
or
Baxial _ a3
Bcentre (a2+,2)3/2
a3
B.I= xB
axia (a2+,1)3/2' centre
Herea=12 em =12x10-2m, r=5 em = 5x10-2m,
Bcentre=0.50x1O-4T
B.= (12 x10-2)3 x 0.50x10-4T
axial[144x10-4+25x10-4]3/2
= (12)3 x 0.50x10-4 = 3.9x10-5T.
169x13
4.15
Example 15.Two identical circular coils of radius 0.1 m,
each having 20turns are mounted co-axially 0.1 m apart. A
current of0.5 Ais passed through both of them (i) inthe
same direction, (ii)in the opposite directions. Find the
magnetic field at the centre ofeachcoil.
Solution. Herea=0.1 m,N=20,r=0.1 m, I=0.5 A
Magnetic field at the centre of each coil due to its
own current is
1\ =~oNI= 41tx10-7x20x0.5 =6.28 x10-5T
2a 2xO.1
Magnetic field at the centre of one coil due to the
current in the other coil is
_ ~o NIa2
Hz-2(a2+,1)3/2
41tx10-7x20x0.5x(0.1)2
2 [(0.1)2 +(0.1)2]3/2
0.628x10-7
[2x(0.1)2]3/2
0.628x10-7=2.22x10-5T.
2..fi x10-3
(i)When the currents are in the same direction, the
resultant field at the centre of each coil is
B=1\+Hz= 6.28x10-5+2.22x10-5
=8.50x10-5T.
(ii) When the currents are in opposite directions,
the resultant field is
B=1\ -Hz= 6.28x10:"'5- 2.22x10-5
=4.06x10-5T.
Example 16.Two coaxial circular loops ~ and ~ of radii
3 cm and4cm are placed as shown. What should be the magni-
tudeand direction of the current in the loop ~sothat the net
magnetic field at the point0be zero? [CBSE SP 08]
;
Solution. For the net magnetic field at the point 0
to be zero, the direction of current in loop ~ should be
opposite to that in loop ~.
Magnitude of magnetic=Magnitude of magnetic
field due to current field due to current
IIin ~ 12in ~···4wx£n~m}r°n4lxv

341. 4.16
Example 17.A long wirehav
ingasemi-circular loop of
radiusrcarries acurrent1,as showninFig.4.28.Find the
magnetic field due toentire wire at thepointO.
R
p T
Fig. 4.28
Solution.Magnetic field due to linear portion. Any
elementdlof linear portions likePQ orSTwill make
-+
angles 0 or11:with the position vector r . Therefore,
field at0due to linear portion is
B= ~1dlsin 8=0
411: ,z
Magnetic field due tosemi-circular portion. Any element
dlon this portion will be perpendicular to the position
-+
vector r , therefore, field due to one such element at
point will be
dB=1-101dlsin11:/2=~1dl
411:•,z 411:,z
Magnetic field due to the entire circular portion is
given by
B=fdB= ~fdl=~ . 11:r =1-101
411:r2 411:,z 4r
:. Total magnetic field at point0=1-101.
4r
Example 18.A straight wire carrying a current of 12Ais
bent into asemicircular arc of radius 2.0emas shown in
-+
Fig. 4.29(a). Whatisthe direction and magnitude ofBat the
centre of the arc? Would your answer change ifthewire were
bent into a semicircular arc of the same radius butinthe
~':t'~~~Fig4~'1
(a) (b)
Fig.4.29
PHYSICS-XII
Solution.(i)Magnetic field at the centre of the arcis
1-101
B=-
4r
Here 1=12 A, r=2.0cm=0.02m,
1-10=411: X10-7 TmA-1
B=411:x10-7x12 = 1.9x10-4T
4x0.02
According to right hand rule, the direction of the
field isnormally into the plane of paper.
(ii)The magnetic field will be ofsame magnitude,
B= 1.9 x10-4 T
The direction of thefieldisnormally out of the plane
of paper.
Example 19.Along wire
isbentas shown inFig. 4.30.
Whatwill be themagnitude
and directionof thefield at
the centre 0of thecircular
portion,ifa current 1is
passed through the wire ?
Assume that thevarious
portions of the wiredo not Fig.4.30
touch at point P.
Solution.The system consists of a straight
conductor and a circular loop. Field due tostraight
conductor at point0is
1-11
E\=_0 -,up the plane of paper
211:r
Field due to circular loop at point0is
1-11
~ =_0_,up the plane of paper
2r
:.Total field at0is
B=E\+ ~=1101(1+.!),up the plane of paper.
2r 11:
Example 20. Figure4.31showsacurrent loop having two
circularsegments andjoined by two radial lines. Find the
magnetic field at the centre O.
5
Fig. 4.31~~~2tuzkyjxo}k2ius

342. MAGNETIC EFFECT OF CURRENT
Solution. Sinc
ethe point 0 lies on lines SP and QR,
so the magneticfield at0due to these straight portions
is zero.
The magnetic field at0due to thecircular segment
PQ is
Here, I=lengthof arc PQ=exa
..1\=~o~,directed normally upward
4na
Similarly, the magnetic field at0due to thecircular
segment SRis
~=~0.!...:!..,directed normally downward.
4n b
Theresultant field at 0is
B=1\ _ ~=~oIex[!_!]
4nab
B= ~0Iex(b-a) .
4nab
or
Example 21 .The wire shown in Fig.4.32carries a current
of10A.Determine themagnitude of themagnetic field at the
centre O.Givenradius of the bent coilis3em.
[Punjab 01;AIIMS 13)
Fig. 4.32
Solution. Ase(rad)= ~
Radius
3n=!..orI=3nr
2r 2
According to Biot-Savart law, magnetic field at the
centre 0 is
B_~oIl_~o 13nr_~o3nI
-4n,z-4n.,z .2-4n2" -;
4nx10-7322 10
----- -x----;:;-
4n 2.7 3x10-2
=1.57x10-3 T.
Example 22. In Fig. 4.33,abedisa circular coil of
non-insulated thin uniformconductor. Conductors pa and
qc are very long straight parallel conductors tangential to
the coil at the points a andc.If acurrent of5Aenters the coil
from ptoa,find the magnetic induction at 0,the centre of
the coil. The diameter of the coilis10em.
4.17
SA
--~~----~~-----p
b
SA
d
--~~-----~------q
Fig.4.33
Solution. HereIabc=lade=2.5 A,
r=Oa=Ob= Oc=Od= 5 em = 5x10-2m.
The magneticinduction at0due to the current in
partabcof the coil is equal and opposite to the
magnetic induction due to the current in partadc.So
magnetic induction at0due to the coil is zero.
Magnetic induction at0due to the straight conduc-
torpa(a half infinite segment) is
1\=..!.~oI= 4nx.1O-7 x5=1O-5T,
22nr 4nx5x10-2
normally out of the plane of paper.
Similarly, magnetic induction at0due to straight
conductorqcis ~
~=~oI=1O-5T,
4nr
normally out of the plane of paper.
Total magnetic induction at0is
B=1\+ ~=10-5 +10-5=2x10-5T,
normally out of the plane of paper.
Example 23. The current-loop PQRSTP formed by two
circular segments of radiiR}andRzcarries acurrent of I
ampere. Find the magnetic field at the common centre O.
Whatwill be the fieldif angleex=90°?
Solution. The magnetic field at0due to each of the
straight parts PQ andRSis zero becausee=0°,for each
of them.
T
Fig.4.34
Magnetic field at the centre 0 due to circular segment
QR of radiusRzis
R=~o ~ I
'14n' ~ 2···4wx£n~m}r°n4lxv

343. 4.18
Here,
12=length ofci
rcularsegmentQR= aRz
~=Il0.!.5:..,directed normally downward
41tRz
Similarly, the magnetic field at0duetothe circular
segment STPis
IlI(21t-a) .
.~=~ ,dIrected normally downward
41tRl
Hence the resultant field at0is
B= ~+ ~=lloI(~+ 21t-a],
41tRz Rl
directed normally downward
If a =90°=1t/2,then
B-!:JL!. (_1t+~]=lloI[~+~]
- 41t 2 Rz2Rl 8 R2 Rl .
Example24.Acurrent1=5.0 Aflows along a thinwire
shaped as shown inFig.4.35.The radius of the curved part
of thewireisequaltoR = 120 mm,the angle 2 ~ = 90°. Find
themagnetic induction of the field at the pointO.
A
o
A
,~,R
, ' '2«jl=900- , ,
, ,
B
Fig. 4.35
Solution.Magnetic induction at0due to the line
segmentABis
~=Ilox I[sin c i-sin e]
41tRcos ~
=Il0 .'!:..!..tan ~ , acting normally downwards
41tR
Magnetic field at0due to the current through arc
segment is
~ =Iloxi(2n-2~), acting normally downwards
4nR
Total magnetic induction at0,
IloI
B= ~+ ~=2n.R[n -~+tan ~]
4nx10-7x5[n nJ
= 2nxO.120n-4+tan4
2x10-7x5x3.356=2.8x10-5T.
0.120
PHYSICS-XII
Example25.Two wires A and B have the same length
equal to 44emand carry a current of 10Aeach.WireAis
bent into acircleandwireBinto a square. (a) Which wire
producesagreater magnetic field at the centre? (b) Obtain
the magnitudes of the fields at the centres of the two wires.
Solution.Given I=10A,
Lengthof each wire=44 cm = 4L (say)
(a)Suppose thewireisbentinto acircle ofradius R.
Then its perimeter 21t R=4 L
:.Magnetic field atthe centre ofthe circular wire is
B_llo I_IloI1t_IloI1t
--------- ...(1)
2R 2nR 4L
Now suppose the wire Bisbentinto a square of
sideL.Weknowthat the magnetic field due to a wire
of finite length whose ends make angles a and 13with
the perpendicular dropped on wire from the given
point at distance rfrom it is given by
dB=IloI(sina+sin 13)
4nr
r
,0
, ,
, ,
,
,
,
"L
,
,
,,
2
"L
-------
L
(/"
,
,,
,, ,
, ,
, ,
,
C
Fig.4.36
:.Magnetic field at0due toconductor ABis
dB= IloI(sin 450 +sin450) =2.J2Ilo I
4n.L/2 4nL
[.:a=I3=45°,r=L/2]
Bysymmetry, magnetic field at 0duetoallthe
foursides of the squarewill be in thesame direction.
Hence totalfieldat0due to the current-carrying
square is
B= 4x2.J2lloI=S.J2lloI
41tL 4nL
...(2)
Comparing equations (1) and (2), we find that the
squarewire produces a greater fieldat its centre.
(b)Magnetic field at the centreofthe circular wire is
B= lloIn =4nx10-7 x10x nT
4L 44x10-2
=0.9x10--4 T [":4L=44cm]~~~2tuzkyjxo}k2ius

344. MAGNETIC EFFECT
OF CURRENT
Magnetic field at the centre of the square wire is
B = 8$x1-101 = 8x1.414x4nx10-7x10 T
41tL nx44x10-2
::.1.0x10-4 T.
Example 26.A straight wire,of length ~ metre, isbent
2
into acircular shape. If the wire were tocarry a current of
5Acalculate themagnetic field, dueto it,beforebending, at
a point distant 0.01 times the radius of the circle formed from
it. Also calculate themagnetic field,at the centre of the
circular loopformed, for the samevalue of current.
[CBSE OD 04C]
Solution. Here 2 nr= ~metre
2
1
r=- =0.25 m
4
Magnetic field due tostraightwire,
B=l-Io1= ~01 4nxlO-7x5
2nr 2nx0.01r2nx0.01x0.25
=4x10-4 T
Magnetic field at the centre of thecircular loop,
B=1-101=4nx10-7 x5 =1.256x10-5T.
2r 2x0.25
<prOblems for Practice
1.Consider a tightly wound100turn coilof radius 10em,
carrying a current of 1 A. What is the magnitude of
the magnetic field at the centre of the coil ?
[NCERTI(Ans.6.28x10-41)
2.Acircular loop of one turn carries acurrent of 5.0 A.
If the magnetic field at the centre is0.20mT, find
the radius of the loop. (Ans. 1.57cm)
3.Whatcurrent has to be maintained in a circular coil
of wireof50turns and 2.54ern in radius in order to
justcancel the effectof earth's magnetic field at a
place where the horizontal component of earth's
field is 1.86 x10-5T? (Ans. 0.015 A)
4.A semicircular arc of radius 20em carries a current
of 10 A. Calculate the magnitude of the magnetic
field at the centre of the arc. [CBSED021
(Ans. 1.57x10-51)
5.An alpha particle moves along a circular path of
radius 1.0 x10-10 m with a uniform speed of
2x106ms-1.Calculate the magnetic field produced
at the centre of orbit. (Ans. 13.4 T)
6.The electron in hydrogen atom moves around the
proton with a speed of 2.2x106ms-1in a circular
orbit of 5.3x10-11m.Calculate(i)the equivalent
4.19
current(ii)equivalent dipole moment and(iii)the
magnetic field at the site of the proton.
[Ans.(i)1.057x10-3A(ii)9.32x10-24 Am-2
(iii)12.5T]
7.A circular coil has 35 turns and a mean radius of
4.0cm.It carries a current of1.2A.Find the
magnetic field(i)at a point on the axis of the coil at a
distance of40em from its centre and(ii)at the centre
of the coil. [Ans. (i)6.5x10-7T(ii)6.6x10-4T]
8.A thick straight copper wire, carrying a current of
10 A is bent into asemicircular arc of radius 7.0 em
as shown in Big.4.37(a).(i)State the direction and
calculate the magnitude of magnetic field at the
centre of arc. (ii)How wouldyour answer change if
the same wire were bent into a semicircular arc of
the same radius but in opposite way as shown in
Fig.4.37(b)? [CBSE Sample Paper 981
[Ans. (i)4.5x10-5T, outside the plene of paper,
(ii)4.5x1O-5T, into the plane of paper]
(a) (b)
Fig. 4.37
9.A long wire is bent as shown in Fig. 4.38. Find the
magnitude and direction of the magnetic field at
the centreaof the circular part, if a current I is
passed through the wire.
[Ans. ~(1--.!)normally into
2R n
the plane of paper]
Fig.4.38 Fig. 4.39
10.Figure 4.39 shows two semicircular loops of radii
~ andRzcarrying current I. Find the magnitude
and direction of the magnetic field at the common
centreO.
[Ans.~oI(2.+2.J,normally downward]
4 ~ Rz···4wx£n~m}r°n4lxv

345. 4.20
11.Aci
rcularsegment of radius 10 ern suends an
angle of 60° at itscentre. A current of9A isflowing
through it. Find the magnitude and direction of the
magnetic field produced at the centre (Fig. 4.40).
(Ans.9.42x10-6T)
Fig. 4.40 Fig. 4.41
12.AcurrentofIampere is flowing through the bent
wire shown in Fig. 4.41.Find the magnitudeand
direction ofthemagnetic fieldat point O.
(Ans.B=!:Jl.!..5:.,directed normally downward)
411: r
13.In Fig. 4.42,the curved portion isasemi-circle and
the straight wires are long. Find the magnetic field
at the point O. [Ans.Il;/(1+;)]
r
d
_+---1
'0
Fig.4.42
14.Twoidentical coils each ofradius Rand having
number of turns Nare lying in perpendicular
planes, such that they have common centre. Find
the magnetic field at the centre of the coils, ifthey
carry currents equal toIand-!3Irespectively.
(Ans.Il0NI/R)
15.A metallicwireisbent
into the shape shown
in Fig. 4.43and carries
a currentI.Ifais the
common centre of all
the threecircular arcs
of radiir,2r and3r,
find the magnetic
field at the point0.
[Ans.51loI8,1
24nr
normally inward
0,\
8\r\_
I \\
\,
,r
\\-
Fig.4.43
PHYSICS-XII
HINTS
1.Asthe coil is tightly wound, so radius of each turn,
r=10cm =O.lm
B=lloNI= 411:x10-7x100xl
2r 2xO.l
=211:x10-4 =6.28x10-4T.
2.Radius r = IloNI= 411:x10-7x1x5.0
, 2B 2xO.20xl0 3
= 1.57x10-2m;=1.57 em.
OTIloNI= B
3.B=BH 2r H
41tx10-7x50xI = 1.86x10-5
2x2.54x102
1.86x2x2.54
I= =0.015 A.
411:x50
or
4.UseB=lloI.
4r
5.Takecharge on a-particle = + 2eandproceed as in
Example 12 on page4.14.
21tr
6.(i)Period of revolution, T= -
v
Equivalent current, I=!..= ~
T 211:r
1.6x10-19x2.2x106 -3
= 2x3.14x5.3xlO 11=1.057x10 A
(ii)Equivalent dipole moment,
m=IA=Ix11:r2
= 1.057x10-3x3.14x(5.3xl0-11)2
=9.32x10-24 Am2.
.. Il0I411:x10-7x1.057 x10-3
(Ill)B=-- = 11 = 12.5 T.
2r 2x5.3x10
7.(i)N=35,I=1.2 A, a=4.0em=0.04 m,
r=40 em = 0.40 m
B _Il0NJa2 411:x10-7x35x1.2x(0.04)2
axial-2(a2+r2)3/2 2[(0.04)2 + (0.40)2]3/2
4x3.14 x10-7x35x1.2x0.0016
2x0.1616x0.402
=6.5x10-7T.
(ii)Bcentre=Il~NI=6.6x10-4T.
8.(i)Magnetic field at the centre of thearc is
B= Ilo I
4r
HereJ=10 A, r=7 em=0.07 m,
Il0=411:x10-7TmA -1°°°3vw}m{lzq«m3kwu

346. MAGNETIC EFFECT OF
CURRENT
41txlO-7xlO 5
B= =4.5x10-T.
4x0.07
The direction of thefield is normally outside the
plane of paper.
(i)B=4.5x10-4T.The field Bwill point normally
into the plane of paper.
9.Magnitude of the magnetic field at0due to the
straight part of thewireis
Il I
~ =.......Q..-, normally outofthe plane of paper
2nR
Magnetic fieldat the centre0due to the current
loopofradiusRis
Ilo1 .
~=--, normally mto the plane of paper
2R
Resultant fieldat0is
normally into the plane of paper.
10.B= ~+~=~oI +1101=1l01(~+~J.
4~ 4~ 4 Rl R2
11.Here8=60°=~rad
3
1 1t
1=!:!.
As 8(rad) =- .. or
r 3r 3
4.21
Magnetic field atadue to the upper straight wire is
1 1101 lloI
~ =2:x21t(d/2)=21td
Similarly, field atadue to lower straight wire is
IloI
~ =2nd
Field at0due to the semicircle of radius d/2is
_1 Ilo I _IloI
~-2:x2(d/2)- 2d
Resultant field at0,
B=~+~+~=lloI[l+~].
2d 1t
14.Magnetic fields produced by the two coils at their
common centre are
~ =lloN1 and ~=lloN..J31
2R 2R
The planes of the two coils are perpendicular to
each other. So the fields ~ and ~ will also be
perpendicular to each other, as shown in Fig. 4.44.
B2- - -- -- -- - - - B
According to Biot-Savart law, magnetic field at theF'
19.4.44
centreais
10-7 x3.14x9=9.42x10-6 T.
3x0.10
-+
12.Anyelementdlon the arc will be perpendicular to
the position vector J7,so the field due to one such
element at the centre 0will be
dB= ~0Idlsin1t/2=Il0Idl
4n r2 41t.r2
Magnetic field due to theentire arc at the centre 0,
B=fdB= 1101 fdl= 1101 .1
41tr2 4nr2
But 1= length of arc=or
B=lloI.ar=~oIa
.. 41t,2 4n r
13.Magnetic fieldat pointadue to any current element
isperpendicular to and points out of the plane of
paper.
The resultant field at the common centre is
B=~~2+ B}
=[("~~Ir+ (~ ~~NIrr
= lloNI(1+3)1/2 =lloNI.
2R R
15.Magnetic field at 0 due to the straight parts of the
wire will be zero. Magnetic fields at 0 due to the
three circular arcs of radiir,2rand3rare '
R __1-101 ~
'"1 acting normally inward
41t.r'
1101 8
~=- -acting normally outward
41t .2r'
lloI8
~ =- - acting normally inward
41t.3r'
Thus the total magnetic field at the centre 0 is
B= ~ _ ~+ ~= ~oI(~ _ ~+ ~)
41tr2r3r
51lI
=_0_ 8facting normally inward.
241tr···4wx£n~m}r°n4lxv

347. 4.22
4.8AMPERE'
S CIRCUITAL LAW AND ITS
APPLICATION TO INFINITELY LONG
STRAIGHT WIRE
9. (a) State Ampere's circuital law and prove it for the
magnetic field produced by a straight current carrying
conductor.
Ampere's circuital law. Just as Gauss's law is an
alternative form of Coulomb's law in electrostatics,
similarly we have Ampere's circuital law as an
alternative form of Biot-Savart law in magnetostatics.
Ampere's circuital law gives a relationship between
the line integral of a magnetic field B and the total
currentIwhich produces this field.
Ampere's circuital law states that the line integral of the
-t
magnetic fieldBaround any closedcircuitisequalto)..I0
(permeability constant) times thetotalcurrentIthreading or
passing through this closed circuit. Mathematically,
fB.dI=)..IoI
In a simplified form, Ampere's circuital law states that if
-t
fieldBisdirected along the tangent toevery point on the
perimeter Lofa closed curve and its magnitude isconstant
alongthecurve,then
BL=)..Io I
where I is the net current enclosed by theclosed
circuit. Theclosed curve is calledAmperean loop
which is ageometrical entityandnotareal wire loop.
Proof for a straight current carrying conductor.
Consider an infinitely long straight conductor carrying
a current I. From Biot-Savart law, the magnitude of the
-t
magnetic field B due to the current carrying conductor
at a point, distantrfrom it is given by
B=)..Io I
21tr
Fig. 4.45Ampere's circuital law.
-t
As shown in Fig. 4.45, the field B is directed along
the circumference of the circle of radiusrwith the wire
PHYSICS-XII
-t
as centre. The magnitude of the field B is same for all
points on thecircle. To evaluate the lineintegral of the
-t
magnetic field B along the circle, we consider a small
current elementlitalong the circle. At every point on
the circle, bothBandlitare tangential to thecircleso
that the angle between them is zero.
-t -t
B .dl= Bdlcos 0° = Bdl
Hence the line integral of the magnetic field along
the circular path is
!-t-t! !)..II
'J'B.dl='J'Bdl=B'J'dl= 2~r .1
=)..IoI.21tr
21tr
This proves Ampere's law. This law is valid for any
assemy of current and for any arbitrary closed loop.
9. (b) Calculate, using Ampere's circuital theorem,
themagnetic field due toan infinitely long wire carrying
acurrent I.
Application of Ampere's law to a straight
conductor.Fig. 4.46 shows a circular loop of radius r
around an infinitely long straight wire carrying current
-t
I.As the field lines are circular, the field B atanypoint
of the circular loop is directed along the tangent to the
Fig. 4.46
circle at that point. Bysymmetry, the magnitudeof
-t
field B is same at every point of the circular loop.
Therefore,
fB.lit=fBdlcos0°= Bfdl= B.21tr
From Ampere's circuital law,
B.21tr=)..IoI
B=)..101
21tr~~~2tuzkyjxo}k2ius

348. MAGNETIC EFFECT OF CURRENT
For Your Knowledge
~Ampere'
s circuital law is not independent of the Biot-
Savart law. It can be derived from theBiot-Savart law.
Its relationship to the Biot-Savart law is similar to the
relationship between Gauss's law and Coulomb's law.
~Both Ampere's circuital law and Biot-Savart law
relate magnetic field to the electriccurrent.
~Ampere's and Gauss's laws relate one physical
quantity (magnetic or electric quantity) on the
boundary or periphery to another physical quantity
(current orcharge), called source, in the interior.
~Ampere's circuital law holds forsteady currents
which do notchange with time.
~Although both Ampere's law and Biot-Savart law are
equivalent in physical content, yet the Ampere's law
is more useful under certainsymmetrical situations.
The mathematics of finding the magnetic field of a
solenoid and toroid becomes much simpler if we
apply Ampere's law.
4.9MAGNETIC FIELD INSIDE A
STRAIGHT SOLENOID
10.Givea qualitative discussion ofthemagnetic field
produced by a straight solenoidly Ampere's
circuital law tocalculate magnetic field inside a straight
solenoid.
Magnetic field of astraight solenoid :A quali-
tative discussion.Asolenoid means aninsulated copper
wire wound closely in the formof a helix. Theword solenoid
comes from a Greek word meaningchanneland was
first used by Ampere. By a longsolenoid, we mean that
the length of thesolenoid is very large ascompared to
its diameter.
Fig 4.47Magnetic field due to a section of a finite solenoid.
Figure 4.47 shows an enlarged view of the magnetic
field due to a section of a solenoid. Atvarious turns of
4.23
the solenoid, current enters the plane of paper at points
marked (8)and leaves the plane of paper at points
marked 0. The magnetic field at points close to a single
turn of the solenoid is in the form of concentric circles
like that of a straight current carrying wire. The
resultant field of the solenoid is the vector sum of the
fields due to all the turns of the solenoid. Obviously
thefields due to the neighbouring turns add up along
the axis of the solenoid but they cancel out in the
perpendicular direction. At outside points such asQ,
the fields of the points marked(8)tend to cancel out the
fields of the points marked 0. Thus the field at interior
midpointPis uniform and strong. The field at the
exterior midpoint Qis weak and is along the axis of the
solenoid with no perpendicular component. Fig.4.48
shows the field pattern of a solenoid of finite length.
Q
Fig 4.48Magnetic field of a finite solenoid.
The polarity of any end of the solenoid can be deter-
mined by usingclockruleor Ampere's right hand rule.
Ampere's right hand rule. Grasp the solenoid withthe
right handsothat the fingers point along thedirection of the
current, the extended thumbwill then indicate theface of the
solenoid that has north polarity(Fig. 4.49).
N-po\e
Right hand
Fig.4.49Ampere's rule for polarity of a solenoid.
Calculation of magnetic field inside a long straight
solenoid. The magnetic field inside a closely wound
long solenoid is uniform everywhere and zero outside···4wx£n~m}r°n4lxv

349. 4.24
.Q
I, I '
I
d,----------«-------, C
: A
B
r-~~~------~p-----------
xxxxxxxxxxxxxxxxxxxxxxxxxx
Fig. 4.50The magnetic field of a very long solenoid.
it.Fig. 4.50 shows the sectional viewofalong solenoid.
At various turns of the solenoid, currentcomes out of
the plane of paper at points marked0and enters the
plane of paper at points marked18>.Todetermine the
-4
magnetic fieldBatany insidepoint, consider a
rectangular closed path abedastheAmperean loop.
According to Ampere's circuital law,
fB.tfl
=110xTotalcurrent throughthe loopabed
-4 b
NowfB.tfl=fB.tfl
c d-4 a
a+fB.tfl+f B.tfl+f B.tfl
bed
c-4 c
fB.!=fBdlcos 90°=0
b b
But
a a
fB.!=fBdlcos 90°=0
d d
d-4
fB.!=0
c
as B=0 forpointsoutside the solenoid.
b
fB.tfl=f s.:
a
b b
fBdlcos0°=Bfdl=Bl
a a
where,
I=length of the sideabof the rectangular loop abe.
Let number of turns per unit length of the
solenoid=n
Then number of turns in lengthIof the solenoid
=nl
Thus the currentIof the solenoid threads theloop
abed, nltimes.
:.Total current threading the loopabed=nll
HenceBl=llonIl or B=llonI
PHYSICS-XII
It can be easily shown that the magnetic field at the
end of the solenoid is just one half of that at its middle.
Thus
1
Bend="211onI
Figure 4.51 shows the variation of magnetic field on
the axis of a long straight solenoid with distancexfrom
its centre.
B
2
B
End of 0 End of
solenoid ~ Distance ~ solenoid
Fig. 4.51 Variation of magnetic field along
the axis of solenoid.
4.10 MAGNETIC FIELD DUE TO A
TOROIDAL SOLENOID
11.Apply Ampere's cireuitallaw tofind the magnetic
field both inside and outside ofatoroidal solenoid
Magnetic field due to a toroidal solenoid. Asolenoid
bent into the form of aclosed ringiscalled a toroidal
solenoi.Alternatively, it is an anchor ring (torous)
around which a large number of turns of a metallic wire
are wound, as shown in Fig. 4.52. We shall see that the
-4
magnetic field B has a constant magnitude everywhere
inside the toroid while itis zero in the open space
interior (point P) and exterior (pointQ)to the toroid.
Fig. 4.52 A toroidal solenoid.
Figure 4.53 shows a sectional view of the toroidal
solenoid. The direction of the magnetic field inside is
clockwise as per the right-hand thumb rule for circular
loops. Three circular Amperean loops are shown by~~~2tuzkyjxo}k2ius

350. MAGNETIC EFFECT OF CURRENT
Fig. 4.53A sectional view of the toroidal
solenoid.
dashed lines. By symmetry, the magnetic fieldshould
be tangential to them and constant in magnitude for
each of the loops.
1. For points in the open space interiortothe
toroi.Let ~ be the magnitude of the magnetic field
along the Amperean loop 1of radiusr1.
Length of the loop 1, ~=2n1J
As the loopencloses no current, soI=0
Applying Ampere's circuital law,
~~ =~oI
or ~x2nr1= ~0x0
or ~=0
Thusthe magnetic field at anypointPin the open space
interior tothe toroid is zero.
2. For points inside the toroidLetBbe the
magnitude of the magnetic fieldalong the Amperean
loop 2 of radius r.
Length of loop 2, ~=2nr
IfNis the total number of turns in the toroid andI
the current in the toroid, then total current enclosed by
the loop 2=NI
Applying Ampere's circuital law,
Bx2nr= ~0xNI
or B=~oNI
2nr
Ifrbe the average radius of the toroid and nthe
number of turns per unit length, then
N=2nrn
B=~onl
3. For points in the open space exteriortothe
toroid. Each turnof the toroid passes twice through the
area enclosed by the Amperean loop 3. But for each
turn, the current coming out of the plane of paper is
cancelled by the current going into the plane of paper.
Thus,I=0 and hence ~=o.
4.25
ForYour Knowledge
~Themagnetic field inside a toroidal solenoid is
independent of its radius and depends only on the
current and the number of turns per unit length. The
fieldinside the toroid has constant magnitude and
tangential direction at every point.
~In ideal toroid, the coils are circular and magnetic
field is zeroexternal to the toroid. In a real toroid, the
turns formahelix and there is a small magnetic field
external to the toroid.
~Toroids areexpected to playa key role in the Tokamak
which acts as a magnetic container for the fusion of
plasma in fusion (thermonuclear) power reactors.
Formulae Used
!--->--->
1.Ampere's circuital law, rBdl=~oI
WhenBis directed along tangent to every point
onclosed curveL,BL= ~0I
2.Magnetic field due to straight solenoid,
(i)At a point well inside the solenoid,B= ~0nI
(ii)At either end of the solenoid, Bend =~~onI
Herenisthe number of turns per unit length.
3.Magnetic field inside a toroidal solenoid, B= ~0nI
Magnetic field is zero outside the toroid.
Units Used
Bis in tesla, currentIin ampere andnin m-1.
Example 27.Asolenoid coil of300 turnslm is carrying a
current of 5A.Thelength of thesolenoid is 0.5 m and has a
radius of 1em.Find the magnitude of the magnetic field
inside the solenoid. [CBSE F 04]
Solution.Here n= 300 tums/m, 1=5A
B=~onI=4nx10-7 x300x5 =1.9x10-3T.
Example 28. Asolenoid of length 0.5 m has a radius oflem
andismadeup of500 turns. Itcarriesacurrent of5A.What
isthe magnitude of the magnetic field inside the solenoid?
[NCERT]
Solution.Number of turns per unit length,
N 500
n=- =-- =1000 turns/m
I0.5 m
Here 1=0.5 m and r=0.01 mi.e.,I»a.So we can
useformula for magnetic field inside a long solenoid.
B=~onI=4nx10-7 x1000x5=6.28x10-3T.···4wx£n~m}r°n4lxv

351. 4.26
Example 29
.A 0.5 m long solenoid has 500 turns and has
aflux density of2.52x10-3T at its centre.Find the current
in the solenoiden /-t0= 41tx10-7Hm-1. [ISCE 95)
Solution. Number of turns per unit length,
N 500
n=- =--=1000 turnsz m
I0.5 m
As B= /-tonI
I=~ = 2.52 x10-3 = 2.0A
/-ton41tx10-7 x1000
Example 30.A copper wire having a resistance of 0.01n
per metre is used to wind a 400 turnsolenoid of radius
1.0 em and length20em.Find the emf of abattery which
when connected across the solenoid wouldproduce a
magnetic field of10-2 T near the centre of the solenoi.
Solution. Length of wire used
= 21trxNo.of turns
= 21tX1.0x10-2x400 m
Resistance per unit length =0.01nm-1
:. Total resistance of wire,
R = 21tx1.0x10-2x400x0.01
= 81tx10-2n
No.of turns per unit length,
n= 400 =2000 m-1
20x10-2
E
B= /-to nI= /-ton-
R
E= BR = 10-2 x81tx10-2 = 1 V.
/-ton41tx10-7x2000
As
Example 31. A solenoid 50cm long has 4layers of
windings of 350 turns each.Theradius of thelowestlayeris
1.4 em. If thecurrent carriedis6.0A, estimate the
-+
magnitude ofB(a)near the centre of the solenoid on itsaxis
and offitsaxis,(b) nearitsendson its axis, (c) outside the
solenoid nearitscentre.
Solution.(a)The magnitude of the magnetic field
at or near the centre of the solenoid is given by
B=/-tonI
wherenis the number of turns per unit length. This
. -+
expression for B can also be used if thesolenoid has
more than one layer of windings because the radius of
the wire does not enter this equation. Therefore,
No. of turns per layerxNo.of layers
n= .
Length of the solenoid
= 350x4 =2800 m-1
0.50
PHYSICS-XII
NowI=6.0A, /-to= 41tx10-7TmA-1, n=2800 m-1
. .B= 41tx10-7 x2800x6T=2.1x10-2T
-+
This value ofB is for bothon andoffthe axis, since
foraninfinitely long solenoid, the internal field near
thecentre is uniform over the entire cross-section.
(b)Magnetic field at the ends of the solenoid is
_ /-tonI _ -2
Bd --- -1.05x10 T.
en 2
(c)The outside field near the centre of a long
solenoid isnegligie compared to theinternal field.
Example32.A coil wrapped around a toroid has inner
radius of20.0emandanouter radius of 25.0 em.If the wire
wrapping makes 800 turns and carries a current of 12.0 A,
what arethe maximum and minimumvalues of the magnetic
field within the toroid?
Solution. Let aandbdenote the inner and outer
radii of the toroid. Then
N
Bmax=/-tonI=/-to21ta I
41tx10-7 x800x12.0
21tx20.0 x10-2
=9.6x10-3T=9.6mT.
B=nl=~1=41tx10-7 x800x12.0
min/-to /-to21tb 21tx25.0x10-2
=7.68x1O-3T=7.68mT.
Example 33. (i)A straight thick long wire of uniform cross-
section of radius'a'iscarrying a steady current I.Use
Ampere's circuital law to obtain a relation showing the
variation ofthemagnetic field (Br) inside andoutside the
wire with distance r, (r ::;a) and (r>a) of the field point from
the centreof its cross-section. Plot a graph showing the
variation offield Bwith distance r.
(ii)Calculate the ratio of magnetic field at a point a12
above the surface of the wire to that at a point a12belowits
surface. What isthemaximum value of thefield ofthis wire?
[NCERT; CBSE D 10)
Fig. 4.54 A steady currentIdistributed uniformly
across a wire of radiusQ.
Solution.(i)Application of Ampere's law to a
long straight cylindrical wire. By symmetry, the~~~2tuzkyjxo}k2ius

352. MAGNETIC EFFECT OF CURRE NT
magneti
clines of force will be circles, with their centres
on the axis of the cylinder andin planes perpendicular
to the axis of the cylinder. So we consider Amperean
loop as acircleof radius r.
Field at outside points. The Amperean loop is a
circle labelled 2 having radiusr>a.
Length of the loop, L=21tr
Netcurrent enclosed bythe loop =I
ByAmpere's circuital law,
BL=1l01
Bx2rrr=1l01
B=1101
21tr
or
or [Forr>a]
i.e.,
1
Boc-
r
[For outside points]
Field at inside points. TheAmperean loop isacircle
labelled 1withr<a.
Length of theloop, L=21tr
Clearly, thecurrent enclosed by loop1islessthanI.
Asthe current distribution is uniform, the fraction ofI
enclosed is
l'=_1_x1t?=I?
1ta2 a2
Applying Ampere's law,
BL=llo I'
I?
or Bx2ttr=11 02
a
or [Forr<a]
i.e., [For inside points]Bocr
Thus the fieldBis proportional to r as we move
from the axis of the cylinder towards its surface and
then it decreases as! .The variation ofBwith distancer
r
from the centre of the wire is shown in Fig. 4.55(a).
B
L-----------------~r
(a) (b)
Fig.4.55 (a)Sketch of the magnitude of the magnetic field for
the long conductor of radiusa.
4.27
(ii)Suppose the point P lies at distancea /2 above
thesurface of the wire and point Q lies at distance a /2
below the surface. [Fig. 4.55(b)]
Magnetic field at point P at distancer=3a/2 from
the axis of the wire is
_ 1101 _1101 _ 1101
Bp - 21tr -2 rt{3/2)a -3rta
Magnetic field at point Q at distance r=a/2from
the axis of the wire is
R..=1l0Ir=1101(~)=1101
\.!2 rta2 2 rta2 2 4rta
Bp 1101 41ta
-=-x-=4:3.
It?3rta 1101
Clearly, Bismaximum on the surface of the wire
i.e.,atr=a.Hence,
B=1101
max 2rta
Example34.Awire of radius 0.5emcarries a current of
100A,whichisuniformly distributed over itscross-section.
Findthe magnetic field (j)at0.1emfrom the axis of the wire,
(ii)at the surface of the wire and(ii)at a point outside the
wire0.2emfrom the surface of the wire.
Solution. HereR= 0.5 em = 0.5 x10-2m,I= 100 A
We use the results of the above example.
(i)B -~ r
inside-21tR2.
41tx10-7x100x0.1x10-2
21tx(0.5x1O-2l
=8.0x10-4T.
(ii)B =1101= 41tx10-7x100
surface 21tR21t x 0.5x10-2
=4.0x10-3 T.
(iii) Herer=0.5+0.2 =0.7 em =0.7 x 10-2 m
B =1101= 41tx10-7x100
outside 2nr21tx0.7x10-2
a
=2.86x10-sT.
rproblems For Practice
1.A long solenoid consists of 20 turns per em. What
current is necessary to produce a magnetic field of
20 mT inside the solenoid? .(Ans. 8.0 A)
2.A long solenoid is made byclosely winding a wire
of radius 0.5 mm over a cylindrical non-magnetic
frame so that successive turns nearly touch each
other. What will be the magnetic field at the centre
of the solenoid if a current of 5 A flows through it ?
(Ans.21tx10-3T)···4wx£n~m}r°n4lxv

353. 4.28
3.T
hemagnetic field at the centre of a 50 em long
solenoid is 4.0 x10-2T when a current of 8.0 A
flows through it.Whatisthe number of turns in the
solenoid? Take rt=3.14. (Ans. 1990)
4.A solenoid is 1.0 m long and 3.0 em in diameter. It
has five layers of windings of 850 turns each and
carries a current of 5.0 A. (i)What isBat its centre?
(i)What is the magnetic flux CPBfor a cross-section
of the solenoid at the centre?
[Ans. (i)2.67x10-2T,(i)1.9x10-5Wb]
5.A solenoid is 2.0m long and 3.0 em in diameter. It has
5layers of winding of 1000 turns each and carries a
current of 5.0 A. What is themagnetic field at the
centre? Use the standard value of Iio[Punjab 97C]
(Ans. 1.57x1O-2T)
6.Atoroid has a core of inner radius 20 em and outer
radius 22 em aroundwhich 4200turns of a wire are
wound. Ifthe current in the wire is 10 A, what is the
magnetic field(i)inside the core of toroid(ii)outside
thetoroid and(ii)in the empty space surrounded
bythetoroid. [Ans. (i)0.04 T(ii)Zero(ii)Zero]
7.Along straight solid conductor of radius 4 em
carries a current of 2 A, which is uniformly distri-
buted over its circular cross-section. Find the
magnetic field atadistance of 3 em from the axis of
theconductor. (Ans.7.5 x10-6T)
HINTS
1.Heren=20cm-1 =20x102m",
B=20mT=20x10-3T
B 20x10-3
Current, I=-= 7 2=B.O A.
Iion41tx10-x20x10
2.Diameter of the wire =2x0.5=1.0mm=10-3m
..Number of turns perunit length,
1 103-1
n=10-3 m= m
Also, I=5A,lio=41tx10-7TmA-1
B=lio nI=4n:x10-7x103x5=21tx10-3 T.
N
3.B=IionI=Ii01I
BI4.0x10-2x0.50
N=-= =1990.
lioI 4x3.14x10-7 x8
4.umberof turns per unit length,
5x850 -1
11=--- =4250m
1.0
(I)B=Ii0nI=4n:x10-7x4250x5.0=2.67x10-2T.
(i)CPB=BA=Bxm2
=2.67 x10-2x3.14x(1.5x10-2)2
=1.9x10-5Wb.
PHYSICS-XII
5.Number ofturns per unit length,
/!=5x1000=2500 m-1
2.0
B=IionI=41tx10-7 x2500x5.0=1.57x10-2T.
6.Meanradius of toroid,
20+22
r=---=21em=0.21m
2
umber ofturns perunit length
4200_42_0_0_ =_10_0_0m-1
Zrr r21tx0.21rt
(i)Field inside the core of the toroid,
-71000
B=lnl=41tx10x -- x10=0.04T.
n:
(ii)Magnetic field outside thetoroid is zero.
(iii)Magnetic field in the empty space surrounded
bytoroidis zero.
7.Current enclosed by the loopofradius r,
, T 2Ir2
1=-2 xnr=-2
n:R R
Using Ampere's circuital law,
BL=lio!'
Ir2 B__liolr
BxZnr=Ii0-2 or 2
R 21tR
41tx10-7x2x3x10-2 -6
------;2...,,-2 -=7.5x10T.
21tX(4x10-)
4.11FORCE ON A MOVING CHARGE
IN A MAGNETIC FIELD
12.Statethefactors on whichthe force acting on a
charge movingina magnetic field depends. Write the
expressionforthis force. Whenisthis force minimum and
maximum?Define magnetic field. Also define the Sf unit
ofmagnetic fiel.
Magnetic force on a moving charge.The electric
chargesmoving in a magnetic field experience aforce,while
there isno such force on static charges. Thisfactwas
first recognized byHendrik Antoon Lorentz, agreat
Dutch physicist, nearly a century ago.
~
Suppose apositive charge qmoves with velocity v
~ ~ ~
in amagnetic field Bandvmakes an angleewith B,
as shown inFig. 4.56. Itisfoundfrom experiments that
~
the chargeqmoving in the magnetic field Bexpe-
~
riences a forceFsuch that
1.the forceisproportional to the magnitude of the
magnetic field, i.e., Fex:B
2.the force isproportional to the charge q,i.e.,Fex:q
3.theJorceis proportional to the component of the velocity
v inthe perpendicular direction of the fieldB,i.e.,
Fex:vsine···4wx£n~m}r°n4lxv

354. MAGNETIC EFFECT
OFCURRENT
Fig.4.56Magnetic Lorentz force.
Combining the above factors, we get
F0:;Bqvsin 8
or F =kqvBsin 8
The unit of magnetic field is so defined that the
proportionality constant kbecomes unity in the above
equation. Thus
F=qvBsin 8
This force deflects the charged particle sideways
and is called magnetic Lorentz force. Asthe direction
~ ~ ~
of F is perpendicular to bothvandB,sowe can
~ ~ ~
express F in terms of the vector product ofvandBas
~ ~ ~
F=q(vxB)
Figure 4.56 shows the relationship among the direc-
;,;~ ~ ~ ~
tions of vectorsF,vandB.VectorsvandBlie in the
~
XY-plane. The direction of Fisperpendicular to this
~
plane and points along +Z-axisi.e.,Factsin the
~ ~
direction ofvxB.
Special Cases
Case1. Ifv=0, then F=0
Thus astationary charged particle does not experience
any force in a magnetic field.
Case2. If 8 =0° or 180°, then F =0
Thus a charged particle moving parallel or antiparallel to
a magnetic field does not experience any force in the
magnetic field.
Case3. If 8 = 90°, then F = q vBsin 90° =q vB
Thus a charged particle experiences the maximum force
when it moves perpendicular to themagnetic field
Rules for finding the direction of force on a charged
particle moving perpendicular to a magnetic field.The
~
direction of magnetic Lorentz force F can be
determined by using either of the following two rules:
4.29
1.Fleming's left hand rule. Stretch the thumb and
the first two fingers of the left hand mutually per-
pendicular to each other.Ifthe forefinger points in
the direction of the magnetic field, central finger in
thedirection of current, then the thumb gives the
direction of the force on the charged particle.
(Fig. 4.57)
z
-->
F
Fig.4.57 Fleming's left hand rule.
2.Right hand (palm) rule. Open the right hand and
placeitsothat tips of the fingers pointinthe
~
direction of the field B and thumb in the direction of
~
velocity vof the positive charge, then the palm faces
towards the forceF,as shown in Fig.4.58.
-->
F
-.[
Fig.4.58Right hand palm rule.
Definition of magnetic field. We know that
B
F
qvsin 8
Ifq=l, v=l,8=90°, sin 90° =1, then B=F
Thus the magnetic field at a point may be defined as theforce
acting on a unit charge moving with a unit velocity at right
anglestothe direction of the field.
SI unit of magnetic field.Again, we use
F
B=---
qvsin 8···4wx£n~m}r°n4lxv

355. 4.30
IfF=lN, q=lC, v=lms
-1, 9=90°,then
Sl unit ofB.e IN
1 C.1 ms-1 .sin 90°
IN
1A.1m
=1 N A-1m-1=1tesla.
Thus the51unit of magnetic field istesla(T).
One tesla is that magnetic field in whichacharge of
1 Cmoving withavelocity of1ms-1atright anglesto
the field experiencesaforce of one newton.
A field of one tesla is a very strong magnetic field.
Very often the magnetic fields are expressed in terms
of a smaller unit, called the gauss(G).
1 gauss = 10 -4tesla
Table 4.1 Some Typical Magnetic Fields
108T
IT
10-2T
10-4 T
1O-12T
Surface of a neutron star
Large field in the laboratory
Field near a bar magnet
Field on the earth's surface
Field in interstellar space
Dimensions of magnetic field.Clearly,
MLr2
[B]= [F]
[q][v][sin 9]
=[Mr2A-1].
AT.Lr1.1
Here A represents current.
4.12 LORENTZ FORCE
13.WhatisLorentz force? Write an expression for it.
Lorentz force.The total force experienced by a charged
particle moving in a region where bothelectric and magnetic
fields are present,iscalled Lorentz force.
A chargeqin an electric fieldEexperiences the
electric force,
-t
This force acts in the direction of fieldEand is
independent of the velocity of the charge.
Themagnetic forceexperienced by the chargeq
-t -t
moving with velocityvin the magnetic fieldBis
given by
PHYSICS-XII
-t
Thisforce acts perpendicular to the plane ofvand
-t -t
Band depends on the velocityvof the charge.
The total force, or theLorentz force, experienced by
thechargeqdue to both electric and magnetic field is
given by
-t -t -;::t
F=Fe+F
nt
or
-t -;::t-t-t
F=q(c+vxB)
For Your Knowledge
~A static charge is a source of electric field only while a
moving charge issource of both electric and magnetic
fields.
~A moving charge produces a magnetic field which, in
turn,exerts a force on another moving charge.
~A stationary source does not produce any magnetic
field to interact with an external magnetic field. Hence
no force is exerted on stationary charge in a magnetic
field.
~An electric charge always experiences a force in an
electric field, whether the charge is stationary or in
motion.
~A charge moving parallel or antiparallel to the
direction of the magnetic field does not experience
any magnetic Lorentz force.
~If in a field, the force experienced by a moving charge
depends on the strength of the field and not on the
velocity of the charge, then the field must be an
electric field.
~If in a field, the force experienced by a moving charge
depends not only on the strength of the field but also
on the velocity of the charge, then the field must be a
magnetic field.
.-..
Formulae Used
Force on a chargeqmoving with velocityvin a
magnetic field at an angleewith it is
F=qvBsine
The direction of the force is given by Fleming's
left hand rule.
Units Used
ForceFis in newton, chargeqin coulom, velocity
vin ms-1 andBin tesla.
Example 35.Aproton entersa magnetic fieldofflux
density2.5Twitha velocity of1.5 x107ms-1at an angle of
30°with the field.Find the forceon the proton.°°°3vw}m{lzq«m3kwu

356. MAGNETIC EFFECT OF CURRENT
Solution. Hereq=e= 1.6x1O-
19C
v=1.5xl07ms-1, B=2.5T,e=30°
Force, F=qvBsin e
= 1.6x10-19x1.5x 107 x2.5x sin30°
=3x10-12N.
Example 36.An alpha particle isprojected vertically
upward with a speed of3x104kms-1 in a region where a
magnetic field of magnitude 1.0T exists in the direction
south to north. Findthatmagnetic force that acts on the
particle.
Solution. Charge on a-particle,
q= +2e=2 x1.6x1O-19C
Herev=3xl04kms-1=3xl07ms-t, B=1.0T,
e=90°.
v
B
90°
Fig. 4.59
Magnetic force on thea-particle is
F=qvBsin e
= 2x1.6x10-19x3x107x1.0xsin 90°
=9.6x10-12 N
According to Fleming's left hand rule, the magnetic
force on the a-particle acts towardswest.
Example 37. An electron ismoving northwards with a
velocity of3.0 x107ms-1 ina uniform magnetic field of10T
directed eastwards. Find the magnitude and the direction of
theforce on the electron.
Solution.q=e=I.6 x1O-19C, v=3.0x107ms ",
B=10T,o=90°.
F
Fig.4.60
The magnitude of magnetic force on the electron is
F=qvBsin e = 1.6 x10-19 x3x107x10xsin 90°
=4.8x10-11 N
4.31
As the electron moves northwards, direction of
current is eastwards. According to Fleming's left hand
rule, the magnetic force on the electron actsvertically
upwards.
Example 38.A positive charge of 1.5/lCismoving with a
speed of2x106ms-1 alongthepositive X-axis. A magnetic
~ " "
field, B= (0.2 j+0.4k ) iesla acts in space.Find the
magnetic force acting on the charge.
Solution. Here q= 1.5 /lC = 1.5 x1O-6C,
~ 61\ 1 ~ 1\ 1\
v=2xlO ims ",B=(0.2j +O.4k)T
Magnetic force on the positive charge is
~ ~ ~
F=q(v xB)
6 6" " "
= 1.5x10-[2x10ix(0.2j+0.4k )1
= 3.0 [0.2 [ xi+0.4[ xk1
= (0.6k-1.2i)N. [.:ixi=i,[xk= -i1
Example 39.A5.0MeVprotonisfalling vertically down-
wardthrough aregion ofmagnetic field1.5T acting horizon-
tally from south to north. Find the magnitude and the direc-
tion ofthe magnetic force exerted on the proton. Takemass of
the proton as 1.6x10-27kg.
Solution. Kinetic energy of the proton is
.!.mv2= 5.0 MeV = 5 x1.6x10-13J
2
v2=2x5x1.6x10-13J
m
10x1.6x}0-13 = 10x1014
1.6x10-27
or
v=3.16x 107ms-1
Force on the proton is
F=qvB sin 90°
= 1.6x10-19x3.16x107x1.5x1
=7.58x10-12N
According to Fleming's left hand rule, the magnetic
force on the proton acts eastwards.
Example 40.A long straight
wireABcarriesacurrent of4A.
Aproton Ptravels at
4x106m/s,parallel to the wire, 4A
0.2m from itand in a direction
opposite tothe current as shown
in Fig. 4.61.Calculate theforce
which the magnetic field of
current exerts on the proton. Also
specify the directionof theforce.
[CBSE OD 02]
B
- _0,2.,--1'
6
4x10m/s
A
Fig. 4.61···4wx£n~m}r°n4lxv

357. 4.32
Soluti
on.Magnetic field at point P due to the
current in wireAB,
B=1101=4nx10-7 x 4 = 4x10-6T
2ttr 2nx0.2
This field acts on the proton normally into the plane
of paper. According to Fleming's left hand rule, a
magnetic force acts on the proton towards right in the
plane of paper. The magnitude of this force is
F=qvBsin900
= 1.6 x10-19x4x106x4x10-6 x1
=2.56x10-18N.
Example 41. Copper has8.0x1028electrons percubic
metre. A copper wire of length1m and cross-sectional area
8.0x10-6rJ carrying a current and lying at right angle to
a magnetic field of strength5x10-3T experiences aforce of
8.0 x 10-2 N. Calculate the drift velocity of free electronsin
the wire.
Solution. n= 8x1028m -3,I= 1m
A=8x1O-6m2, e=1.6 x 10-19 C
Total charge contained in the wire,
q=Volume of wirexne=Alne
= 8 x 10-6 x 1 x 8x1028x 1.6 x 10-19 C
= 102.4 x 103 C
Ifvdis the drift speed of electrons, then
F=qvdBsin900=qV dB
F 8.0x10-2 -1
V=-= ms
•.dqB102.4 x103x5x 10-3
=1.56x 10-4ms-1•
<prOblems For Practice
1.Anelectron moving with a velocity of5.0x107ms-1
enters a magnetic field of1.0Wb m-2at an angle of
300•Calculate the force on the electron.
(Ans.4.0 x10-12N)
2.An a-particle of mass6.65x10-27kg and charge
twice that of an electron but of positive sign travels
at right angles to a magnetic field with a speed of
6x105ms-1. The strength of the magnetic field is
0.2T.(i)Calculate the force on the a-particle.
(ii)Also calculate its acceleration.
[Ans.(i) 3.84 x 10-14 N(ii) 5.77 x 1012 ms "]
3.An electron is moving northwards with a velocity
of107ms-1 in a magnetic field of 3 T, directed
downwards. Calculate the instantaneous force on the
electron. (Ans. 4.8x10-12N, vertically upwa:ds)
4.A solenoid, of length 1.5m, has a radius of1.5em
and has a total of1500turns wound on it. Itcarries a
current of 3 A. Calculate the magnitude of the axial
magnetic field inside the solenoid. If an electron
were to move with a speed of 2x104ms-1 along
the axis of thiscurrent carrying solenoid, what
would be the force experienced by this electron ?
[CBSED08Cl (Ans. 0.38T,0)
5.An electron is moving at 106ms-1 in a direction
parallel to a currentof5A, flowing through an
infinitely long straight wire, separated by a perpen-
dicular distance of10cm in air. Calculate the
magnitude of the force experienced by the electron.
[CBSED 99](Ans.1.6x10-18N)
6.A proton ofenergy 3.4MeVmoves vertically down-
wards through a horizontal magnetic field of 3T which
acts from south to north. What is the force on the
proton? Mass of protonis1.7x10-27kg;charge on
proton is1.6x10-19 C. (Ans.12.15x10-12N)
HINTS
1.q=e= 1.6 x10-19 C,V= 5.0x107ms-1
B=1.0Wbm-2,e=300
Force,F=qvBsine
= 1.6x10-19 x5.0 x 107 x 1.0 xsin300
=4.0x10-12N.
2.(i)Herem= 6.65x1027kg,
q=+2e= 2x1.6x1O-16C, B= 0.2T,
v=6x105ms-I,e=900
F=qvBsin 900
= 2x1.6x10-19x 6x105x 0.2 x1N
=3.84x10-14N
F3.84x10-14 12_2
a=-= 27= 5.77x10ms
m6.65x10
3.F=qvBsin900= 1.6 x10-19 x107 x3x1
=4.8x10-12N
According to Fleming's left hand rule, the force acts
vertically upwards. I
lloNI4nxlO-7x1500x3T T
4.B=-1- = 1.5 x102 = 0.38
FOI;ce,F=evBsin00""0.
$.Magneticfieldof toe straight wire carrying a current
ofZ A,at a distance of10cmor 0.1 m from it is
B=110I=4nx10-7 x5= 10-5 T
2n r 2n x0.1
This field, acts perpendicular to the direction of the
electron. So magnetic force on the electron is
F=qvBsin900
= 1.6x10-1~ x106x 10-5 xI=1.6x10-18 N.
fl.Proceed as in Example39,on,page4.31.
PHYSICS-XII···4wx£n~m}r°n4lxv

358. MAGNETIC EFFECT OF CURRENT
4.13WORK DONE
BY A MAGNETIC FORCE
ON A CHARGED PARTICLE IS ZERO
14.Show that the work done by a magnetic field on a
moving charged particle isalwayszero.
Work done by a magnetic force
~ ~
particle. The magnetic force F=q(vx
on acharged
~
B)always acts
~
perpendicular to thevelocity vor the direction of
motion of charge q.Therefore,
~~ ~ ~ ~
F.v=q(vx B).v=O
According toNewton's second law,
~
~ ~
F =ma =m--
dt
or
~
dv ~
m--. v=0
dt
!!!.[d-; .-;+s.d-;j=0
2dt dt
md ~ ~
--(v. v)=O
2dt
~(..!.mv2)=0
dt2 ~~ 2
[v.-v =v]
or
or
or
dK=0
dt
or K=constant
Thusa magnetic forcedoesnot change thekinetic energy
of the charged particle. This indicates that thespeed ofthe
particle does notchange. According to thework-energy
theorem, the change in kinetic energy is equal to the work
done on the particle by the net force. Hence the work
doneon the charged particleby the magnetic forceiszero.
4.14MOTION OF A CHARGED PARTICLE
IN A UNIFORM MAGNETIC FIELD
15.Discuss the motion of a charged particle in a
uniform magnetic field with initial velocity(i)parallel to
the field, (ii)perpendicular to the magnetic field and (ii)
at an arbitrary angle with the field direction.
Motion of a charged particle in a uniform magnetic
field. When a charged particle having chargeqand
~ ~
velocityventers a magnetic fieldB,it experiences a
force
~ ~ ~
F=q(vxB)
The direction of this force is perpendicular to both
~ ~ -
vandB.The magnitude of this force is
F=qvBsine
4.33
Followingthreecases are possie:
1. When the initial velocityisparalleltothe
magnetic fieldHeree=0°,soF=qvBsin 0°=O.
Thus the parallel magnetic field does not exert any
force on the moving charged particle. The charged
particle will continue to move along the line of force.
2.When the initial velocityisperpendiculartothe
magnetic fieldHeree=90°,soF=qvBsin 90° =qvB=a
maximum force. As the magnetic force acts on a particle
perpendicular to its velocity, it does not do any work
on the particle. Itdoesnot change the kinetic energy or
speedoftheparticle.
~
Figure 4.62shows a magnetic fieldBdirected
normally into the plane of paper, as shown by small
crosses. A charge+qis projected with a speedvin the
plane of the paper. Thevelocity is perpendicular to the
x x x x xxx
x ~x x x
x x
x xx
x xx
q
-7
x xB
x x x x
x x x x
xx x x x
x x x x x x x
Fig. 4.62 A positively charged particle moving in a magnetic
field directed into the plane of paper.
magnetic field. A forceF=qvBactson the particle
~ ~
perpendicular to bothvandB.This force conti-
nuously deflects the particle sideways withoutchanging
its speed and the particle will move along a circle
perpendicular to the field. Thus the magnetic force
provides the centripetal force. Letrbe the radius of the
circular path. Now
2
Centripetal force, mv=Magnetic force,qvB
r
or
mv
r=-
qB
Thus the radius of the circular orbit is inversely
proportional to the specific charge (charge to mass
ratioq/m)and to the magnetic field.
P.dfrevoluti Circumference
enoarevo ution=----::-----:---
Speed
T _2ttr _2remv_2rem
---;---;'qi3-qs
or···4wx£n~m}r°n4lxv

359. 4.34
Clearly
, the time period is independent ofvandr.If
the particle moves faster, the radius is larger, it has to
move along a largercircle so that the time taken is the
same.
Thefrequency of revolution is
f.=!=~
cTTttm
This frequency iscalledcyclotron frequency.
3.When the initial velocity makes an arbitrary
angle with the field direction.A uniform magnetic
--->
field B is set up along +ve X-axis.A particle of chargeq
---> --->
and mass menters thefield Bwithvelocity vinclined
--->
at angle 8 with the direction of the field B,as shown in
Fig.4.63.
y
z
B
Fig. 4.63Helical motion of charged particle in a magnetic field.
--->
Thevelocity vcan be resolved into two rectangular
components:
1.Thecomponent vIIalong the direction of the field
i.e.,alongX-axis. Clearly
vII=vcos 8
The parallel component remains unaffected by the
magneticfield and so the charged particle continues to
move along the fieldwith a speed ofvcos 8.
2. The componentv1.perpendicular to the direction
of the fieldi.e.,in the YZ-plane. Clearly
v1.=Vsin 8
Due to this component of velocity, the charged
particle experiences a force F =qv1.B which acts
--->
perpendicular to bothv1.and B. This force makes the
particle move along a circular path in the YZ-plane.
The radius of thecircular pathis
r=mv1.=mvsin 8
qB qB
PHYSICS-XII
The period of revolution is
T_2ttr _2rtmvsin 8 _ 2tim
- ~-vsin8·qB -qB
Thus a charged particle moving in a uniform
magnetic field has two concurrent motions : a linear
--->
motion in the direction ofB(along X-axis) and a
--->
circular motion in a plane perpendicular to B(in
YZ-plane). Hence the resultant path ofthe charged
particle will be a helix, with its axisalong thedirection
--->
of B.
Thelineardistance travelled by the charged particle
in the direction ofthemagnetic field during its period
of revolution is called pitch of the helical path.
. Tnm 2mllvcos8
pitch =VII xT=vcos8x --=----
qB qB
4.15MOTION OF A CHARGE IN
PERPENDICULAR MAGNETIC AND
ELECTRIC FIELDS
x
16.Electric and magnetic fields are applied mutually
perpendicular to each other. Show that a charged particle
will follow a straight line path perpendicular to both of
thesefields,ifits velocity isE/Binmagnitude.
Velocity selector. Suppose a beam ofcharged
particles, say electrons, possessing arange of speeds
passes through aslit 51 and then enters a regionin
which crossed (perpendicular) electric and magnetic
--->
fields exist. As shown in Fig. 4.64,the electric field E
actsin the downward direction and deflects the
electrons in the upward direction. The magnetic field
--->
B acts normally intothe plane of paper and deflects
theelectrons in the downward direction.
Electron
••
I I -,,_
'-X~~'X,XiA~-~,'_,
\ E I '"
, X X / "
, - -.: \De~;c:on '
Region of
crossed fields
v
Fig. 4.64Motion of an electron in a region of
crossed magnetic and electric fields.
Only those electrons will pass undeflected through
theslit52on which the electric and magnetic forces are···4wx£n~m}r°n4lxv

360. MAGNETIC EFFECT OF CURRENT
equal and
opposite. The velocity vof the undeflected
electrons is given by
E
V=-
B
Such an arrangement can be used toselectcharged
particles of a particular velocity out of a beam in which
the particles are moving with different speeds. This or
arrangement is calledvelocity selector orvelocity
filter. This method was used by IIThomson to
determine the charge to mass ratio (e/m)of an electron.
eE=evB or
Examples based on
Motion of Charges in Electric
and Magnetic Fields
Formulae Used
1.Electric force on a charge, ~ =qE
2.Magnetic force on acharge, F,n=qvBsin8
3.Ina perpendicular magnetic field, thecharge
follows acircular path.
InV2 mv
qvB=-- orr= -
r qB
T=2nInandf=3..!!.-
qB 2nm
--> -->
4.Whenvmakes angle 8 with B, the charge follows
helical path.
mv.l mvsin8 21(r 2nm
r=--= ; T=-=--
qB qB v.l qB
2nmvcos8
Pitch of helix, h=viiT=vcas8.T=----
qB
5. K.E. gainedbyan electron when accelerated
through a potential differenceV,
~mv2=ev :.v=~2:::
Units Used
Eis in Vm-1orNe-I, Bin tesla, vinms-I, rin metre.
Example 42. An electron moving horizontally witha
velocity of4x104m/sentersaregionof uniform magnetic
field of10-5T acting vertically downward as shown in
Fig. 4.65(a). Draw its trajectory and find out the time it takes
to come out of the region of magnetic field [CBSE F 15]
I
:XX X
I
:XXB X
•
I
-e
I
'XX X
I
I
:XX X
Fig.4.65(a)
4.35
Solution. The electron moves along semicircular
trajectory insidethe magnetic field and comes out, as
shown in Fig.4.65(b).Radius r of the path is given by
mv2
-=qvB
r
mv 9.1x10-31x4x104
r=- = m
qB 1.6x10-19 x10-5
= 9.1x 4x10-3m =22.75 x1O-3m
1.6
I
:XX X
I
e:XXBX
:3
X X
e I
:xX X
Fig. 4.65(b)
Time taken to come out of the region of magnetic
field,
ttr22x22.75x10-3
t=-= s
v 7x4x104
= 17.875 x10-7 S::::.1.8x10-65•
Example 43. Anelectron travels ina circular path of
radius 20emin a magnetic field2x10-3T.(i)Calculate
the speed of the electron. (ii)Whatisthe potential difference
through which theelectron must be accelerated to acquire
this speed?
Solution.Herer=20 em =20xla-2m,
B=2x10-3T,e= 1.6x10-19C,m=9.1x10-31 kg
(i)Magnetic force on the electron
=Centripetal force on electron
mv2
evB=--
r
eBr
:.Speed,v=-
m
1.6x10-19 x2x10-3x20x10-2
9.1x10-31
=7.0x107ms-t.
(ii)IfVis the p.d. required to give speedvto the
electron, then
eV=1mv2
2
V=mv2= 9.1x10-31 x(7.0x107)2
2e 2x1.6x1O-19
or
= 13.9x103V::::.14kV.···4wx£n~m}r°n4lxv

361. 4.36
Example 44.An
electron after being accelerated through a
potential difference of104 Venters a uniform magnetic field
of 0;04 T perpendicular to its direction of motion. Calculate
the radius of curvature of its trajectory.
Solution. HereV= 104 V,B=0.04 T,
e=1.6 x10-19 C,m=9.1x 10-31 kg
Anelectron accelerated through a p.d. Vacquires a
velocityvgiven by
l' ~2ev
- mv2 =eVorv=--
2 m
Asthe electron describes a circular pathof radius of
rinthe perpendicular magnetic fieldB,therefore,
mv2
--=evB
r
or r= mv =m~2eV=J2iileV
eB eB III eB
_~2x9.1x10-31 x1.6x10-19 x104
- 1.6 x 10- 19x0.04
5.4x10-23
1.6x10-19x0.04
=8.43 x 10-3m = 8.43 mm.
Example 45.If a particle of charge qismoving with or
velocityvalong the z-axisand the magnetic fieldBisacting
-+ -+-+
along the x-axis, use the expression F=q (vxB )to find
the direction of the forceFacting on it.
A beam of proton passesundefleciedwith a horizontal
velocity v, through a region ofelectric and magnetic fields,
mutually perpendicular to each other and normal tothe
direction of the beam. Ifthe magnitudes of the electric and
magnetic fields are 100 kVlm and 50mT respectively,
calculate:(i)velocity v of the beam. (ii)force with which it
strikes a target on a screen, if theproton beam current is
equal to 0.80 mA. [CBSE 00 08]or
~ ~ ~ 1\ 1\
Solution.F=q (vxB)=q(vjxBk )
= qvB!xk=qvBi
Thus the force Facts on the charge qalong the +ve
x-direction.
(i)For undeflected proton beam,
qvB=qE
E100 kVm-1
v=-=----
B 50mT
100x103Vm-1
50x1O-3T
PHYSICS-XII
(ii)Current carried by proton beam,
1=0.8mA =8x10-4A
Number of protons striking the screen per second,
n=!.=8x10-4 =5x1015s-1
e1.6x10-19
mp= 1.675x1O-27kg
Force with which aproton beam strikes a target on
the screen,
F=dp=m nv
dt p
= 1.675x10-27 x5x1015x2x106N
=1.675x10-5 N.
Example 46. An electron beam passes through a mag-
netic field of 2x10-3Wb«?and an electric field of
3.4x104Vm-t, both acting simultaneously. If thepathof
the electron remainsundeoiaied,calculate thespeed of
the electrons. If theelectric fieldisremoved, what will be
the radius of the circular path?Massofan electron
=9.1 x10-31 kg.
Solution.Here B=2x10-3Wb m-2,
E=3.4x104Vm-1
Magnetic force on the electron
=Electric force on the electron
evB= eE
Velocity of electrons,
E3.4x104
v= - = ms-1 =1.7x 107ms-1
B 2x10-3
When electric field has been removed,
Force exerted by the magnetic field on an electron
=Centripetal force on an electron
mv2
evB=--
R
R=mv=9.1x10-31 x 1.7x107
eB1.6 x 10-19 x2x10-3
=4.8x10-2m= 4.8cm.
Example 47. In a chamber a uniform magnetic field of8.0G
(1G=10-4T)ismain tained. Anelectron with a speed of
4.0x106ms-1enters the chamber in a direction normal to
thefiel.
(i)Describe thepath of the electron.
(ii)Whatisthefrequency of revolution of theelectron?
(iii)What happens to the path of the electron if it
progressively loses its energy due to collisions with
the atoms or molecules of the environment?
[NCERT]
i.e.,www6notesdrive6com

362. MAGNETIC EFFECT OF C
URRENT
Solution.(i)The path of the electron is a circle of
radiusrgiven by
mv
r=-
eB
HereB;"B.OG=8.0 x1O-4T, v=4.0x106ms ",
e=1.6x10-19 C,m=9.1 x10-31 kg
9.1 x10-31x4.0x106
r=------~--------~
1.6 x10-19 xB.Ox10-4
=2.Bx 10-2 m=2.8 em.
The sense of rotation of the electron in its orbit can
be ascertained from the direction ofthe centripetal
-4 -4-4
forceF= -e(vxB).Thus if we look alongthe
-4
direction ofB, the electron revolves clockwise.
(b)The frequency of revolution of the electron in its
circular orbit is
f=~ =1.6 x10-19 x8.0 x10-4
2nm 21tx9.1x1O-31
= 0.22 x108Hz=22 MHz.
(c)In successive collisions, electron loses its speed
progressively. If after collision its velocity vector
remains in thesame planeof the initial circular orbit,
the radiusof the circular orbitwill decrease in
proportion to the decreasing speed. Otherwise, the path
of the electronwill be helical between two collisions.
Example48.A monoenergetic electron beam of initial
ener81J18keV movinghorizontally issubjected to a
horizontal magnetic field of 0.4Gnormal toitsinitial
direction. Calculate the vertical deflection of the beam over a
distance of30em. [CBSESample Paper 98]
Solution. Under the action of the magnetic field,
the electrons will move along a circular path.
..Centripetal force on an electron
=Magnetic force on an electron
mv2
--- =evB
r
mv ~2m.1/2 mv2
or r=-= -'-----------
eB eB
Herem=9.11x 10-31kg,e=1.6x 10-19C,
B=0.40 G=0.40x10-4T
K.E. =.!.mv2= 18 keY = 18x1.6x10-16J
2
~2x9.11x10-31 x18x1.6x10-16
r= -'------------:;-;;,--------...,----
1.6x10-19x0.4x10-4
= 11.32 m
4.37
Thus the electron moves in a circle of radius
11.32 m, as shown in Fig. 4.66.As it covers a distance
PQ =30 ern, it goes down through a vertical distance
Fig. 4.66
equal to PA.Ifeis the angle suended by arc PQ at the
centre0,then
PA=OP-OA=r-rcose=r(1 - cose)
Now e=~ = 30 x10-2= 0.02650 rad
Radius 11.32
= 0.02650 x180=1.52°
.n
cose= cos 1.52 °= 0.99965
Hence P A= 11.32 (1 - 0.99965)
= 3.9744x10-3m ::.4mm.
Example49.A beam of protons enters a uniform magnetic
fieldof0.3T witha velocityof4x105ms-1at an angle of
60°to the fieldind the radius of the helical path taken
by the beam.Also find the pitchof the helix (distance
travelled by a proton parallel to the magnetic field during
oneperiod ofrotation). Mass of proton is1.67x10-27kg.
[lIT 86]
Solution. The components of the proton's velocity
parallel and perpendicular to the magnetic field are
VII=vcos 60° = 4x105x~ =2x105ms-1
v.l=vsin 60° = 4x105x.J3=3.464x105ms-1
2
The componentVIImakes the electron move along
the fieldBwhilev.lmakes the proton move along a
circular path. Hence the path of the proton is a helix.
The radiusrof the helix is given by
mv2
qv.lB=__.l
r
or
mv.l 1.67x10-27 x3.464x105
r= -- =----------:;=------
qB 1.6x10-19 x0.3
= 12x10-3m = 1.2 em.···4wx£n~m}r°n4lxv

363. 4.38
Period of
revolution of the electron is
T= 21tr= 2x3.14x12x10-3
v.l 3.464x105
= 21.75 x10-8s
Pitch of the helix is
p= vIIxT=2x105x21.75x10-8
= 43.5 x1O-3m = 4.35 cm.
Example50.A proton projected in a magnetic field of
0.02T travels along a helical path of radius5.0emand pitch
20em.Find the components of the velocity of the proton
along and perpendicular to the magnetic field. Take the mass
of the proton= 1.6 x10-27kg.
mv
Solution. Radius of helical path, r=_.l_
qB
rqB5x10-2 x1.6x10-19 x0.02
v=-
.l 111 1.6x10-27
=1.0x105ms-1
Period of revolution,
T= 21t r = 21tX5x10-2 = 1tx10-6 S
v.l 1.0x105
_ Pitch _20x10-2 _637104 -1
vII- - -. x ms
T 1tx10-6
~roblems For Practice
1.An electron entering a magnetic field of10-2T with
a velocity of107ms-1 describes a circle of radius
6x10-3m. Calculateelmof the electron.
(Ans.1.67x1011C kg-I)
2.An electron after being accelerated through a poten-
tial difference of100Venters a uniform magnetic
field of0.004T perpendicular to its direction of
motion. Calculate the radius of the path described
by the electron. [CBSE 00 92]
(Ans. 8.4mm)
3.A particle having a charge of100J.lC and a mass of
10mg is projected in a uniform magnetic field of
25mT with a speed of10ms-1 in a direction perpen-
dicular to the field. What will be the period of revolu-
tion ofthe particle inthe magnetic field ?
(Ans.25s)
4.An electron having a kinetic energy of100eV
circulates in a path. of radius10cm in a magnetic
field. Find the magnetic field and the number of
revolutions made by the electron per second.
(Ans.3.4x10-4T,9.4x106rps)
PHYSICS-XII
5.An electron beam passes through a magnetic field
of2x10-3Wb m-2and an electric field of
1.0x104V m-1, both acting simultaneously. If the
path of the electrons remains undeviated, calculate
the speed of the electrons. If the electric field is
removed, what will be the radius of the circular
path? (Ans.5x106ms ", 1.43ern)
6.An electron moving perpendicular to a uniform
magnetic field completes a circular orbit in10-6s.
Calculate the value of the magnetic field. Mass of
electron= 9x10-31 kg. (Ans.3.5x10-3T)
7.Find the flux density of the magnetic field to cause
62.5eV electron to move in a circular path of radius
5 em. Givenme=9.1x10-31 kg ande=1.6x10-19 C.
(Ans.5.335x10-4T)
8.An electron of energy2000eV describes a circular
path in a magnetic field of0.2T.What is the radius
of path? Takeme=9x10-31 kg,e=1.6x10-19C.
(Ans.0.75mm)
9.What should be the minimum magnitude and
direction of the magnetic field that must be produced
at the equator of earth so that a proton may go
round the earth with a speed of1.0x107ms-1 ?
Earth's radius is 6.4x106m.
(Ans.1.63x10-8T, perpendicular to the
equator in a horizontal direction)
10.A stream of charged particles possessing a range of
speeds enters region I after passing through a slit ~
(Fig.4.67). In region I there exist crossed (perpen-
~
I I
J
E==:==:===;:
IB I
I I,,
I
II
-J
1000ms
Fig. 4.67
dicular) electric and magnetic fields. The electric
field has magnitude100Vm-1.We want the
particles emerging from slit ~ into region II to have
a fixed velocity of1000ms-1.What should be the
value of the uniform magnetic field in region I ?
(Ans. 0.1T)
11.A proton, a deutron and an alpha particle, after
being accelerated through the same potential
difference, enter a region of uniform magnetic fieldwww6notesdrive6com

364. MAGNETIC EFFECT OF CURRE N
T
~ ~
B,in a direction perpendicular toB.Compare their
kineticenergies.Iftheradius of proton's circular
pathis5 em, what will be the radii of the paths of
deutronandalpha particle?
(Ans.1:1:2,rd= 7.07em, ra= 10 ern)
12.A particle having a charge of 5.0 IlC and a mass of
5.0x10-12 kg is projected with avelocity of
1.0 km s-1 in a magnetic field of magnitude 5.0 mT.
The angle between the magnetic field and the
velocity is sin -1(0.90). Show that the path of the
particle will be a helix. Find the diameter ofthe
helix anditspitch. (Ans.36ern,55ern)
HINTS
1.Use~=~.
m rB
2.Proceed as in Example 45 on page 4.36.
21tm 2n x10x10-6
3.T=--= = 25 s.
qB100x10-6x25x10-3
~2meV 21tm
4.Use B= andT=-.
er eB
5.AseE=evB
E1.0x104 6-1
, ,v=-= 3= 5x10 ms
B 2x10-
When electric field is removed, electrons follow
circular path.
mv2
--= evB
r
mv 9.1x10-31x5x106
or r------""'----;;-
-eB-1.6x1019x2x10-3
= 1.43x10-2m = 1.43 em.
6.AsT=21tm:.B=21tm
eB eT
7.Here2mv2= 62.5eV=62.5x1.6x10-19J
2
=10-17J
v=~
2x10-17
I------,=_ = 4.69x106ms-1
9.1x10-31
mv 9.1x10-31 x4.69x106
B =-e-r=-1-.-6-x-1-0~1;n9""'x-5-x-1-0-""'2;-
= 5.335x10-4 T.
B..2mv2=eVorv=~2ev
2 m
r=mv=m!2eV=~
.. eBeBYm eB
4.39
~2x9x10-31x1.6x10-19x2000
1.6x10-19x0.2
=7.5x10-4m = 0.75 mm.
mv 1.67x10-27 x107 8
9.B =-= =1.63x10-T.
qr1.6x10-19 x6.4x106
10.For the particles to go undeflected,
Force due to electric field=Force due to magnetic
field
qE= qvB
E100 Vm-1
or B = -= 1= 0.1 T.
v1000 ms
11.For a given p.d., the kinetic energy of a charged
particle is proportional to its charge.
..Kp :Kd :Ka=e :e :2e=1:1:2
Radius of thecircular path of any particle of kinetic
energy K,
r=;;=;.f!=~~~K
~2mpKp
..For proton,rp= = 5ern
eB
~2m K 12x2mxK
Fordeutron, r= ddV P P
d eB eB
=.firp= 1.414x5ern= 7.07 em.
~ 12x4xmp x2K
For a-particle, r= ad=V P
a 2eB .2eB
=2rp=10 em.
12.Hereq= 5.0 IlC = 5x10-6C,m= 5x10-12kg,
v= 1.0 kms-1 =103ms-1, B= 5.0 mT = 5 x10-3T
As 8=sin-1(0.90), so sin8=0.90
cos 8= ~1- sin28 =~1-0.B1= ~0.19 ~ 0.436
vJ..=vsin 8 = 103 x0.90=0.9x103ms-1
VII=vcos8 = 103 x0.436=4.36x102ms-l
Velocity componentVIImoves the electron along the
field andv1.along circular path. Hence the motion
ishelical. .
. 2mvJ.. 2x5xlO-12xO.9x103
DIameter =2r=-- = -6 3
qB 5x10 x5x10-
= 0.36 m = 36 em.
21tr3.14x0.36 -3
T=-= 3= 1.25x10 s
vJ..0.9x10
Pitch =VIIxT=4.36x102x1.25x10-3
=0.55 m = 55 em.···4wx£n~m}r°n4lxv

365. 4.40
4.
16CYCLOTRON
17. Whatisa cyclotron?Discuss the principle, con-
struction, theory and working of a cyclotron. Whatisthe
maximum kinetic energy acquired by the accelerated
charged particles?Give the limitations and uses of a
cyclotron.
Cyclotron.Itis a device used to accelerate charged
particles like protons, deutrons, a-particles, etc., to very high
energies.It was invented byE.O.LawrenceandM.S.
Livingstonin 1934 at Berkeley, California University.
Principle. Acharged particle can be accelerated to very
high energies by making it pass through a moderate electric
field a number of times. This can be done with the help of a
perpendicular magnetic field which throws the chargedparticle
into a circular motion, thefrequency of which does not depend
on the speed of the particle and the radius of the circular orbit.
Construction.As shown in Fig. 4.68, a cyclotron
consists of the following main parts:
1.It consists of two small, hollow, metallic
half-cylinders01and02'calleddeesas they are
in the shape of D.
2.They are mounted inside a vacuum chamber
between the poles of a powerful electromagnet.
3. The dees are connected to the source of high
frequency alternating voltage of few hundred
kilovolts.
4. The beam of charged particles to be accelerated
is injected into the dees near their centre, in a
plane perpendicular to the magnetic field.
5. The charged particles are pulled out of the dees
by a deflecting plate (which is negatively charged)
through a windowW.
6. The whole device is in high vacuum (pressure
- 10-6 mm of Hg) so that the air molecules may
not collide with the charged particles.
Theory. Let a particle of chargeqand massmenter
~ ~
a region of magnetic field B with a velocityv,normal
~
to the field B. The particle follows a circular path, the
necessary centripetal force being provided by the
magnetic field. Therefore,
Magnetic force on chargeq
=Centripetal force on chargeq
mv2 mv
orqvBsin90°=-- orr=-
r qB
Period of revolution of the charged particle is
given by
PHYSICS-XII
High ~
frequency
oscillator
(a)
Magnetic field into
theplane ofpaper
rv--.~,,~r-~=r~~~X~
X
X
Deflection
plate-s-,
Proton
source
Vacuum
chamber
(b)
Fig.4.68Cyclotron(a)Front view(b)Section diagram.
Hence frequency of revolution of the particle will be
f.=:!.=~
cT21tm
Clearly, this frequency isindependent of both the
velocity of the particle and the radius of the orbit and is
calledcyclotron frequencyormagnetic resonance
frequency.This is the key fact which is made use of in
the operation of a cyclotron.
Working.Suppose a positive ion, say a proton,
enters the gap between the two dees and finds dee01
to be negative. It gets accelerated towards dee01'As it
enters the dee01'it does not experience any electric
field due to shielding effect of the metallic dee. ThewwwLnotesdriveLcom

366. MAGNETIC EFFECT
OF CURRENT
perpendicular magnetic field throws it into a circular
path. At the instant the proton comes out of dee01'it
finds dee01positive and dee02negative. It now gets
accelerated towards dee02.It moves faster through02
describing a larger semicircle than before. Thus if the
frequency of the applied voltage is kept exactly the
same asthe frequency of revolution of the proton, then
every time the proton reaches the gap between the two
dees, the electric field is reversed and proton receives a
push and finally it acquires very high energy. This is
called the cyclotron'sresonance condition. The proton
follows a spiral path. The accelerated proton is ejected
through a window by a deflecting voltage and hits the
target.
Maximum K.E. of the accelerated ions. The ions
will attain maximum velocity near the periphery of the
dees. Ifvais the maximum velocity acquired by the
ions androis the radius of the dees, then
mv2
__0=qVoB or
ro
The maximum kinetic energy of the ions will be
Ko=!mv~=!m(qBro)2
22m
q2wd
Ko=-_o
2m
Limitations of cyclotron:
1. According to the Einstein'sspecial theory of relativity,
the mass of a particle increases with the increase in its
velocity as
or
m= 1110
~1":'"v2/?
where1110is the rest mass of the particle. At high
velocities, the cyclotron frequency (fc=qB/ 2rcm)will
decrease due to increase in mass. This will throwthe
particles out of resonance with the oscillating field.
That is, as the ions reach the gap between the dees, the
polarity of the dees is not reversed at that instant.
Consequently the ions are not accelerated further.
The above drawback is overcome either by increasing
magnetic field as in asynchrotronor by decreasing the
frequency of the alternating electric field as in a
synchro-cyclotron.
2. Electrons cannot be accelerated in a cyclotron. A
large increase in their energy increases their velocity to
a very large extent. This throws the electrons out of
step with the oscillating field.
3. Neutrons, being electrically neutral, cannot be
accelerated in a cyclotron.
4.41
Uses of cyclotron:
1.Thehigh energy particles produced in a cyclotron
are used to bombard nuclei and study the resulting
nuclear reactions and hence investigate nuclear
structure.
2. The high energy particles are used to produce
other high energy particles, such as neutrons, by
collisions. These fast neutrons are used in atomic
reactors.
3.It is used to implant ions into solids and modify
their properties or even synthesise new materials.
4.It isused to produce radioactive isotopes which
are used in hospitals for diagnosis and treatment.
ForYour Knowledge
~As the magnetic force on a charged particle acts
perpendicular to the velocity, it does not do any work
on the particle. As a result, the kinetic energy or the
speed of the particle does not change due to the
magnetic force.
~When a charged particle is projected into a uniform
magnetic field with its initial velocity perpendicular
to the field, the magnetic force acts on the charged
particle perpendicular to both the magnetic field and
its direction of motion. This force produces
centripetal force to make the particle move in a circle
in a plane perpendicular to the magnetic field.
~When a charged particle moves perpendicular to a
uniform magnetic field: (i)its path is circular in a
plane perpendicular to the magnetic field and its
direction of motion,(ii)the radius of the circular path
is proportional to its momentum,(iil)the kinetic
energy and speed of the particle do not change,
(iv)the force acting on the particle is independent of
the radius of the circular orbit but is proportional to
its speed.e.,Focr"and Focvand(v)the period of
revolution of the charged particle is independent of
its speed and the radius of its circular orbit.
~When a charged particle is projected into a uniform
magnetic field at an arbitrary angle with the field, the
component of the initial velocity parallel to the
magnetic field will make the particle move along the
direction of the field while the perpendicular
component will compel it to follow a circular path. As
a result, the particle will follow a helical path with its
axis parallel to the field.
~In a cyclotron, it is the electric field which accelerates
the charged particles. The magnetic field does not
change the speed, it only makes the charged particle
to cross the same electric field again and again by
making it move along a circular path.···4wx£n~m}r°n4lxv

367. 4.42
Exam l
es based on
Cotron
Formulae Used
For the accelerated charged particle,
. qBr
1.Velocity,v=-
m
2. Period of revolution,T=21tm
qB
3.Cyclotron frequency,Ie=qB
21tm
2B2R2
4.Maximum kinetic energy, ~ax=-q--
2m
whereRis the radius of the dees.
Units Used
.Bis in tesla,vin ms-1,rin metre,Tin second and
I,in Hz.
Example 51 .Deutrons are accelerated ina cyclotron that
has an oscillatoryfrequency of107 Hzand a dee radius of50em.
(i)Whatisthe strength of the magnetic field needed to acce-
leratethe deuirons ?(ii)Whatistheenergyofdeutrons emerging
from the cyclotron. Mass of a deutron=3.34x10-27kgand
charge of a deutron=1.6x10-19C.
Solution. v = 107 Hz,R= 50 em = 0.50 m,
m=3.34 x10-27 kg, q=1.6xl0-19 C
(i)Cyclotron frequency, fe=~
21tm
B=21tmfe = 2x3.14x3.34x 10-27x 107
. . q 1.6x10-19
=1.3 T.
q2WR2
(ii) T(
"max=2m
(1.6x10-19)2 x(1.3)2x(0.50)2
2x3.34x10-27
= 1.62 x10-12J=1.62x10-12 MeV
1.6x10-13
=10.125 MeV.
Example 52.Acyclotron's oscillatorfrequencyis10MHz.
What should bethe operating magnetic field for accelerating
protons? If theradius of the 'dees' is 60em,what is the kinetic
energy of the ~roton beam produced by the accelerator ?
(e= 1.60x10-9C,mp= 1.67 x10-27kg). Express your
answerinunits of MeV(1MeV= 1.602x10-13J).
[CBSE OD 05; CERT]
Solution. Herefe= 10 MHz=107.Hz,
e=1.6xl0-19 C, R =60 em =0.6 m,
mp= 1.67x10-27kg
PHYSICS-XII
The operating magnetic field for accelerating
protons is
B=21tmpfe =2x3.14x1.67x10-27x107
e 1.6x10-19
=0.66 T.
Kinetic energy of the emerging beam will be
e2WR2(1.6x10-19)2 x (0.66lx(0.6)2
~ax=2mp= 2 x1.67x10-27
=1.2x10-12J= 1.2 x10-12 MeV
1.602x10-13
=7.4 MeV.
Example 53. In a cyclotron, a magnetic induction ofl.41
is used to accelerate protons. How rapidly should the electric
field between the deesbereversed? The mass and charge 0,
proton are 1.67x10-27kgand1.6x10-19Crespectively.
Solution. HereB= 1.4 T, m= 1.67x10-27 kg,
e=1.6 x10-19 C
The time required by a charged particle to complete
a semicircle in a dee is
t=1tm= 3.14x1.67x10-27=2.34x10-8s
eB 1.6xl0-19x1.4
Thus the direction of electric field should reverse
after every 2.34x10-8s.
The frequency of the applied electric field should lx
1 1 7
fe=-= 8= 2.14x10Hz.
2t2x2.34x10-
Example 54. If themaximumvalueof accelerating potentia
provided by a radio frequency oscillator be 20kV,findth.
number of revolutions madeby a protoninacyclotron t(
achieve one fifth of thespeed of light. Mass of a proton
= 1.67 x 10-27 kg.
Solution. In a cyclotron, a proton gains energy eV
when it crosses a region of potential differenceV.Ir
one revolution, the particle crosses the gap twice. S(
the energy gained in each revolution =2eV.
Suppose the particle makesnrevolutions befon
emerging from the dees. The gain in its kinetic energ~
will be
mv2
n=--
4eV
c3x108 8-1
V= - =--- =0.6x10 ms
55'
m=1.67x10-27kg
1.67x10-27x(0.6x108)2 .
..n= 19 3=470revolutions.
4x1.6x10-x20x10
..!.mv2=2neV
2
or
Given~~~2tuzkyjxo}k2ius

368. MAGNETIC EFFECT OF CURRENT
rproblems For Practice
1.Anelec
tronof energy 10,000eV describes a circular
path in a plane at right angles to a uniform
magnetic fieldof0.01T.(a)What is the radius of the
circular orbit? (b)What is thecyclotron frequency?
(c)What is the period of itsrevolution? (d)What is
thedirection of revolution as viewed byan
observer looking in the direction of the field ?
(Ans. 3.4x1O-2m, 2.8x108s-1,
3.6x10-9s,clockwise sense).
2.The protons are accelerated by a cyclotron, when a
magneticfieldof2.0Tisapplied perpendicular to
theplane of the dees. Calculate the energy of the
proton in MeV, if the circular path of the protons
hasa radius of 40cm before the protons leave the
cyclotron. Given mass of a proton=1.67x10-27kg.
(Ans.30.6MeV)
3.A cyclotron has an oscillatory frequency of 12MHz
and a dee radius of50ern. Calculate the magnetic
field re~uired to accelerate deutrons of mass
3.3x10-2 kg and charge1.6x10-19C.What is the
energy of the deutrons emerging from the cyclotron?
(Ans.1.56T,14.7MeV)
4.Alpha particles of mass 6.68x10-27kg and charge
3.2x10-19Careaccelerated in a cyclotron in which
a magnetic field of1.25T is applied perpendicular
tothe dees. How rapidly should the electric field
between the dees be reversed ? What are the
velocity and kinetic energy of an alpha particle
when it moves in a circular orbit of radius25cm?
(Ans.9.5x 106 Hz,1.5x107ms ", 7.5x10-13 J)
HINTS
_ q2B2R2_ (1.6 x10-19)2 x(2.0)2x(0.40)2
2.~ax -~ - 2x1.67x10-27 J
=(1.6x1O-19l x(2.0)2x(0.40)2MeV
2x1.67x10-27x1.6x10-13
=30.6MeV.
3.AsJ.=qB
c2nm
B=2nm/r,=2x3.142x3.3x10-27 x12x106
q 1.6x10-19
=1.56 T.
_ q2tfR4 _(1.6x1O-19l x(1.56)2x(0.50)2J
~ax -2m- 2x3.3x10-27
=(1.6 X10-19)2 x(1.56)2 x(0.50)2 MeV
2x3.3x10-27x1.6x10-13
=14.7MeV.
4.43
n m3.14x6.68x10-27 -8
4.t=-= 19 =5.25x10s
qB3.2x10x1.25,
Direction of electric field should be reversed after
every5.25~10-8s.
Applied frequency,
116
J.=-= 8=9.5x10 Hz
c2t2x5.25x10-
As
mv
r=-
qB
rqB0.25x3.2x10-19 x1.25
v=-=-------;;;:;----
m 6.68x10-27
=1.5x107ms-1
K=~mv2
2
=~x6.68x10-27x(1.5x107)2
2
=7.5x10-13J.
4.17FORCE ON A CURRENT CARRYING
CONDUCTOR IN A MAGNETIC FIELD
18. Describe an experiment to illustrate that a current
carryingconductor experiences a force in a magnetic
field ?
Force on a current carrying conductor in a magnetic
field.When a conductor carrying a currentis placed in
an external magnetic field, it experiences a mechanical
force. To demonstrate this force, take a small
aluminium rodAB.Suspend it horizontally by means
of connecting wires from a stand, as shown in Fig. 4.69.
Fig. 4.69Force on a current in a magnetic field.
Place a strong horse-shoe magnet in such a way that
the rod is between the two poles with the field directed
upwards. Now, if a current is passed through the rod
fromAtoB,the rod gets deflected to the right. If we
reverse the direction of current or interchange the···4wx£n~m}r°n4lxv

369. 4.44
poles of
the magnet, the deflection of the rod is also
reversed.The direction of force is perpendicular to
both the current andthe magnetic field and is given by
Fleming'sleft hand rule.
Cause of the force on a current carrying conductor
in a magnetic field.A current is an assemy of moving
charges and a magnetic field exerts a force on a moving
charge. That is why a current carrying conductor when
placed in a magnetic field experiences a sideways force
asthe force experienced by the moving charges (free
electrons) is transmitted to the conductor as a whole.
19.Derive an expression for the force experiencedbya
current carrying straight conductor placed in a magnetic
field, isthisforce
(i)zero and (ii) maximum?
Statethe rule used to determine the direction of this
force?
Expression for the force on a current carrying
conductor in a magnetic field.As shown in Fig. 4.70,
consider a conductor PQof lengthI,area of cross-
sectionA,carrying current Ialong+veY-direction. The
~
fieldBacts along+veZ-direction. The electrons drift
~
towardsleft with velocityvd'Each electron expe-
riences a magnetic Lorentz force along +veX-axis,
which is given by
Ifnis the number of free electrons per unit
Z
-->
B
Area=A
Q
Pr---~-->--------~------------~
y,_+- V-'d:...- --''-{ -)-------------1 -+-t~- Y
-->
II
---------------1
x
Fig. 4.70Forceon a current in a magnetic field.
volume, then total number of electrons in the con-
ductor is
N=nxvolume=nAI
Totalforce onthe conductor is
~ ~ ~~
F=Nf=nAI[-e(vdx B)]
~ ~
=enA[-lvd xB ]
PHYSICS-XII
IfIIrepresents a current element vector in the
~ ~
direction of current, then vectorsIandvdwillhave
opposite directions and we can take
~ ~
-Ivd=vdl
;::t ~ ~
t:=enAvd (IxB)
ButenAvd=current,I
;::t ~ ~
Hence t:=I(IxB)
Magnitude of force.Themagnitude of the force on
the current carrying conductor is given by
F=llBsine
whereeis the angle between the direction of the
magnetic field and the direction offlow of current.
Special Cases
(i)Ife=0°or180°, then
F=IlB(O)=O
Thus a current carrying conductor placed parallel to the
direction ofthe magnetic field does not experience any force.
(ii)Ife=90°, then
F=llBsin90°=IlB
or Fmax =llB
Thus acurrent carrying conductor placedperpendicular to
the direction of a magneticfield experiences a maximum force.
Direction of force.The direction of force on a
current carrying conductor placed inaperpendicular
magnetic field is givenbyFleming's left hand rule.
Stretch the thumb and thefirst two fingers of the left handin
mutually perpendicular directions. If theforefinger points in
thedirection of the magnetic field,central finger inthe
direction of current, then the thumb gives the direction of
~
force on the conductor. In Fig.4.70,the field B is along
+Z-direction, the current Ialong+Y-direction and so
~
the force F acts along +X-direction.
Us d
--> -->-->
1. F=I (IxB) 2. F= IIBsine
3. Fmax =IIB
U,...Used
Force F is in newton, current Iin ampere, lengthI
in metre and field B in tesla.~~~2tuzkyjxo}k2ius

370. MAGNET
IC EFFECT OF CURRENT
Example 55.Awire oflength Icarries a currentIalongthe
~ 1\ 1\ 1\
X-axis. A magnetic field B=1b(i+j+k )tesla exists in
space.Findthe magnitude of the magnetic force on the wire.
Solution. Asthe wire carries current Ialong the
~"
X-axis, so I=Ii
Also,
~ 1\ 1\ 1\
B=1b(i+j+k)tesla
Magnetic force on the wire is
----t -) ~ 1\ 1\ 1\ 1\
F=I ( 1xB)=I[lixIb(i+j+k )]
=1bII[ix (i+i+k )]
~ 1\ 1\ 1\ 1\
=BoI1(0+k-j)=(k-j )BoI 1
Magnitude of the magnetic force is
F=~12+(_1)21bII=.fiBolInewton.
Example 56. Thehorizontalcomponent of the earth's
magnetic field at acertain place is 3.0x10-5Tand the
direction of thefieW.isfrom the geographic south tothe
geographic north. Avery longstraight conductoriscarrying
a steady current of1A.Whatisthe force per unit length on
itwhenit isplaced on a horizontal table and the direction of
the currentis(a) east to west, (b) south to north?
[NCERll
Solution.Theforce on a conductorof lengthI
placed in a magnetic field B,and carrying current I,is
F=lIBsine
The force per unit length will be
f=£=IBsine
I
whereeisthe angle that the conductor makes with the
~
directionof B .
(a)When the current flows east to west,e=90°.
f=IBsin 90° =1x3.0x10-5x1
=3.0x10-5Nm-1
According to Fleming's left hand rule, thisforce
actsvertically downwards.
(b)When the current flows fromsouth to north,
e=0°
f=IIsin0°=0
Thus the force per unit length of the conductor is
zero.
Example 57. A current of2A enters at the corner 'a' of a
squareframe of side20cm and leaves at opposite corner' c'. A
magnetic field of B = 0.25 Tactsina direction perpendi-
4.45
cular totheplaneof paper, as shown inFig.4.71.Find the
magnitude and direction of themagnetic forces on thefour
sides oftheframe.
____ .- __Ta ~d
b c
Fig. 4.71
Solution. By symmetry, thecurrent through each
of the four sides will be1A.Also,
1=20 ern =0.20 m, B =0.25 T
Magnitude of force on each side is
F =IlBsin 90°
= 1x0.20x0.25x1 = 0.05 N
ByFleming's left hand rule, forces onabanddewill
betowards left and onadandbedownward.
Example 58. Amagnetic fieldof1.0Tisproduced by an
electromagnetina cylindrical region ofradius 4.0 em,as
shown in Fig. 4.72.Awire, carrying
current of 2.0A,isplaced perpen-
diculartoand intersecting theaxis
ofthe cylindrical region. Find the
magnetic force acting onthe wire.
Solution. Clearly, the mag-
neticfieldacts vertically down-
wards while the current flows
horizontally, so
e=90°.
Length of the wire in thecylindrical region
=2r=2 x4.0em=0.08m
F =IlBsin 90° =2.0x0.08x1.0xl =0.16N
Ifsi
Fig.4.72
Thisforce acts on the wire normally intothe plane
of paper.
Example 59.Astraight wire ofmass 200gandlength
1.5m carriesacurrent of2 A.Itissuspended inmi-air by a
~
uniform horizontal magnetic fieldB.Whatisthemagnitude
ofthemagnetic field ? [NCERT; CBSE F15]
Solution. Suppose that a wire ABcarries a current
of 2 A in the direction asshown in Fig. 4.73. The weight
mgof the wire acts vertically downwards. Therefore,
according to Fleming's left hand rule, the magnetic field
~
B must act perpendicularly into the plane of paperso
~
that the magnetic force F on the wire acts vertically
upwards.···4wx£n~m}r°n4lxv

371. 4.46
x
x
X
A
X
X
x
Fig
.4.73
XF X X
PHYSICS-XII
X X
X
X
B
X
X
~'b"'~~
X
X X XX
X X IX
X X XX
X XX X
xlIlg X XX
For mid-air suspension,
Magnetic force on the wire=Weight of the wire
IZBsin 90° =mg
B=mg
II
Butm=200 g =0.2 kg,g=9.8 ms-2, 1=1.5 m,
1=2A
or
B=0.2x9.8=0.65 T.
1.5x2
Examp e60.Whatisthe force on a wire of length4.0em
placed inside a solenoid near its centre, making an angle of
60°with its axis?The wire carriesacurrent of12Aand the
magnetic field due to thesolenoid has a magnitude of0.25T.
Solution. The force on a conductor of lengthI
placed in a magnetic field B, and carrying current I,is
F=IZBsine
whereeis the angle that the conductor makes with the
-+
direction ofB.
Since the field due to a solenoid near its centre is
along its axis, soe= 60°.
AlsoI= 12 A, I= 4.0 ern = 0.04 m, B = 0.25 T
F= 12x0.04x0.25 sin 60° = 0.10 N.
Examp e61.On a smooth plane inclined at30°with the
horizontal, a thin current-carrying metallic rodisplaced
parallel to the horizontal ground. The plane islocated in a
uniform magnetic field of 0.15T in the vertical direction. For
whatvalueof current can the rod remain stationary?The
mass per unit length of the rodis0.03kgm-l. CERT]
Solution. Suppose a rod PQ is placed horizontally
on an inclined plane as shown.in Fig. 4.74. Various
forces acting on the current carrying rod PQ are
(i) its weightMgacting vertically downwards; and
-+
(ii)horizontal forceBIIdue to the magnetic fieldB.
In order that the rod remains stationary, the compo-
nent of the weight of the rod along the inclined plane
must be balanced by the component of the forceBIZ
along the inclined plane,i.e.,
Mgsine=BIZcose
p
Fig.4.74
Ifmisthe mass per unit length of the rod, then
M=ml
mlgsine=BIZcose
1=mgtane= 0.30x9.8xtan 30°
B 0.15
= 0.30x9.8x0.5774 = 11.32 A
0.15
Example62.Ashortconductor oflength 5.0emisplaced
parallel to a long conductor of length 1.5m near its centre.
The conductors carrycurrents 4.0 Aand3.0 Arespectively
in thesame direction. What is the total force experienced by
the long conductor when they are 3.0emapart? [NCERTI
Solution. As the two conductors have different
lengths, the longer conductor may be considered to be
of infinite length. Therefore, magnetic field produced
byit at a distance of 3 cm (0.03 m)isgiven by
B=!-to 12= 41tx 10-7x3T =2x10-5T
21tr 2 1tx0.03
or
Force on the short conductor due to this magnetic
field will be
F=1111B = 4x5x1O-~x2x10-5N
=4xlO-6N
According to Newton's third law, the longer
conductor will also experience a forceof reaction equal
to 4.0x10-6 N.As the currents are in the same
direction, the force is attractive.
Examp e63.Asolenoid60cm long and of radius 4.0em
has3layers ofwindings of300turns each. A 2.0emlong
wireof mass 2.5gliesinside the solenoid near its centre
normaltoitsaxis;both the wireand the axis of the solenoid
arein the horizontal plane. Thewireis connected through
two leads parallel totheaxisof the solenoid to anexternal
batterywhichsupplies acurrent of6.0Ainthe wire. What
valueofcurrent inthe windings of the solenoid cansupport
the weight of the wire? g=9.8 ms-2. CERTJwww6notesdrive6com

372. -+ -+-
+
F3=IRPxB
F3=n~sin 120° =.J3liB
2
This force acts normally out of the plane ofpaper. Fig.4.76
MAGNETIC EFFECT OF CURRENT
Solution. Let I be the current in the windings of the
solenoid which cansupport the weight of the wire. The
magnetic field inside thesolenoid along its axis will be
B=J.lonI
Total number of turns
Now n=---------
Length of thesolenoid
---
= 300x3 = 1500turns m-1
60x10-2
B = 4n:x10-7x1500xI = 6n: x10-4Itesla
This field actsperpendicular to thecurrent carrying
wire, therefore, the magnetic force on thewirewill be
F=I'IB= 6 x(2x10-2) x6n:x10-4 I newton
The current Iwould support the wire if the above
force equals theweight of the wire,
i.e.,6x2x10-2 X6n:x10-41 =2.5 x10-3x9.8
1=2.5x10-3 x9.8 A = 108.3· A
72x3.14x10-6
or
Example64.Figure4.75showsa triangular loop PQR
carrying currentI.The triangle isequilateral withside equal
to1.If auniform magnetic field Bexists parallel to PQ, then
find the forcesacting on the three wires separately.
R
p
B-.
Fig.4.75
-+
Solution. As BIIPQ so force on wire PQ is
or
-+ -+-+
Fl=I PQxB
Fl= IxPQx BxsinOo=O
Force onwireQR,
-+ -+-+
F2=IQRxB
F2=lIBsin 120° =.J3nB
2
Byright hand rule, this force acts normally into the
plane of paper.
Force on wireRP,
or
or
4.47
~roblems For Practice
1.A current of 1 A flows in a wire of length 0.1 m in a
magnetic field of 0.5 T. Calculate the force acting on
thewirewhen the wire makes an angle of(a)90°(b)
0°, with respect to the magnetic field.
(Ans.0.05N,0)
2.A current of 5.0 A is flowing upward in a long
vertical wire placed in a uniform horizontal
northward magnetic field of 0.02 T. How much
force and in what directionwill the field exert on
0.06 m length of the wire?
(Ans.6x10-3N,towards west)
3.What is the magnitude of force ona wireof length
0.04 m placed inside a solenoid near its centre,
making an angle of 30° with its axis ? The wire
carries a current of 12 A and the magnetic field due
to the solenoid is of magnitude 0.25 T.
[CBSE OD 90 C)
(Ans.0.06N)
4.A longstraight conductorPcarrying a current of
2 A is placed parallel to ashort conductor Qof
length 0.05 m carrying a current of 3 A. Thetwo
conductors are 0.10 m apart.
Calculate
(a)the magnetic field due to PatQ
(b)the approximate force on Q.
(Ans.4x10-6T,6x10-7 N)
5.A straight wire1 m long carries a current of10 A at
right angles to a uniform magnetic field of 1 Wb m-~.
Find the mechanical force on the wire and the
power required to move it at 15 ms-1 in a plane at
right angles to the field. (Ans.10 N, 150W)
6.A wireABmaking an angle of 30° with a horizontal
issupported by a magnetic field of 0.65 T, directed
normally into the plane of paper. If the wire carries
a current of 5 A, determine its mass per unit length.
(Ans.0.2872 kg m-I)
A
B···4wx£n~m}r°n4lxv

373. 4.48
7.A
horizontal wire 0.1 m long carries a current of
5 A. Find the magnitude and direction of'the
magnetic field which can support the weight of
wire assuming that its mass is 3x10-3kg m-1.
(Ans. 5.88x10-31)
8.A conductor of length 10 em is placed per-
pendicular to a uniform magnetic field of strength
100 oersted. If a charge of 5 C passes through it in
5 s, find the force experienced by the conductor.
(Ans. 10-3N)
9.A conductor of length 20 cm is placed(i)parallel
(ii)perpendicular(iii)inclined at an angle 30°, to a
uniform magnetic field of 2 T. If a charge of 10 C
passes through it in 5 s, find the force experienced by
the conductor. [Ans. (i)zero(ii)0.8N(iii)0.4N]
10.A current of 5.0 A exists in the loop shown in
Fig.4.77. The wireABhas a length of 50emand lies
in a magnetic field of 0.20 T. What is the magnetic
force acting on the wire ?
(Ans. 0.50 N, towards the inside of the loop)
XOX x
x x x xX
A B
X X xl X X
Fig.4.77
11.A horizontal wire 0.1 m long having mass 3 g
carries a current of 5 A. Find the magnitude of the
magnetic field which must act at 30° to the length of
the wire in order to support its weight?
(Ans. 0.1176 T)
12.Find the magnitude of the magnetic force on the
segment PQ placed in a magnetic field of 0.25 T, if a
current of 5 A flows through it, as shown in
Fig. 4.78. (Ans. 0.32N)
p
B----i ••.
•
Q
Fig.4.78
HINTS
3.F=IlBsina=12x0.04 x 0.25 sin 30° =0.06N.
4.(a)Magnetic field due to P at Q is
11 0I 4nx10-7 x2 6T
B=--= =4x10- .
2na 2nxO.10
PHYSICS-XII
(b)Force onQ,F=IlBsina
=3x0.05x4x10-6xsin 90°
=6.0xl0-7T.
5.F=IlBsin 90° =10x1x1x1=10 N
P=Fv=10 x 15=150 W.
6.Force on wireAB,F=IlBsin 90° =IlB
Component of the force in the vertically upward
direction=Fcos 30° =IlB.J3
2
Ifmis the mass per unit length of wire, then its
weight=mlg
J3
mlg=llB.-
2
IB.J35x0.65xJ3
m=--=-----
2g 2x9.8
=0.2872 kg m-1 .
7.In equilibrium,
Magnetic force on wire=Weight of wire
IlBsin 90° =mg
mg3x10-3x9.8 3
B=-.-= =5.88xl0- T.
II 5
or
or
or
8.F=IlBsina=!1..IBsin 90°
t
5 x 0.10x100x10-4xl -3
-------~=10 N.
5
9.Proceed as in Proem 8 above.
10.F=lIBsina=5.0x0.50x0.20xsin 90° =0.50N.
11.F=IlBsina=mg
. B _mg _ 3x10-3x9.8_ T
..---- -0.1176 .
Ilsina0.lx5xsin30°
12.F=IlBsina=5x0.30x0.25 sin (180° -120°)
=5 x 0.30x0.25xsin 60°=0.32N.
4.18 FORCES BETWEEN TWO PARALLEL
CURRENT-CARRYING CONDUCTORS
20. How will you show experimentally the existence
of(i)attractive forces between parallel currents and
(ii) repulsive forces between anti-parallel currents ?
Forces between two parallel current-carrying
conductors. Itwas first observed byAmpere in 1820
that two parallel straight conductors carrying currents in
the same direction attract each other and those carrying
currents in the opposite directions repel each other.
Experiment1.As shown in Fig. 4.79, the upper
ends of two wires are connected to the -ve terminal of
a battery and their lower ends are connected to the +ve
terminal of the battery through a mercury bath. Whenwww6notesdrive6com

374. MAGNETIC EFFECT OF CURRENT
the circuit is completed, the current flows in the two
wires in the same direction. The two wires are found to
be closer to each other, indicating a force of attraction
between them. "
Fig. 4.79Attractive force between parallel currents.
Experiment 2.As shown in Fig. 4.80, two wires are
connected to a batterythrough a mercury
bath in such
a way that current flows in them in succession. When
the circuit is closed, the currents in the two wires flow
in opposite directions. The two wires move away from
each other, indicating a force of repulsion between
them.
Fig. 4.80Repulsive force between
anti parallel currents.
21. Derive an expression for the force per unit length
between two infinitely long straight parallel current
carrying wires. Hence define one ampere. Also define
coulomb in terms of ampere.
Expression for the force between two parallel
current-carrying wires.As shown in Fig.4.81(a), con-
sider two long parallel wiresABand CD carrying
currentsIIand12,Letrbe the separation between
them.
The magnetic field produced by currentIIat any
point' on wire CD is
_110II
B---
I21tr
B
4.49
B 0
I
I
I
I
I...•
II 12
Bl
o
,.£ - - - ...•
: F2
'- - - - - - - -y"
,
,
,--- - - - - -y''
II
...•
12II 12 B2
.£- - -- --~-
,...• ...•
r, r,
,
------
--y' "'II' - ___
...•
I
I
B2
I I
I I I I
:+--r-+: :+--r--+:
I I I I
A C A C
Fig. 4.81(a)Parallel currents attract,
(b)Antiparallel currents repeL
This field acts perpendicular to the wire CD and
points into the plane of paper. It exerts a force on
current carrying wireCD.The force acting on lengthI
of the wire CD will be
IlIIlII
F. = IIRsin 90°= I I.~ = ~. I
2 2 '1 2 21tr 21tr
Force per unit length,
f=F2=1101112
I 21tr
According to Fleming's left hand rule, this force
acts at right angles toCD,towardsABin the plane of
the paper. Similarly, an equal force is exerted on the
wireABby the field of wireCD.Thus when the
currents in the two wires are in the same direction, the
forces between them are attractive. It can be easily seen
that
As shown in Fig.4.81(b),when the currents in the
two parallel wires flow in opposite directions (anti-
parallel), the forces between the two wires are repulsive.
Thus,
Parallel currents attract and antiparallel currents repel.
Definition of ampere.
WhenII=12= 1 A andr= 1 m, we get
f=110=2x10-7Nm-1
21t···4wx£n~m}r°n4lxv

375. 4.50
One ampereisthat
value of steady current, which on
flowing ineach of the two parallel infinitely long
conductors of negligible cross-section placed in vacuum at
adistance of 1m from each other, produces between them a
force of2 x10-7newtonper metre of their length.
Definition of coulomb interms of ampere.If a
steady current of 1 ampere is set up in a conductor,
then the quantity of charge thatflows through its
cross-section in 1 second is called one coulomb.
1C=l As
E I b d•
Forces betwen Parallel Current-Carrying
Wires
Formulae Used
~ II
1. Force per unit length, f=~
21tr
2.Force onlength Iof one ofthe wires,
F=~oII121
21tr
Units Used
Force is innewton, currents in ampere, distance r
in metre and field Bin tesla.
Constant Used
~a= 41tx10-7TmA -1.
Example 65. A current of 5.0 A flows through each of two
parallel long wires. The wires are 2.5em apart. Calculate the
forceacting per unit length of each wire. Use thestandard
value of constant requirede of the
force,ifboth currents flow inthesame direction?
[Punjab 99]
Solution. HereII=12=5A,
r=2.5ern =2.5x10-2m, ~o=41tx 10-7TmA-I
Force acting perunit length ofeachwire,
f=1101112=41tX10-7x5x5
21tr21tx2.5x10-2
=2x10--4Nm-1
As the currents in both the wires flow in the same
direction, the force will be attractive.
Example 66.A longhorizontal wire Pcarries a current of
50 A. Itisrigidly fixedire Qisplaced
directly above and parallel toP.Theweight of the wireQ is
0.075 Nm-I and it carries, acurrent of 25A.Find theor
position of the wireQfrom the wirePso thatQremains
suspended due to the magnetic repulsion. Also indicate the
direction of current inQwithrespect to P. [Roorkee 96]
PHYSICS-XII
Solution. The magnetic force per unit length on the
wire Q due to the current in wire P is
_110II 12
F--.-
21tr
F
I25A
Q-----< --+ .•-T
P W__ -i~_A_ ~
Fig. 4.82
The currents in PandQmust have opposite
directions, only then Q will experience a repulsive
force which would balance the weight of Q.
F=110II 12=W
21tr
r=110 .II 12 = 2x10-7 x50x25'
21t W 0.075
=3.33x10-3m=3.33mm.
or
Example 67.Acurrent balance (or ampere balance)isa
devicefor measuring currents. The current to be measured is
arranged to go through two long parallel wires of equal
length in opposite directions one of which islinked to the
pivot of the balance. The resulting repulsive force on the wire
is balanced by putting a suitable massin thescalepan
hanging on the other side of thepivot,In one measurement,
themassin thescalepanis 30.0g,thelength ofthe wires is
50.0em each, and the separation between them is10.0 mm.
What is the value of the current being measured ?Take
g=9.80ms-2and assume that the arms of the balance are'
equal. [NCERT]
Solution.m=30.0g=0.03 kg, I=50em=0.50 m,
r=10.0 mm =0.01 m,g=9.8 ms-2
Force per unit length between two parallel conductors,
f=110.1112
21tr
:. Force on a conductor of length I,
F=110 .1112I
21tr
When the pan is balanced,
Weight in scale pan = Balancing force
mg=110 .Ix1.1
21t r
12 = 21tmgr= 21tx0.03x9.8x0.01
110I 41tx10-7x0.50
.i.e.,
= 29400
I=.J29400=171.46 A.°°°3vw}m{lzq«m3kwu

376. MAGNETIC EFFECT
OF CURRENT
Example 68. Arectangular loop of sides 25emand10cm
carrying acurrent of 15Aisplaced with its longer side
parallel to a long straight conductor 2.0 emapartcarrying a
current of 25A.What isthenet force on the loop?
[CBSEOO 05]
Solution. Consider the rectangular loop ABCD
placed near a long straight conductor XY,as shown in
Fig. 4.83. The armABwill get attracted, while CD will
get repelled. Forces on arms BC andAD,being equal,
opposite and collinear, will cancel each other.
0
CT
r2=12cm
AI
1'~15A
18l_yr;-I2 em
X
t
I,= 25 A
Fig. 4.83
Current through the rectangular loop, II=15A
Current through the long wireXY, 12=25A
Force onAB,
1.12II
Fl=-.!L .--.L1. xlength of conductorAB
4n r1
10-7x2x15x25x25x10-2
2.0x10-2
=9.375 x 10-4N (Attractive)
ForceonCD,
1.12II
F2=-.!L .--.L1. xlength of conductor CD
4n r2
10-7x2x15x25x25x10-2
12.0x10-2
= 1.5625 x10-4 (Repulsive)
Net force on the loop,
F = Fl - F2 =9.375 x10-4-1.5625x10-4
=7.8125xl0-4 N"'" 7.8x10-4 N (Attractive)
Thusthe force on the loopwill act towards the long
conductor(attractive) if the current in its closer side is
in the same direction as the current in the long
conductor, otherwise it will be repulsive.
Example 69. In Fig. 4.84,thewires AB,CDandEFare
long and have identical resistances. The separation between
the neighbouring wiresis1.0em. The wires AE and BFhave
negligible resistance and the ammeter reads 15A.Calculate
themagnetic force per unit length of AB and CD.
Solution. By symmetry, current through each of
the wires AB,CD and EF is 5 A.
4.51
Fig.4.84
Force per unit length of ABdue to current in CD is
fl=1.10II 12=4nx10-7x5x5
2nr 2nx 1.0x10-2
= 5.0x10-4 m",directed downward
Force per unit length ofABdue to current in EF is
f=4nx10-7x5x5
22nx2.0x10-2
= 2.5x10-4Nm-1,directed downward
Total force per unit length ofABis
f=fl+f2
= 7.5 x10-4Nm-1,directed downward
Force per unit length of CD due to current inABis
f-4nx10-7 x5x5
1-2nx1.0x10-2
=5.0x10-4Nm-1,directed upward
Force per unit length of CD due to current in EF is
4nx10-7 x5x5
f2=2nx1.0x10-2
=5.0x10-4Nm-1,directed downward
Totalforce per unit length of CD=zero.
cproems For Practice
1.A long horizontal rigidly supported wire carries a
current of 100 A. Directly above it and parallel to it
is a fine wire that carries a current of 200 A and
weighs 0.05 Nm -1.How far above the wire should
the second wire be kept to support it by magnetic
repulsion? (Ans.8 em)
2.A wireABis carrying a steady current of 12 A and
is lying on the tae. Another wire CD carrying 5 A
is held directly aboveABat a height of 1 mm. Find
the mass per unit length of the wire CD so that it
remains suspended at its position whenleftfree.
Give the direction of the current flowing in CDwith
respect to that inAB.
[Take the value ofg=10 ms-2]. [CBSE 0013]
(Ans. 1.2x10-3kg m-1,in the opposite direction)
3.Two very long, straight, parallel wiresAand B
carry currents of 10 A and 20 A respectively, and···4wx£n~m}r°n4lxv

377. 4.52
A
are at
a distance 20 cm apart, as shown in Fig. 4.85.
If a third wire C (length 15 ern) having a current of
10 A is placed between them, how much force will
act on C?The direction of current in all the three
wires is same. (Ans. 3.0x10-5N,towards B)
C B
A B
2.
~ ~
D C
-20cm+-
Fig.4.85 Fig.4.86
4.In Fig. 4.86,ABCDis a rectangular loop made of
uniform wire. The lengthAD=BC=1cm. The
sidesABand DC are much longer thanADor BC
Find the magnetic force per unit length acting on
the wire DC due to the wireABif the ammeter
reads 10 A. (Ans.5x10-4Nm-1,attractive)
5.A rectangular loop of wire of size 2crnx5 em carries
a steady current of 1 A. A straight long wire
carrying 4 A current is
kept near the loop as
shown in Fig. 4.87. If the
loop and the wire are
coplanar, find(i)the
torque acting on the loop
and(ii)the magnitude
and direction of the force
on the loop due to the
current carrying wire.
[CBSE 0 121
I=4A
2cm
§
LO
I+--
1an
Fig.4.87
[Ans.(i)r=0(ii)F=2.671lN,towards
the straight wire1
6.A square loop of side 20 em carrying current of 1 A
is kept near an infinite long straight wire carrying a
current of 2 A in the same plane as shown in
Fig. 4.88. Calculate the magnitude and direction of
the net force exerted on the loop due to the current
carrying conductor. [CBSE 00 15C1
(Ans.5.33xlO-7N)
2A
Fig.4.88
PHYSICS-XII
HINTS
1.Force of repulsion per unit length,
41tx10-7x100x200 1
I= =0.05Nm-
21txr
4x10-3
r= m=8ern.
0.05
lA
Weight per unit length of upper wire
= Magnetic force per unit length
W F=>mg=1101112
[[ [21tr
Mass per unit length
m1101112 41tx10-7 x12x5
=I=21trg= 21txlO3x10
= 1.2x10-3kgm-1
The direction of current in CD must be opposite to
that of current inABso that the force between the
two wires is repulsive.
3.Force on C due toA,
41tx10-7x10x10x0.15
E ------:-----
1 - 21tx0.10
= 3.0x10-5N,towardsA
Force on C due to B,
41tx10-7x20x10x0.15
E--------
2 - 21txO.10
= 6.0x10-5N, towards B
Net force on C
=F2 -1i= 3.0x10-5N, towards B.
f=1l01J2= 41tx10-7 x5x5
4.
21tr 21tx0.01
= 5x10-4 Nm -1,attractive
5.(i)As direction of the magnetic field due to the
straight conductor is parallel to the area vector
(both normal to the plane of the loop), so
torque r=O.
(ii)Proceed as in Example 68 on page 4.51
6.F=1101} 12[
2nr
:.Net force on sidesabandcd
=110/112I [~_~]
21t1.r2
= 41tx10-7 X2X1X20X10-2[ 1 l]N
21t 10x10 230x10 2
=4x10-7 X20[~]N =5.33x10-7N
lOx30
This force is directed towards the infinitely long
straight wire.···4wx£n~m}r°n4lxv

378. MAGNETIC EFFECT OF
CURRENT
4.19TORQUE EXPERIENCED BY A CURRENT
LOOP IN A UNIFORM MAGNETIC FIELD
22. Derive anexpression for the torque acting on a
current carrying loop suspended in a uniform magnetic
field.
Torque on a current loop in a uniform magnetic
field. Asshown in Fig. 4.89(a), consider a rectangular
~
coilPQRSsuspended in a uniform magnetic field B,
with its axis perpendicular to the field.
(a)
--+
F
lup
,e
,
,---------
asine
--+
F
(b)
Fig.4.89(a)A rectangular loopPQRSin a uniform magnetic
fieldif.(b)Top view of the loop. magnetic dipole moment;rt
is shown.
LetI=current flowing through the coilPQRS
a,b=sides of the coilPQRS
A=ab=area of the coil
~
8=angle between the direction of B and
normal to the plane of the coil.
According to Fleming's left hand rule, the
magnetic forces on sidesPSand QR are equal, \..
4.53
opposite and collinear (along the axis of the loop), so
their resultant is zero.
The sidePQexperiences a normal inward force
equal toIbBwhile the side RS experiences an equal
normal outward force. These two forces form a couple
which exerts a torque given by
r=Forcexperpendicular distance
=IbBxasin 8=lBAsine
If the rectangular loop has N turns, the torque
increases N times i.e.,
t=NIBAsine
ButNIA=m,the magnetic moment of the loop, so
't=mBsin8
~
Invector notation, the torquetis given by
~~ ~
't=InxB
~
The direction of the torquetis such that it rotates
the loopclockwise about the axis of suspension.
Special Cases
(i)When 8 =0°, t=0,i.e.,the torque ismtntmum
when the plane of the loop is perpendicular to
the magnetic field.
(ii)Whene=90°,t=NIBA,i.e.,the torque is
maximumwhen the plane of the loop is parallel
to the magnetic field. Thus
'tmax=NIBA
For Your Knowledge
~The expression for torque(t=NIBAsine)holds for a
planar loop of any shape. Thus thetorqueon a planar
current loop depends on current, strength of magnetic
field and area of the loop. It isindependent of the shape
of the loop.
~For a planar current loop of a given perimeter sus-
pended in a magnetic field, the torque is maximum
when the loop is circular in shape. This is because for
a given perimeter, a circle has maximum area.
.-+ -+ -+
~The expresslOn"t=mx B for the torque on a current
loop in a magnetic field is analogous to the expression
:[' =Pextfor the torque on an electric dipole in an
electric field. This supports the fact that acurrent loop
isa magnetic dipole. )···4wx£n~m}r°n4lxv

379. 4.54
~Theto
rque on a currentloop in a magnetic field is the
operating principle of the electric motor andmost
electric meters used for measuring currents and
voltages, called galvanometers.
~Ifthedirection ofthe magnetic field makes an angle u
with the plane of the currentloop,then
I}+u= 90°or I}= 90° -n
r=NlBAsin(90°-u)=NlBAcasn.
~Inauniform magnetic field, the net magnetic force on
a current loop is zero but torque acting on it may be
zero or non-zero.
~Inanon-uniform magnetic field, the net magnetic
force on a current is non-zero buttorque acting on it
maybe zero or non-zero.
Examples based on
~------
Formulae Used
Torque on a current loop in a magnetic field,
,=NlBAsinI}=mBsine
wherem=NIA=magnetic dipolemoment of the
current loop.
-> -> ->
Invector form" =mxB.
Units Used
Current Iis inampere, areaAinm2,field Bin tesla,
torque tin Nm and magnetic moment min Am2.
Example 70. Themaximum torque acting on a coil of
effective area0.04 ~is4x10-8Nmwhen the current in it
is100J.lAFind the magnetic induction inwhich itiskept.
Solution. A=0.04 m2, 'max= 4x10-8Nm,
I=100J.lA=10-4 A,N=1
As 'tmax=NIBA
:.Magnetic induction,
B='tmax
NIA
4x10-8
1x10-4 x0.04
Example 71. Calculate the torque on a 100 turn
rectangular coil of length 40cmand breadth 20em, carrying
acurrent of10A,when placed making an angle of 60°with a
magnetic field of 3T.
PHYSICS-XII
Solution. Here I= 10 A, N =100, B=3T,
A= 40 emx20 em =800 em2 =8x10-2m2
e= 90° -600 =300
~
=Angle between Band the normal to
the plane ofthe coil
:.Torque,
't=NIBAsine
=100x10x3x8x10- 2xsin 300
=120 Nm.
Example 72. Given auniform magnetic fieldof100Gin
aneast to west direction and a44em long wire with a
current carrying capacity of almost10A.Whatisthe shape
and orientation of theloop made of this wire which yields
maximumturning effect on the loop?Whatisthemagni-
tudeof themaximum torque?
Solution. B= 100G= 100x10-4T,I= 10 A
The torque on the planar loop will be maximum if
itsarea is maximum. Since for a given perimeter, a
circle encloses maximum area, therefore, the wire should
be bent into a circular loop of radius rgiven by
21tr= 44
44 44x7
r=-=-- =7 em=0.07 m
21t2x22
:.Area ofthe circular loop,
A= 1t,2=22x(0.07)2 = 0.0154 m2
7
Again, for maximum torque, the loop must be
oriented with its plane in N-S direction.
Then
'tmax =IBA=10x100x10-4x0.0154 Nm
=1.54x10-3Nm.
Example 73. Acircular coil of25 turns and radius 6.0 em,
carrying acurrent of 10A,issuspended vertically in a
uniform magnetic field of magnitude 1.2T.The field lines
run horizontally in the planeof the coil. Calculate the force
and the torque on the coil due to the magnetic field. In which
direction should a balancing torque be applied to prevent the
coilfrom turning?
Solution. Consider any element dtofthe wire.
Force on this element is IdtxB.For each elementdt,
there is another element - dton the current loop.
Forces on each pair of such elements cancel out.
Hence net force on the coil in a uniform magnetic field
is zero.···4wx£n~m}r°n4lxv

380. Thedir
ection ofn;isnormal to areaItof the coil from
S-pole to N-pole. Magnitude of torque onthe coil is
r=mBsin 8
~ ~
For maximum torque, mmust be perpendicular to B .
Therefore,
tmax ';"mB=1.92x0.2 = 0.384 Nm
Thus the torque on the coil is maximum whenever
the X-axis lies in the plane of the coil. Fig.4.92
MAGNETIC EFFECT OF CURRENT
Fig.4.90
In Fig. 4.90,~is a unit vector normal to the plane of
the loop, directed outward. The angle between ~and
~
B is 90°. The magnitude of the torque acting onthe
loop is
t=NIBAsin8
= 25x10x1.2x1t(0.06)2xsin 90°
=3.4 Nm
This torque acts in the vertically upward direction
producing turning effect in the direction of curved
arrow. To prevent the coil from turning, a balancing
torque r'=rmust be applied.
Example74.A rectangular coil ofsides8cm and 6em
having 2000 turns andcarrying acurrent of 200 mA is
placed in a uniform magnetic field of 0.2 Tdirected along the
+veX-axis.
(i) Whatisthe maximum torque thecoil canexperience?
In which orientation does itexperience the maximum
torque?
(ii) For which orientations of the coil isthetorque zero?
When isthis equilibrium stable and whenis it
unstable? [NCERT]
Solution.I=8 em =0.08m,b=6 em =0.06m
N=2000, I=200mA=0.2 A,B=0.2T
Themagnitude of the magnetic dipole moment is
given by
m=NIA= 2000x0.2x(0.08x0.06) = 1.92 Am 2
4.55
~
The torque on the coil is zero when mis parallel or
~
antiparallel to B, i.e.,when it lies in the YZ-plane. The
~
coil will be instae equilibrium whenmis parallel to
~ ~
Band in unstae equilibrium whenmis antiparallel
~
toB.
Example75.A lOa-turns coil kept ina magnetic field
B= 0.05Wbm-2,carries acurrent of 1 A, as shown in
Fig.4.91.Findthetorque acting on the coil. [MNREC 97]
Pits
JGJ[
Q R
1+-15cm -----+I
Fig.4.91
Solution. Here the angle between the axis of
~
rotation of the coil and the magnetic field B is 90°.
N = 100,I= 1 A,
A= 15 emx15 cm =225x10-4m2,
B=O.5 Wbm-2, 8 =90°
Torque,
r=NIBAsin 8
= 100x1x0.5x225x10-4xsin90°
=1.125 Nm
As the force onthe arm PQ actsupwards and that
on SR downwards, so the torque acts anticlockwise.
Example76.A parallelogram-shapedcoil PQRS of sides 0.7
m and 0.5 m carries a current ofl.5 A, as shown in Fig. 4.92.
~
Itisplaced in a magnetic fieldB=40T parallel to PS. Find
(i)forcesonthe sides of the coil and(ii)torque on the coil.
1.5A Q
~-----.--------~
P
/i1
R
10------ 0.7m I~~~2tuzkyjxo}k2ius

381. 4.56
->
Solu
tion. As the magnetic field B acts parallel to
sides PS andQR,no forces act on these sides.
Force on side PQ is
F=nBsine = 1.5 x0.7x40xsin 60°
= 1.5 x0.7x40x0.866 =36.37N
According to Fleming's left hand rule, the force F
will act normally upward.
Similarly, force on side SR will also be 36.37 N, but
it will be directed normally inward.
(ii)As the forces on the sides PQ and SR are equal,
opposite and parallel, they form a couple which exerts
a torque.
1= Forcex.1distancebetween the two forces
= FxPS sin 60° = 36.37x0.5x0.866
=15.75 Nm.
Example 77.A100turnclosely woundcircular coil of
radius 10emcarries acurrent of3.2A.(i)What isthefield
at the center of the coil?(ii) Whatisthemagnetic momen t of
this arrangement?
The coilisplaced in a vertical plane andisfree to rotate
about a horizontal axis which coincides with its diameter. A
uniform magnetic field of2Tinthe horizontal direction or
exists such that initially the axis of the coil isin thedirection
of the field. Thecoil rotates through an angle of90°under
the influence of the magnetic field. (ii)What are the
magnitudes of the torques on the coil in the initial and final
position? ()Whatisthe angular speed acquired by the coil
when it has rotated by900? The M.I.of the coil isO.1 kg ~.
[NCERT]
Solution.(i)Here N = 100, I=3.2 A,
r=10em =0.1 m
Magnetic fieldat the centre of the coil,
B=J.loNI=4nx10-7x100x3.2[.:1tx3.2 =-10]
2r 2xO.1
4x10-5x10 =2x10-3T
2x0.1
The direction ofthe field is given byright hand
thumbrule.
(i)Magnetic moment associated with the coil,
m=NIA=NIxn,z= 100x3.2x3.14x(0.1)2
=10 Am2
->
The direction ofmis given by right hand rule.
(iii)Torque,1=mBsine
Initially, e= 0°
PHYSICS-XII
:.Initial torque,
1=mBsin 0° = 0
Final torque,
1=mBsin 90° = 10 x2 x 1 =20 Nm.
(iv)ByNewton's second law,
1=Ia=IdO)=IdO) .de=I.dO)•0)
dt de dt de
But 1=mBsin e
I.dO) .0)=mBsine or
de
IO)ds»=mBsin ede
When e changes from0 tort /2, suppose the angular
speed changes from0 to0).Integrating above equation
within these limits of e and0),we get
w rr/2
If0)ds»=mBfsin ede
o 0
I[~2I=mB[- cose]~/2
~ I0)2 =-mB[cos ~ -cos 0] =nzB
0)=~2mB= ~2x10x2=20rads-1.
I 0.1
Example78.Asolenoid of length 0.4mand having
500turns of wire carries acurrent of3A.Athin coil having
10turnsofwireand ofradius0.01m carriesa current of0.4A.
Calculate the torque required to hold the coil in the midle of
thesolenoidwithitsaxisperpendicular tothe axis of the
solenoid(J.lo=4nx10-7 V-s /A -m). [Roorkee 90]
Solution. For solenoid, 1=0.4m, Nl =500,II=3 A
Forcoil,N2=10, r=O.Olm, I2=0.4 A
Field inside the solenoid,
J.loNlI . .
B = 1,along the aXISof solenoid.
I
Magnetic moment of coil,
m= N2 I2 A=N2I2n,z,along the axis of coil.
As the axis of thecoil is perpendicular to the axis of
-> ->
solenoid,mand B will be perpendicular to each other.
Required torque,
l=mBsine
=N2I2n,z. J.loNlII.sin 90°
I
= 10x0.4X3nx(0.01)2 x4nx10-7x500x 3 x1
0.4
=6n2x10-7= 6x9.87x10-7= 5.92x10-6Nm.~~~2tuzkyjxo}k2ius

382. MAGNETIC EFFECT OF CURRENT
rprOblems For
Practice
1.What is the maximum torque on a rectangular coil
of area 5 emx12 em of 600 turns, when carrying a
current of 10-5A in a magnetic field of 0.10 T ?
(Ans. 3.6x10-6Nm)
2.What torque acts on a 40 turn coil of 100 em2 area
carrying a current of 10 A held with its axis at right
angles to a uniform magnetic of 0.2 T ?
(Ans. 0.8 Nm)
3.A square shaped plane coil of area 100em2of
200 turns carries a steady current of SA. It is placed
in a uniform magnetic field of 0.2 T acting
perpendicular to the plane of the coil. Calculate the
torque on the coil when its plane makes an angle of
60°with the direction of the field. In which
orientation will the coil be in stae equilibrium?
[CBSE 00 15C]
(Ans. 1 Nm)
4.A rectangular coilPQRSis placed in a uniform
. -->
magnetic fieldB,as shown in Fig. 4.93. Find the
torque on the coil when it carries a current of 2.0 A.
The magnitude of the fieldBis 2.0x10-2T.
(Ans. 4.0x10-3Nm)
r-------40em---II
Fig. 4.93
5.A rectangular coil of 100 turns has length 5 cm and
width 4 cm. It is placed with its plane parallel to a
uniform magnetic field and a current of 2 A is
passed through the coil. If the torque acting on the
coil is 0.2 Nm, find the magnitude of the magnetic
field. (Ans. 0.5 T)
6.A circular coil of radius 2.0 ern has 500 turns and
carries a current of 1.0 A. Its axis makes an angle of
30°with the uniform magnetic field of 0.40 T that
exists in the space. Find the torque acting on the
coil. (Ans. 0.13 m)
7.A circular coil of 200 turns and radius 10 em is
placed in a uniform magnetic field of 0.5 T, normal
to the plane of the coil. If the current in the coil is
3.0 A, calculate the (a)total torque on the coil.
(b)total force on the coil. (c)average force on each
electron in the coil, due to the magnetic field.
4.57
Assume the area of cross-section of the wire to be
10-5m2and the free electron density is 1029/m 3.
[CBSE 0008]
[Ans. (a)zero(b)zero(c)1.5x10-24N]
HINTS
1.A=5cm x12cm =60x10-4m2
'tmax =NIBA
= 600x10-5x0.10x60x10-4
=3.6xlO-6Nm.
2.'t=NIBA sine= 40x10x0.2x100x10-4sin 90°
=0.8 Nm.
3.T=NIBAsine= 200x5xO.2x100x10-4sin (90°-60°)
=lNm.
4.T=IBAsine= 2.0x2.0x10-2x0.40x0.25xsin 90°
=4.0xlO-3 Nm.
5.B=__T__
NIAsine
0.2
100x2x0.05x0.04xsin 90°
=0.5T.
6.T=NIB(nr2)sine
= 500x1.0x0.40x3.14x(0.02)2xsin 30°
= 0.1256""0.13 Nm.
7.Proceed as in Exercise 4.25 on page 4.104.
4.20MOVING COIL GALVANOMETER
23.Describe the principle, construction and working
of a pivote-type moving coil galvanometer. Define its
figure of merit.
Moving coil galvanometer.A galvanometer isa device
to detect current in a circuit. The commonly used moving
coil galvanometer is named so because it uses a
current-carrying coil that rotates (or moves) in a
magnetic field due to the torque acting on it.
In aD'Arsonval galvanometer,the coil is suspended
on a phosphor-bronze wire. It is highly sensitive and
requires careful handling. InWeston galvanometer, the
coil is pivoted between two jewellel bearings. It is rugged
and portae though less sensitive, and is generally
used in laboratories. The basic principle of both types
of galvanometers is same.
Principle. A current carrying coil placed ina magnetic
field experiences a current dependent torque, which tends to
rotate thecoil and produces angular deflection.
Construction. As shown in Fig. 4.94, a Weston
(pivoted-type) galvanometer consists of a rectangular
coil of fine insulated copper wire wound on a light
non-magnetic metallic (aluminium) frame. The two
ends of the axle of this frame are pivoted between two
jewelled bearings. The motion of the coil is controlled
by a pair of hair springs of phosphor-bronze. The inner···4wx£n~m}r°n4lxv

383. 4.58
ends of the springs are soldered to the two ends of the
coil and the outer ends are connected to the binding
screws. The springs provide the restoring torque and
serve as current leads. A light aluminium pointer
attached to the coil measures its deflection on a
suitae scale.
Scale
Pointerc--s Permanent
magnet
1
N
Uniform r
adial
magnetic field
(a)
Pointer
TerminalsConcave~~~IIIIr-"1111
pole
Soft-iron - /
cylinder /<!:
Hair
spring
(b)
Fig. 4.94(a)Top view(b)Front view of a
pivoted-type galvanometer.
The coil is symmetrically placed between the
cylindrical pole pieces of a strong permanent horse-
shoe magnet.
A cylindrical soft iron core is mounted symme-
trically between the concave poles of the horse-shoe
magnet. This makes the lines of force pointing along
the radii of a circle. Such a field is called aradial field
The plane of a coil rotating in such a field remains
parallel to the field in all positions, as shown in
Fig.4.94(a).Also, thesoftiron cylinder, due to its high
permeability, intensifies the magnetic field and hence
increases the sensitivity of the galvanometer.
PHYSICS-XII
Theory and working.In Fig.4.95(a), we have
I=current flowing through the coilPQRS
a,b=sides of the rectangular coilPQRS
A=ab=area of the coil
N=number of turns in the coil.
->
F
->
B
->
F
(b)
Fig. 4.95(a)Rectangular loopPQRSin a uniform
magnetic field.(b)Top view of the loop.
Since the field is radial, the plane of the coil always
---+
remains parallel to the fieldB .The magnetic forces on
sidesPQandSRare equal, opposite and collinear, so
their resultant is zero. According to Fleming's left rule,
the sidePSexperiences a normal inward force equal to
NIbBwhile the sideQRexperiences an equal normal
outward force. The two forces onsidesPSandQRare
equal and opposite. They form a couple and exert a
torque given by
t=ForcexPerpendicular distance
=NIbBx asin90°=NIB (ab)=NIBA
Heree=90°,because the normal to the plane of coil
->
remains perpendicular to the fieldBin all positions.
The torquetdeflects the coil through an anglea.A
restoring torque is set up in the coil due to the elasticity
of the springs such that
trestoring a:a r restoring =k«
or
wherekis thetorsion constantof the springsi.e.,
torque required to produce unit angular twist. In
equilibrium position,
Restoring torque=Deflecting torque
ka=NIBA
a=NBA. I
k
or
or aocI···4wx£n~m}r°n4lxv

384. MAGNETI
C EFFECT OF CURRENT
Thus the deflection produced in the galvanometer
coil is proportional to the current flowing through it.
Consequently, the instrument can be providedwith a scale
with equal divisions along a circular scale to indicate
equal steps incurrent. Suchascaleis calledlinear scale.
k
Also, I=--. a=Ga
NBA
The factor G=k / NBA is constant for a galvano-
meter and is calledgalvanometer constant orcurrent
reduction factorof the galvanometer.
Figure of merit of a galvanometer.Itisdefined as the
current which produces adeflection of one scale division in
the galvanometer and isgiven by
G=i=_k_
aNBA
4.21SENSITIVITY OF A GALVANOMETER
24. When isagalvanometer said to be sensitive ?
Define current sensitivity and voltage sensitivity of a
galvanometer. State thefactors on which the sensitivity of
amoving coil galvanometer depends. How can we
increasethe sensitivity of a galvanometer?
Sensitivity ofagalvanometer. A galvanometer issaid
tobe sensitive ifit shows large scale deflection even when a
smallcurrent ispassedthrough itora smallvoltage is
appliedacrossit.
Current sensitivity.Itisdefinedasthedeflectionproduced
in the galvanometer when a unitcurrent flows through it.
I_!!:._ NBA
Current sensitivity, s - -
I k
Voltage sensitivity. Itisdefinedasthedeflection produced
inthe galvanometer when a unit potential difference is
applied across its ends.
a
Voltage sensitivity, Vs=-
V
aNBA
-=--
lR kR
Clearly,
I ... Currentsensitivity
votagesensitivity =----R---"--
Factors on which the sensitivity of a moving coil
galvanometer depends:
1.Numberofturns Nin its coil.
2.Magnetic fieldR
3. Area Aofthecoil.
4.Torsion constant kof the spring andsuspension
wire.
Factors by which the sensitivity of a moving coil
galvanometer can be increased :
1. By increasing thenumber of turns Nofthecoil.But
the value of Ncannot be increased beyond a
certain limit because that will make the
galvanometer bulky and increase its resistance R.
4.59
2.By increasing the magnetic field B.This can be
done by using a strong horse-shoe magnet and
placing a soft iron core within the coil.
3. By increasing thearea A of the coil.However,
increasingAbeyond a certain limit will make
the galvanometer bulky and unmanageae.
4.By decreasing the value of torsion constant k.The
torsion constant kis made small by using sus-
pension wire andsprings of phosphor bronze.
25. Give theadvantages and disadvantages of using a
moving coil galvanometer.
Advantages of a moving coil galvanometer:
1.Asthe deflection of the coil is proportional to
the current passed through it, so a linear scale
can be used to measure the deflection.
2.A moving coil galvanometer can be made highly
sensitive byincreasingN, B, Aand decreasingk.
3.As the coil is placed in a strong magnetic field of
a powerful magnet, its deflection is not affected
byexternal magnetic fields. This enaes us to
use the galvanometer in any position.
4.Asthe coil is wound over a metallic frame, the
eddy currents produced in the frame bring the
coil to rest quickly.
Disadvantages of a moving coil galvanometer:
1.The main disadvantage is that its sensitiveness
cannot be changed at will.
2.All types of moving coil galvanometers are
easily damaged by overloading. A current greater
than that which the instrument is intended to
measure will bum out its hair-springs or
suspension.
For YourKnowledqe
~If the radial field were not present in a moving coil
galvanometer, for example, if the soft iron cylinder
were removed, then the torque would be NBAlsin9
andIwould be proportionala/sin9.The scale would
then benon-linear and difficult to calibrate or to read
accurately.
~Phosphor-bronze is usedfor suspension or hair
springs because of several reasons:
1.It is a good conductor of electricity.
2.It does not oxidise.
3. It is perfectly elastic.
4.It has very Iittle elastic after effect.
5.It is non-magnetic.
6.Of all materials, it has the minimumvalue for
rest?ring torque per unit twisti.e.,smallest)
torsion constantk. "···4wx£n~m}r°n4lxv

385. 4.60
Formulae Used
1. Inamoving
coil galvanometer,
k
Current, I=-- .U
NBA
NBA
Deflection produced, U=-k- .I
I k
2.Figure of merit, G=-;;: =NBA
... a NBA
3.Current sensitivity, Is=I=-k-
U U NBA
4.Voltage sensitivity, Vs=V=IR=kR
Units Used
Current Iisin ampere, area Ainm2,field Bin
tesla, angle aindegrees, torque rinNrn,
resistance Rinohm,potential difference Vinvolt,
torsion constant kinNm deg-I.
Example79. Arectangular coilof area
5.0x10-4nr-and60turnsispivoted about one of its
verticalsides. The coilis ina radial horizontal field of 90 G
('radial' heremeans thefield lines are intheplaneof thecoil
for any orientation). Whatisthe torsional constant of the
hair-springs connected to the coilifa current of 0.20 mA
produces an angular deflection of 18°? [NCERT]
Solution. B= 90G= 90x10--4T,
A= 5.0x1O-4m2, 1=0.20mA=0.20x 10-3A,
N=60,u=18°
Torsional constant of the hair spring is given by
k=NlBA
U
60x0.2x10-3x90x10--4x5x10--4Nd-1
m eg
18
=3.0x10-9Nm deg-1.
Example80.A rectangularcoil havingeach turn of length
5 em and breadth2emissuspended freely ina radial
magnetic field of induction 2.5x10-2Wbm-2,torsional
constant ofthe suspensionfibreis1.5x10-8Nm rar1. The
coil deflects through an angle of 0.2 radian when a current of
2 ~Aispassed through it.Find the number of turns of the
coil.
Solution.k=5emx2em=10x10-4m2 =10-3m2
B=2.5x 10-2Wbm-2, k=1.5x1O-8Nm rad-1
9=0.2 rad, I=2 ~A= 2x10-6A
PHYSICS-XII
As
k
I=--.u
NBA
k
N=--.u
IBA
1.5x10-8x0.2 =60.
2x10-6x2.5x102x103
Example81.Thecoilof amoving coil galvanometer has an
effective area of5x10-2nr-.Itissuspended ina magnetic
field of2x10-2Wbm:2.If the torsional constant of the
suspension fibre is4x10-9Nm deg-\finditscurrent
sensitivityindegree per-microampere.
Solution.HereN =1, A= 5x10-2m2,
B=2x1O-2Wb m-2,k= 4x10-9Nm deg-1
Current sensitivity
NBA 1x2x10-2x5x10-2
=--=--------,,---
k 4x10-9
= 0.25x106deg A-I
=0.25 deg~A-I [.:1 A =106 ~Al
Example 82.A current of 200~A deflects the coil of a
moving coil galvanometer through 30°.What should be the
currenttocausethe rotation through 1t/10radian? What is
the sensitivity of the galvanometer?
Solution. Here II=200~A,91=30°,
9 =...::.. rad= 18° I =?
210 '2
I__k_u and 12=_k_.U2
1 -NBA' 1 NBA
or
12=u2
IIu1
u 18
12=~ .II=- x200 =120~A
u1 30
Current sensitivity
_ u2 _18deg_015 d A-I
- - -. eg ~ .
12120~A
Example83.Agalvanometer needs50mV for afullscale
deflection of50divisions. Find its voltage sensitivity. What
must beitsresistance if its current sensitivity is1division/
~A?
Solution.Voltage sensitivity,
u50divisions
Vs=V=50mV
=50divisions =103div V-I
50x10-3 Vwww4}~trsq»vvr4p~{

386. MAGNETIC EFFECT
OFCURRENT
Resistance of galvanometer,
R= ~1div I-lA -1
gVs103divV-I
106divA-l
3 1 =1000n.
10divV-
Example84.A moving coil meter has the following
particulars: Number of turns,N=24;Area of the coil,
A=20xlO-3n?; Magnetic field strength, B=0.20T;
Resistance of the coil,R=14n.
(i)Indicate a simple wayto increase the current
sensitivity of the meter by25%.(Itisnoteasy to
change A or B).
(ii)If in doing so, the resistanceofthe coil increases to
21n,isthe voltage sensitivity of the modified meter
greater or less than the original meter? [NCERT)
Solution.(i)Current sensitivity,
Is=~ =NBA
I k
Since itis easier to change Nthank,AorB,so the
currentsensitivity can be increased by increasing N.To
increase it by 25%,.Nshould be increased from24to30.or
(ii)Voltage sensitivity,
Vs=~=~= NBA
VRI kR
Ask, A, Bare same in the two cases, we need to
compare only NIRratio.
N= 24=1.7
R14
N=30=1.4
R21
Thusthe modified meter has lessvoltage sensi-
tivity than the original meter. By increasing the
number ofturns, ithasgained incurrent sensitivity but
lostin voltage sensitivity.
Example85.To increase the current sensitivity of a
moving coil galvanometer by50%,its resistance isincreased
sothatthenew resistance becomes twice itsinitial resistance.
By what factor does its voltage sensitivity change?
[CBSE 0001)
Fororiginal meter,
For modified meter,
Solution.Current sensitivity,
a
Is=i
Voltage sensitivity,
V=a=~=~
SVIRR
Newcurrentsensitivity,
I~=Is+ ~Is=~[s
100 2
4.61
New voltage sensitivity,
3
T.-Is3.
V~=_s_=_2_ =-Vs=0.75Vs
2R 2R 4
Thus new voltage sensitivity becomes75%of its
initial valuei.e.,it decreases by25%.
Example86.Thecoil of agalvanometer is0.02mx0.08m
It consists of200turnsoffine wire andisina magnetic field
of0.2tesla.The restoring torque constant of the suspension
fibreis10-6Nmdeg-1.Assuming the magnetic field to be
radial, (a) what isthemaximumcurrent thatcan be measured
by this galvanometer, ifthe scale can accommodate 30°
deflection?(b)Whatisthe smallest current that can be
detected, iftheminimumobservable deflectionis0.1degree?
[CBSE 00 13C)
Solution.HereA=0.02x0.08m2= 1.6x10-3m2
N =200,B=0.2 T,k=10-6 Nmdeg "
(a)The maximum current (Imax) that can be
measured is given by
NBAImax =kamax
[ =kamax= 10-6 x30 A
max NBA 200x0.2x1.6x10-3
=4.69x10-4A.
(b)The smallest current(Imin )that can be detected
is given by
I.=kamin
mm NBA
10-6 x0.1 A=1.56x10-6A
200x0.2x1.6x10-3
~roblems For Practice
1.A rectangular coil of area 100cm2 and consisting of
100turns is suspended in a magnetic field of
5x10-2T. What current should be made to pass
through it in order to keep equilibrium at an angle
of45°with the field? Given that torsion constant of
the suspension fibre is 10-8Nm deg ".
(Ans. 9xlO~ A)
2.The coil of a galvanometer consists of250turns of
fine wire wound on a2.0emx1.0em rectangular
frame. It is suspended in a uniform radial magnetic
field of strength2,000G. A current of10-4A
produces an angular deflection of30°in the coil.
Find the torsional constant of its suspension.
(Ans.1.9x10~Nm rad-1 )
3.A moving coil galvanometer is placed in a radial
magnetic field of0.2T. The galvanometer coil has
200turns and area of1.6x10-4m2. The torsion
constant of the suspension fibre is 10-6 Nm deg".···4wx£n~m}r°n4lxv

387. 4.62
Determine the maximum current that can be
measured by this galvanometer if its scale can
accommodate a deflection of 45°. (
Ans.7x10-3A )
4.The coil of moving coil galvanometer is 40 mm long
and 25 mm wide. It has 100 turns and is suspended
in a radial magnetic field of 10-2 T.Ifthe sus-
pension fibre has a torsional constant of
10-8Nm deg-1,find the current sensitivity of the
moving coil galvanometer. (Ans. 0.1 deg ~A-1)
5.A coil of a moving coil galvanometer twists
through 45° when a current of 1 micro-ampere is
passed through it. Ifthe area of the coil is 10-5 m2
and it has 1000 turns, find the magnetic field of the
magnet of the galvanometer. The restoring torque
per unit twist of the galvanometer coil is
1O-4Nm deg ". (Ans.45 T)
6.The coil of a pivoted-type galvanometer has 50 turns
and encloses an area of 6 m2.The magnetic field in
the region in which the coil swings is 0.01 T and is
radial. The torsional constant of the hair spring is
1.0 x 10-8Nm deg ", Find the angular deflection of
the coil for a current of 1 mA. (Ans.300)
7.A rectangular coil of area 8x10-4m 2is suspended
freely in a radial magnetic field of induction
2x10-2Wb m-2.When a current of 5 ~A is passed
through the coil, it deflects through 60°. The torsional
constant of the suspension is 3.821 x 10-9 Nm rad-I.
Find the number of turns of the coil. (Ans.50 turns)
8.A galvanometer needs 25 mV for a full scale def-
lection of 50 divisions. Find its voltage sensitivity.
What must be its resistance if its current sensitivity
is 1 div/20 ~A ? (Ans.2 x103div V-I, 25 Q)
9.Ifthe current sensitivity of a moving coil galvano-
meter is increased by 20%, its resistance also increases
by 1.5 times. How will the voltage sensitivity of the
galvanometer be affected? [CBSE00 99]
(Ans.Decreases by 20%)
10.Compare the current sensitivity and voltage
sensitivity of the following moving coil galvano-
meters: MI and Mz :
N;=30, ~=1.5x10-3m2,
~ = 0.25 T, R = 20Q.
N2= 35, Az = 2.0x10-3m 2
~ = 0.25 T, R = 30Q.
You are given that the springs in the two meters have
the same torsional constants. (Ans.9: 14, 27 : 28)
HINTS
I=_k_ u =-10-8x45
1. NBA.100x5 x 10-2 x 100x10-4
= 9x10-6 A.
PHYSICS-XII
2.c.= 30° = ~ rad = 0.523 rad
NIBA250x10-4x0.2x2x10-4
k=-- =---------
u 0.523
=1.9x10-6 Nm rad-I.
I =kumax
3.max NBA
10-6 x45 3
-------.=7x10- A.
200x0.2x1.6x10-4
C ... NBA
4.urrent sensitivity =-k-'
k k
5.AsI=-- . U,soB=-- .U
NBA NIA
NBA
6.UseU=--.1.
k
7.UseN=_k_.u.
IBA
u50 divisions 50 divisions
8.Vs=V= 25 m V = 25 x10-3V
=2x103 div V-I
R=~ = 1div / 20~A
Vs2x103div V1
~x 106div A-I
- 20 -2S!1.
- 2 x 103 div V1 -
9.Current sensitivity,Is=!"!..
I
Voltage sensitivity,
V _u _ u _ Is
s-V-IR-"R
New current sensitivity,
Is=Is+ ~Is= ~Is
100 5
New voltage sensitivity,
6
I:-Is 4
Vs=_5_ =_5_ =-Vs=0.8Vs
1.5R1.5R5
Thus the new voltage sensitivity becomes 80% of its
initial valuei.e.,it decreases by 20%.
10.Proceed as in Exercise 4.10 on page 4.99.
4.22MEASUREMENT OF CURRENT AND
VOLTAGE
Introduction. A galvanometer is a basic instru-
ment for electrical measurements. It is a sensitive
current detector. It produces a deflection proportional
to the current flowing through it. It can be easily
converted into an ammeter for measuring current and
into voltmeter for measuring voltage.~~~2tuzkyjxo}k2ius

388. 26. Explain how can we convert a galvanometer into
an ammeter of given range.
Conversion of a galvanometer into an ammeter.An
ammeterisa device used to measure current through a
circuitelement. To measure current
through a circuit
element, an ammeter is connected inserieswith that
element so that the current which is to be measured
actually passes through it. In order to ensure that its
insertion in the circuit does not change the current, an
ammeter should have zero resistance: So ammeter is
designed to have very small effective resistance. In
fact,an ideal ammeter should have zero resistance.
An ordinary galvanometer is a sensitive instru-
ment. It gives full scale deflection with a small current
of few microamperes. To measure large currents with
it, a small resistance is connected in parallel with the
galvanometer coil. The resistance connected in this
way is called ashunt.Only a small part of the total
current passes through the galvanometer and remai-
ning current passes through the shunt. The value of(
shunt resistance depends on the range of the current
~Since an ammeter is a parallel combination of the
required to be measured. galvanometer and the shunt resistance, so its
resistance is even less than that of the shunt
resistance. Moreover, R A< <G.
~Because of its very small resistance, an ammeter
placed in a series circuit does not practically change
the current in the circuit to be measured.
MAGNETIC EFFECT OF CURRENT
Following essential requirements should be met
while converting a galvanometer into ammeter or
voltmeter:
1.Ammeter or voltmeter should be accurate,
reliae and sensitive.
2. The use of these devices in a circuit mustnotor
alterthe current in the circuit or the potential
difference across any element in the circuit.
4.23CONVERSION OF A GALVANOMETER
INTO AN AMMETER
Ammeter
r------------------~
: 5 :
I
I
I...... I
I
I
I
I
I (1- Ig) (I - Ig)
I
I
I
I
I
I
I
I
I
G
I
I
I
Fig.4.96
LetG=resistance of the galvanometer
Ig=the current with which galvanometer
. gives full scale deflection
o -I ~ the required current range of the ammeter
5=shunt resistance
I -Ig=current through the shunt.
4.63
As galvanometer and shunt are connected in
parallel, so
P.D. across the galvanometer=P.D. across the shunt
IgG=U-1g)5
I
5=--g-xG
I-I
g
So by connecting a shunt of resistance 5 across the
given galvanometer, we get an ammeter of desired
range. Moreover,
5
I=--x I
gG+5
The deflection in the galvanometer is proportional
to Igand hence to I. So the scale can be graduated to read
the value of current I directly.
Hencean ammeterisa shunted or low resistance
galvanometer.Its effective resistance is
G5
RA=--<5
G+5
27. What is a shunt? Mention its important uses.
Shunt. A shuntisa low resistance whichisconnected in
parallel with a galvanometer (or ammeter) to protect it from
strong currents.
Uses of shunt:
1.To prevent a galvanometer from being damaged
due to large current.
2. To convert a galvanometer into ammeter.
3. To increase the range of an ammeter.
For YourI<no-w-I-e-dg-e- -----1
~The resistance of an ideal ammeter is zero.
~Higher the range of ammeter to be prepared from a
given galvanometer, lower is the value of the shunt
resistance required for the purpose.
~The ammeter of lower range has a higher resistance
than the ammeter of higher range.
~The range of an ammeter can be increased but it
cannot be decreased.
4.24CONVERSION OF A GALVANOMETER
INTO A VOLTMETER
28. Explain how can we convert a galvanometer into a
voltmeter of given range.°°°3vw}m{lzq«m3kwu

389. 4.64
Conversion of
a galvanometer into a voltmeter.A
voltmeterisa device for measuring potential difference
across any two pointsina circuit. It is connected in
parallel with the circuit element across which the
potential difference is intended to be measured. As a
result, a small part of the total current passes through
the voltmeter and so the current through the circuit
element decreases. This decreases the potential
difference required to be measured. To avoid this, the
voltmeter should be designed to have very high
resistance. In fact,an ideal voltmeter should have infinite
resistance.
___C~o~~e~e~ _ ~
R
o---+ ...•-fGl--"'IV'v---t---o
A BI
I
•.._----------
+---v---
Fig. 4.97
A galvanometer can be converted into a voltmeter
by connecting a high resistance in series with it. The
value of this resistance is so adjusted that onlycurrent
Igwhich produces full scale deflection in the galvano-
meter, passes through the galvanometer.
Let
G=resistance of the galvanometer
Ig=the current with which galvanometer
gives full scale deflection
o -V=required range of the voltmeter, and
R=the high series resistance which restricts
the current to safe limitIs:
:.Total resistance in the circuit=R+G
By Ohm's law,
I=Potential difference=~
g Total resistanceR+G
or
V
R+ G=- or
Ig
V
R=--G
Ig
So by connecting a high resistance R in series with
the galvanometer, we get a voltmeter of desired range.
Moreover, the deflection in the galvanometer is
proportional to currentIand hence toV.The scale can
be graduated to read th~ value of potential difference
directly.
Hencea voltmeterisa high resistance galvanometer. Its
effective resistance is
Rv=R+G» G
PHYSICS-XII
For Your Knowledge
~Since a voltmeter is a series combination of a galvano-
meter and a high resistanceR,so its resistance is much
higher than that of the galvanometer.
~An ideal voltmeter should have infinite resistance.
~A voltmeter is placed in parallel with the circuit
element across which thevoltage is to be measured.
Because of its high resistance, it draws a verysmall
current and hence the potential difference across the
element remains practically unaffected.
~Higher the range of voltmeter to be prepared from a
given galvanometer, higher is the value of series high
resistance required for the purpose.
~The voltmeter of lower range has a lower resistance
than the voltmeter of higher range.
~The range of voltmeter can be both increased or
decreased.
/es based on
•
•
•
Formulae Used
1.For conversion of a galvanometer into ammeter,
the shunt resistance,
I
R=--g - xR ;Here
SI - I g
g
RgRs
2.Resistance of an ammeter,RA=--"---
Rg+Rs
R
I= S xI
gRg+l\
3. For conversion of a galvanometer into a
voltmeter, the value of high series resistance,
V V
R=- - Rg; HereI=---
Ig gRg+R
4.Resistance of a voltmeter,Rv=Rg+R
5.For a galvanometer,Ig=nk
wheren=no. of divisions on the galvanometer
scale
k=current required to produce deflection of one
scale division or figure of merit of the galvanometer.
Units Used
All resistances are in ohm(n)and current in
ampere (A).
Example 87. A galvanometer with acoilof resistance
12.0nshows fullscaledeflection for a current 2.5mA.How
will youconvert the meter into:
(i) an ammeter of range 0to7.5A,
(ii)a voltmeter of range0to10.0V?~~~2tuzkyjxo}k2ius

390. MAGNETIC EFFECT OF CURRENT
Determine thenet resistan
ce of the meter ineach case.
When an ammeter is put in a circuit, does it read (slightly)
less or more than the actual current in the original circuit?
When a voltmeter is put across a part of thecircuit, does it
read (slightly) less or more than the original voltage drop?
Explain. [NCERT; CBSE D 05]
Solution. (i)For conversion into ammeter:
Rg= 12n.Ig=2.5mA=0.0025A,I=7.5A
R= ~ x R 0.0025 x12
sI-I g7.5 -0.0025
g
2.5x12x10-3=4.0x10-3n
7.4975
So by connecting a shunt resistance of4.0x10-3n
in parallel with the galvanometer, we get an ammeter
of range 0 to7.5A.
Net resistance RAisgiven by
1 1 1 3001
-=-+ --
RA 124 x10-3 12
R=~n =4x 10-3n
A3001
or
When an ammeter is put in a circuit, it reads
slightly less than the actual current in the original
circuit because a very small resistance is introduced in
the circuit.
(ii)For conversion into voltmeter:
Rg= 12o.Ig=2.5x10-3 A,V=10V
R=~- R 10 -12
Ig g2.5x10-3
= 4000 -12 =3988n
So by connecting a resistance of3988 nin series
with the galvanometer, we get a voltmeter of range 0 to
10V.
Net resistance,Rv= (3988 + 12)n=4000n
Because voltmeter draws small current for its deflec-
tion,so itreads slightly less than the original voltage drop.
Example88.Agalvanometer with ascaledivided into 100
equal divisions hasacurrent sensitivity of 10divisions per
mA and a voltage sensitivity of 2divisions per mV. What
adoptions are required to read (i)5A for fullscale and
(ii)1division per volt ? [lIT]
Solution.Asthe current sensitivity is10divper
mA and there are 100divisions on the scale, so current
required for full scaledeflection is
I= ~x100mA=10mA=10x 10-3A
g10
=0.01 A
4.65
As voltage sensitivity is 2 div per m V, so voltage
required for full scale deflection is
V=.!x100 mV= 50 mV= 50x1O-3V
g2
Galvanometer resistance,
R=Vg50x10-3= 5 n
gI10x10-3
g
(i)Forconversion into an ammeter. 1=5A
R=~x R=~x5=~ o
sI -I g5 - 01 499
g .
So a shunt of5/ 499 nshould be connected across
the galvanometer to read5A for full scale deflection.
(ii)For conversion into voltmeter. For reading1div
per volt, the voltage range should be 100Vbecause
there are100divisions.
V 100
R=- -R=- -5 =10000 -5 =9995n
'« g0.01
So a resistance of9995 nshould be connected in
series with the given galvanometer to read1div pervolt.
Example 89.An ammeter of resistance 0.80 ncan measure
currents upto1.0A(i)What must be the shunt resistance to
enable the ammeter to measure current upto5.0A?(ii)
Whatisthe combined resistance of the ammeter and the
shunt? [NCERT; CBSE D 13]
Solution. The given ammeter can be regarded as
the galvanometer.
Ig=1.0A, Rg=0.80n
(i)Total current in the circuit,I=5.0A
The required shunt resistance,
', 1.0
Rs=-- xRg x0.80 =0.20n.
I - I 5.0 -1.0
g
(ii)The combined resistanceRAofthe ammeter and
the shunt is given by
1 1 1 1 1 1 + 4 25
-=-+-=-+-=--=-
RA Rg Rs0.8 0.2 0.8 4
or RA=4/25=0.16n.
Example90.In the circuit (Fig.4.98),thecurrentisto be
measured. Whatisthevalue of the currentiftheammeter
shown (a)isa galvanometer with a resistance Rg = 60.00 n;
.--_.--(A
3.00n
3.00V
Fig. 4.98···4wx£n~m}r°n4lxv

391. 4.66
(b)isa galvanom
eter described in(a) butconverted to an
ammeter by a shunt resistance Rs=0.02 0;and(c)isan
ideal ammeter with zero resistance ? [NCERT]
Solution.(a)Total resistance in the circuit
=R+ 30 = 60 + 3 = 630
g
3V
Current, 1=-- = 0.048 A.
630
(b)Resistance of the galvanometer converted to an
ammeter is
R=RgRs = 60x0.02==-0.020
ARg+Rs60 + 0.02
Total resistance in the circuit
=RA +30 =0.02 + 3 =3.02 0
3V
1=-- =0.99 A.
3.020
Current,
(c)As the ideal ammeter has zero resistance, so
3.00V
Current, I =-- = 1.00 A.
3.000
Example 91. In agalvanometer there isa deflection of
10 divisions per mA. The internal resistance of the galvano-
meteris60O.If a shunt of2.50isconnected to the galvano-
meter and there are50divisions in all,on the scale ofgalvano-
meter, what maximum currentcanthis galvanometer read?
[CBSE D OIC]
Solution. As the galvanometer has 50 divisions,
current required to produce fullscale deflection is
I =~x50 mA = 5 mA = 0.005 A
g10
Rg=600 and e;=2.50
Let I be the maximum current that the galvano- or
meter can read.
R
I= s xI
gRg+Rs
1=(Rg +Rs)Ig = (60+2.5) 5 =125mA.
« 2.5
Example 92.A galvanometer having a resistance of500
givesafull deflection for a current of 0.05ACalculate the
length of the shunt wire of2mm diameter required to
convert the galvanometer to an ammeter reading current
upto5 ASpecific resistance for the material of thewireis
5x10-70m [Punjab 96]
Solution.Rg=500, .Ig=0.05 A, I=5A
R=~ xR 0.05x50 = 50
sI - I g5 - 0 05 99
g .
Then'
or
PHYSICS-XII
Now R = 50 0, P = 5x10-70 m, r= 1 mm = 10-3 m
99
AsR=p..!.-.,solength of required shunt is
A
RA 50x'Itx(l0-3l
1=-= m
p 99x5x10-7
= 10x3.142x10m=3.17m.
99
Example 93.Amoving coil galvanometer when shunted
with a resistance of 50gives a full scaledeflection for
250 mA and when a resistance of1050 0isconnected in
series,itgivesafullscaledeflection for25volt.Find the
resistance of the galvanometer and the current required to
produce afullscaledeflection when it isused alone.
Solution. With shunt, current required to produce
full scale deflection is given by
I=Rs .I
gR+R
g s
5x250x10-3
I= ...(1)
g R+5
g
When a resistance Ris connected inseries,
or
I=_V_= 25
gRg+RRg+ 1050
From equations (1) and (2),
1.25 25
Rg+ 5Rg+ 1050
1.25Rg+ 1312.5 = 25 Rg+125
23.75 Rg = 1187.5
R= 1187.5 = 50 O.
g23.75
...(2)
or
Example 94. When a galvanometer having 30divisions
scale and100 0resistance isconnectedin series to a battery
ofemf3V through a resistance of 200 0, shows full scale
deflection. Find thefigure of merit of the galvanometer in IlA
Solution. Here n= 30, Rg= 100 0, E.=3V,
R=2000, k=?
The current required to produce fullscale deflec-
tion in the galvanometer is
I=_E._= 3 =_1_A=104 A
gRg+R100 +200 100 Il
AsIg=nk,therefore, the figure of merit is
I 104 A
k=.J..= Il=333.31lA div-t.
n30 divisions~~~2tuzkyjxo}k2ius

392. MAGNETIC EFFECT OF CURRENT
Example95.Th
edeflection produced inagalvanometer is
reduced to 45divisions from55whena shunt of8nisused.
Calculate the resistance of the galvanometer.
Solution.Withoutshunt, Ig=1= 55k
where kisthefigure of merit of the galvanometer.
Withshunt,
I~=(55-45)k=10k
I~=10k=~
Ig55k 11
1'=~I=~I
g11g11
or
When shunt of 8nis used,
r=Rs x I
gR+R
g s
8 2
-- x 1=- Ior 88=2Rg+16
Rg+8 11
R = 72 =36n.
g2
Example96.Agalvanometer of resistance 'G'can be
converted intoavoltmeter of range (0-V)volts by
connecting a resistance 'R'inseries with it. How much
resistance will be required to change its range from 0to
V/2? [CBSE 00 lSC]
or
or
V
Solution.In first case, R=--G
Ig
I=~
gR+G
LetR'be the required resistance
range from 0 toV/2.So insecondcase,
I=V /2
gR'+G
V V /2
-- --
R+G R'+G
2R'+2G=R+G
Hence, R'=R-G.
2
tochange the
or
,Example97.Agalvanometer can be converted into a
voltmeter ofcertain range byconnecting a resistance of
980 n in series with it. When the resistance is 470 n
connected inseries, therange is halved. Find the resistance of
the galvanometer.
Solution.The current for full scale deflection of a
voltmeter is given by
I =_V_
gR+R
g
4.67
I= V
gRg+980
I=V/2
gRg+470
V V
Rg+980 2(Rg+470)
or 2Rg+940 =Rg+980
or Rg=40n.
Example98.A voltmeter reads5.0V at fullscale
deflection and isgraded according to its resistance per volt at
fullscaledeflection as5000 nV-I.How will you convert it
into a voltmeter that reads 20Vat full scale deflection?Will
it still begraded as5000 nv'?Will you prefer this volt-
meter to one thatisgraded as2000 nV-1?
[NCERT;CBSE 0 OlC]
Solution.Resistance per volt is another way of speci-
fying the current at full scale deflection. The grading of
5000 nv- 1at full scale deflection means that the
current required for full scale deflection is
I=_I_A=0.2 mA
g5000
In first case,
In second case,
In order to convert it into a voltmeter of range 0 to
20 V, a resistance R has to be connected in series with
it.Then on applying an extra P.D. of 15 V (20 V-5 V),
the current through it is again 0.2 mA at full scale
deflection.
Rx0.2x10-3= 15
or R = 15 3n=75,o00n
0.2x10-
Thus(i)to convert the givenvoltmeter(0 - 5 V
range) into a voltmeter of range 0 to 20 V, a resistance
of 75,000 n should be connected in series with the
given meter.
(ii)Original resistance of voltmeter
= 5000ov'x5 V = 25,000 n
:.Total resistance after conversion
= 25,000+75,000 = 100,000 n
Resistance per volt of new meter
= 100,000 = 5,000 nv'
20
i.e.,it has the same grading as before.
(iii) The higher the 'resistance per volt' of the meter,
the lesser is the current it draws from the circuit and
the better it is. So this meter is more accurate than the
one graded as 2000 n V-1.
Example 99. Agalvanometer having 30divisions has a
currentsensitivity of 20llA/division. It has a resistance of···4wx£n~m}r°n4lxv

393. 4.68
20 O.How w
illyou convert itinto anammeter measuring
upto1ampere ?Howwill you convert this ammeter into
voltmeter reading upto1volt? [Roorkee 87]
Solution. Here n= 30,Rg= 200
Current sensitivity, k= 20 flA/div
:.Current required for full-scale deflection is
I=nk=30x20 =600 flA =6x1O-4A =0.0006 A
g
(i)Forconversion into ammeter. I =1 A
R=~x R0.0006x20
sI-I g1-00006
g .
=25x6=0.150
9994
i.e.,a shunt of 0.150 should be connected across the
galvanometer.
(ii)For conversion of resulting ammeter into voltmeter.
The resistance of the ammeter formed is
R'=RgRs=20x0.15 =0.0150
gRg+Rs20+0.15
Current for fullscale deflection, I~= 1A
Voltage range, V= 1 V
Required series resistance,
R= ~-R'=! - 0.015 =0.985O.
I' g1
g
Example100.A voltmeter V of resistance400 0 is used to
measure thepotential difference acrossa1000resistorin
the circuit shown in Fig. 4.99.
(i)Whatwillbe the reading on the voltmeter?
(ii)Calculate the potential difference across1000resistor
beforethe voltmeter isconnecte. [CBSE 0 98)
84V
iooo 200n
Fig. 4.99
Solution. (i)Resistance of the parallel combi-
nation of voltmeter V(4000) and 1000 resistance
R' = 400x100 =800
400+.100
Total resistance in the circuit,
R = R'+200 = 80+200 = 280 0
PHYSICS-XII
Current in thecircuit,
I=!=~=~A
R 280 10
Reading on thevoltmeter =P'D.acrossR'
= ~x80 =24 V.
10
(ii) Total resistance before the voltmeter is
connected = 100+200 = 3000
84V 7
Current,I=-- =-A
3000 25
r.o.across 100 0 resistor='!-.x100 = 28O.
25
Example101.Ad.c. supply of 120Visconnected to a
largeresistance X.A voltmeter of resistance 10k0placedin
seriesinthecircuitreads4V. What isthe value of X?What
do you thinkisthepurpose in using avoltmeter, instead of
an ammeter, to determine thelargeresistance X?[NCERT)
Solution. Current through voltmeter,
I=V= 4V =4x10--4A
R10x1030
xn 10 kn
r--"J\I\r---{ V
~
+j-1---...-.(.
120V K
Fig. 4.100
or
I= total e.m.f.
total resistance
4x10-4= 120
X+104
4x10-4X+4 = 120
X= 116 40 = 29x1040 = 290 k0
4x10-
Also
or
As the current in the circuit is very small, the
ammeter's reading will be too small to be measured
accurately. This is an unusual use of voltmeter for
measuring very high resistance.
Example102.(a) A battery of emf9V and negligible
internal resistance isconnected to a 3k0resistor. The
potential drop across a part of the resistor(between points A
andBinFig.4.101) ismeasured by(i) a20k0voltmeter;
(ii) a1k0voltmeter. In(iii)boththe voltmeters are
connected across AB. In which case would you get the
(1)highest, (2)lowest reading?···4wx£n~m}r°n4lxv

394. MAGNET
ICEFFECT OF CURRENT
(b)Doyour answers to this problem alterifthe potential
dropacross the entire resistorismeasured? What ifthebattery
has non-negligible resistance? [NCERT]
BB A
!jV 9V 9V
Fig. 4.101
Solution.(a)The voltmeter, which has maximum
resistance, will draw minimum current and allow
maximum current to flowthrough resistor AB.Conse-
quently, therewill be maximum potential difference
across AB.
Incase(i),resistance of voltmeter,
Rv=20 kO
Rv=1kO
11 1 1+20
-=-+-=--
Rv20 1 20
R=20 kO
v21
Incase (ii),
In case (ii),
or
(1) Astheresistance of voltmeter is maximumin
case (i),itwillshow maximum reading.
(2)Asthe resistance of voltmeter is minimum in
case(iii),it will show lowest reading.
(b)In all cases, the voltmeter reading will besame if
thebattery hasnegligie internal resistance. But ifthe
internal resistance is non-negligie, then the answers
will be similar to those in (a).
Example 103.You are giventworesistors XandYwhose
resistances are to be determined using an ammeter of
resistance0.50 and a voltmeter of resistance 20kO.Itis
knownthatXisin the range of a few ohms, while Yisthe
range of several thousand ohms. In each case, which of the
two connections shown in Fig. 4.102wouldyouchoosefor
resistance measurement ?Justify your answer quanti-
tatively. [NCERT]
L-._-( v}-----'
(a) (b)
Fig.4.102
4.69
Solution. In circuit(a),the voltmeter V will measure
the sum of the potential drops across the resistance and
the ammeter. Thevalue of the resistance determined
from these calculations will include the resistance of
theammeter. This will not be desirae if resistance is
very small. So the circuit(a)is suitae only for
measurement of large resistanceY.
Incircuit(b),the ammeter will read thesum of
currents flowing through the resistance and the
voltmeter V. The value of the resistance oained by
these calculations will be less than the actual value.
The difference will increase with the increase in the
value of the resistance. So the circuit(b)is suitae only
forthemeasurement of the small resistance X.
We can justify the above arguments quantitatively
as follows:
(i)Measurement of X.LetX=5 O.In circuit (a), the
ammeter reading isIIand the voltmeter reading is
II(X+0.5).
Voltmeter reading =II(X+0.5) =II(5+0.5) = 5.5 O.
Ammeter reading II II
L----IV l---...J
20kQ
(a)
V
20kQ
(b)
Fig. 4.103
With circuit(a),the error in the measurement ofXis
0.50.
Incircuit (b),theammeter reading is Iand the
voltmeter reading isXII(=20,000 12),
Clearly,
I -XII_5I
2-20,000-20,000 1
Voltmeter reading=XII=~
Ammeter reading I II+12
511
I+_5_1
120,000 1
= 5x20,000 = 4.99870
20,005
With circuit(b),the error in the measurement ofX
is0.0013 O. This error is much smaller than that
oained by using circuit(a).Hence for measuring a
resistance of few ohms, the circuit(b)should be used.···4wx£n~m}r°n4lxv

395. 4.70
(i)Measurement afY.Let
Y =20,000 O. In circuit(a),
we get
Voltmeter reading = 20,005 11= 20,005 0
Ammeter reading 11
The error in the measurement of Y is 5 O.
In circuit(b),we get
Voltmeter reading = 20,000 12= 20,000 12
Ammeter reading I 11 +12
20,00012
12+12
[.:20,000II=20,00012]
=10,0000
The error in the measurement of Y is 10,0000,
which is much larger than error oained by using
circuit(a).Hence for measuring large resistance of
several thousand ohms, the circuit(a)should be used.
jOroblems For Practice
1.A galvanometer has a resistance of 960 and it is
desired to pass 4% of the total current through it.
Calculate the value of shunt resistance.
(Ans.40)
2.A galvanometer has a resistance of 50O. A resis-
tance of 5 0 is connected across its terminals. What
part of the total current will flow through the
galvanometer? [Haryana01]
(Ans.1/ 11)
3.A galvanometer coil has a resistance of 30 0 and the
meter shows full scale deflection for the current of
2.0 mA. Calculate the value of resistance required
to convert it into an ammeter of range 0 to 1 A. Also
calculate the resistance of the ammeter.
[CBSE Sample Paper 98]
(Ans. 0.060 in parallel, 0.059880)
4.How will you convert 1 mA full scale deflection
meter of resistance 100 ohms into an ammeter to
read 1 A (full scale deflection) and into a voltmeter
to read 1 volt (full scale deflection).
[CBSE D 93C ; Punjab 97]
(Ans. 0.10 in parallel, 9000 in series)
5.A moving coil galvanometer of resistance 100
produces full scale deflection, when a current of
25 mA is passed through it. Describe showing full
calculations, how will you convert the galvano-
meter into(i)a voltmeter reading upto 120V and
(ii)an ammeter reading upto 20 A.
[ISCE 94]
(Ans.47900 in series, 0.01250 in parallel)
PHYSICS-XII
6.A galvanometer of resistance 200 gives a deflection
of one division when a potential difference of 4 mV
is applied across its terminals. Calculate the
resistance of the shunt if the current of 10 A is to be
measured by it. The galvanometer has 25 divisions.
(Ans. 0.010)
7.A galvanometer of resistance 400 gives a deflec-
tion of 5 divisions per mA. There are 50 divisions
on the scale. Calculate the maximum current that
can pass through it when a shunt resistance of 2 0 is
connected. [lIT]
(Ans.210 mA)
8.It is intended to measure a maximum current of 25 A
with an ammeter of range 2.5A and resistance 0.9O.
How will you do it ? What will be the combined
resistance? (Ans. 0.10 in parallel, 0.090)
9.A galvanometer has a resistance of 300 and a
current of 2 mA is needed to give full scale deflec-
tion. What is the resistance neededand how is it to
be connected to convert the galvanometer(i)into an
ammeter of 0.3 A range and(ii)into a voltmeter of
0.2 V range? [Roorkee 92]
(Ans. (i) ~0 in parallel(ii)700 in series)
149
10.A galvanometer has a resistance of 100O. A dif-
ference of potential of 1.0 V between its terminals
gives a full scale deflection. Calculate the shunt
resistance which will enae the instrument to read
upto 2 A. (Ans. 0.5 0)
11.A resistance of 9000 is connected in series with a
galvanometer of resistance 100O. A potential diffe-
rence of 1V produces a deflection of 100 divisions
in the galvanometer. Find the figure of merit of the
galvanometer. (Ans. 10-5A div-1.)
12.A galvanometer has a sensitivity of 60 divisions per
ampere. When a shunt is used, its sensitivitybecomes
10 divisions per ampere. What is the value of the
shunt used if the resistance of the galvanometer is
200? (Ans.40)
13.A galvanometer of resistance 36630 gives full scale
deflection for a certain currentIt:Calculate the
resistance of the shunt which when joined to the
galvanometer coil will result in 1/34 of the total
current passing through the galvanometer. Also
find the total resistance of the galvanometer and the
shunt. (Ans. 1110, 107.70)
14.A shunt of 60 is connected across a galvanometer
of resistance 294O. Find the fraction of the total
current passing through the galvanometer.
(Ans.1/50)°°°3vw}m{lzq«m3kwu

396. MAGNETIC EFFECT OF CURRE
NT
15.A galvanometer has currentsensitivity of
5 divisions/mA and a voltage sensitivity of
2 divisions/m V. If the instrument has 30divisions,
howwillyou use it to measure (i)acurrent of 3A
and(ii)a voltage of 15 V ?
[Ans.(i)0.005nin parallel(ii)2497.5n in series1
16.It is required to pass only one-tenth of the main
current through a galvanometer having a resistance
of 27 n. Calculate the length of the wire ofspecific
resistance 48xlO-6n cm and areaof cross-section
0.2 mm2required to make ashunt for this purpose.
(Ans.1.25m)
17.A galvanometer gives afullscale deflectionwith a
current of 1 A. It isconverted into ammeter of range
10 A.Find the ratio of the resistance of the ammeter
tothe resistance of the shunt used. (Ans.9: 10)
18.A galvanometer has a resistance of 8 n. It gives a
full scale deflection for a current of 10 mA. It isto be
converted into an ammeter of range 5 A. The only
shunt resistance availae is of 0.02 n, which is not
suitae for this conversion. Find thevalue of
resistanceRthat must be connected in series with
the galvanometer (Fig. 4.104) to get ammeter of
desired range. (Ans.1.98n)
5
R R
G}-J\f\>/\r-'--
Fig. 4.104 Fig. 4.105
19.The circuit shown in Fig. 4.105 is used to measure
theresistanceRThe ammeter reads 0.13 A and the
voltmeter reads 117 V. The resistance of the
ammeter is 0.015 n and that of the voltmeter is
9000 n. Find the value of R (Ans 1000 n)
20.The scale of a galvanometer is divided into 150 equal
divisions. The galvanometer has the current sensi-
tivity of 10 divisions per mA and thevoltage sensi-
tivity of 2 divisions per mY. Howthe galvano-
meter can be designed to read(i)6 A division-1and
(ii)1 V per division -1? [Roorkee 99]
[Ans.(i)8.33x10-5n in parallel
(ii)9995oinseries1
21.Two resistance coils of 100n and 200 n respectively
are connected in series across 100 V. A moving coil
voltmeter of 200 n is connected in turn across each
coil.What will it read ineach case ?(Ans. 25 V,50 V)
22.A battery of emf 12 V and internal resistance 1.2n
·supplies a current through a coil of resistance 48 n.
Avoltmeter of resistance 72 n is used to measure
the potential difference acrossthe coil. Whatwould
be the reading on the voltmeter? (Ans. 11.52 V)
4.71
HINTS
IgRg 0.04Ix96
1.1\ = --= = 4n.
I - Rg I -0.04I
IR I
2AsR=~ or...1..
SI-I J
g
3.Shuntresistance,
R=~ x R= 0.002x30 = 0.002x 30=0.06 n.
''s 1-Ig g1-0.002 0.998
Netresistance of theammeter
5
50+ 5
1
11
Rg1\ 30x0.06 = ~ =0.OS988n.
Rg+Rs30+0.06 167
4.(i)Forconversion into ammeter:
Rg= 100n.Ig= 1 mA = 0.001 A, I= 1 A
R=~ x R= 0.001x100 = 0.001x100
SI - Ig g 1-0.001 0.999
=o.i n .
(ii)Forconversion into voltmeter:
Rg= 100n.Ig= 0.001 A, V= 1 V
V 1
R=- -Rg=-- -100=900n.
Ig 0.001
5.Forconversion into voltmeter:
R=~-R= 120 - 10 = 4790n.
I g25x10-3
g
For conversion into ammeter :
IgRg 25x10-3x10
Rs=-- = 3=0.012Sn.
I-Ig20 - 25x10-
Vg4mVx25 3
6.I=-= = 5 mA = 5x10-A
gRg 20n
IgRg 5x10-3 x20
R=-- 3=O.Oln.
SI-Ig 10-5xlO-
50
7.Ig=5"= 10 m A, Rg= 40 n, 1\= 2o
:.Maximum current,
I=Rg+1\xI= (40 + 2) x 10 =210 mA.
R g 2
g
8.Toconvert an ammeter of lower current range to
higher current range, a shunt has to be connected
across it. The shunt resistance is
Ig 2.5x0.9
Rs= -- xR= = O.ln
I - Ig g 25-2.5
Combined resistance,
RgRs
R=--"--
Rg+1\
0.9x0.1
---=0.09n.
0.9+ 0.1···4wx£n~m}r°n4lxv

397. 4.72
9.Proceed
as in Example 88on page 4.65.
Vg 1
10.I=-=-=O.OlA
gRg 100
Ig 0.01x100
1\=-- xR= = 0.50.
I - Ig g 2-0.01
Vg 1
11.R=- -R or 900=- -100
Ig g Ig
orI=_1_ A=10-3A
g1000
Currentsensitivity
I10-3A
=...!=--- =10-5Adiv-1
n100div
12.Here Ia:60 andIgCf:.10
_Ig__10 1
.. andRg=200
I60 6
ButIg 1\
I Rg +Rs
.!=~
620+Rs
or 6 Rs = 20+ ~orRs =4O.
I
13.HereI=-orI=34Ig
g34
1
~=-g-xR
I-I g
g
Igx3663
-"--- =1110
34Ig-Ig
Combined resistance,
RgRs
R=---'~-
Rg+ ~
= 3663 x111 =107.70.
3663+111
14.~=1\= __6_=~ or 2%ofthe
I Rg +Rs294+650
current passes through thegalvanometer.
15.Proceed as in Example 89 on page 4.65.
IgRg (I110)x27
16.~ = ~ = I _ I 110=30
g
_ 1\A _30x0.2x10-6m2_
I- -- 8 -1.25m.
p 48x10-Om
IgRg 1xRgRg
17.1\ =~= 10-1 =90
g
total
PHYSICS-XII
Resistance of the ammeter formed,
v; (RgI9)Rg Rg
R - - --
A -1\+Rg- Rg/9+Rg-10
R RgI10
~=--=9:10.
1\ Rg I9
18.P.D. across the series combination ofGand R
= P.D. across the shunt 5
1g(Rg+R)=(I - 1g)Rs
0.01(8+R)=(5-0.01) x0.02 = 4.99x0.02
or8+R = 4.99x0.02 = 9.98
0.01
or R= 9.98 - 8 =1.980.
19.Current through the voltmeter
117V
=--=0.013A
90000
Currentthrough the resistance
R=0.13 - 0.013 =0.117 A
Resistance,
117V
R=--=10000.
0.117 A
20.1= ~x150 mA = 15x10-3A
g10
V=.!x150 mV = 75x10-3V
g2
Vg75x10-3
Rg=T=15x10-3= 5O.
g
(i)Required. current range, 1=6x150=900 A
IgRg 15x10-3 x5 -5
.. 1\ =--= 3=8.33x10O.
I-Ig900-15x10-
(ii)Required voltage range,V=1x150=150 V.
R=~- R=150 -5=99950.
.. I g15 x103
g
21.Proceed as in Example 100 on page 4.68.
22.Heree= 12 V,r= 1.20, R = 480, Rv= 720
The combined resistance of the parallel combi-
nationofRand Rvwillbe
R=RxRv= 48x72 = 28.80
pR+Rv 48+72
Current inthe circuit,
I=_e_= 12 =O.4A
Rp+r28.8+1.2
Reading of the voltmeter
=IRp= 0.4 ,x28.8=11.52 V.~~~2tuzkyjxo}k2ius

398. GUIDELINES To NCERT EXERCISES
4.1.A circular coil of wire consisting of 100
turns each of
radius 8.0 cm carries acurrent of 0.40 A. What is the
....
magnitude of the magnetic field Bat the centre of thecoil?
Ans.GivenN= 100, r= 8 em = 0.08 m,I=0040A
~0NI41tx10-7x100x0.40
..B=-- =--------
2r 2x0.08
= 1tx10-4= 3.1x10-4 T.
4.2.A long straight wire carries acurrent of35A.What is
....
the magnitude of the fieldBat a point20em from the wire?
Ans.Here I= 35 A, r= 20 cm = 0.20 m,
~o= 41tx10-7T mA-1
~oI 41txlO-7 x35 5
B =-= = 3.5x10- T.
21tr21tx0.20
4.3.A long straight wire in the horizontal plane carriesa
current of50A in the north tosouth direction. Givethe
....
magnitude and direction ofBat a point 2.5m east ofthe wire.
Ans.HereI= 50 A, r=2.5 m
~ 0I41tx10-7x50 6
B =-- = =4x10-T
21tr 21tx2.5
Applying right hand thumb rule, we find that the
magnetic field will act in the vertically upward direction
at the point 2.5 m east of the wire.
4.4.Ahorizontal overhead power line carries a current of
90 A inaneastto west direction. Whatisthe magnitude and
direction ofmagnetic fieldduetothe current 1.5m below the
line?
Ans.Here 1=90 A,r=1.5m,~o =41tx10-7Tm A-1
B=~oI= 41txlO-7x90T=1.2x10-ST
21tr 21tx1.5
Applying right hand thumb rule, we find that the
direction of the field B will be towards south at a point
below the power line.
4.5.Whatisthemagnitude of magnetic force per unit
length on a wirecarrying acurrentof8 Aandmakingan angle
of30°with the direction of a uniform magnetic field of 0.15 T?
Ans.Given1=8 A,e= 30°, B = 0.15 T
As F=IlBsine
Force per unit length,
f=£=IBsine
I
= 8x0.15xsin 30° = 0.6 Nm-1.wwwmnotesdrivemcom

399. MAGNETIC EFFECT OF CURRENT
4.6.A 3.0
cm wire carrying acurrent of10Aisplaced
insideasolenoid perpendicular to its axis. The magnetic field
inside the solenoid isgiventobe0.27T.Whatisthemagnetic
force on the wire ?
Ans.Given I = 3.0cm = 0.03m,1=10A,
S = 90° rB=0.27T
F=IlBsin8= 10x0.03x0.27xsin 90°
= 8.1x10-2 N
Thedirectionofthe force is given by Fleming's left
hand rule.
4.7.Twolong and parallel straight wires A andBcarrying
currents of 8.0 A and 5.0 A in the same direction are separated
by adistance of 4.0em.Estimate the force on a10cmsection of
wire A.
Ans. Force per unit length of each wire is
f=fl0II 12= 41t x10-7x8x5 = 2x10-4Nm-1
21tr 21t x4x10-2
Force on 10emsection of wire Ais
F=f1=2x10-4 x 10 x 10-2 =2x10-5 N.
4.8.A closely wound solenoid 80cm long has5layers of
windings of400turns each. 771ediameter ofthe solenoid is1.8em.
.....
Ifthe currentcarried is8.0 A, estimate themagnitude of B
inside the solenoid near its centre.
Ans.Number of turns per unit length of the solenoid
is
Numberofturns per layerxNumber of layers
n=------------~----~------------~--
Lengthofsolenoid
= 400 x5 = 2500 m-1
0.80
Magnetic field inside the solenoid is
B =fl0nI= 41tx10-7 x2500x8 = 81tx10-3T
=2.5x10-2T.
4.9.A square coilof theside10cm consists of20turns and
carries a current of12A.The coil issuspended vertically and
normal to the plane of the coil and makes an angle of30°with
thedirection of a uniform horizontal magnetic field of magnitude
0.80 T. Whatisthemagnitude of torqueexperienced by the coil?
Ans.GivenA= 0.10 mx0.10 m = 0.01 m 2,N = 20,
I= 12 A, 8 = 30°, B = 0.80 T
Magnitude of torque is
T=NIBAsin 8
= 20x12x0.80x0.01xsin 30°= 0.96 Nm.
4.10.Two moving coil galvanometers MlandM2havethe
following particulars:
1)= 10n,Nl= 30, ~ = 3.6x10-3 m2, ~ = 025T
Rz= 14o.N2= 42,.Az= 1.8 x1O-3m2, ~ = 050T
The spring constants are identical for the two springs.
Determine the ratio of (i)current sensitivity and (ii)voltage
sensitivity of Mz andMI'
4.99
Ans.Let torsion constant for each meter = k
For a galvanometer, we have
NIBA =ko:
Its current sensitivity is defined as the deflection
produced per unit current,i.e.,
a.NBA
I k
..Current sensitivity ofM2=N2~Az/k=N2~Az
Current sensitivity of ~Nl ~ ~ /kNl ~ ~
42x0.50x1.8x10-37
----------.. =-=1.4.
30x0.25x3.6x10-35
Voltage sensitivity of a galvanometer isdefined as the
deflection produced per unitvoltage, i.e.,
a. a.NBA
--=--
VIRkR
Voltage sensitivity of M2=N2~Az / kRz
Voltage sensitivity of Ml Nl ~ ~ /k1)
=N2~Az x..&.=?x10=~
Nl~~ Rz 5 14
4.11.In a chamber, a uniform magnetic field of 6.5 G (1 G=
10-41) is maintained. An electron isshotinto the field with a
speed of4.8x106ms-1 normal to thefield. Explain why the
path ofthe electron isa circle. Determine the radius of the
circular orbit. Given thate= 1.6x10-19C,me=9.1x10-31kg.
Ans. The perpendicular magnetic field exerts a force
on theelectron perpendicular to its path
continuously deflects the electron from its path and
makes it move along a circular path.
:.Magnetic force on the electron = Centripetal force
mv2
evBsin 90°= _e_
r
or
r=mev
eB
Now B = 6.5 G = 6.5 x10-4T, v=4.8x106ms-1
r= 9.1x10-31 x4.8 x 106 = 4.2x10-2m = 4.2 em.
.. 1.6x10-19 x6.5x10-4
4.12.In Exercise 4.11,obtain the frequency of revolution of
the electron in itscircular orbit. Does the answer depend on the
speed of the electron?Explain.
Ans.Frequency of revolution of the electron in its
circular orbit,
f= ~ = 1.6 x10-19 x6.5xlO-4
21tm2x3.14x9.1xlO 31
= 18.18x106Hz. =18 MHz.
No,the frequencyfdoes not depend on the speedvof
the electron.
4.13.(a) Acircularcoil of30turns and radius 8.0 em
carrying a current of 6.0 A issuspended vertically in a uniformwww7notesdrive7com

400. 4.100
horizontal magnetic field of magnitude 1.0 T.The field lines
make
an angle 60°withthe normalto the coil. Calculate the
magnitude of the counter torque that must be aplied to prevent
the coil from turning.
(b) Would your answer change if the circular coil in (a) were
replaced by a planar coil of some irregular shape that encloses
the same area? [CBSE 00 98C)
Ans.(a)N= 30,r= 8.0cm = 0.08 m,I= 6.0 A,
B=1 T,e= 60°
Magnitude of counter torque
=Magnitude of deflecting torque
= NIBAsine
= 30 x 6 x 1 x (3.14 x 0.08 x 0.08) sin 60°
=30 x 6 x 3.14 x64 x 10-4 x 0.866 =3.1Nm.
(b)No,the answer would not change because the
above formula forthetorque is true for a planar loop of
any shape.
4.14.Two concentric circular coilsXandYof radii16em
and10em respectivelylie inthe same vertical plane containing
the north-south direction. CoilXhas20turns and carries a
current of16A ; coilYhas25turns and carries a current of
18 A. The sense of the current inXisanticlockwise, and inY
clockwise, for an observerlooking at the coils facing west. Give
themagnitude and direction of the net magnetic field due to the
coils at their centre.
N
Observer
W~.~~----++-+-r~------.E
Coil Y
Coil X
5
Fig. 4.160
Ans.For coil X:rx= 16 cm= 0.16 m ,Nx=20,Ix= 16 A
Magnetic field at the centre of coil Xis
B=1l0IxNx 41t x10-7x16 x20 T
x 2rx 2 0.16
=41tx1O-4T
Asthe current in the coilXisantic1ockwise, the field is
directed towards east.
ForcoilY:ry=lOcm""O.lOm, Nx=25,I=18A
:.Magnetic field at the centre of coil Yis
By11oIyNy 41tx10-7 x18x25T=91tx10-4T
2ry 2 0.10
PHYSICS-XII
As the current in the coil Y is clockwise, the field Byis
directed towards west. Since By>Bx'therefore, the net
field is directed towards westand its magnitude is
B=By-Bx= 51tx 10-4 =1.6 x10-3 T.
4.15.A magnetic field of100 G(l G =1O-4T) isrequired
whichisuniform in a region of linear dimension about10em
and area of cross-section about 10-3m2• The maximum
current-carrying capacity of a given coil of wireis 15A and the
number of turns per unitlength that can be wound round a core
isat most1000turns m-1. Suggest some apropriate design
particulars of a solenoid for the required purpose. Assumethe
coreisnot ferromagnetic.
Ans.Here B = 100 G = 10-2 T,I= 15 A,
n= 1000 turns m-1
Magnetic field inside a solenoid,
B=1l0nI
B 10-2
. . nI=-= = 7955 "" 8000
11041tx 10-7
We maytakeI=10A,thenn=800
The solenoid mayhave length 50 cm and area of cross-
section 5 x 10-2m 2(five times the given value) so as to
avoid edge effects, etc.
4.16.For a circular coil of radius Rand N turns carrying
current I, the magnitude of the magnetic field at a point on its
axis at a distance x from its centre isgiven by,
110 IR2N
B=~+- ---=
2(x2 +R2)3/2
(a) Show that this reduces to the familiar result for field at
the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal
radiusR,and number of turns N, carrying equal currentsinthe
same direction, and separated by a distance R.Show that the
field on the axis around the mid-point between the coils is
uniform over a distance that issmall as compared toR,andis
given by
11NI
B = 0.72_0_, approximately
R
Such an arrangement used to produce a nearly uniform
magnetic field over a small region isknown as Helmholtz coils.
Ans.(a)Given
110 IR2 N
B=----'-+--:-~
2(x2 +R2)3/2
At the centreof thecoil,x= 0, so
B=110IR2N=110IN
2R3 2R
This is the standard result for the field at the centre of
the coil.www6notysxr•vy6wom

401. MAGNE
TIC EFFECT OF CURRENT
(b)Asshown in Fig.4.161, consider asmall region of
length2dabout the midpoint°between the two coils.
Coil1 Coil 2
d d
••..-•••--01
----..--..- -..•..- - - -
Q0 P
IR R
--"2----------"2--
I+-----------R -----------01
Fig.4.161
IlTNR2
GivenB ------.<'0'------"--,,..,.,
- 2(R2 +x2)3/2
Therefore, the magnetic field at the point Pdue to coil 1,
lloIR2N .
f\= 3/2,acting alongP02
2[R2+(~ +dr]
Magnetic field atthe point P dueto coil2,
lloIR2N .
Bz= 3/2,actmg alongP02
2 [ R2 +(~ -dr]
Total magnetic field atthe point P will be
B=f\+Bz
~ "OI~'Nr
1
IlOIR2Nr 1 11
= 2 (5R2 )3/2+(5R2 )3/2
-+ Rd --- Rd
4 4
[Neglecting d2,asd«R]
or
[Expanding bybinomial theorem and
neglecting higher powers ofd/R]
B:::.O.72lloTN .
2R
4.101
Magnetic field will also be same at the point Q.In fact,
it will be uniform over the small region of length 2 d
around the midpoint 0.
4.17.A toroid has a core(non-ferromagnetic) of inner
radius25em and outer radius 26em around which 3500turns
of a wire are wound. If the current in the wire is11A, whatis
themagnetic field (a) outside the toroid (b) inside the core of the
toroid (c)intheempty space surrounded by the toroid?
Ans.Here,I=11A, total number ofturns=3500
Mean radius of toroid,
r=25+26=25.5ern=25.5x10-2m
2
Total length (circumference) of the toroid =21tr
=21tx25.5x10-2 =51.0x10-21t m
Number of turns per unit length,
3500
n=------,,-
51.0 x10 21t
(a)The field outside the toroid is zero.
(b)The field inside the core of the toroid,
-7 3500
B=Il°nI=41tX10x 2x11
51.0x10-1t
=3.02x10-2T .
(c)The field in the empty space surrounded by the
toroid is alsozero.
4.18.Answer the following questions:
(a) A magnetic field that varies inmagnitude from pointto
point but has aconstant direction(east to west) isset up in a
chamber. Acharged particle enters thechamber and travels
undeflecied along a straight path with constant speed. What can
you say about the initial velocity of the particle ?
(b) Acharged particle enters an environment of a strong
and non-uniform magnetic field varying from point to point
bothinmagnitude and direction and comes out ofitfollowing a
complicated trajectory. Woulditsfinal speed equal the initial
speed ifitsuffered no collisions with the environment?
(c) Anelectron travelling west to east enters a chamber having
a uniform electrostatic fieldina northtosouth direction. Specify
thedirection in which a uniform magnetic field should be set up
to prevent the electron from deflecting from its straight line path.
Ans. (a)The force on a charged particle moving in a
magnetic field is given by
F=qvBsine
The force on a charged particle will be zero or the
particle will remain undeflected if
sine=0ore=00,1800
i.e.,initial velocity 1!is either parallel or aniiparallel toB.
(b)Yes,a magnetic fieldexerts force on a charged
particle in adirection perpendicular to its direction of
motion and hence does no work onit.Sothe charged
particle will have its final speed equal to its initial speed.www7notesdrive7com

402. 4.102
(c)The electron travelling west to east experiences a
force towards north due to the electrostatic field. It will
remain undeflected
if it experiences an equal force
towards south due to the magnetic field. According to
Fleming's left hand rule, the magnetic field must act in the
vertically downward direction.
4.19.Anelectronemitted by a heatedcathode and
accelerated through a potential difference of 2.0 kV, enters a
region with uniform magnetic field of 0.15 T.Determine the
trajectoryof the electron if the field (i) istransverse to itsinitial
velocity, (ii)makesan angle of30°with the initialvelocity.
Ans. V= 2.0 kV = 2x103V,B= 0.15 T,
e= 1.6x10-19C,m= 9.1x10-31 kg
Potential differenceVimparts kinetic energy to the
electron given by
1mv2 =eV
2
or, Velocity gained by electron,
;----~----;;-
v=~2ev = 2x1.6xlO-19x2xl03ms_1
m 9.1x10-31
= 2.65x107ms-1
(i)When fieldBis transverse to the initial velocity1!,
mv2
e vBsin 90° =--
r
mv9.1x10-31 x2.65x107
..r--- m
-eB-1.6x10-19 x0.15
.=10-3m =1 mm.
Thus the electron follows a circular trajectory of radius
1 mm normal to the fieldB.
~
(ii)When fieldBmakes an angle of 30° to the initial
velocity1!,
v.l=vsin 30° = 2.65x107x~ = 1.33xl07 ms-1
VII=vcos30°= 2.65x107x0.866 = 2.3x107ms-1
The radius of the helical path is given by
mv.l mvsin30° 9.1xlO-31x1.33xl07
r=--=----
eB eB 1.6x10-19xO.15
= 50.4 x10-5m =0.50mm.
4.20.A magnetic field set up using Helmholtz coils is
uniform in a small region and has a magnitude of 0.75 T. In the
same region, a uniform electrostatic fieldismaintained in a
direction normal to thecommon axis of the coils. A narrow beam
of (single-species) charged particles all accelerated through
15kVenters this region in a direction perpendicular to both the
axis of the coils and the electrostatic field. If the beam remains
undeflected when the electrostatic field is9.0 x 105 Vm-1,make
a simple guess as to what the beam contains. Why is the answer
not unique?
Ans.B= 0.75 T, E= 9.0x105Vm-1,
V= 15 kV = 15 x 103 V
PHYSICS-XII
For undeflected beam, velocity ofcharged particles
must be
E9.0x105
V= ms-1=12xl05ms-1
B 0.75
But the kinetic energy of the charged particles is
given by
1mv2 =qV
2
!L=!. L=!.)12 x105)2 Ck-1
m2.V22 15x103 g
=4.8x107C kg-1
Now fordeutrons,
1.6x10-19
q= =4.8xl07C kg-1
m2x1.67x10-27
which means that the particles may be deutrons, each of
which contains one proton and one neutron. The answer
is not unique because we have determined only the ratio
of charge to mass. Other possible answers are He2+ and
Li3+,etc.
4.21.A straight horizontal conducting rod oflength 0.45 m
and mass 60gissuspended by twovertical wires atitsends. A
current of 5.0 A isset up in the rod through the wires.
(a) What magnetic field should be set up normal to the
conductorin order that the tension in thewiresiszero?
[CBSE DISC)
(b)What will be the total tension in the wires if the direction
of currentisreversed, keeping the magnetic field same as
before? (Ignore the massof the wires)g=9.8ms-2.
Ans.HereI=0.45m,m= 60g= 0.06 kg,
1= 5.0 A, g= 9.8 ms-2
(a)Tension in the supporting wireswill bezero when
the weight of the rod is balanced by the upward forceliB
of the magnetic field.
i.e., IIB=mg
B=mg= 0.06 x9.8 T=0.26T
11 5x0.45
T
mg
liB
T
I I
---~-r------------1----
According to Fleming's left hand rule, the magnetic
field should be applied normally into the plane of paper
so as to exert an upward magnetic force on the rod.
(b)If the direction of current is reversed, the magnetic
force will act in the downward direction. Hence the total
tension in the wires will be
T= 2xthe weight of the rod
= 2x0.06x9.8 N = 1.176 N.www7notesdrive7com

403. MAGNETIC EFFECT OF CURRENT
4.22.The wir
es which connect the battery of anautomobile
to its starting motor carry acurrent of300A (for a short time).
Whatistheforce per unit length between the wires if they are 70em
longand1.5emapart? Istheforceattractive or repulsive?
[Haryana 01]
Ans. II=12=300 A, r=1.5em=1.5x10-2m,
1=70 em=0.70m
Theforce per unit length between the wires is
f=1l0III2 =41txlO-7 x300x300 Nm-I =1.2 Nm-1
21tr 2nx1.5x10-2
Total force between the wires,
F=fxI=1.2x0.70=0.84 N
As the currents in the two wires are in opposite
directions, the force is repulsive.
4.23.Auniform magnetic fieldof1.5Texists in a
cylindrical region of radius 10.0 em, itsdirection being parallel
totheaxis along east to west.A wire carrying current of 7.0 A
inthe north to south direction passes through this region. What
isthemagnitude and direction of the force on the wire if
(i)the wire intersects the axis,
iii)the wire isturned from N-Sto north east ornorth
west direction,
(iii)the wire intheN-S direction islowered from the
axis by adistance of 6.0 em?
Ans.Here B=1.5 T, I=7.0A
(i)Asshown in Fig. 4.163,
lengthofwire in cylindrical region
=diameterABof cylindrical region
=20 em=0.20 m
,
,
,
,
,
I
',8
5
------
Fig.4.163
As the wire lies inN-Sdirection and field acts along
E-Wdirection, soe=90°
..Force on wire,
F=IBIsine =7x1.5x0.20x1=2.1 N
ByFleming's left hand rule, thisforce acts in the
vertically downward direction.
(ii)When the wire turns from N-StoN-EorN-W
direction, suppose it makes angleewith field B, asshown
in Fig. 4.164. Then length of wire in magnetic field,
A'B'=I'say.
Clearly
I.e
-=sm or
I'
1'=_1_
sine
4.103
A
8
I-
I
IAxis
Wj------
I 9
I
I
I
I
Fig.4.164
Force onwire,
F=I1'sine=1._1_. Bsine=IlB=2.1 N
sino
This force acts in the vertically downward direction.
(iii)As shown in Fig. 4.159, when the wire is lowered
by6.0 ern, length of the wire in the magnetic field=2x
But x=~102- 62=8em=0.08 m
.. 2x=0.16,e=90°
Force onwire, F=lIB=7x0.16x1.5=1.68N
This force also acts in the vertically downward direction.
4.24.Auniform magnetic fieldof 3000 Gisestablished
along thepositive Zdirection. A rectangular loop of sides 10em
?-~---.Y
y
x x
(a) (b)
z
y
X X
(c) (d)
Z Z
8f 8f
Y
X
[
X
[
(e) If)
Fig. 4.165www7notesdrive7com

404. 4.104
and5em carries a curr
ent of12 A. Whatis the torque on
the loop in the different cases shown in Fig. 4.165?What
is the force on each case ?Which case corresponds to
stableequilibrium?
Ans.HereB= 3000 G = 3000 x10-4= 0.3 T,
A= 10x5 = 50cm2 = 50x10-4m2, I= 12 A
Magnetic moment,
m=IA=12x50x10-4=0.06 Am 2
We apply right hand rule tovarious current loops to
decide the direction ofn:.
4 A 2 A
(a)Herem= 0.06iAm,B=0.3kT
4 4 4
't=mxB
1\ 1\ 21\
=0.06i xO.3k=-1.8xlO- jNm
Thus atorque of1.8x10-2Nm acts along negative
Y-axis.
----. f'I 2~ 1\
(b)Herem= 0.06iAm rB=0.3kT
4 4
Clearly, mandBaresame as in case (a).In this case
also,atorque ofl.8 x10-2Nm acts along negative Y-axis.
-4 1\ 2 -4 1\
(c)Herem=- 0.06 jAm, B= 0.3 kT
4 4 4
t=mxB
=-0.06ix0.3k=-1.8x10-2iNm
Thus atorque of1.8x10-2Nm acts along negative
X-axis.
(d)This caseissimilar tocase(c).But here the direction
of the torqueis 60° anticlockwise with negative
X-directioni.e.,240°with positive X-direction.
-4 1\ 2-4 1\
(e)Herem= 0.06 kAm,B=0.3kT
-4 -4 -4 1\ 1\
't=mxB=0.06kx0.3k=O.
-4 1\ 2--1- 1\
if)Herem=-0.06kAm rB=0.3kT
---+ -t -t 1\ 1\
r.=mxB=-0.06kx0.3k=O.
The netforce on the loop is zero in eachcase.
Case (e)correspondstostable equilibrium, because
4 4
heremis parallel toB.
Case(j)correspondstounstable equilibrium, because
heren:isantiparallel toB.
4.25.A circularcoil of20turns andradius 10emisplaced
ina uniform magnetic field of0.10 Tnormal to the plane of the
coil.If the current inthe coilis5.0 A, whatisthe
(a) total torque on the coil,
PHYSICS-XII
(b) total force on the coil,
(c) average force on each electron inthe coil due to the
magnetic field?
(Thecoilismade ofcopper wire of cross-sectional area
10-5m2,andthe free electron density incoperisgiven tobe
about(1029m-3 ).
Ans.N= 20,r= 10 cm=0.10m,B=0.10T,
I=5.0 A,e=0°
(a)Torque on the coil,
r=NIBAsine=0 [.:e=0°]
(b)Magnetic forces on the opposite arms of coilare
equal and opposite, andact in thesame plane; hence the
totalforce on the coil is zero.
(c)Force on each electron is
F=evB=.!!!..
nA
[.:1=enAv]
For given wire,
n= 1029m-3,A=10-5m2
0.1x5 -25
F=29 5N=5x10 N.
10 x10
4.26.A solenoid60em longandof radius 4.0 em has 3
layersof windings of300 turns each. A 2.0 em long wire of mass
2.5gliesinside the solenoid near itscentre normal to its axis;
both the wire and the axis of the solenoid are in the horizontal
plane. The wireisconnected through two leads parallel to the
axis of the solenoid to anexternal battery which suplies a
current of6.0 Ainthewire. Whatvalue ofcurrent (with
appropriate sense ofcirculation) inthe windings of the solenoid
can suporttheweight of the wire? g =9.8ms-2.
Ans. Let 1 be the current in the windings of the
solenoid which cansupport the weight of the wire. The
magnetic fieldinside the solenoid along its axis will be
B=l1on1
Total numberofturns
n=---------
Length of the solenoid
300x3 -1
= 2= 1500 turns m
60x10
B=4nx10-7 x1500x1
=6nx10-41 tesla
Here,
This field acts perpendicular to the current carrying
wire, therefore, the magnetic force on the wire will be
F=l'lB
=6x(2x10-2)x6nx10-4 Inewton
The current 1 wouldsupport thewireif the above
forceequals the weight ofthe wire,
i.e.,6x2x10-2 X6nx10-4I=2.5x10-3x9.8
2.5x10-3x9.8
1=72x3.14x10-6A=108.3 A.
orwww7notesdrive7com

405. MAGNETIC EFFECT OF CURRENT
4.27.A galvanometer coil has a resistance of120and meter
shows full scale defle
ction for a current of3mA.How will you
convert the meter into a voltmeter of range0to18V?
Ans.HereRg=120,Ig=3 mA=3 x10-3A,V=18V
R=~-R 18 -12
I g3x10-3
g
=6000-12=59880
Byconnecting a resistance of5988in series with the
given galvanometer, we get a voltmeter of range 0 to
18 V.
4.105
4.28.A galvanometer has a resistance of150and the meter
shows full scale deflection for a current of4mA.How will you
convert the meter into an ammeter of range0to6A?
Ans.HereRg=150,Ig=4 mA=0.004 A,I=6 A
R=--.!.LxR=0.004x15
SI-Ig g 6-0.004
=0.0100=10m0
Byconnecting a shunt of resistance10mO across the
given galvanometer, we get an ammeter of range 0 to
6A.www7notesdrive7com

406. "YPE A : VERY
SHORT ANSWER QUESTIONS (1mark each)
1.State Oersted's observation.
2.StateBiot-Savart's law. [ISCE94]
3.Mathematically, Biot-Savartlaw may be expressed as
dB=K. Idls~8
r
Write the value ofKin SI units.
4.What is the SI unit offl 0?
5.What is the value of41t / fl 0?
6.State the rule that is used to find the direction of
magnetic field acting at a point near a current
carrying straight conductor.
[CBSE98;Pb.97, 98]
7.Write an expression for the magnetic field
produced by an infinitely long straight wire
carrying a currentI,at a short distanceafrom itself.
[ISCE98]
8.Show the magnetic lines of force around a straight
current carrying conductor. [Punjab97C]
9.What is the nature of the magnetic field associated
with the current in a straight conductor?
10.Where is the magnetic field due to current through
circular loop uniform ?
11.Where is the magnetic field of a current element
(i)minimum and(ii)maximum?
12.Figure 4.166shows a circular
loop carrying a currentI.
Show the direction of the
magnetic field with the help
of lines of force.
[CBSE D 04]
Fi. 4.166
13.Which physical quantity has the unit Wb m-2? Is it
a scalar or a vector quantity? [CBSE D 04]
14.An electric current is flowing due south along a
power line. What is the direction of the magnetic
field at a point(a)above it and(b)below it ?
15.How does a current carrying coil behave like a bar
magnet? [CBSE D11]
16.Draw the magnetic field lines due to a current
carrying loop. [CBSE D 13C]
17.How much is the flux densityBat the centre of a
long solenoid ? [ISCE95,97]
18:What is a toroid ?
19.What is magnetic Lorentz force? [Punjab01]
20.Write the expression, in a vector form, for the
....•
Lorentz magnetic forceFdue to a charge moving
with velocityZ;in a magnetic fieldB.What is the
direction of the magnetic force? [CBSED 14]
21.A particle of chargeqmoves with a velocityvat
an angle 8 to a magnetic field B. What is the force
experienced by the particle?[ISCE 93; CBSEF91]
22.What is the force experienced by a stationary
charge in a magnetic field ? [Himachal02]
23.What is the work done by magnetic field on a
moving charge? [Haryana94]
24.Write down the expression for the Lorentz force on
a charged particle.[Himachal99; Punjab99,99C]
25.What is the force on a charge moving along the
direction of the magnetic field ?[CBSED 94C]
26.State Fleming's left hand rule.[Pb01;CBSED 94C]
27.An electron beam is moving vertically downwards.
If it passes through" a magnetic field which iswwwDnotesdriveDcom

407. 4.106
directed from south to north in a horizontal plane
,
then in which direction the beam would be
deflected? [eBSED 96C]
28.What will be the path of a charged particle moving
perpendicular to a uniform magnetic field ?
[eBSE D 93]
29.What will be the path of a charged particle moving
in a uniform magnetic field at any arbitrary angle?
[eBSE F 93]
30.What will be the path of a charged particle moving
along the direction of a uniform magnetic field ?
[eBSEOD95]
31.When a charged particle moving with a velocity :;
~
is subjected to a magnetic fieldBthe force acting on
it is non-zero. Would the particle gain any ener?
[eBSE F13]
.32.An electron with speed venters at right angle in a
regionof uniformmagneticfieldB.Writethe expression
for the radius of the path it follows. [eBSEF 1995]
33.An electron beam projected along +X-axis,
experiences a force due to a magnetic field along
the+Y-axis. What is the direction of the magnetic
field ? [eBSEDOS]
34.An electron and a proton moving with the same
speed enter the same magnetic field region at right
angles to the direction of the field. For which of the
two particles will the radius of circular path be
smaller? [eBSEOD98]
35.A charged particle moving in a uniform magnetic
field penetrates a layer of lead and thereby loses
one-half ofits kinetic ener. How does the radius
of curvature of its path change?[eBSE FlOC]
36.Write the condition under which an electron will
move undeflected in the presence of crossed
electric and magnetic fields. [eBSEF 13;OD14C]
37.A straight conductorABof a circuit lies along the
X-axis fromx=-a /2 tox=+a/2 and carries a
currentI.What is the magnetic field due to this
conductorABat a pointx=+a?
38.State the principle of a cyclotron. [Punjab01]
39.Does the time spent by a proton inside the dees of a
cyclotron depend on(i)the speed of the proton and
(ii)the radius of its circular path ?
40.In a field, the force experienced by charge depends
upon its velocity and becomes zero, when it is at
rest.Is the field electric or magnetic in nature?
41.In a field, the force experienced by a charge
depends only upon the magnitude of the field and
does not depend upon the velocity. What is the
nature of the field?
PHYSICS-XII
~
42.What is the force that a conductor dlrcarrying a
currentIexperiences when placed in a magnetic
~
fieldB.Whatisthe direction of theforce?
[eBSEOD90]
43.An electron beam ismoving horizontally in a tube.
Thevertical component of earth's magnetic field is
directed downwards. In which direction will the
electron beam be deflected ?
44.A charged particle moves inauniform magnetic
field at right angles tothe direction of the field.
Which of-thefollowing quantities will change: speed,
velocity, momentum, kinetic ener, displacement?
45.Inwhich orientation is the force experienced by a
current-carrying conductor placed in amagnetic
field maximum ?
46.A current carrying conductor does not tend to
deflect in a magneticfield.What conclusion canbe
drawn from it?
47.ame the rule that gives the direction of force on a
current-carrying conductor placed perpendicular
to the magnetic field.
48.Write an expression for the force between two
parallelshortwires carrying currents.
49.Two current elements are placed a certain distance
apart but not parallel to each other. Do they exert
equal andopposite forces on each other?
50.Whatisthedirection of force between two parallel
wires carrying currents inoppositedirections?
51.The force existing between two parallel current
carrying conductors is F.If the current in each
conductor is doubled, what is the value of the force
between them?
52.Is the force between two parallel current-carrying
wires affected by the nature of the dielectric
medium between them ?
53.What isthevalueofnet force acting on a current
carrying (i)rectangular and(ii)circular loop, placed
in a uniform magnetic field ? What do you expect
about the torque in eachcase?
54.Write anexpression for the torque acting on a current
carrying coil located in auniform magnetic field.
55.Write an expression for the magnitude of the torque
acting on a current carrying coil placed in a uniform
radial magnetic field.
56.Under what circumstances will acurrent carrying
loopnot rotate in the magnetic field ?
57.State the principle of working of a moving coil
galvanometer. [eBSED 15]
58.What doyou mean bythe figure of merit of a
galvanometer?www6notesdrive6com

408. MAGNETIC EFFECT OF CURRENT
59
.Define the current sensitivity of a moving coil
galvanometer and state its S1unit.
[CBSEOD 13, 13C]
60.Write two factors by which the current sensitivity
of a moving coil galvanometer can be increased.
[CBSED01;F08]
61.Define voltage sensitivity of amoving coil
galvanometer. Give its S1unit. [Punjab91]
62.Write two factors by which voltage sensitivity of a
moving coil galvanometer can be increased.
[CBSED 01;F 08]
63.What is the nature ofthe magnetic field in a moving
coil galvanometer? [CBSED 96; OD96]
64.State two properties of the material of thewireused
for suspension of the coil in a moving coil
galvanometer. [CBSEOD01, 06C]
65.The current sensitivity of a moving coilgalvano-
meter is 5division/rnA and voltage sensitivity is 20
division/volt. Find the resistance of thegalvano-
meter.
66.An electron and a proton, having equal momenta,
enter a uniform magnetic field at right angles to the
fieldlines. What will be the ratio of curvature of
their trajectories? [CBSESamplePaper05]
67.An electron is moving with a velocityv,along the
axis of a long straight solenoid, carrying a current 1.
What will be the force acting on the electron due to
the magnetic field of the solenoid ?
[CBSESample Paper 05]
68.Among alpha, beta and gamma radiations, which
get deflected by the magnetic field? [CBSEF 04]
69.A solenoid coil of 300tums/m is carrying a current
of 5 A. The length of the solenoid is 0.5 m and has a
radius of 1 ern. Find themagnitude of the magnetic
field inside the solenoid. [CBSEF04]
70.What is the resistance of an ideal ammeter?
71.Whatisthe resistance of an ideal voltmeter?
72.Why should anammeter havea high current carrying
capacity ?
73.Why should a voltmeter have a low current
carrying capacity ?
74.What is the effective resistance of an ammeter if a
shunt of resistance Rsis used across theterminals of
a galvanometer of resistance Rg?
75.Suppose a shunt of resistance O.OHl isconnected
across a galvanometer, what can besaid about the
resistance of the resulting ammeter?
76.A student wants to increase the range of an
ammeter from 1mAto 5mA.What should be done
to the shunt resistance?
4.107
77.What is the direction of the force acting on a
charged particleq,moving with a velocity1/in a
->
uniform magnetic field B ? [CBSED08]
78.Twoidenticalcharged particlesmoving with same
speed enter a region of uniform magnetic field. If
one of these enters normal to the field direction and
the other enters along a direction at 30° with the
field, what would be the ratio of their angular
frequencies? [CBSESamplePaper08]
79.An a-particle and a proton are moving in the plane
of the paper in a region where there isauniform
->
magnetic field(B ) directed normal to the plane of
the paper. If the two particles have equal linear
momenta, what will be the ratio of the radii of their
trajectories in the field ? [CBSESamplePaper08J
80.Why should the spring/suspension wire in a
moving coil galvanometer have low torsional
constant? [CBSE0008]
81.The coils, in certain galvanometers, have a fixed
core made of a non-magnetic metallic material.
Why does the oscillating coil come to rest so quickly
in such a core? [CBSED 08C]
82.A long straightwire carries a currentIalong the
positive y-direction. A particle of charge +Qis
->
moving with a velocity valong thex-axis. Inwhich
direction will the particle experience a force ?
[CBSEF 13]
83.Two particles Aand B of masses mand2mhave
chargesqand2qrespectively. Both these particles
moving with velocitiesl1.andv2respectively in the
same direction enter the same magnetic field B
acting normally to their direction of motion. If the
two forcesFAandFBacting on them are in the ratio
1:2, find the ratio of their velocities.
[CBSED 11C]
84.A beam of a particles projected along +x-axis,
experiences a force due to magnetic field along the
+y-axis.What is the direction of the magnetic field?
(Fig.4.167) [CBSEOD 10]
x
1
• 0.particle
z
y
Fi. 4.167www5notesdrive5com

409. 4.108
85.A beam of
electrons projected along+x-axis,
experiences a force due to a magnetic field along
the+y-axis. What is the direction of the magnetic
field? (Fi. 4.168) [CBSEOD 10]
x
ee
z
y
Fig.4.168
PHYSICS-XII
their common direction of motion. What would be
the ratio of the radii of the circular paths, described
by the protons and deuterons?
[CBSEF11]
87.A square coil, OPQR, of sidea,carrying a currentI,
is placed in theY-Zplane as shown in Fi. 4.169.
Find the magnetic moment associated with this
coil. [CBSESamplePaper13]
'Z
RI--__----,Q
r---I-~---- y
po
86.A narrow stream, of protons and deuterons, having
the same momentum values, enter a region of a
uniform magnetic field directed perpendicular toFi.4.169
Answers
x
•
1.A magnetic needle brought close to a straight
current-carrying wire aligns itself perpendicular to
the wire, reversing the direction of current reverses
the direction of deflection.
2.According to Biot-Savart law, the magnetic field
~
due to a current elementIdlat the observation
~
point whose position vector isris given by
-> ~
dB=llol.~
4n r3
where110is the permeability of free space.
3.K=llo=10-7TmA-1.
41t
4.S1unit of permeability(110)=TmA-lor WbA -1m-1.
5.41t=107r1m-1A.
110
6.The direction of magnetic field due to a straight
conductor can be determined with the help ofright
hand thumb rule.According to this rule if we grasp
the conductor in the right hand so that the thumb
points in the direction of the current, then the
magnetic field will be in the direction of the curl of
the fingers.
7.B=1l01
2na
8.See Fi. 4.8.
9.The magnetic field consists of concentric circular
lines of force with the conductor at their centre and
in a plane perpendicular to the conductor.
10.At the centre of the current loop.
11.(i)Magnetic field is minimum (zero) along the
axis of a current element.
(ii)Magnetic field due to current element is
maximum in a plane passing through the
element and perpendicular to its axis.
12.See Fi. 4.25.
13.Wb m-2is the S1unit of magnetic fieldBwhich is a
vector quantity.
14.According to right hand rule, the direction of the
field is(a)towards west above the wire and
(b)towards east below the wire.
15.A current carrying loop behaves as a bar magnet
because
(i)it possesses a magnetic dipole moment
(m=lA),and
(ii)it experiences a torque in an external magnetic
field. This torque tends to align the axis of the
loop along the direction of the field.
16.See Fi. 4.25.
17.The magnetic field well inside a long solenoid
having n turns per unit length and carrying current
Iis B=Ilonl.
18.An anchor ring around which a large number of
turns of a metallic wire are wound is called a toroid.
19.The force experienced by a charged particle while
moving through a region of magnetic field is called
magnetic Lorentz force.
-> ->->
It is given byF=q(vxB ) .www6notesdrive6com

410. MAGNETIC EFF
ECT OF CURRENT
20.F=q(1!xB).The direction ofthe force is perpen-
~ ~
dicular to the plane containing vectors vand B.
F=qvBsine.
For astationary charge, v=O.
Therefore, F=qvBsine=q(0)Bsine=o.
Zero, because a magnetic force acts perpendicular
tothe direction of velocity or the direction of
motion of the charged particle.
-+- 4 4 4
F=q(E+vxB).
Force on a charge moving alongthedirection of the
magnetic field is zero F=qvBsin0°=O.
Fleming's left handrule gives the direction of force
ona charged particle moving in amagnetic field :
Stretch the thumb and the first two fingers of the
left hand so that they are perpendicular to each
other. If the forefinger points in the directionof
magnetic field, central finger in the direction of
current, then the thumb givesthedirection of force
on the charged particle.
Towards west. . 28.Circular path.
Thepath of the particle will be a helix with itsaxis
along the field B.
Thecharged particle will move along a straight line
path.
The magnetic force acts perpendicular to the
direction of motion of the charged particle. No
work is done by the magnetic force on it. The
particle does not gain any energy.
Magnetic force on electron = Centripetal force
2
evBsin 90°= mv :.Radius, r=mv.
r eB
According to Fleming's left hand rule, the magnetic
field acts in the +Z-direction.
Rdi mv.
aIUS,r=- t.e., rocm
eB
As electron has smaller massthan proton, so it will
circulate in a circular path of smaller radius.
Radius of curvature,
r=mv=!!!..{2K=.J2mK i.e.;r «J]:
qB qBV-;;;qB
:.If the kinetic eneris halved, radius ofcurvature
isreduced to1/12times its initial value.
36.For undeflected beam,
21.
22.
23.
24.
25.
26.
27.
29.
30.
31.
32.
33.
34.
35.
F,,,=F.,or evBsin900=eE
E
orV=-.
B
37.Zero, because the observation point liesontheaxis
of the straight conductor.
4.109
38.Refer answer to Q.16on page4.34.
39.Timespent by a proton inside the dees of a
cyclotron is independent of both its speed and
radius of itscircular path.
40.The field is magnetic in nature.
41.The field is electric in nature.
~ -:t ~
42.d F=I(dl xB)
~
The direction of forcedFisperpendicular to the
~ ~
plane ofdland Bandwill point in the same
direction inwhich a right-handed screw, when
~ ~
rotated fromdlto B, will advance.
43.Towards west.
44.Only velocity, momentum and displacement will
change as they are allvectors.
45.When the conductor is held perpendicular to the
magnetic field, it experiences a maximum force.
46.Thismeans that no force is acting on the current
carrying wire due to the magnetic field. Thisis
possible when the conductor is parallel to the
direction of the magnetic field.
47.Fleming's left hand rule.
48.The force between two parallel shortwires of
lengthsd~andd12,separated by distance rand
carrying currents IIand12respectively, is given by
dF=!::Q..III2d~ d12
41t ,2
49.Like other forces, theseforces also obey Newton's
third lawof action and reaction and are, therefore,
equal and opposite.
SO.The direction of force is perpendicular to the two
wires and is outwards, so that the two wires repel
each other.
51.The value of force is4F.This is because force
between two parallel current carrying conductors is
proportional to the product of the currents through
them.
52.No.Thisinteraction is between the magnetic fields
produced by the two wires which does not depend
on the nature ofthe dielectric medium.
53.Ineach case the net force, is zero but torque is
non-zero.
.54.If a coil of area A,turns Nand carrying currentIis
heldin a uniform magnetic field B, it experiences a
torque given by
~
, =NIBA sine,whereeistheangle between B and
the normal totheplane of the loop.www5notesdrive5com

411. 4.110
55. 1"=NIBA.
56
.If the current carrying loop is placed inamagnetic
field, with its plane perpendicular to the field, then
it will not rotate.
57.Refer to point 26 of Glimpses.
58.The figure of merit of a galvanometer is defined as
the amount of current required to produce one
scale deflection in the galvanometer.
It is given by
59.The current sensitivity of a galvanometer is defined
as the deflection produced in the galvanometer on
passing unit current through it.
C ... aNBA
urrent sensitivity =-=--
I k
The S1unit of current sensitivityis radian ampere ".
60.The current sensitivity of a moving coil galvano-
meter can be increased by(i)increasing the number
of turns in the galvanometer coil. (ii)decreasing the
torsion constant of its suspension fibre.
61.The voltage sensitivity of amoving coil galvano-
meter is defined as the deflection produced in the
galvanometer when a unit voltage is applied across
its coil.
V I . .. a a NBA
o tage sensitivity = - = - =--
V IR kR
The S1unit of voltage sensitivity is radian volt-I.
62.The voltage sensitivity of a moving coil
galvanometer can be increased by
(i)increasing the number of turns of the galvano-
meter coil
(ii)decreasing the torsion constant of the suspension
fibre.
63.Radial magnetic field is used in a moving coil
galvanometer.
64.The material used for the suspension wire of a
moving coil galvanometer should have following
properties :
(i)Small torsion constantkwhich makes the
galvanometer highly sensitive.
(ii)High tensile strength so that even thin wire
does not break under the weight of the
suspension coil.
65.HereIs=5div mA-I =5x103div A-I,
Vs= 20div V-I
I5x103
R=~=-·-=2500.
gVs 20
66.As r=mv i.e., ra:mv..re: rp =1 :L
eB
PHYSICS-XII
67.F=evBsin0°=O.
68.Alpha and beta radiations are deflected by the
magnetic field.
69.B=flonI= 41tx10-7x300x5 = 1.9x10-3T.
70.Zero.
71.Infinite.
72.Due to high current carrying capacity, an ammeter
is not damaged by excessive currents.
73.Due to low current carrying capacity, the
voltmeter will draw only a small part of the total
current. The potential difference(V=IR)to be
measured will not be much different from the
actual value.
74.Effective resistance of ammeter
RsR
R= g
AR+R .
s g
75.The resistance of the resulting ammeter will be less
than 0.010.
76.The value of shunt resistance should be reduced so
that more current may pass through it.
77.The force Facts in the direction of the vector
-> ->
vxBi.e.,perpendicular to the plane of the vectors
-> ->
vandB.
78.Angular frequency, 00=qB.It is independent of
m
angle8.
:.Ratio of the angular frequencies, ~:002= 1 :L
79.Radius, r=mv=L
qBqB
(
rqp e1
For samepandB, .1!.=-=-=-= 1 : 2
Ypqa2e2
80.Low torsional constant of the suspension wire ensures
high sensitivity of the moving coil galvanometer.
81.The eddy currents set up in the metallic material
oppose the motion of the coil in the magnetic field
and hence bring it to rest at once.
->
82.The fieldBdue to currentIacts along -ve
z-direction. By Fleming's left hand rule, the charge
+Qwill experience a force along +ve y-direction.
83.FA=q'1Bsin 90° =.!
FB2qv2Bsin 90° 2
5..=1: 1.
v2
84.By Fleming's left hand rule, the magnetic field acts
along=z-axis.
85.By Fleming's left hand rule, the magnetic field is
directed along +z-axis.www6notesdrive6com

412. 86.Radius,r=mv=L
qB qB
F d Brp __qd _-_e
_-1 .'1.
or same pan,
rdqpe
"YPE B :SHORT ANSWER QUESTIONS (2 or 3 marks each)
1.
MAGNETIC EFFECT OF CURRENT
Briefly describe Oersted's experiment leading to
the discovery of magnetic effect of current. State
Ampere's swimming rule.
2.State Biot-Savart's law in vector form expressing
->
the magnetic field due to an elementdlcarrying
currentIat a distance"? from the element. How will
you find the direction of the magnetic field ?
[ISCE 93; CBSED02 C; OD14C]
3.A wire of length Lis bent into a semi-circular loop.
Use Biot-Savart law to deduce an expression for the
magneitc field at the centre due to currentIpassing
through it. [CBSED11 C]
4.Using Biot-Savart's law, derive the expression for
the magnitude of the magnetic field at the centre of
a circular loop of radiusrcarrying a steady current
1.Draw the field lines due to the current loop.
[ISCE 96; CBSED01C;OD14 C]
5.State Biot-Savart law. Deduce the expression for the
magnetic field at a point on the axis of a current
carrying circular loop of radius'R,distant'x'from
the centre. Hence write the magnetic field at the
centre of a loop. [CBSED05;ODOS,15]
6.State Ampere's circuital law and prove this law for
a circular path around a long current carrying
conductor.' [Himachal 98;Haryana 98C 01]
7.Using Ampere's circuital theorem, calculate the
magnetic field due to an infinitely long wire
carrying currentI. [CBSEOD90]
8.A long solenoid with closely wound turns hasn
turns, per unit of its length. A steady currentIflows
through this solenoid. Use Ampere's circuital law
to obtain an expression, for the magnetic field, at a
point on its axis and close to its mid point. Draw its
field lines. [CBSED 04C,14C]
9.(a)How is a toroid different from a solenoid?
(b)Use Ampere's circuital law to obtain the
magnetic field inside a toroid.
(c)Show that in an ideal toroid, the magnetic field
(i)inside the toroid and(ii)outside the toroid
at any point in the open space is zero.
[CBSEOD08,14C]
10.(a)State Ampere's circuital law, expressing it in the
integral form.
(b)Two long coaxial insulated solenoids,qandSz
of equal lengths are wound one over the other as
4.111
87.By right hand rule, the direction of
moment will be along +ve X-direction.
-. 1\ 21\
m=IA(+i)=Iai
magnetic
shown in the figure. A steady current"I"flows
through the inner solenoidqto the other end B,
which is connected to the outer solenoidSzthrough
which the same current"I"flows in the opposite
direction so as to come out at endA.If 11and ~ are
the number of turns per unit length, find the
magnitude and direction of the net magnetic field
at a point(i)inside on the axis and(ii)outside the
combined system. [CBSED14]
Fi. 4.170
11.A long straight wire of a circular cross-section of
radius'a'carries a steady current' I'. The current is
uniformly distributed across the cross-section.
Apply Ampere's circuital law to calculate the
magnetic field at a point'r' in the region for(i)r<a
and(ii)r>a. [CBSED10]
->
12.Write anexpression for forceFacting on a chargeq
moving with a velocity7!in the region, where
->
magnetic induction B is uniform. How does the
speed change, as the charge moves ? Under what
->
circumstances the forceFshall be zero?[ISCE96]
13.Write an expression for the force on a charge
moving in a magnetic field. Use this expression to
defirie the SI unit of magnetic field.
[CBSE D 08C]
14.Consider the motion of a charged particle of mass'
mand charge 'q 'moving with velocity7!in a
->
magnetic field B .www4yztp•o·tvp4nzx

413. 4.112
--> -->
(a)Ifvis perpendicular
to B, show that it
describes acircular path having angular
frequency 00=qB /m:
•-->
(b)Ifthe velocity vhas a component parallel to the
-->
magnetic field B, trace the path described
by the particle. Justify your answer.
[CBSED 14C)
-->
15.A uniform magnetic field B is set up along the
positive x-axis. A particle of charge' o'and mass' mr
-->
moving with a velocity venters the field at the origin
inX-Y plane such that it has velocity components
bothalong and perpendicular to the magnetic field
-->
B . Trace, giving reason, the trajectory followed by
the particle. Find out the expression for the distance
moved by the particle along the magnetic field in
one rotation. [CBSEOD 15]
16.A steady magnetic field cannot change thekinetic
energy of a movingcharged particle, it can deflect
thecharged particle sideways. Comment.
17.A charged particle moving withauniform velocity
-->
venters a region where uniform electric and
--> -->
magnetic fields E and B are present. It passes
through the region without anychange in its velocity.
What can we conclude about the (i)relative
-4' ----. ~ ----+
directions of B, u,and B ?(ii)magnitudes of E
-->
and B. [CBSESamplePaper08]
18.A hydrogen ion of mass 'niand charge'c(travels
with aspeed'dalong a circle ofradius 'r' in a
uniform magnetic fieldof flux density 'H.Obtain
the expression for the magnetic force on the ion
and determine its time period. [CBSED 03C ;OD04]
19.Show that the frequency of revolution, of acharged
particle (in theX-Yplane), in a uniform magnetic
-->--> A
field B (B=Bk ),is independent of its speed.
Which practical machine makes use of this fact?
What is the frequency of the alternating electric
field, used in this machine? [CBSED 09C]
20.Draw a schematic sketch of a cyclotron. Explain
clearly the role of crossed electric and magnetic
fields in accelerating the charge. Hence derive the
expression for the kinetic ener acquired by the
particles. [CBSEOD 13]
21.Derive an expression for the maximum force
experienced by a straight conductor of length I,
carrying currentIand keptin a uniform magnetic
field, B. [CBSED 06C)
PHYSICS-XII
22.Briefly describe an experiment to show the
existence of arepulsive forcebetween two parallel
conductors carrying currents inopposite
directions.
23.Derive the expression forforce per unit length
between two long straight parallel current carrying
conductors. Hence define oneampere.
[CBSED 01, 09]
24.Derive a formula for theforcebetween two parallel
straight conductors carrying current inopposite
directions and write the nature of the force. Hence,
define anampere. [CBSEOD98]
25.Derive an expression for the torque on a
rectangular coil of area A,carrying a current Iand
placed in a magnetic field B.The angle between the
direction of Band vector perpendicular to the plane
of the coil ise.Indicate the direction ofthe torque
acting on the loop. [CBSEF 09]
26.Arectangular coilofsides 'I' and 'b'carrying a
-->
currentIis subjected to a uniform magnetic field B
acting perpendicular to its plane. Obtain the
expression for the torque acting on it.
[CBSED 14C)
27.A rectangular loop of area A,having N turns and
carrying a current ofIampere is heldina uniform
magnetic field B. (i)Write the expression forthe
maximum torque experienced by the loop. (ii)In
which orientation, will the loopbeinstable
equilibrium ? [CBSEOD98C)
28.State the principle of amoving coil galvanometer.
Show that the current passing through the coil is
directly proportional to the deflection of the coil.
[Haryana02]
29.Amoving coilgalvanometer consists of a rectan-
gular coil of N turns, each of areaA,suspended in a
radial magnetic field of flux density B. Derive the
expression for the torque on the coil,when current I
passes through it.Draw suitable labelled diagram.
[CBSED 93C]
30.A moving coil galvanometer of resistance Ggives a
fullscale deflection for acurrent Ig'Usethe suitable
circuit diagram to convert it into an ammeter of
range 0 toI (l>19)'Deduce the expression for the
shunt requried for this conversion. Hence write the
expression for the resistance of the ammeter thus
obtained. [Punjab2000; CBSED 09C]
31.Explain how will you converta galvanometer into a
voltmeter to readamaximum potential difference
ofVvolts. Can one use avoltmeter to measure the
emf of a cell? Justify your answer.
[CBSEOD97C, F 98]www4yztp•o·tvp4nzx

414. MAGNETIC EFFECT OF CURRENT
Answers
4.113
•
1.Refer answer toQ.2
on page 4.1.
2.Refer answer toQ.3 on page 4.2.
3.Refer to solution of Example 17 on page 4.16.
Here L=1trorr=L/1t
B=~oI
4r
_ ~ ol1t_1t ~ 01
-4·r:-4L
4.Refer answer toQ.7 on page 4.12. See Fig. 4.25.
5.Refer answer to Q. 8 on page 4.13. See Fi. 4.25 on
page 4.14.
6.Refer answer to Q. 9 on page 4.22.
7.Refer to solution of Example 33 on page 4.26.
8.Refer answer toQ.10 on page 4.23.
9.Refer answer toQ.11 on page 4.24.
10.(a)fB.di=~ol
(b)(i)Magnetic field inside solenoid ~ '
1\=~o'\I
Magnetic field inside solenoid ~,
~ =~o11z1
As the currents in the two solenoids are oppositely
directed, so direction of ~ is opposite to that of1\.
The net magnetic field at any inside point along
the axis,
B=1\ - ~=~o('\-11z)I
(ii)Outside the combined system, net magnetic
field=O.
11.Refer to the solution of Example33(i)on page 4.26.
---> --->--->
12.Force, F=q (vx B ). The speed of the charge is not
--->
affected in the magnetic field. Force F will be zero
if:;=0 or if:; is parallel or antiparallel toB .
13.Refer answer toQ.12 on page 4.28.
14.(a)Magnetic force acts on the charged particle in a
---> --->
direction perpendicular to bothvand B and
provides centripetal force.
Magnetic force,qvBsin 90°
mv2 mv
=Centripetal force,-- or r= -
r qB
.. 00=!!. =qB
rm
(b)Refer answer toQ.15(3) on page 4.34.
15.Refer answer toQ.15(3) on page 4.34
16.A magnetic field exerts force in a direction perpen-
dicular to the direction of motion of the charge. No
work is done by the magnetic force on the moving
charge. So the kinetic ener of the charged particle
is not affected. The perpendicular magnetic force
only deflects the charged particle sideways.
---> --->
17.(i)The directions ofEand B are perpendicular to
each other and also perpendicular to the direction
of:; so that electric and magnetic forces are in
opposite directions.
---> --->
(ii)The magnitudes ofEand B should be such that
qE= qvB
E
or v=-.
B
18.In the uniform ,?agnetic field, the magnetic force
on the hydrogen ion acts perpendicular to bothv
andB.
F=evBsin 90° =evB
or
Magnetic force on=
the hydrogen ion
mv2
evB=--
r
Centripetal
force
mv
or r=-
eB
Time period of hydrogen ion,
T _21tr_21tmv_Znm
--;---:;.-;a-d3.
19.Refer answer toQ.15(2) on page 4.33.
We obtain,L=qB
21tm
A cyclotron makes use of this fact in which
alternating electric field of frequencyIeis applied.
20.Refer answer toQ.17 on page 4.40.
21.Refer answer toQ.19 on page 4.44.
22.Refer answer to Q. 20 (Experiment 2) on page 4.49.
23.Refer answer toQ.21 on page 4.49.
24.Refer answer toQ.21 on page 4.49.
25.Refer answer to Q. 22 on page 4.53.
26.Magnetic moment associated with the current
carrying coil is
---> " "
m=IAn= Ilbn
where ~ is a unit vector perpendicular to the plane
of the coil.www5notesdrive5com

415. 4.114
4441\4
44
Torque,1"=mxB=IlbnxB=lIb0=0
This is because ~ andBare either parallel or
antiparallel vectors.
27.Refer answer toQ.22 on page 4.53. The loop will
be in stable equilibrium whenntis parallel toB.
28.Refer answer toQ.23 on page 4.57.
PHYSICS-XII
29.Refer answer toQ.23 on page 4.57.
30.Referanswer toQ.26 on page 4.63.
31.Refer answer toQ.28 on page 4.64.No, avoltmeter
cannot be used to measure the emf acell.A
voltmeter requires a small current forits operation.
It measures p.d. in a closedcircuit, which is less
than the emf of the cell.
~YPE C :LONG ANSWER QUESTIONS (5 marks each)
1.(a)Writeany two important points of similarities
and differences each between Coulomb's law
for the electrostatic field and Biot-Savart's law
for the magnetic field.
(b)Use Biot-Savart's law to find the expression for
the magnetic field due to acircular loop of
radius'r' carrying current' I "at its centre.
[CBSE F 15)
2.State Biot-Savart law expressing it in vector form.
Use it to obtain the magnetic field, at an axial point,
distant r from the centre of a circular coil of radius a
carrying a currentI.Hence, compare the magnitudes
of the magnetic field of this coil at the centre and at
an axial point for which r=j3a. [CBSE SP 08)
3.(a)Use Biot-Savart law to derive the expression
for the magnetic field due to a circular coil of
radiusRhavingNturns at a point on the axis
at a distance' x'from its centre.
Draw the magnetic field lines due to this coil.
(b)A current' I'enters a uniform circular loop of
radius' Kat pointMand flows out atNas
shown in the figure.
Obtain the net magnetic field at the centre of
the loop. [CBSE DISC]
Fig.4.171
4.(a)Write, using Biot-Savart law, the expression for
~
the magnetic field elementdlcarrying current
Iat a distancer"from it in a vector form.
Hence derive the expression for themagnetic
field due to a current carrying loopof radius R
atapointPdistantxfromits centre along the
axis of the loop.
(b)Explain howBiot-Savart law enables one to
express the Ampere's circuital law inthe
integral form,viz.,
fiLii=f.loI,
whereIis the totalcurrent passing through
thesurface. [CBSE OD 15)
5.(a)State Ampere's circuital law connecting the
~
line integral ofBover aclosed path tothenet
current crossing the area bounded by thepath.
(b)Use Ampere's law to derive theformula for the
magnetic field due to an infinitely long straight
current carrying wire.
(c)Explain carefully why thederivation as in(b)is
not valid for magnetic field inaplane normal
toacurrent-carrying straight wire of finite
length and passing through the midpoint of
the axis.
6.(a)Show how Biot-Savart law can be alternatively
expressed in the form of Ampere's circuital
law.Use this law toobtain the expression
for the magnetic field inside a solenoid of
length'1',cross-sectionalarea'A'having' N'
closely wound turnsandcarrying a steady
current T',
(b)Sketch the magnetic field lines for a finite
solenoid. Explain whythe field atthe exterior
midpointis weak while at theinterior it is
uniformandstrong. [CBSED06C, 15C]
7.(a)State Ampere's circuital law. Use this law to
obtain the expression for the magnetic field
inside an air cored toroid ofaverage radius' r',
having' n'turns per unit length andcarrying a
steady currentI.Show thatthe magnetic field
in the openspace inside andexterior to the
toroid is zero.www4yztp•o·tvp4nzx

416. MAGNETIC EFFECT OF CURRENT
(b)An observ
er to the left of a solenoid of N turns
each of cross-section area'A'observes that a
steady currentIin it flows in theclockwise
direction. Depict the magnetic field lines due
to thesolenoid specifying its polarity and
show that it acts as a bar magnet of magnetic
momentm=NIA. [CBSEOD 13, D 15]
A/'-;j)'-;jJ'
I I I I I
i I I I I
'T \ \ \ \
\, \ \ \
N ...
Fig.4.172
8.(a)Using Ampere's circuital law, obtain the
expression for the magnetic field due to a long
solenoid at a point inside the solenoid on its
axis.
(b)In what respect is a toroid different from a
solenoid ? Draw and compare the pattern of
the magnetic field lines in two cases.
(c)How is the magnetic field inside a given
solenoid made strong? [CBSEOD 11]
9.Derive a mathematical expression for the force
acting on a current carrying straight conductor kept
inamagnetic field. State the rule used to determine
thedirection of this force. Under what conditions is
this force (i)zeroand(ii)maximum ?
[CBSED 97C, 98]
10.Drawaschematic sketch of a cyclotron. Explain
briefly how itworks and how it is used to accelerate
thecharged particles.
(i)Show that time period of ions in a cyclotron is
independent of both the speed and radius of
circular path.
(ii)What is resonance condition? How is it used to
accelerate thecharged particles?
[CBSED 08;OD09]
11.With the helpof a labelled diagram, state the
underlying principle of a cyclotron. Explainclearly
how it worksto accelerate the charged particles.
Showthatcyclotron frequency is independent of
energy of the particle. Is there an upper limit on the
ener acquired by the particle? Give reason.
[CBSED11, 14,14C]
12.(a)Derive an expression for the force between two
long parallelcurrent carrying conductors.
(b)Use this expression to define SIunit of current.
(c) A long straight wireABcarries a currentI.A
protonPtravels with a speedv,parallel to the
wire, at a distancedfrom it in a direction
4.115
opposite to the current as shown in Fig. 4.173.
What is the force experienced by the proton
and what is its direction?[CBSED 06; OD 10]
B
A
P
d--+.Proton
1
v
I
Fig.4.173
13.Derive an expression for the torque acting on a loop
of N turns, areaA,carrying currentI,when held in
a magnetic field B. With the help of a circuit diagram,
show how a moving coil galvanometer can be
converted into an ammeter of given range. Write
the necessary mathematical formula.[CBSED 04]
14. (a)Two straight long parallel conductors carry
currents/1and12in the same direction.
Deduce the expression for the force per unit
length between them.
Depict the pattern of magnetic field lines
around them.
(b)A rectangular current carrying loopEFGHis
kept in a uniform magnetic field as shown in
Fi.4.174.
N
ill
H G
5
Fi. 4.174
(i)What is the direction of the magnetic
moment of the current loop?
(ii)When is the torque acting on the loop (A)
maximum, (B) zero?[CBSEOD 05, 09]
15.(a)With the help of a diagram, explain the principle
and working of a moving coil galvanometer.
(b)What is the importance of a radial magnetic
field and how is it produced?
(c) Why isit necessary to introduce a cylindrical
soft iron core inside the coil of a galvanometer?
(d)"Increasing the current sensitivity of a
galvanometer may not necessarily increase its
voltage sensitivity". Justify this statement.
[CBSED 06, 13C;OD 14,14C, 15]www5notesdrive5com

417. 4.116
Answers
PHYS
ICS-XII
•
1.(a)Refer answer toQ.4on page4.3.
(b)Refer answer toQ.7on page4.12.
2.Referanswer to Q. 8 on page 4.13.
J.!Iaz
B -_-"...0"--...- •..••
axial - 2 (,z + aZ)3/Z
Z
B = J.!oIa =J.!oI
r;J'3a 2(3az+aZ)3/Z 16a
J.!oIaz J.!°I
B - - [Put, =0]
centre -2(Oz + aZ)3/Z -2a
Bcentre _16a _8
..Br;J'3a-z;;--.
3.(a)Refer answer to Q.8on page4.13.See Fig. 4.25.
(b)At point M, let the current Ibedivided into two
parts :IIalong the smaller part andIzalong the
larger part of the loop.
Fi. 4.175
Field due toIIat 0,
B-.!.J.!OII normally into thepaper.
1-4'2R'
Field due to 1z at 0,
~3J.!I
~ =-.~, normally out of the paper
42R
->-> ~
Net field at 0, B= ~+~
IIiI=.!.J.!011 _ ~J.!°Iz
4 2R 4 2R
As the resistance of the larger part is 3 times the
resistance of the smaller part, soII=3Iz
~
Hence, IBI=O.
4.(a)Refer answer toQ.80-1page4.13.
(b)Biot-Savart law can be expressed as Ampere's
circuital law by considering the surface to be made
up of a large number of loops. The sum of the
tangential components of the magnetic field
multiplied by the length of all such elements, gives
the resultfB.di=J.!01.
5.(a)Refer answer to Q. 9 on page4.22.
(b)Refer tosolution of Example 33on page4.26.
(c) A straight conductor of finite length cannot by
itself form a complete steady currentcircuit.
Additional conductors are necessary to close
the circuit. ThesewiII spoil thesymmetry of
the problem. The difficulty disappears if the
conductor is infinitely long.
6.(a)Refer answer to Q.9 on page 4.22and Q. 10on
page4.23.
(b)The magnetic field due to the neighbouring
turns add up along theaxisof thesolenoid and
tend to cancel out in the perpendicular
direction. Thus the field at the exterior
mid-point is weak and at the interior, it is
uniform and strong.
7.(a)Referanswer to Q. 9. on page 4.22and Q. 11on
page 4.24.
(b)
~0-;~'~--;~.
-=====---~~--_~I~(-~ __~~---------~~~~~
, \ \ \
...
Fig.4.176
Thesolenoid consists of N loops, each ofareaA
and carrying a current 1. Each loop acts as a
magnetic dipole of dipole moment m=1A.As
the magnetic momentsofall loops arealigned
along thesame direction, so the net magnetic
moment ofthesolenoid isNIA.
8.(a)Refer answer to Q.10on page 4.23.
(b)Asolenoid bent into the form of aclosed ring
iscalled atoroidal solenoid. The field pattern of
solenoid is similar to that of a bar magnet. The
field lines inside a toroid are circular loops and
the field is uniform everywhere inside the toroid.
See Fi.4.48on page4.23and Fi. 4.52on page4.24.
(c) The field inside a solenoid can be increased by
(i)inserting an iron core inside it
(ii)increasing number of turns per unit length, and
(iii)increasing the current through the solenoid.
9.Refer answer to Q. 19 on page4.44.
10.Refer answer to Q.17on page4.40.
11.For cyclotron, refer answer to Q.17on page 4.40.
Cyclotron frequency, f.=~
cZnmwww4yztp•o·tvp4nzx

418. MAGNETIC EFFECT OF CURRENT
AsIeis
independent of velocity v,soIeis inde-
pendent of the kinetic energy of the particle.
According to Einstein's special theory of relativity,
the mass of a particle increases with its velocity. At
high velocities the cyclotron frequency will
decrease due to increase in mass. This will throw
the particle out of resonance with the oscillatory
field. Hence the particles are not accelerated
further.
12.(a),(b)Refer answer toQ.21 on page 4.49.
(c)The field due to currentIat point Pis,
B=iloI
2nd
Thisfieldactsnormally into the plane of paper.
According to Fleming's left hand rule, a force acts
on the proton in a direction away from wire AB.
F=evBsin90o=eviloI =ilolev.
2nd Znd
4.117
13.Refer answer toQ.22 on page 4.53.
14.(a)Refer answer to Q.21on page 4.49.
(b)(i)According to right hand thumb rule, the
direction of the magnetic moment of the
current loop will be normally into the plane
of the paper.
(ii)Torque acting on the loop ismaximum
when its plane isparallel to the magnetic
field. Torque acting on the loop iszero
when its plane isperpendicular to the
magnetic field.
15.(a)Refer answer to Q.23 on page 4.57.
(b)Refer to the solution of Problem 29 on page 4.85.
(c) A soft iron core makes the field radial. It also
increases the strength of the magnetic field and
hence increases the sensitivity of the
galvanometer.
(d)Refer to the solution of Problem 33 on page 4.86.
""YPE D :VALUE BASED QUESTIONS (4 marks each)
1.Dimpi's classwasshown a video on effects of
magnetic field on a current carrying straight
conductor. She noticed that the force on thestraight
current carrying conductor becomes zero when it is
oriented parallel to the magnetic field and this force
becomes maximum when it is perpendicular to the
field. Sheshared this interesting information with
hergrandfatherin the evening. The grandfather
could immediately relate it to something similar in
real life situations. He explained it to Dimpi that
similar things happen in real life too. When we
align and orient our thinking and actions in an
adaptive.and accommodating way, our lives
become more peaceful and happy. However, when
weadopt an unaccommodating and stubborn
attitude, life becomes troubled and miserable. We
should therefore always be careful in our response
todifferent situations in life and avoid unnecessary
conflicts. [CBSESamplePaper 15]
Answer thefollowing questions based on above
information:
(a)Express the force acting on a straight current
carrying conductor kept in a magnetic field in
vector form. State the rule used to find the
direction of this force.
(b)Which one value is displayed and conveyed by
(i)grandfather as well as
(ii)Dimpi ?
(c)Mention one specific situation from your own
lifewhich reflects similar values shown by you
towards your elders.
2.Deepak was performing an experiment on poten-
tiometer in his practical period. Unfortunately, a
galvanometer fell from his hands and broke. He
was sad and his friend advised him not to tell the
teacher about that incident. But Deepak went to his
teacherand narrated the incident. The teacher
heard him patiently and on finding that it was not a
Deepak's fault but just an accident, did not scold
him. Instead, he used the broken galvanometer to
explain its internalconstruction to the entire class.
Based on the above paragraph, answer the following:
(a)What were the values displayed by Deepak ?
(b)State the basic principle of a moving coil
galvanometer.
3.Kamal's uncle was advised by his doctor to
undergo an MRlscan test of his chest and gave him
an estimate of the cost. Not knowing much about
the significance of this test and finding it to be too
expensive he first hesitated. When Kamal learnt
about this, he decided to take help of his family,
friends and neighbours and arranged for the cost.
He convinced his uncle to undergo this test so as to
enable the doctor to diagnose the disease. He got
the test done and the resulting information greatly
helped the doctor to give him proper treatment.
[CBSEF13]www5notesdrive5com

419. 4.118
Based on
the above paragraph, answer the following:
(a)What according to you, are the values
displayed by Kamal and her family, friends
and neighbours?
(b)Assuming that the MRI scan of her uncle's
chest was done by using a magnetic field of
,
Answers
PHYSICS-XII
0.1 T, find the maximum and minimum values
atforce that this magnetic field could exert on
a proton (charge=1.6x10-19 C) moving with a
speed of 104 m/s.State the condition under
which the force can be minimum.
•
-> ->->
1.(a)F=J(IxB )
The direction of the force is given by Fleming's
left hand rule. For statement, refer to point 11
of Glimpses on page 4.118.
(b)(i).Adaptation to different situations and
flexible and adjustable attitude. (ii)Sharing
excitement in classroom learning with family
members.
(c) Avoiding unnecessary arguments in con-
flicting situations ineveryday life.
2.(a)Courage to tell the truth and gratitude to the
teacher for his patience and tolerance.
(b)The basic principle of moving coil galvano-
meter is that a current-carrying coil placed in a
magnetic field experiences a torque, the
magnitude of which depends on the strength
of current.
3.(a) (i)Presence of mind;
High degree of general awareness;
Ability to take prompt decisions;
Concern for her uncle;
(ii)Empathy;
Helping and caring nature.
(b)Maximum force
=qvB= 1.6x10-19 xl04xO.l
=1.6x10-16N
->
Force is maximum when-;.LB.
Minimum force=0
->
Force is minimum when-;11B.www4yztp•o·tvp4nzx

420. Magnetic Effect of Current
GLIMPS
ES
1.Oersted observation.A compass needlesuffers a
deflection when it is placed near a wire carrying
an electric current. When the direction of
current is reversed, the direction of deflection of
the needle also reverses. This conclusively
proves that a current carrying conductor
produces a magnetic field around it. This is
calledmagnetic effect ofcurrent.
2.Biot-Savart law. According to this law, the
magnetic field due to a current carrying element
dzcarrying currentIat a point P at distance r
from it is given by
dB=110.Idlsin8
4n-?
Y I
->
dB
P
->
r
x
Invector notation, dB=110I.dzxr
4n r3
Here 8 isthe angle betweendzandrand
110=4nx10-7Tm A-1,isthe permeability of
,
free space. The direction ofdBis same as that of
dzxrand is given by right-hand screw rule.
3.Magnetic field due to straight current carrying
conductor.The magnetic field at a point at
perpendicular distance' a'from a straight con-
ductor carrying currentIisgiven by
B=110I(sin<1\+sin<12)
4na
where <1\and<12are the angles which the
perpendicular from the observation point to the
conductor makes with the lines joining the ends
of the conductor to the observation point.
The magnetic field due to a straight conductor
of infinite length (<1\=<12= ~)is given by
B=110I
2na
4.Rules for finding the direction of magnetic field
due to straight current carrying conductor.
(i) Right hand thumb rule. If we holdthe
straight conductor in the grip of our right
hand insuch a way that the extended
thumb points in the direction of current,
then the direction of curl of fingers will give
the direction of the magnetic field.
(ii)Maxwell's cork screw rule. If a right
handed screw be rotated along a wire so
that it advances in the direction of current,
then the direction in which the thumb
rotates gives the direction of the magnetic
field.
(4.119)wwwnnotesdrivencom

421. 4.120
5.Magnetic field of a circular current loop.The
magnetic field of a circular current loop of
radiusacarrying currentIis
(i)At the centr
eof theloop: B=!-t0I
2a
(ii)Atan axial point at distance rfrom the centre:
B !-to Ia2 .
2(?+a2)3/2 •
6.Rules for finding the direction of magnetic field
due to circular current loop.
(i)Right hand thumb role.If we curl the
fingers of our right hand around the
circular wire with the fingers pointing in
the direction of the current, then the
extended thumb gives the direction of the
magnetic field.
(ii)Clock role.This rule gives the polarity of
anyface ofthe circular current loop. Ifthe
current round any face of the coilisin
anticlockwise direction, it behaves like a
north pole. If the current flows in the
clockwise direction, it behaves like a south
pole.
7.Ampere's circuital law.The line integral of the
~
magnetic field B around any closed path is
equal to!-t0times the totalcurrentIthreading
the closed path.
fB.1z=!-toI
In a simplified form, the law states that if fieldB
is directed along the tangent to every point on
the perimeterLof a closed curve andits
magnitude is constant along the curve, then
BL=!-toI.
8.Magnetic field of a straight solenoid.A solenoid
is a long insulated wire wound in the form of a
helix. Its length is very large as compared to its
diameter. The magnetic field of a straight
solenoid carrying currentIand havingnturns
per unit length is given by
(i)B=!-to nI
(At a point well inside the solenoid)
(ii) 1
Bend =-!-to nI
2
(At either end of the solenoid)
9.Magnetic fi~ld of a toroidal solenoid.A sol~noid
bent into the form of a closed ring is called a
PHYSICS-XII
toroidal solenoid. The magnetic field inside the
toroidal solenoid has a constant magnitude and
tangential direction. It is given by
B=!-to nI
whereIis the current in the windings and nis
the numberofturnsperunit length. The field
lines are concentric circles.
10.Force on a charge moving in a magnetic field.A
charge qmoving with velocity 7!atan angle 8
~
with the magnetic field B experiences the
magnetic Lorentz force,
F=qvBsin8
~ ~
In vector notation, F =q(7!xB)
The directionofthis force is perpendicular to
~
bothVand B, and work done by it is zero. This
force ismaximum when the charged particle
moves perpendicular to the direction of the
field (8 =90°) and minimum when the charged
particle moves alongthe field (8 =0°).
11.Rules for finding the direction of force on a
charge moving perpendicular to a magnetic
field.
(I)Fleming's left hand role. Stretch the thumb
and the first two fingers of theleft hand
mutually perpendicular to each other. If the
forefinger points in thedirection of the
magnetic field, central finger in the direction of
current, then the thumb gives the direction of
force on the charged particle.
(ii)Right hand palm role. Open theright hand
and place it so that tips of thefingerspoint in
~
the direction of the field B and thumb in the
direction of velocity 7!of positive charge, then
~
the palm faces towards the force F.
12.Definition of magnetic field.The magneticfield
at a point may bedefined as the force acting on
a unit charge moving with aunitvelocity at
right angles to the directionofthe field.
13.SI unit of magnetic field is tesla.One tesla is that
magnetic field inwhich a charge of 1 coulomb
moving with a speed of I ms-I at right angles to
the field experiences a force of 1 newton.
1 tesla (T)=1 N A-lm-1,
1 gauss (G) =10-4T.www=notesdrive=com

422. MAGNETIC EFFECT OF CURRENT (Competition Section)
14.Lorentz force. The total force
,called Lorentz
force, acting on a chargeqmoving with velocity
--+ --+
vin an electric fieldEand magnetic fieldBis
--+ --+ --+
F=q(E+itx B)
15.Motion of charge inside an electric field.If a
potential differenceVis applied between two
parallel plates separated by distanced,then
electric field set up between the plates is
E=V
d
The chargeqof massmexperiences the electric
force, Fe=qE
Acceleration produced in the charge,a=qE
m
The moving charge follows a parabolic path
inside the electric field.
16.Motion of charge inside a magnetic field.
--+
(i)Whenit1.B, the magnetic force on the charge
makes it move along acircular path of radius,
mv
r=-
qB
--+
(ii)Whenitmakes angleewith B, the
perpendicular component :v1.=vsineof the
initial velocity makes the charge move along a
circular path of radius,
mvsine
r=---
qB
The parallel component :vII=vcoseof the
initial velocity makes it move along the
direction of the magnetic field. The resultant of
the two components makes the charge move
along a helical path of pitch,
h=2rrmvcose .
qB
17.Cyclotron.It is a device used to accelerate
charged particles like protons, deutrons,
a -particles, etc. to very high energies. Here
charged particles move along a spiral path
under the action of a perpendicular magnetic
field and gain energy as they cross an
alternating electric field again and again.
18.Cyclotron frequency.In a cyclotron, the
frequency of the applied alternating electric
field is equal to the frequency of revolution of
the charged particle. This frequency is called
cyclotron frequency.
4.121
It is given by, f.=~
c2rrm
wheremisthe mass andqis the charge of the
positive ion. The cyclotron frequency is
independent of both the velocity of the particle
and the radius of its orbit.
19.Maximum energy gained by positive ions.IfVo
androare the maximum velocity and maximum
radius of the circular path of the positive ions in
a cyclotron, then
mv2 _qBto
_0_=qvoBorVo ---
'0 m
2W2
M· ki ti 1 2qro
:. aximum me c energy =2mvo=2m
IfVis the accelerating potential of the high
frequency oscillator and the charged particle com-
pletes nrevolutions before leaving the dees, then
Maximum kineticenergy =2nqV.
20.Force on a current carrying conductor in a
magnetic field.A conductor of lengthIcarrying
currentIheld in a magnetic fieldIfat an anglee
with it, experiences a force given by
F=IlBsine
--+ --+--+
In vector notation, F=I (IxB )
--+ --+
The direction ofFis perpendicular to bothI
--+
and B and is given by Fleming's left hand rule.
The force is maximum whene=90° and zero
whene=0°or 180°.
Fmax=IlB
21.Force between two parallel infinitely long
current carrying conductors. The force per unit
length between two long parallel conductors
carrying currentsIIand12and separated by
distance, is given by
f=110 .II 12
, 2rr ,
The force is attractive when the currents are in
the same direction and repulsive when the
currents are in opposite directions.
22.51 unit of current is ampere.One ampere"is that
strength of current which, when flowing in two
parallel infinitely long conductors of negligible
cross-section placed in vacuum at a distance of
1 m from each other, produces between them a
force of 2x10-7newton per metre length.wwwAnotesdriveAcom

423. 4.122
23.Force between two current elements.Theforce
between
two parallel current elementsdl1and
dl2carrying currents 11 and12and separated by
distance r much greater than their length is
given by
_ J..lo II12dl1dl2
dF- .--2-
411: r
24.Relative sizes of electric and magnetic fields.
Under similar conditions, the magnetic forces
are much smaller than the electric forces.
F v v
~n=VIv2(J..loeo)=?2
e
10-5x10-5=-10- 27
(3x108)2
HereVIandv2are drift speeds of the electrons
in the two conductors.
25.Torque on current carrying coil in a magnetic
field.Arectangular coil of area A,carrying
currentIand capable of rotation about an axis
~
perpendicular to the field Bexperiences a torque,
r=N1BAsin 9=mB sin 9
where N=number of turns in the coil,
m=NIA= magnetic dipole moment, 9=angle
which the normal to the plane of the coil makes
~
with the field B.
.~ ~ ~
Invector rotation, r=mxB
Torque is minimum when the plane of the coil is
perpendicular to the magnetic field (9 =0°).
Torque is maximum when the plane of the coil
isparallel to the magnetic field(9 =90°).
'max=NIBA.
26.Moving coil galvanometer. It is a device used to
detect current in a circuit. It is based on the
principle that a current carrying coil placed in a
magnetic field experiences a current dependent
torque, which tends to rotate the coil and
produces angular deflection. It consists of a coil
of wire of areaAand N turns carrying currentI
to be measured. It is suspended in a radial
magnetic field so that its plane always remains
~
parallel to B (sin 9=1) by a suspension fibre of
torsion constantk.In equilibrium position,
Restoring torque=Deflecting torque
orka=NIBA ora=NBA.1
k
i.e.,Deflection of coilex:Current in the coil.
PHYSICS-XII
27.Figure of merit of a galvanometer. It isthe
current which produces a deflection of one scale
division in the galvanometer. It isgiven by
G=i=_k_.
aNBA
28.Current sensitivity of a galvanometer.Itisthe
deflection produced in a galvanometer when a
unitcurrent flows through it.
C .... aNBA
urrent sensitivity=-=-- .
I k
29.Voltage sensitivity of a galvanometer. Itisthe
deflection produced ina galvanometer when a
unit potential difference is applied across its
ends.
V I ... a a NBA
o tage sensitivity = V=lR =kR
30.Conversion of a galvanometer into an ammeter.
Anammeteris an instrument usedto measure
electric current ina circuit. A galvanometer of
resistance Gcan be converted into anammeter
ofrange 0 - Iby connecting a small resistance S
in parallel with it. Thevalue of small resistance,
called shunt, is given by
I
S=-g-x G
1-I
g
whereIgis the current with which galvano-
meter gives full-scale deflection.
Total resistanceof anammeter,
R-~
A -G+S
An ammeteris alow resistancedevice and is
connected in series ina circuit.
31.Conversion of galvanometer into a voltmeter. A
voltmeter is used to measurepotential dif-
ference between any twopoints of acircuit. A
galvanometer of resistance G can be converted
into a voltmeter of range 0 -Vbyconnecting a
large resistance R inseries with it.Thevalue of
R is given by
R=~-G
Ig
Total resistance ofavoltmeter,
Rv=G+ R
Avoltmeteris ahighresistance device and is
always connected parallel tothe conductor
across which p.d. istobe measured.www4notzsyrivz4xom

424. C H APT E R
MAGNETISM
"""" ~.,., • • •• ~ ~ " > ~
. .
- >, .
5.
1
INTRODUCTION
1. What do you mean by the terms magnet and
magnetism?What are natural magnets?What is the
origin of the word magnetism? o
Magnets and magnetism.A magnet is a material that
has both attractive and directive properties.It attracts small
pieces of iron, nickel, cobalt, etc. This property of
attraction is calledmagnetism.When suspended frely,
a thin long piece of magnet comes to rest nearly in the
ggraphical north-south direction.
As early as the 6th century B.C., the Grs had
some knowledge ofnatural magnets. Thales of Miletus
kn that pieces of a naturally occurring iron ore,
lodestone or magnetiteor black iron oxide Fe304 had the
property of attracting small pieces of iron. The word
magnetismoriginates from the place-Magnesia -a
province in.the upper part of Grece -where this ore
was found. Later on the Chinese discovered that thin
long pieces of lodestone, if suspended horizontally and
frely with a string, would naturally orient themselves
roughly in the ggraphical north-south direction. In
fact, the word lodestone meansa leading stonewhich
expresses this directional property, valuable to
travellers. By about A.D. 1000, the Chinese were using
the magnetic compass for navigation. It wasWilliam
Gilbertwho, in his book'De Magnete'of 1600, first
suggested that the earth itself was a huge magnet,
causing the alignment of compass nedles.
In the early part of the ninetenth century,Oersted
discovered that moving charges or currents are the
sources of magnetic fields. However, the science of
magnetism was known long before the ninetenth
century.
5.2ARTIFICIAL MAGNETS
2. What are artificial magnets?What are their
common shapes?
Artificial magnets.Generally, the natural magnets
are not strong enough magnetically and have
s -:;.._-
/s
Bar magnet Magnetic nedle
S N
€t=!======--_@
Horse shoe
magnet
Ball ended magnet
Fig. 5.1
(5.1)
Different forms of artificial magnets.
Nwww4°±trsp€vvr4o±~

425. 5.2
inconvenient shapes.
The pieces of iron and other
magnetic materials can be made to acquire the properties of
natural magnets. Such magnets are called artificial
magnets.The main advantage of these magnets is that
they can be made much more stronger than the natural
magnets and also of any convenient shape and size
They are generally available in the following forms:
1.Barmagnet. It is a bar of circular or rectangular
cross-section.
2.Magnetic nedle
nedle having pointed ends and is pivoted at its centre
sothat it is fre to rotate in a horizontal plane.
3.Horse shoe magnet. It has the shape of a
horse-shoe and thus it hasben named so.
4.Ball-ended magnet. Itis a thin bar of circular
cross-section ending in two spherical balls.
5.3BASIC PROPERTIES OF MAGNETS
3. State the important properties of magnets.
Basic properties of magnets :
1. Attractive property. A magnet attracts small
pieces of iron, cobalt, nickel, etc. When a magnet is
brought near a heap of iron filings, the ends of the
magnet show the greatest attraction. These ends, where
the magnetic attraction is the maximum, are called
polesof the magnet. Thusevery magnet has two poles.
Fig. 5.2Polesofabar magnet.
2.Directive property. When a magnet is suspended
or pivoted frely, it aligns itself in the ggraphical
north-south direction. The pole of the magnet which
points towards the ggraphical north is called the
north-seeking or north (N) pole
points towards the ggraphical south is called the
south-seekingor south (S) pole of the magnet.
Fig.5.3 Amagnet points north-south when
freely suspended.
PHYSICS-XII
3. Like poles repel and unlike poles attract. If the
N-pole of a magnet is brought near the N-pole of a
suspended magnet, the poles are found to repel each
other. Two S-poles also repel each other. In contrast, N-
and S- poles always attract each other. This action can
be described by thelaw of magnetic poleswhich states
thatlikemagnetic poles repel, and unlike magnetic poles
attract each other.
(~
Repulsio~
Fig.5.4 Likepoles repel and unlike poles attract.
4. Magnetic poles always exist in pairs. If we try to
isolate the two poles of a magnet from each other by
breaking the magnet in the middle, each broken part is
found to be a magnet with N- and S-poles at its ends. If
we break these parts further, each part again is found
to be a magnet. So unlike electric charges, magnetic
monopoles do not exist. Every magnetexists as a dipole.
IN 5I
IN 571 N 51
IN5f/fC}7IN 5fiN5I
Fig. 5.5Polesalways exist in pairs.
5. Magnetic induction.A magnet induces mag-
netism in a magnetic substance placed near it. This
phenomenon is calledmagnetic induction.When N-pole
of a powerful magnet is placed close to a soft iron bar,
the closer end of the bar becomes S-pole and the farther
end N-pole
bar. Thusinduction precedes attraction.
For Your Knowledge
~Repulsion is the surer test of magnetism. A magnet
can attract another magnet. Also it can attract
magnetic substances like iron, nickel, cobalt, etc.
However, amagnet can repel another magnet only.
Sorepulsion is the surer testof magnetism.
5ASOME IMPORTANT DEFINITIONS
CONNECTED WITH MAGNETISM
4. Define the terms magnetic field, uniform magnetic
field, magnetic poles, magnetic axis, magnetic equator
and magnetic length with reference to a bar magnet.www4°±trsp€vvr4o±~

426. MAGNETISM
Some important definitions conn ected with
ma
gnetism:
1. Magnetic field.The space around a magnet within
which its influence ambeexperienced iscalledits magnetic field.
2. Uniform magnetic field.Amagnetic field in a
regionissaid to be uniform if it has same magnitude and
direction at all points of that region. At a given place, the
magnetic field of the earth can be considered uniform.
Thefield due to a bar magnet is not uniform.
A uniform magnetic field acting in the plane of paper
is represented by equidistant parallel lines [Fig. S.6(a)].
A uniform magnetic field acting perpendicular to the
paper and directed outwards isrepresented by dots
[Fig.S.6(b)].A uniform magnetic field acting perpen-
dicular to the plane of paper and directed inwards is
represented by crosses[Fig.S.6(c)].
~••••
XX X X
~••••••
X X X X
B
X X X X~••••
~• •••
X X X X
(a) (b) (c)
Fig. 5.6Representations of a uniform magnetic field.
3. Magnetic poles.These aretheregions of apparently
concentrated magnetic strength inamagnet where themagnetic
attraction ismaximum. The poles of a magnet lie some-
what inside the magnet and not at its gmetrical ends.
: Magnetic
: Iaxis
I0-- - - -- - --- -:--j --------0I
N ' 5
:Magnetic
:equator
Fig. 5.7 A bar magnet.
4. Magnetic axis.The line passing through the poles of
amagnet iscalled the magnetic axisof themagnet.
5. Magnetic equator.Theline passing through the
centre of themagnet and at right anglesto the magnetic axis
iscalled the magnetic equator of the magnet.
6. Magnetic length.Thedistance between the two
polesofa magnet iscalledthemagnetic length of the magnet.
Itisslightly less than the geometrical length of themagnet.
It is found that Magnetic length =0.84
Gmetrical length
I-- Magnetic length ---+!
oN 50
I<--- Geometrical length---I
Fig. 5.8 Magneticand gmetricallengths of a magnet.
5.3
5.5COULOMB'S LAW OF MAGNETIC FORCE
5. State Coulomb's law of magnetic force. Hence define a
unit magnetic pole.
Coulomb's law of magnetic forc.This lawstatesthat
theforceof attractionorrepulsion between two magnetic
polesisdirectly proportional to the product of their pole
strengths and inversely proportional to the square of the
distance between them.
Ifqnandqare the pole strengths of the two
I1 nI2
magnetic poles which are distance r apart, then the
force betwen them is given by
qnI1qm2
Fcc?-
or
wherekis a proportionality constant which depends
on the nature of the medium as well as on the system
of units chosen. For SI units and for poles invacuum,
!-Lqm1qnI2
F=-.Jl.
47t?-
wher!-Lois the permeability of fre space and is equal
to47tx10-7henry/metre
pole from Coulomb's law :
Ifq=q= 1unit; r= 1m,then
nil nI2
F=!-Lo=10-7 N
47t
Hence a unit magnetic pole may be defined as that
pole which when placed invacuum at a distance of one
metre from an identical pole repels itwith a force of10-7
newton.
5.6MAGNETIC DIPOLE AND MAGNETIC
DIPOLE MOMENT
6. Whatisa magnetic dipole? Give some examples.
Magnetic dipoleIn electricity, the fundamental or
simplest structure that can exist is apoint charge. Here
two equal and opposite charges separated by a small
distance constitute anelectric dipole,which is described
~
by an electric dipole momentp .In magnetism, isolated
magnetic poles do not exist. Here the simple structure
that can exist is themagnetic dipolewhich is described
~
bya magnetic dipole momentm.
Anarrangement of two equal and opposite magnetic
poles separated byasmall distanceiscalleda
magnetic dipole.···3wx£n~m}r°n3lxv

427. 5.4
Every bar
magnet is a magnetic dipole. A current
carrying loop behaves as amagnetic dipole. Even an
atom acts as a magnetic dipole due to the circulatory
motion of the electrons around itsnucleus.
7.Define magnetic dipole moment. Isitascalar or
vector quantity? Give its SI unit.
Magnetic dipole moment. Themagnetic dipole
momentof amagnetic dipole is defined as the product
of its pole strength and magnetic length. It isavector
quantity, directed from S-pole to N-pol.
-4
where qmisthepole strength and 2 Iisthe magnetic
length of the.dipole measured in the direction 5 -to
N-pole.
We shall se later on that the51unitof magnetic
dipole momentisampere metre2 (Am2)or joule per
tesla (JT - 1).
For Your Knowledge
~ Basic d pnbetwen elect 'icity and m gnetism.
In electricity a point charge is thesimplest sourceof
electricity and can be used as a test object also.By
measuring the force on a test charge at various places
in a given electric field, we can map out the entire
field. In magnetism, on the other hand, isolated
magnetic poles do not exist. Amagnetic dipoleis the
simplest source of a magnetic field. It can be used as a
test object for mapping a magnetic field. The mapping
isdone by measuring the torque experienced by a test
magnet oramagnetic dipole atvariouspoints in the
magnetic field.
In short, a small magnet is described by avectoriit
while an electric charge by thescalarcharge qon it. A
magnet experiences a torque in a magnetic field while
an electric charge experiences a force in an electric
field.
~The pole strengthqmis also calledmagnetic charge.
Thus we assign magnetic charge +qmtothe north
pole and-qmto the south pole.
~The direction of magnetic dipole moment iitisfrom
S-pole to N-poles to the direction of
the electric dipole momentpof an electric dipole
from negative charge to positive charge.
~When a magnet of pole strengthqrnis cut into two
equal parts.
(i)along its axis (longitudinally), the pole strength of
each half becomesq rn/2
(il)perpendicular to its axis (transversely), the pole
strength ofeach half still remains qrn•
PHYSICS-XII
Examples based on
'::,'/~:"~'~C9~loWp~~s1-~W.~~ri~~QJR~let.: -~;v:'~1
. .' .... "Momentofa ';iagnet<__ 't:'~';o:,if]
Formulae Used
1. Magnetic dipole moment,m=qrnX21
110q,"J«:
2. Coulomb's law, F=_ . 2
41t r2
Units Used
Pole strength is in Am, force in nton, distance
in metre.
Constant Used
110=41tx10-7TmA-1.
Example 1. Two magnetic poles, one ofwhichisfour times
stronger than the other,exert aforceof5gfoneach other
when placed at a distance of 10em.Find the strength of each
pole.
Solution. Let the pole strengths of the two dipoles
beqmand4qm'
Here F=5 g f =5x10-3kgf =5x10-3x9.8 N,
r=lO ern=0.1m
Using Coulomb's law of magnetism,
F=110qmlqm2
41t .r2
5x10-3x9.8 10-7 xqmX4qm
(0.1)2
q2=5x9.8x(0.1)2 x104=25x49
III 4
or
or qlll=5x7=35Am
4qm=4x35=140Am.and
Example 2.Two similar magnetic poles,having pole
strengths in the ratio 1: 2are placed 1m apart. Find the
pointwhere a unit pole experiences no net force duetothe
two poles.
Solution. Let the pole strengths of the two magnetic
poles beqmand 2qm'Suppose the required point is
located at distance xfrom the first pole at this
point,
Force on unit pole due to first pole
=Force on unit pole due to second pole
i-loqmx1 _i-lo2qlllX1
or 41t--.;z--41t'(1-xl
2x2=(I-xl
1
x=~=0.414m.
1+,,2
or or.fix=I-x
orwww4°±trsp€vvr4o±~

428. MAGNETI
SM
Example3.Calculate theforce acting between two
magnets of length 15emeach and pole strength 80Ameach
when the separation between their northpolesis10emand
that between south poles is40em.
Solution. Thesituation is shown in Fig. 5.9. Here
qllll=q11l2=80 Am
I+-- 15em --1--10 em+-- 15em --.j
I!•..•-------40 em--------+i·1
Fig.5.9
Force of repulsion betwen polesN1andN2is
F=110.qmlqlll2=10-7x80x80 =0.064 N
1 4lt r2 (0.10)2
Force of repulsion betwen poles 51 and 52 is
F= 10-7 x80x80 =0.004 N
2 (0.40)2
Force of attraction betwen N1and 52 is
F=10-7x80x80 =0.010 N
3 (0.25)2
Force of attraction between N2and51 is
F4=F3=0.010 N
Resultant force betwen the two magnets is
F=F1+F2-F3-F4
=0.064+0.004-0.010-0.010
=0.048N(repulsive).
Example4.A magnetic dipole of length 10em has pole
strength of20 Am.Findthemagnetic moment ofthedipole.
Solution. Here21=10 em=0.10m,qm=20 Am
:. Magnetic moment,
m=qmX21=20x0.10 Am2 =2.0 Am 2.
Example5.A bar magnet of magnetic moment 5.0Anl
haspoles20em apart. Calculate thepole strength.
[CBSE D92C]
Solution. Here m= 5.0 Am 2, 21= 20em= 0.20 m
As m=qmx21
Polestrength,
«;=~=~~~=25Am.
Example6.A steel wire of lengthIhas amagneticmoment
m.Itisbent into a semicircular arc. What isthe new
magnetic moment ?
Solution. Pole strength, qm=!!!.
I
5.5
When the wire is bent into a semicircular arc, the
separation betwen the poles changes fromIto2r,
whereristhe radius of thesemicircular arc. Thus
I
1=ttrorr=-
1t
. m 212m
N magnetic moment = qx2r=-x-=-.
111 I 1t rr
j2)roblems ForPractice
1.Two magnetic south poles are located 4.0 cm apart.
Ifthepoles ofeach magnet have a strength of
8.0Am and are 20.0 ern apart, find the force exerted
by one south pole on the other. (Ans.4.0x1O-3N)
2.Twoequal andunlike poles placed 5 cm apart inair
attract each other with a force of 14.4 x10-4N.How
far from each other should they be placed so that
the force of attraction will be 1.6 x10-lN?
(Ans.0.15 m)
3.Two magnetic poles, oneof which is 10 times as
strong as the other, exert on each other a force equal
to 9.604 mN, when placed 10 cm apart inair.Find
the strength of the two poles.
(Ans. 9.8 Am, 98 Am)
4.Twolike magnetic poles of strengths 5 Amand
20 Am are situated 1.0 m apart. At what pointon
thelinejoining the two poles, will the magnetic
field be zero?
(Ans.0.33m from 5 Ampole towards 20 Am pole,
1 mfrom 5 Am pole away f~om20 Am pole)
5.Two bar magnets of length 0.1 m and pole strength
75 Ameach, are placed on the same line
distance betwen their centres is0.2 m. What is the
resultant force due to one on the otherwhen(i)the
north pole of one faces the south pole of the other
and(ii)the north pole of one faces thenorth pole of
the other? [Ans. (i)3.4x10-2N (attraction),
(ii)3.4x10-2N(repulsion)]
6.A magnetic dipole of length 15 ern has a dipole
moment of 1.5 Am 2.What is the pole strength?
(Ans. 10 Am)
7.A magnetised steel wire 31.4 ernlong has a pole
strength of0.2Am. It is bentin the form of a
semicircle. Calculate the magnetic moment of the
stel wir. (Ans. 0.04 Am2)
8.Twothin bar magnets of pole strengths 25 Am and
48 Am respectively and lengths 0.20m and 0.25 m
respectively are placed at right angles to each other
with theN-poleof firsttouching the S-pole of the
second. Find the magnetic moment of the system.
(Ans.13 Am 2)···3wx£n~m}r°n3lxv

429. 5.6
HINTS
I!0qm
l qm210-7 x8x8 -3
1.F=-.--2- = 2 2N = 4.0x10N.
4n, (4x10 )
_1.10 qn, q~ . 1
2.F----2-I.e., Focz
4n, r
2
.1i_'2
"--2"
F2'1
~1 14.4x10--4 -2
'2=E2·'i= --4x5xl0 =0.15m.
1.6x10
F=1.10q,"Jqn'2
4n· ?
10-7xqxlOq
.. 9.604x10-3 = m m
(0.1)2
or q;=96.04
or qm=9.8Am
and 10 q",=98Am.
4.Let the magnetic field be zeroat distancexfrom the
pole of strength 5 Am.
3.As
I+-- x--.••·1•...' --- 1-x---+1'1
•• •
o
Fig.5.10
Magnetic field at0due to 5 Am pole
=Magnetic field at0due to 20 Am pole
1.1051.10 20
4n.x2=4n.(1-x)2
or4x2=(I-x)2
or 2x=±(1 -x)
or x= -1m,+0.33m.
5.Proced asin Example 2 on page 5.4.
m1.5Am2
6.q= - = = 10Am.
m21 0.15m
7.Proced asin Example 6 on page 5.5.
8.Heren;=25x0.20=5.0 Am2
~ =48x0.25=120 Am2
Resultant magnetic moment of the two magnets
placed perpendicular to each other is
m=~n;2+~2=~52+122=13 Am2.
5.7MAGNETIC FIELD LINES
8. What are magnetic lines of force?Givetheir
important properties.
Magnetic lines of forceMichael Faraday, the cele-
brated physicist of London (1791-1867) introduced the
PHYSICS-XII
concept of the magnetic lines of force to represent a
magnetic fieldvisually. Magnetic lines of force do not
really exist but they are quite useful in describing
many different magnetic phenomena.
Amagnetic line of forcemaybedefinedasthe
curve the tangenttowhich at any point gives the
direction of the magnetic field at that point. It may also
bedefinedasthe path along whichaunit north pole
would tend to move if free todoso.
Properties of lines of force :
1.Magnetic lines of force are closed curves which
start in air from the N-pole and end at the S-pole
and then return to the N-pole through the
interior of the magnet.
2. The lines of force never cross each other. If they
do so, that would mean there are two directions
of the magnetic field at the pointof intersection,
which is impossible
3. They startfromand endonthe surface of the
magnet normally.
4. The lines of force have a tendency tocontract
lengthwise andexpand sidewise. This explains
attraction between unlike poles and repulsion
betwenlike poles.
5. The relativecloseness of the lines of force gives
a measure of the strength of the magnetic field
which is maximum at the poles.
9. Describe a method for plotting themagnetic field of
a bar magnet.
Plotting magnetic field of a bar magnet. The
magnetic field around a magnet can be traced with
the help of a magneticcompass needle. Itnsists of a
small and light magnetic
nedle pivoted at the centre
of a small circular brass case
provided with a glass top, as
shown in Fig. 5.11. The
north pole of the magnetic
nedle is generally paintedFig. 5.11Compass nedle
black or red.
Fig. 5.12To plot magnetic field of a bar magnet
with a compass nedlewww4°±trsp€vvr4o±~

430. Imagine aunit north-pole placed at point
P. Then
Therefore, the strength of the magnetic fieldBat from Coulomb's law of magnetic forces, the force exerted
point P is by the N-pole of the magnet on unit north-pole is
BaxI·aJ=Force experienced bya J..lq
FN= ~ .-1!!.2' along NP
unit north - pole at point P 41tx
- F - F-J..loqm[_1 1_] Similarly, the force exerted by the S-pole of the
- N S - 41t (r_1)2(r+1)2 magnet on unit north-pole is
_ J..loqm 4rl
-4;-.(r2 _12)2
MAGNETISM
To plot the magnetic field of a magnet, the magnet
NS is placed on a white paper. The compass nedle is
placed near its N-poJe, as shown in Fig5.12.The
positions of the two ends S and N of the nedle are
marked by pencil dots on the paper. The nedle is now
displaced to the n position so that its south pole
exactly comes over the mark previously made against
the north pole and again the n position of the north
pole is marked. The process is repeated till the south
pole is reached. The various dots are joined together by
a smooth curve which gives a line of force.
Similarly, other lines of force are drawn. A complete
pattern of the magnetic field around a bar magnet is
shown in Fig.5.24on page5.18.
5.8MAGNETIC FIELD OF A BAR MAGNET
AT AN AXIAL POINT
10.Derive an expression for the magnetic field
inieneitvat a point on the axis of a bar magnet. Whatis
the direction of the field?
Magnetic field of a bar magnet at an axial point
(end-on position).Let NS be a bar magnet of length 21
and of pole strengthqm.Suppose the magnetic field is
to be determined at a point P which lies on the axis of
the magnet at a distancerfrom its centre, as shown in
Fig.5.13.
~I'-------r+l------~-I
f.suuu+O-u--N .•..}--u----- ~ p
1--1 '" 1'1' r-1~
I' r -I
Fig. 5.13Magnetic field of a bar magnet at an axial point.
Imagine aunit north pole placed at point P. Then
from Coulomb's law ofmagnetic forces, the force exerted
by the N-pole of strengthqmon unit north pole will be
F =Ilo.J«: alongNP
N 41t.(r_1)2'
Similarly, the force exerted by S-pole on unit north
pole is
F=llo ~
s41t.(r+1)2 '
~
along PS
5.7
Butqm.21=m, is the magnetic dipole moment, so
B .=Ilo 2mr
axial4n .(?_12)2
For a short bar magnet,I< <r,therefore, we have
1102m ~
Baxial =4n ."?" 'alongNP ...(1)
Clearly,the magnetic field at any axial point of magnetic
dipoleis inthesame direction as that ofitsmagnetic dipole
moment i.e., from S-pole to N-pole, so we can write
~
B.=J..lo2m
axial41t· r3
5.9MAGNETIC FIELD OF A BAR MAGNET
AT AN EQUATORIAL POINT
11. Derive an expression for the magnetic field
intensity at a point on the equatorial line of a bar magnet.
Whatisthe direction of this field?
Magnetic field of a bar magnet at an equatorial
point (broadsid-on position).Consider a bar magnet
NS of length21and of pole strengthqm.Suppose the
magnetic field is to be determined at a point P lying on
the equatorial line of the magnet NS at a distancer
from its centre, as shown in Fig.5.14.
Q
Rt+--~
,'T
,
,
x .
,
,
,
,
,
,
.:e
,-t-
\~
\~'"',
,",...."
\--''=?
,
,
,
,
,
r
5 N
Fig.5.14Magnetic field of a bar magnet at
an equatorial point.
F -J..loqmalong PS
s-41t· x2 ' r···3wx£n~m}r°n3lxv

431. 5.8
As the magn
itudes ofFNandFsare equal, so their
vertical components get cancelled while the horizontal
components add up along PR.
Hence the magnetic field at the equatorial point Pis
Bequa=Net force on a unit N-pole placed at point P
=FNcas8+Fscas8
=2 FN cas 8 r.FN=Fsl
=2 ~o ~ I
.41t .x2 . x
orB =~o m
equa 41t'(1+12)3/2
wherem=qm.21,is the magnetic dipole moment.
Again for a short magnet, I< <r,so we have
~om
Beq=-. 3"' along PR ...(2)
ua41t r
Clearly, the magnetic field at any equatorial point of a
magnetic dipoleisin the direction opposite to that of its
magnetic dipole moment i.e., from N-pole to 5-pole
we can write
~
~_ ~o m
Bequa--4nr3
On comparing equations (1) and (2), we note that
the magnetic field at a point at a certain distance on the axial
line of a short magnet is twice of that at the same distance on
its equatorial line.
Formulae Used
Magnetic field of a bar magnet of length 21and
dipole momentmat a distance rfrom its centre,
1.B .I= ~o./mr2 2 (onthe axial line)
axia41t (r-I)
2.B -~o m
equa - 4lt.(r2+12 )3/2
Fora short magnet, I< <r,so
_ ~o2m
3.Baxial -4lt .7 (on the axial line)
4.B-~0m (on the equatorial line)
equa-41t . r3
(on the equatorial line)
Units Used
Magnetic field Bisin tesla, distances randIin
metre and magnetic moment in JT-1 or Am2.
Example 7. What is the magnitude of the equatorial and
axial fields due to a bar magnet of length5cm at a distance of
50em from the midpoint? The magnetic moment of the bar or
magnet is0.40A~. [NCERT]
PHYSICS-XII
Solution.Herem=0.40 Am2, r = 50 em =0.50 m,
21=5.0 cm
Clearly, the magnet is a short magnet (I«r).
(i)B = ~.m= 10-7 x0.4 =3.2x10-7T.
equa4 1t r3 (0.5)3
(ii)B.I= ~ .23m=6.4x10-7T.
axia4 1t r
Example 8. A bar magnet of length10cm has a pole
strength of10Am.Calculate the magnetic field at a distance
of 0.2 m from its centre at a point on its(i)axial line and
(ii) equatorial line.
Solution. Here21= 10 cm or1=5cm =0.05 m,
qm=10 Am, r=0.2 m
Magnetic moment,
m=qmX21=10x0.1 =1 Am2
(i) Magnetic field on axial line is
B.= ~02mr 10-7x2x1x0.2
axial 41t.(1_12)2(0.22 _0.052)2
10-7x0.4T 5
=2.84x10-T.
(0.0375)2
(ii)Magnetic field on equatorial line is
B=~o m 10-7xl T
equa 41t'(1+12)3/2 (0.22+0.052)3/2
10-7 10-7
-----::-= T= T
(0.0425)3/2 8.76x10-3
=1.14x10-5T.
Example 9. Two small magnets are placed horizontally,
perpendicular to the magnetic meridian. Their north poles
are at30em east and20em west from a compass needle. If
the compass needle remains undeflected, compare the magnetic
moments of the magnets.
I_S N_~20 ~~£--30-c;; -~'__N s__1
Fig.5.15
Solution.The compass nedle atClies on the axial
line of thetwo magnets. As itremainsun deflected, the
fields ofthetwomagnets at C must be equal and opposit.
Ilo2~ _~o 2111z
-'-3---'-3-
41tr1 41t r2
orwww4°±trsp€vvr4o±~

432. MAGNETISM
Example 10.Two
short magnets PandQareplaced one
over another with their magnetic axesmutually perpen-
dicular to each other. It is found that the resultant field at a
pointon the prolongation of themagnetic axis of Pisinclined
at30°with this axis.Compare themagnetic moments of the
two magnets.
Solution. LetAbe any point on the prolongation of
-7 -7
the axis of magnet P. Let~and ~ be the fields of the
magnets P andQrespectively at the pointA.Let
-7 -7
m1andrrIzbethe magnetic moments of the two magnets.
I+---r--l
Qrr52
I
I
I
I
P :
~---------~'-+_-.ml 1--A",--.-L,;...;."..---.,'S1
N1
Fig 5.16
As pointAlies on the axial line of P, therefore,
_1102I~
Bl--'-3-
4rcr
The pointAlies on the broad-side-on position of Q,
therefore,
~=110 111.;,
4rc.r3
But the resultant field ~ is inclined at 30° with ~,
so
Hence
~ =tan300=~
~ 13
1_rrIz
13-2~
or
Example 11.Two identical magnetic dipoles of magnetic
moments1.0Anteach are placed at a separation of2 m with
their axes perpendicular to each other. What is the resultant
magnetic field at~point mid-way between the dipoles?
Solution. The situation is shown in Fig. 5.17.
Fig. 5.17
5.9
The magnetic fields of thetwo magnets at the mid-
point Pare
B=110 2m=1O-7x/X1=2 xlO-7T,
14rc' r3 1
in horizontal direction
~=110n;= 10-7 T, in vertical direction
4rc r
BR=~B/+~2=.J5 x10-7T
Ifthe resultant field BRmakes angle8with ~,then
R 10-7
tan8=~= =0.5
~ 2x10-7
8=26.57°
~rOblems For Practice
1.A bar magnet is 0.10 m long and its polestrength is
12 Am. Find the magnitude of the magnetic field at
apoint onits axis at a distance of 20 ern from it.
(Ans.3.4x10-51')
2.Calculate the magnetic field due to a bar magnet
2 cm long and having a pole strength of 100Am at a
point 10 cm from each pole. (Ans.2x10-4 T)
3.A bar magnet has a length of 8 cm. The magnetic
field at a point at a distance of 3emfrom the centre
in the broad-side onposition is found to be 4x10-6T.
Calculate the pole strength of the magnet.
(Ans.6xl0-5 Am)
4.The magnetic moment of a current-loop is
2.1x10-25Am2.Find the magnetic field on the axis
of the loop at a distance of 1.0Afrom the loop.
(Ans.4.2x10-2T)
5.If the earth's magnetic field has a magnitude of
3.4x10-5T at the magnetic equator of the earth,
whatwould be its value at the magnetic poles of the
earth? (Ans.6.8x10-5T)
6.Theintensities of magnetic field at two points on
theaxis of a bar magnet at distances 0.1m and 0.2 m
from itsmiddle point are in the ratio 12.5 :1.
Calculate the distance betwen the poles of the
magnet. (Ans.0.1 m)
7.Two short magnetsaandbof magnetic moments
0.108Am2 and 0.192Am2 are placed along
mutually perpendicular straight lines meting at a
point P. Find the magnitude and direction of
magnetic field at point P, ifit lies at distances 30 ern
and 60 cm respectively from the centres of the two
magnets.
(Ans.8.24x10-7T,at 14° with the axis of a)···3wx£n~m}r°n3lxv

433. 5.10
8.Two short magnets
of magnetic moments aandbof
magnetic moments 32 Am2and 27 Am 2 are placed
on a table, as shown in Fig. 5.18. Find the magni-
tude and direction of the magnetic field produced
5
40em
----------~P
I
I
:30cm
a
N
Fig. 5.18
bythese magnets at point P situated at theequa-
torial lines of boththe magnets at distances 40 cm
and 30 cm respectively from the centres of the two
magnets. (Ans.1.12x 10-4T, at an angle
of 26.57° withthe equatorial line of a)
HINTS
110 2mr 1102xqmx2/xr
1.Baxia1=41t .(,1 _/2)2 = 41t'(r2 _12)2
10-7x2x12xO.10xO.20
(0.22_ 0.052)2
=3.4x10-5T.
B _110 m_llo qmx21
2.equa-41t . ,3 - 41t .-,3-
107x100x0.02 -4
----;;-3-- = 2x10 T.
(0.10)
3.UseB 110 In
equa=41t.(,2_12 )3/2
110 qmx21
= 41t . (,2 _12)3/2 .
4.The current loopis short magnetic dipole. So
B. =110 2m=1O-7x2x2.1xlO-25
axial41t.r3 (1.0x10-10)3
=4.2x10-2T.
5.At the equator,
110 m -5
B=Bequa= -.3"=3.4x10T
41tr
Atthe poles,
Br=B'aJ= ~2m=2 B
aXI 41t',3
=6.8x10-5T.
7.As shown in Fig. 5.19,the pointPlies onthe axial
line of both the magnets aandb.
PHYSICS-XII
Bb B
/-30 em
Is N~-
IP
B.a I
I
I
I
I
I
I
60 em
I
I
1
N
b
5
Fig.5.19
B=1102n;= 1O-7x2xO.108
a41t1.3 (0.30)3
=8x10-7T (along the axisofa)
R _1102~_10-7x2xO.192
"'b- • 3- 3
41tr2 (0.60)
= 2x1O-7T (along the axis of b)
The resultant field at Pis
B=~B/+1),2= 10-7 ~82+22 =8.24x10-7T
If the field Bmakes angleewith the direction ofBa'
then
tane=1),= ~ = 0.25 or
Ba8
8.As shown in Fig. 5.20, the point Plies on the equa-
torialline ofboth the magnets aandb.
a5
B. B
40em.--1(':
----------~:
P, B
I b
:30 em
I
I
I
N
51
Fig. 5.20 b
B- ~n; _10-7x32-0 5 10-4T
.. a- '3- 3-'x
41t 1. (0.40) (antiparallel to "1 )
1),=110.1= 10-7 x;7=lO-4T
41tr2 (0.30) (antiparalleltoInz)
The two fields are perpendicular to each other. So
the resultant field at point Pis
B= ~B;+~= 10-4 ~(0.5)2 +12=1.12 x10-4T
If the fieldBmakes an angleewith the directionof
Bv,then
B0.5x10-4
tane=-1!...= 4= 0.5 or e=26.57°
1), 10www3«°tssr•wvs3q°~

434. MAGNETISM
5.10TORQUE ON A
MAGNETIC DIPOLE
IN A MAGNETIC FIELD
12. Derive an expression for the torque on a magnetic
dipole placed in a uniform magnetic field. Hence define
magnetic dipole moment.
Torque on a magnetic dipole in a uniform magnetic
field.Consider a bar magnetNSof length21placed in a
--7
uniform magnetic field B. Letqmbe the pole strength
of its each poleet the magnetic axisof the bar magnet
--7
make an angle 8 with the field B, asshown in
Fig.S.21(a).
--7
Force on N-pole=qmB; along B
--7
Force on S-pole=qmB,opposite to B
••
It!
q",B
B
••
q"'B•..•-~
------~~------------------~ ..~~ ~ ~
T=m xB
(b)(a)
Fig. 5.21(a)Torque on a bar magnet in a magnetic field.
(b)Relation betwen the directions of t>, ;;,B.
The forces on the two poles are equal and opposite
They form a couple
given by
T=Forcexperpendicular distance
=qmBx21sin8=(q",x21)Bsin8
or T=mBsin8
wherem=qmX2/,isthe magnetic dipole moment of
the bar magnet. Invector notation,
--7 --7 --7
T=mxB
--7
The direction of the torque1is given by the right
hand scr rule as indicatedin Fig. S.21(b).The effect
--7
of the torque1is to make the magnet align itself
--7
parallel to the field B . That is why a frely suspended
magnet aligns itself in the north-south direction
because the earth has its own magnetic field which
exerts a torque on the magnet tending it to align along
the field.
5.11
Special Cases
1. When the magnet lies along the direction of the
magnetic field,
8=0°, sin 8=0, 1=0,
Thus the torque is minimum.
2. When the magnet lies perpendicular to the
direction of the field,
8 = 90°, sin 8 = 1, 1 =mB
Thus the torque ismaximum
1max=11IB
Definition of magnetic dipole moment. If in Eq.(1),
B=l, 8 =90°, then
1=m
Hencethe magnetic dipole moment may be defined as
the torque acting on a magnetic dipole placed perpendicular
to a uniform magnetic field of unitstrength.
SI unit of magnetic moment. As
1
m=-----
Bsin8
SI.f 1Nm
units0m=---
IT.1
=NmT-1 orJT-1orAm2.
5.11 POTENTIAL ENERGY OF A MAGNETIC
DIPOLE IN A MAGNETIC FIELD
...(1)
13.Deriveanexpression for the potentialenergy of a
dipole placed in a uniform magnetic field at an anglee
with it. When will the magnetic dipole be in the positions
of stable and unstableequilibrium?
Potential energy of a magnetic dipoleAs shown in
Fig.S.21(a),when a magnetic dipole is placed in a
--7
uniform magnetic field B at angle8with it, it exp-
riences a torque
1=mBsin 8
This torque tends to align the dipole in the
--7
direction of B.
...(2)
If the dipole is rotated against the action of this
torque, work has to be dones work is stored as
potential energyof the dipole
The work done in turning the dipole through a
small angled8is
dW=1d8=mBsin 8d8
If the dipole is rotated from an initial position 8=81
to the final position 8=82,then the total work done
will be···3wx£n~m}r°n3lxv

435. 5.12
92
W
=fdW=fmBsin 9d9=mB [- cos 9] ~
~
= - mB(cos 92 -cos 91)
This work done is stored as the potential energy U
of the dipole
:. U=-mB(cos92-cos91)
The potential energy of the dipole is zero when
~~
m.1B. So potential energy of the dipole in any orien-
tation 9can be obtained by putting 91 =90° and 92 = 9
in the above equation.
U= - mB(cos 9-cos 90°) ~
~ ~
or U = -mBcos 9=-m.B
Special Cases
1.When 9 =Ob, U =- mBcosO° =-mB
Thus the potential energy of a dipole is
~ ~
minimumwhenmisparallelto B. In this state,
the magnetic dipole is in stable equilibrium.
2.When 9 =90°, U = -mBcos 90° =0.
3.When 9 = 180°, U = -mBcos 180° =+mB.
Thus the potential energy of a dipoleismaximum
~ ~
whenmisaniiparallel to B . In this state, the magnetic
dipole is inunstable equilibrium.
5.12 CURRENT LOOP AS A MAGNETIC DIPOLE
14.Show that a current carrying loop behaves as a
magnetic dipole. Hence write an expression for its
magnetic dipole moment.
Current loop as a magnetic dipoleWe know that
the magnetic field produced at a large distance rfrom
the centre of a circular loop (of radiusa)along its axis is
given by
or
whereIis the current in the loop andA=na2is its
area. On the other hand, the electric field of an electric
dipole at an axial point lying far away from it is given
by
E=_l_. 2: ...(2)
4nEOr
wherepis the electric dipole moment of the electric
dipole.
PHYSICS-XII
On comparing equations (1) and(2),we note that
both BandEhave same distance dependence(:3).
Morver, they have same direction at any far away
point, not just on the axis. Thissuggests that a circular
current loop behaves as a magnetic dipole of magnetic
moment,
m= IA
In vector notation,
~ ~ A
m=IA=IAn
This result is valid for planar current loop of any
shap.Thusthemagnetic dipole moment of any current
loopisequal totheproduct of the current and its loop area.
Its direction is defined to be normal to the plane of
the loop in the sense given by right hand thumb rule.
Right hand thumb rul. Ifwe curl the fingers of the
right handin the direction ofcurrent in theloop,then the
extendedthumb gives thedirection of the magnetic moment
associated with the loop.
A
11
N-Face Area A
S-Face
Fig. 5.22Current loop as a magnetic dipole.
...(1)
It follows from the above rule that the upper face of
the current loop shown in Fig. 5.22.hasN-polarity and
the lower face has S-po)arity. Thusa current loopbehaves
likea magnetic dipole.
If a current carrying coil consists of Nturns, then
m= NIA
The factorNIis calledamperes turnsof current loop.
So,
Magnetic dipole moment of current loop
=Ampere turnsxloop area
Clearly, dimensions of magnetic moment
=[A] [L2] =[AL2]
51 unit of magnetic dipole moment is Am2.Itis
defined as themagnetic moment associated with one turn
loopof areaonesquare metre when a current of one ampere
flowsthrough it.···3wx£n~m}r°n3lxv

436. MAGNET
ISM
Table5.1 Analogy between electric
and maetic dipoles
~;..~.
1
Freespaceconstant
Dipole moment
-+
m
Axial field
--_.-
Equatorialfield
Torque in external field
-+ -+ -+ -+
pxE mxB
---r----------
-+-+
-m.B
-+-+
-p.EP.E.in external field
5.13MAGNETIC DIPOLE MOMENT OF A
REVOLVING ELECTRON
15.Deriveanexpression for the magnetic dipole
moment ofanelectron revolving around a nucleus.
DefineBohrmagneton andfindits value.
Magnetic dipole moment of a revolving electron.
According to Bohr model of hydrogen-like atoms,
negatively charged electron revolves around the posi-
tively charged nucleus. This uniform circular motion
of the electron is equivalent to a current loop which
possesses a magnetic dipole moment =fA.Asshown in
Fig. 5.23, consider an electron revolving anticlockwise
around anucleus in an orbit of radius rwith spedv
andtime periodT.
1
e
+-
I
I
-
e
Fig. 5.23Orbital magnetic moment of a revolving electron.
Equivalent current,
f_Charge _e_ e _ ev
- -------
TimeT 21tr /v21tr
Area of the current loop,A=n,z
5.13
Therefore, theorbital magnetic moment (magnetic
moment due to orbital motion) of the electron is
ev 2
III=fA=-.1tT
2nr
evr
1l1=-
2
...(1)or
Asthe negatively charged electron is revolving
antic1ockwise, the associated current flows clockwise.
According to right hand thumb rule, the direction of
the magnetic dipole moment of the revolving electron
will be perpendicular to the plane of its orbit and in the
downward direction, as shown in Fig. 5.23
Also, the angular momentum of the electron due to
its orbital motionis
1=mevr ...(2)
The direction of ris normal to the plane of the
electron orbit and in the upward direction, as shown in
Fig. 5.23.
Dividing equation (1)by(2), we get
!:1.=evr /2=_e_
Imevr2me
The above ratio is a constant called gyromagnetic
ratio. Its value is 8.8 x1010C kg-I. So
e
III=-1
2me
Vectorially,
-+ e-t
Il--- I
1-2m
e
The negative signshows that the direction ofris
-t
opposite to that ofIlrAccording toBohr'squantisation
condition, the angular momentum of an electron in any
permissible orbit is integral multiple ofh /2n, whereh
is Planck's constant, i.e.,
I=nh h 123
2n'w eren=, " .....
1l1=n[~)
4nme
Thisequation gives orbital magnetic moment of an
electron revolving innth orbit.
Bohr magneton.Itis defined as the magnetic moment
associated with an electron due to its orbital motion in the
first orbitof hydrogen atom. It is the minimumvalue ofIII
which can be obtained by puttingn=1in the above
equation. Thus Bohr magneton is given by
eh
IlB=(Ill)rnin=-4--
nme···3wx£n~m}r°n3lxv

437. 5.14
Putting thevalues
ofvarious constants, we get
1.6x10-19Cx6.63x10-34 JS
IlB=
4x 3.14x9.11x10-31kg
=9.27x10-24Am2.
->
Besides theorbital angular momentum I ,an
->
electron has spin angular momentum 5dueto its
spinning motion.Themagnetic moment possessed by an
electron due to its spinning motion iscalled intrinsic mag-
netic moment or spin magnetic moment. It isgiven by
-> e->
11=-- 5
5 m
e
The total magnetic moment of the electron isthe
vector sum of these two momenta. It is given by
->->-> e->->
11=11/+115=--(1 +25)
2me
Examples based on
•
-> -> ->
1.Torque,'t=mBsin8orr=mxB
2. Work done in turning the dipole or P.E. of a dipole,
W=U= -mB(cas82-cos ~)
3. If initially the dipole is perpendicular to the field,
U=-mBcas8
(1') -> ->
When m is parallel to B, 8=0°, U= -mB
Potential energy of the dipole is minimum. It
is in a state of stable equilibrium.
(ii)WheniitisperpendiculartoB,e=90°,U=O.
(iii)When1:;;isantiparalleJtoB,e=180°, U =+mB
Potential energy of the dipole is maximum. It
is in a state of unstable equilibrium.
4. Magnetic moment of a current loop, m=NIA
5. Orbital magnetic moment of an electron innthorbit,
11 /=~r=2:eI=n(4::J
6. Bohr magneton is the magnetic moment of an
electron in first(n=1) orbit.
ell
11 B=(11/).=-- .
nun4nme
Units Used
Torque1:is in Nm, magnetic moment m in JT-1or
Arn2, field B in tesla, potential energy U in joule
PHYSICS-XII
Example12.A magnetised needle of magnetic moment
4.8x10-2JT-1is placedat30°with the direction of uniform
magnetic field of magnitude 3x10-2T.What isthe torque
acting onthe needle ? [CBSE D OIC]
Solution.Here m= 4.8x10-2JT-1,e= 30°,
B=3xlO-2 T
:.Torque, r=mBsin8
= 4.8x10-2x3x10-2xsin30°
=7.2x10-4J.
Example13.Ashort bar magnet placed with its axis at 30°
to a uniform magnetic field of 0.2T experiences a torque of
0.06Nm.(i)Calculate the magnetic moment ofthe magnet.
(ii)Find out what orientation of the magnet corresponds to
its stable equilibrium in the magnetic field. [CBSEOD02]
Solution.(i)Here B=0.2T, 8=30°, T=0.06 Nm
Magnetic moment,
r 0.06
m=---=----
B sin8 0.2 sin 30°
0.06 =0.6 Am2.
0.2x0.5
(ii)TheP.E. of a magnetic dipole in auniform
magnetic field is
U=-mBcos 8
Instable equilibrium, theP.E. is minimum. So
cosOeIor 8=0°
Hence the bar magnet will be in stable equilibrium
->
when its magnetic moment mis parallel to the
->
magnetic field B.
Example14.In an iron bar(5cmx1cmx1em) the
magnetic moment of an atomis1.8x10-23Ant-. (i)What
willbemagnetic moment of the bar in the state of magnetic
saturation ?(ii)Whattorque will have to be applied to keep
the bar perpendicular to an external magnetic field of15,000
gauss?Densityofiron=7.8g cm-3, its atomic mass=56.
Solution. (i)Mass of iron bar =volume xdensity
= 5 cm3 x7.8 gcm-3 =39 g
Number of atoms in 56g ofiron= 6.02 x1023
:.Number ofatoms in 39g ofiron
6.02x1023x39= 4.19 x1023
56
Magnetic momentof each atom
=1.8x10-23 Am2
Magnetic moment ofthe iron bar in the state of
magneticsaturation is
m= 1.8x10-23 x4.19x1023=7.54 Am2.
(ii)Here 8 =90°, B=15,o00G =15000x10--4T
Required torque, 't=mB sin 8
=7.54x15000x10-4xsin 90° =11.3 Nm.···3wx£n~m}r°n3lxv

438. MAGNETISM
Example15
.A planar loop of irregular shape encloses an
area of7.5 x10--4nl-andcarries acurrent of12A.Thesense
of flow of current appears to be clockwise to an observer.
Whatisthemagnitude and direction of the magnetic
moment vector associated with the current loop?
[NCERT]
Solution. Here A= 7.5 x10-4m2,I= 12 A
Magnetic moment associated with the loop is
m=IA=12x7.5x10--4
=9.0x10-3JT-1
Applying right hand rule, the direction of magnetic
moment is along the normal to the plane of the loop
away from the observer.
Example16.A current of5Aisflowing through a10turn
circular coil of radius 7em.Thecoil liesinthe x-y plane.
What isthemagnitude and direction of the magnetic dipole
moment associated with it?
If this coil wereto beplaced ina uniform external
magnetic field directed along the x-axis, inwhich plane
would the coil lie, when in equilibrium? (Take1t=22/7)
[CBSE Sample Paper 03]
Solution. Magnetic dipole moment,
m=NIA=NIx1tr2
22(7)2
=10x5x --;;-x 100
=0.77 Am2
The direction of magnetic dipole moment is
perpendicular to the plane of the coil. Hence it is along
z-axis.
Torque on the current loop of magnetic moment
mis
t= mBsina
whereais angle betwenrr;andB.For stable
~
equilibrium torque is zero, soa=0°.For this B should
be perpendicular to the plane of the coil. Hence the coil
will lie iny-zplane in the condition of stable
equilibrium.
Example 17. A bar magnetwith poles25cm apart and of
pole strength 14.4Am rests withitscentre on africtionless
pivot.Itisheldinequilibrium at60°to a uniform magnetic
field of induction0.25T by applying a forceF,at right
angles toitsaxis,12emfromitspivot. CalculateF.What
will happen if the force Fisremoved? [lIT]
Solution. Here m=qmX21= 14.4 x0.25 = 3.6 Am2
9=60°, B=0.25T, r=12cm=0.12m
Torque, t=Fr=mBsin 9
5.15
F =mBsin 9 = 3.6x0.25xsin 60°
r 0.12
= 3.6x0.25x0.866 = 6.5 N
0.12
When the forceFis removed, the magnet aligns
itself in the direction of field B.
Example18.An electroninan atom revolves around the
nucleus inan orbit of radius0.5A.Calculate the equivalent
magnetic moment if the frequency of revolution of the
electronis1010MHz. [lIT 88, CBSE D 98]
Solution. The electron revolving around the
nucleus in a circular orbit is equivalent to a current
loop. Its magnetic moment ism=IA=evx1t,z
Heree= 1.6x10-19Cv=1010MHz =1016Hz,
r=0.5A=0.5x1O-lOm
:.m=1.6x 1O-19x1016x3.14x(0.5xlO-1O)2
=1.2S6x 10-23Am2.
Example19.An electron moves around the nucleus in a
hydrogen atom of radius0.51Awith avelocity oj2x106m/s.
Calculate thefollowing:
(I)theequivalent current due toorbital motionofelectron
(ii)the magnetic field produced at the centre of the nucleus
(iii)themagnetic moment associated with the electron.
[CBSEOD08]
Solution. Here r =0.51x 1O-lOm, v=2x105ms-1
(i)I=~=~= 1.6x10-19x2x105= 10-4 A
T21tr 21tx0.51x10-10
41tx10-7x10-4
------.;;- =1.23 T.
2x0.51x10-10
(iii)m=IA=~ xif=evr
21tr 2
1.6x10-19 x2 x 105 x0.51x10-10
2
=8.16x10-25 Am2.
Example 20.Two magnets of magnetic moments mand
.J3marejoined to form a cross(+ ~The combinationis
suspended freely ina uniform magnetic field. In equilibrium
position, the magnet of magnetic moment mmakes an angle
9with the field. Find9.
Solution.When the magnet of moment inmakes
angle 9 with the field B, the other magnet of moment
.J3mwill make angle (90° - 9) with the field B. In the
equilibrium position,
Torque experienced by first magnet
= Torque experienced by second magnet
or mBsin 9 =.J3mBsin (90° - 9)···3wx£n~m}r°n3lxv

439. 5.16
or sin8=.J
3cas 8
sin8=.J3 or tan8=.J3
cas8
or
8=60°.
Example 21 .Abar magnet having a magnetic moment of
1.0x104JT-1isfree to rotate in a horizontal plane. A
horizontal magnetic field of4x10-5Texistsinspace. Find
the work done in rotating the magnet slowly from a direction
parallel to the field to a direction 60°from the field.
Solution. Herem=1.0x104JT-I,B=4x10-5T,
81=0°, 82=60°
Work done,
W=-mB(cos82-cas81)
=-1.0 x104x4x10-5(cas60°-cas0°)
=1.0x104x4 x10-5 x.!=0.2J.
2
Example22.Acurrent of 7.0Aisflowingina plane
circular coil of radius1.0 cm having 100 turns. The coil is
placed in a uniform magnetic field of0.2Wbm-2. Ifthecoil
isfree to rotate, what orientations would correspond to its (i)
stable equilibrium and (ii) unstable equilibrium? Calculate
the potential energy of the coil in these cases. [CBSE 00 92)
Solution. HereN=100,A= 7.0A,
r=1.0cm=1.0x10-2m, B=O.2 Wbm-2
Magnetic moment associated with the coil is
m=NI A=Nlx1<?
= 100x7.0x22x(1.0x10-2)2=0.22Am2
7
-t
(i)The stable equilibrium corresponds tomparallel
-t
to B. The potential energy is then minimum.
Umin= -mB cos0°=-0.22x0.2xl=-0.044J.
-t
(ii)Theunstable equilibrium corresponds to m anti-
-t
parallel to B. Thepotentialenergy is then maximum.
Umax =-mBcas180°=-0.22x0.2x(-1)
=+0.044J.
Example23.Ashort bar magnet placed with itsaxisat30°
experiences a torque of 0.016 Nm in an external field of800 G.
(a) What is the magnetic moment of the magnet? (b)Whatis
the work done by an external force in movingitfrom its most
stable to most unstable position? (c) Whatisthe work done
by the force duetothe external magnetic field in the process
mentioned in part (b)?(d)TIlebar mTetisreplaced by a
solenoid of cross-sectional area 2x101~and 1000turns,
butthe same magnetic moment. Determine the current
flowingthrough the solenoid. [NCERT)
PHYSICS-XII
Solution. (a)Here8 = 30°, B= 800 G = 800X10-4T,
t=0.016Nm
Magnetic moment,
1:
m=---
Bsin8
0.016 = 0.40 Am2.
800x10-4xsin30°
(b)For most stable position,8=0°and for most
unstable position 8= 180°. So the required work done
bythe external force,
W=-mB(cos180° -cos0°)=2mB
= 2x0.40x800x10-4=0.064J.
(c)Here the displacement and the torque duetothe
magnetic field are in opposition. Sothe work done bythe
magnetic field due to the external magnetic field is
W8=-0.064J.
(d)HereA= 2x10-4m2,N=1000
Magnetic moment of solenoid,
nls=m=0.40 Am2
But ms =NlA
m 0.40
•.Current, 1=_5= =2A
NA1000x2x10-4
~rOblems For Practice
1.Ashort bar magnet of magnetic moment 0.9 JT -1is
placed with its axis at30°toauniform magnetic
field. It experiences atorque of 0.063 J.(i)Calculate
the magnitude of the magnetic field. (ii)In which
orientation will the bar magnet be in stable
equilibrium in the magnetic field ?[CBSE F 121
...• -t
(Ans.0.14 T, mIIBfor stable equilibrium)
2.A circular coil of300turns and diameter 14 em
carries a current of 15 A. What is the magnitude of
magnetic moment associated with the loop?
[Haryana OIl(Ans. 69.3JT-1 )
3.Calculate the magnitude of the torque required to
hold a bar magnet of magnetic moment 200 Am2
along a direction making an angle of30°with the
direction of a uniform magnetic field of 0.36 G.
(Ans.3.6x10-3Nm)
4.Calculate the torque acting on a magnet of length
20em and of pole strength 2x10-5 Am,placed in
earth's magnetic field of flux density 2 x10-5T,
when(i)magnet is parallel to the field (ii) magnet is
perpendicularto the field.
[Ans.(i)Zero (ii) 0.8 x10-10Nm]···3wx£n~m}r°n3lxv

440. MAGNETISM
5.The magnetic dipole moment of the earth is
6.4x1021Am 2. If
we consider it to be due to a
current loop wound round the magnetic equator of
the earth, then what should be the magnitude of the
current? Assume the earth to be sphere of radius
6400 km. (.6..s5x107A)
6.A straight solenoid of length 50 ern has 1000 turns
and a mean cross-sectional area of 2x10-4m 2.It is
placed with its axis at 30°, with a uniform magnetic
field of 0.32 T. Find the torque acting on the
solenoid when a current of 2 A is passed through it.
(\ns.0.064 Nm)
7A current of 3 A flows through a planecircular coil
of radius 4 cm and having 20 number of turns. The
coil has ben placed in a uniform magnetic field of
0.5 T. Find(i)dipole moment of the coil
(ii)potential energy of the dipole
(Ans0.3 Am 2, - 0.15 J)
8.A bar magnet placed in a uniform magnetic field of
strength 0.3 T with its axis at 30° to the field, exp-
riences a torque of 0.06 Nm. What is the magnetic
moment of the bar magnet? [ISeE 98]
(Ans.0.4 Am 2)
9.A short bar magnet placed with its axis at 30° with a
uniform external magnetic field of 0.16 T exp-
riences a torque of magnitude 0.032 J.
(a)Estimate the magnetic moment of the magnet.
(b)Ifthe bar were fre to rotate, which orienta-
tionswould correspond to its
(i)stable, and(ii)unstable equilibrium?
What is its potential energy in the field for
cases(i)and(ii)?
[Ans.(a)0.40 JT-I,(b)(i) -0.064 J(ii)+0.064 JJ
10.Calculate the work done in rotating a magnet of
magnetic moment 3.0 J T-1 through an angle of 60°
from its position along a magnetic field of strength
0.34x10-4T. (Ans.5.1x10-5J)
11.A bar magnet of magnetic moment 2.5 Am 2 is fre
to rotate about a vertical axis through its centre
magnet is released from rest from the east-west
direction. Find the kinetic energy of the magnet as it
aligns itself in the north-south direction. The
horizontal component of earth's magnetic field is
0.3 G. (!1S.75JlJ)
HINTS
1.Here m=0.9JrI, 9=30°, ,=0.063J
(i)B=-'- = 0.063 = 0.14 T
msin 9 0.9 xsin 30°
(ii)When 9 = 0°, U =-mBcosO° =-mB
5.17
The P.E. of the magnet is minimum. Hence the bar
magnet will be in stable equilibrium whenn;is
->
parallel to B.
2HereN= 300,I= 15 A, r= 7 em = 7x10-2m
:. m=NIA=NIx1tr2= 300x15x3.14x(7x10-2)2
=69.3 JT-1.
5.Use 111=LA=Ix1tr2.
6.r>I11Bsin 9 =NIABsin 9
= 1000x2x2x10-4x0.32xsin 30°
= 0.064 Nm.
7.(i)m=NIA=NIx1tr2= 20x3x3.14x(0.04)2
=0.3 Am2
(ii)In equilibrium position,
U= -mB= - 0.3 x 0.5 = - 0.15 J.
, Q06 2
8.111=---= = 0.4 Am.
Bsin 9 0.3 sin 30°
9.(a)Here 9 = 30°, B = 0.16 T, r= 0.032 J
Magnetic moment,
111=_'_ = 0.032 =0.40 JT-1.
B sine0.16xsin 30°
(b)Potential energy of the dipole in a magnetic field
->
B is given by
-> ->
U= - 111. B = -mBsin 9
(i)The bar will be in stable equilibrium when its
-> ->
magnetic momentmis parallel to B (9 = 0°).
Its potential energy is then minimum and is
given by
Umin= - I11Bcos 0° =-mB
= - 0.40x0.16 = - 0.064J.
(ii)The bar will be in unstable equilibrium when
n;is antiparallel toB(9 = 180°). Its potential
energy is then maximum and is given by
Umax= -mBcos 180° =+mB=+0.064J
10W=-I11B(cos92-cos~)
= - 3.0 x0.34x10-4(cos 60° - cos 0°)
= 5.1x10-sJ.
11.Here ~ = 90°, 92 = 0°,111= 2.5 Am 2,
B= 0.3 G= 0.3x10-4 T
Kinetic energy = Loss in P.E.
=Ui-Uf= - mBcos 90° +mBcos 0°
=2.5x0.3x10-4J=75JlJ.···3wx£n~m}r°n3lxv

441. 5.18
5.14BARMAGNET AS AN EQUIVALENT
SOLENOID
16. State som
e similarities between a current carrying
solenoid and a bar magnet.
Similarities betwen a current carrying solenoid
and a bar magnet. When a current is passed through a
solenoid, it behaves like a bar magnet. Some observa-
tions of similar behaviour are as follows:
1.A current carrying solenoid suspended frely
always comes to rest in north-south direction.
2.Two current-carrying solenoids exhibit mutual
attraction and repulsion when brought closer
to one another. This shows that their end
faces act as N-and S-poles like that of a bar
magnet.
3. Figure 5.24 shows the lines of force of a bar
magnet while Fig. 5.25 shows the lines of force
of a finite solenoid. The two patterns have a
striking resemblance
Fig. 5.24Field lines of a bar magnet.
If we move a small" compass nedle in the
neighbourhood of the bar magnet and the
current carrying finite solenoid, we shall find
that deflections of the nedle are similar in the
two cases. This again supports the similarity
betwen the two fields.
+
Fig. 5.25Field lines of a current carrying finite solenoid.
PHYSICS-XII
4.The magnetic fields of both the bar magnet and
current carrying solenoid at any far away axial
point are given by the same expression:
B"=1102m
axial 41t . r3
Thus a bar magnet and a solenoid produce
similar magnetic fields.
17. Explain how isa current carrying solenoid
equivalent to a bar magnet.
A solenoid as an equivalent bar magnet. A solenoid
can be regarded as a combination of circular loops
placed side by side, as shown in Fig. 5.26(a). Each turn
of the solenoid can be regarded as a small magnetic
dipole of dipole moment fA.Then the solenoid
becomes an arrangement of small magnetic dipoles
placed in line with each other, as shown in Fig.5.26(b).
The number of such dipoles is equal to the number of
turns in the solenoid. The north poleof one touches the
south of the adjacent one. The opposite poles
neutralise each other except at the ends. Thus, a
current carrying solenoid can be replaced by justa
single south pole and a single north pole, separated by
a distance equal to the length of thesolenoid. Hence a
current carrying solenoid is equal to a bar magnet as
shown in Fig. 5.26(c).
14---------L --------~
(a)
E:BE:BE:BE:BE:BEJ
snsn5nsns "5n
(b)
Is. .NI(c)
14--- ----- L-- -- -- --~
Fig. 5.26A solenoid as an equivalent bar magnet.
A bar magnet and a finite solenoid produce similar
magnetic field patterns, as shown in Fig. 5.24 and
Fig. 5.25 respectively. It may be noted that the
magnetic field inside the solenoid is in direction
opposite to that we expect on the basis of the above
pole model (N ~5).
18. Explain howisa bar magnet equivalent to a
current carrying solenoid.
A bar magnet as an equivalent solenoid.We can
explain this byAmpere's hypothesisaccording to
whichall magnetic effects are produced by current-loops.
The electrons in an atom kep on revolving around its
nucleus and hence set up electric currents. These atomic···3wx£n~m}r°n3lxv

442. MAGNETISM
currents are equivalent
to small circular current-loops.
Inamagnet, these current-loops are arranged parallel
to each other and have currents in thesame sense.
Figure 5.27 shows theatomic current loops in a
cross-section of a cylindrical bar magnet. At any point
inside the magnet, the currents from the adjacent loops
cancel eachother and hence thenetcurrent is zero. But
thereis anet current on the surface. Due to this surface
current, the bar magnet is equivalent to aclosely-
wound, current carrying solenoid. Hence a bar magnet
produces a magnetic field similar to the solenoid.
Fig. S.27A bar magnet as an equivalent solenoid.
It may benoted here that at theendsof the magnet,
thecurrent loops behave differently from those inside
themagnet. As aresult, the magnetic poles are located
slightly inside the bar magnet. That is why the
magnetic length of abar magnet is slightly lessthan its
gmetrical length.
5.15GAUSS'SLAWIN MAGNETISM
19.State Gauss'slaw in magnetism. What are its
important consequences ?
Gauss's law in magnetism. Gauss's lawin elec-
trostatics states that the surface integral of the
-4
electrostatic field E over aclosed surface 5 isequal to
1/EOtimes the total chargeqenclosed by the surface 5,
i.e.,
fE.is=_5...
s EO
Suppose that theclosedsurface 5 encloses an
electric dipole which consists of two equal and oppo-
site charges. Then the total charge enclosed by 5 is zero
so that thesurface integral of the electrostatic field of a
dipoleover theclosed surface is also zero, i.e.,
f~ipo)e' dS=0
5
Now a magnetic field is produced only bya
magnetic dipole because isolated magnetic poles do
not exist, so the above equation for a magnetic field can
be written as
fB.dS=0
5
5.19
ThisisGauss's law in magnetismwhich states that
the surface integral ofa magnetic fieldover aclosedsurface is
always zero. But the surface integral of a magnetic field
over asurface gives magnetic flux through thatsurface.
So Gauss's law in magnetism can also bestated as
follows:
The net magnetic flux through a closed surface iszero.
ConsequencesofGauss's law:
1.Gauss's law indicates thatthere are no sources or
sinks ofmagnetic field inside aclosed surface. So
there isnopoint at which the fieldlines start or
there is no point at which the field lines
terminate. In other words, there are nofre
magnetic charges. Henceisolated magnetic poles
(also called monopoles) donotexist.
2. The magnetic poles always exist as unlike pairs
of equal strengths.
3.If a numberof magnetic lines of force enter a
closed surface,then anequal number oflines of
force must leave thatsurface.
Closed
surface
Fig. 5.28Magnetic field lines never terminate.
\
For Your Knowledge
•Gauss's law of magnetism formally expresses the fact
thatmagnetic monopoles do not exist.Hence the most
elementary magnetic element is a magnetic dipole or
a current loop. All magnetic phenomena can be
explained in terms of an arrangement of magnetic
dipoles and/or current loops.
>Basic difference betwen electric and magnetic lines
of force
magnetic monopoles do not exist is that magnetic
lines of force are continuous and form closed loops.
They do not start or end at a point. In contrast, the
electric lines of force start from a positive charge and
end on a negative charge or they fade out at infinity in
case of isolated charges.
5.16MAGNETIC FIELD OF THE EARTH
20. Give some experimental evidences which support
the existence of earth's magnetic field.···3wx£n~m}r°n3lxv

443. 5.20
Magnetic field of the earth. Earth is a
powerful
natural magnet. Its magnetic field is present every-
where near the earth's surface. TIUsfield can be approxi-
mated to the field of a magnetic dipole of dipole
moment8.0x1022Am2 assumed to be located at the
centre of the earth. The axis of the dipole makes an
angle of about20°with the axis of rotation of the earth.
Themagnetic north pole Nmof the earth lies somewhere
near thegeographic south poleSwhile themagnetic south
poleSmlies somhere neargthegeographic north pole
Ng'The magnitude of the magnetic field on theearth's
surface istypically about lO-4T whichis equal to
1gauss(G).Agauss isalso often called an oersted. Thus
theearth's magnetic field is of theorder of 1 oersted.
The branch of physics that deals with the study of
earth's magnetismiscalledterrestrial magnetism
orgeomagnetism.
Experimental evidences in support of earth's
magnetism:
1. A freely suspended magnetic needle comes torest
roughly innorth-south direction. TIUs suggests
that the earth behaves as a huge magnet with its
south pole lying somhere near thegeographic
north pole and its north pole lying somhere
near the ggraphic south pole.
2. An iron bar buried inthe earth becomes weak magnet
aftersome times induced by
earth's magnetic field.
3. Existence of neutral points near a bar magnet indi-
cates the presence of earth's magnetic field. At
these points, the magnetic field of the magnet is
cancelled by the earth's magnetic field.
5.17ORIGIN OF EARTH'S MAGNETIC FIELD
21.Givea brief account of different theories regarding
thesource ofearth's magnetism.
Origin of earth's magnetic field.The magnetic field
of the earth is approximately like that ofagiant bar
magnetembedded dep inside the earth. Many thries
have ben proposed about the cause of earth's
magnetism from time to tim. Someofthese are
mentioned below :
1. In1600,William Gilbert in his book'DeMagnete'
first suggested that the earth behaves as a bar magnet
and its magnetism is due to the presence of magnetic
material at its centre, which could be a permanent
magnet. However, the core of the earth is so hot thata
permanent magnet cannot exist ther.
2.Prof Blackett suggested that the earth's
magnetism is due to the rotation of the earth about its
own axis. Everysubstance is made of charged particles
PHYSICS-XII
such as protons andelectrons. Astheseparticles rotate
along with the earth, theycause circulating currents
which, in turn, magnetise the earth.
3.Cosmic rays causetheionisation of gases in the
earth's atmosphere the earth rotates, strong electric
currents areset up due tothe movement of the charged
ions.These currents maybe the source of earth's
magnetism.
4.According toSirE.Bullard (U.K.) and w.M.
Elaster(U.S.A.), there are large deposits of ferro-
magnetic materials like iron, nickel, etc. in the core of
the earth. The coreof the earth is veryhot and molten.
The circulating ions in the highly conducting liquid region
of the earth's coreform current loops and hence produce a
magnetic field. At present, this hypothesis seems most
probable because our moon, which has no molten core,
has no magneticfield. Venus, which has a slower rate
of rotation, has a weaker magnetic fieldwhile Jupiter,
withafasterrateof rotation has a stronger magnetic
field.
The changes in the earth's magnetic field are so
complicated and irregular that the exact cause of
earth's magnetism is yettobe known.
5.18SOME DEFINITIONS IN CONNECTION
WITH EARTH'S MAGNETISM
22.Define theterms geographic axis, magnetic axis,
magnetic equator, magnetic meridian and geographic
meridian inconnection with geomagnetism.
S( me definitions in connection with earth's
magnetism,Fig.5.29showsthe magnetic lines of force
around the earth.
Magnetic
equator
Ggraphic G hi
north Olel .grap ~c
.:::~tt."on
/ Earth's
~magnetic
.Yt:::::---\-_ south pole
Earth's
magnetic --
north pole / / / //7
Ggraphic
south pole
Fig. 5.29Magnetic field of the earth.
1. Ggraphic axis.The straight line passingthrough
the geographical north and south poles of the earth iscalled
its geographic axis. It isthe axis ofrotation of the earth.www3«°tssr•wvs3q°~

444. MAGNETISM
2.Magnetic
axis.The straight line passing throughthe
magnetic north and south poles of the earth is called its
magnetic axis.
The magnetic axis of the earth makes an angle of
nearly 20° with the geographic axis. At present, the
magnetic south pole Smis located at a point in
Northern Canada at a latitude of 70.5°N and a
longitude of 96°W. The magnetic north pole NIII is
located diametrically opposite to Smi.e.,at a latitude of
70.5°5 and a longitude of84°E. The magnetic poles are
nearly 2000 km away from the ggraphic poles. The
magnetic equator intersects the ggraphic equator at
longitudes of 6°W and 174°E.
3. Magnetic equator.Itisthe great circle onthe earth
perpendicular to the magnetic axis.
4. Magnetic meridian. The vertical plane passing
through the magnetic axis of a freely suspended small
magnetiscalled magnetic meridian. 171eearth's magnetic
field actsinthe direction of the magnetic meridian.
5. Ggraphic meridian. The vertical plane passing
through the geographic north and south polesiscalled
geographic meridian.
5.19ELEMENTS OF EARTH'S
MAGNETIC FIELD
23.What are the elements of earth's magnetic field?
Explain their meanings. Show these elements ina labelled
diagram and deduce various relations between them.
Elements of earth's magnetic field. The earth's
magnetic field at a place can be completely described by three
parameters which are called elements of earth's magnetic
field.They are declination, dip and horizontal component
of earth'smagnetic field.
1. Magnetic declination.171eangle between the
geographical meridian and the magnetic meridian at a place
iscalled the magnetic declination(a)at that place.
Magnetic declination arises because themagnetic
axis ofthe earth does not coincide with its ggraphic axis.
To determine magnetic declination at a place, set
up acompass nedle that is fre to rotate in a
horizontal plane about a vertical axis, as shown in
Fig.5.30. The angleathatthis nedle makes with the
ggraphic north-south (Ng -Sg)direction is the
magnetic declination. By knowing declination, we can
determine the vertical plane in which the earth's
magnetic field lies. In India, the value ofais small. It is
0°41' Efor Delhi and 0°58' W forMumbai. Thismeans
that the N-pole of a compass nedle almost points in
the direction of ggraphic north.
5.21
Declination
Ng
Fig. 5.30Determination of declination at a plac.
2. Angle of dip or magnetic inclination.Theangle
~
made by the earth's total magnetic fieldBwith the horizontal
direction inthemagnetic meridian iscalled angle of dip(8)at
any place.
Vertical
-+
B Sill
Fig. 5.31Determination of dip at a place.
Theangle ofdip is different at different places on
thesurface of the earth. Consideradip needle, whichis
just another compass nedle but pivoted horizontally
sothatit isfre torotatein a vertical plane coinciding
with the magnetic meridian. It orients itself so that its
N-pole finally points exactly in thedirection of the
~
earth'stotal magnetic field B.The angle betwen the
horizontal and the final direction ofthedip nedle
gives theangleof dip at the given location.
At the magnetic equator, thedip nedle rests
horizontally so that the angle ofdipiszero at the magnetic
equator.The dip needle restsvertically at the magnetic
polessothattheangleofdip is90°atthemagnetic poles.At
allother places, thedip angle liesbetwen 0° and 90°.
3.Horizontal component of earth's magnetic
field. Itisthe component ofthe earth's total magnetic
~
fieldBinthe horizontal direction inthe magnetic···3wx£n~m}r°n3lxv

445. 5.22
meridian.If 8
is the angle of dip at any place, then the
~
horizontal component of earth's field B at that place is
given by
BH=Bcos 8
Atthe magnetic equator, 8=0°,BH=B cos0°=B
Atthemagnetic poles, 8=90°, BH=Bcas90°=0
Thus the value of BHisdifferent atdifferentplaces on
the surface of the earth.
Ggraphic
meridian
Magnetic
meridian
Fig. 5.32Elementsof earth's magnetic field.
Relations betwen elements of earth's magnetic
field. Fig.5.32 shows the thre elements of earth's
magnetic field. If8is the angleof dip at any place, then
the horizontal and vertical components of earth's
~
magnetic field B at that place will be
and
BH·= Bcos 8
Bv=Bsin 8
Bv=Bsin 8
BH Bcos 8
Bv=tan 8
~
or
Also
BH2+Bv2=W(cos28+sin28)=W
or B=~BH2+Bv2 ...(3)
Equations (1),(2)and (3) are the different relations
betwen the elements of earth's magnetic field. By
knowing the thre elements, we can determine the
magnitude and direction of the earth's magnetic field
at any plac.
PHYSICS-XII
ForYour Knowledge
~ Magnetic maps.These arethe detailed charts which
indicate on the world map the linespassingthrough all
such places where oneof the threemagnetic elements has
thesame value. Thre types of lines are drawn on such
maps. These are
1.Isogonic lines. The lines joining theplaces of equal
declination are called isogonic lines. The line of
zero declination is calledagonic line.
2.Isoclinical lines. Thelinesjoining. theplaces of
equal dipor inclinationare called isoclinical lines.
The line of zero dip is calledaclinic line or
magnetic equator. The points of 90° dip are called
magnetic poles. The magnetic equator crosses
the ggraphic equator twice once in Atlantic
and then in Pacific ocean.
3.Isodynamic lines. The lines joining the places
having the same value of the horizontal component
of earth's magnetic fieldare called isodynamic lines.
The horizontal component is zero at poles and
maximum at the magnetic equator.
5.20GLOBAL VARIATIONS IN THE EARTH'S
MAGNETIC FIELD· .
...(1)
24.Describe the variations of earth's magnetic field
from place to place.
Global variationsinthe earth's magnetic field.
Earth's magnetic field changes both in magnitude and
direction from place to place
global variations are as follows:
1.The magnitude of the magnetic field on earth's
surface is small, nearly 4x10-5T.
2.Still smaller is the background field of our own
galaxy, the Milky Way, being about 2 pTi.e.,
2x10-12T.
3. If we assume that the earth's field is due to
dipole of 8.0x1022Am2 located at its centre,
then the earth's magnetic field will be less than
1 u'T (l0-6T) at a distance of 5 times the radius of
the earthi.e.,at about 32,000 km. Upto this
distance, the magnetic field is entirely governed
by the earth.
4. At distances greater than32,000 km, the pattern
of the earth's magnetic field gets severely
distorted by thesolar wind.
5.Solar wind causes ionisation of atmosphere
near the magnetic poles of the earth. This in
turn causes beautiful displays of colours high
up in the sky and is known asaurora.
...(2)www4°±trsp€vvr4o±~

446. MAGNETISM
25. Whatissolar wind? Ho
wdoesit affect earth's
magnetic field?
Solar wind. Thesolarwindisastream of hot charged
ions, composed of equal numbers of protons andelectrons
continuously flowing radially outward from the SUIlwith a
speed of approximately 400kmls. A long magneto tail
stretches out forseveral thousand earth diameters in a
direction away from the sun.
At distances greater than 32,000 km, thedipole
field pattern of the earth's magnetic field gets severely
distorted by the solar wind, asshowninFig.5.33.
Solar
wind
Fig.5.33Distortion of earth's magnetic field by
the solar wind.
26. What are aurora borealis and auroraaustralis ?
Can this effect be seen anywhere in India?
Aurora borealis and aurora australis.This is a
spectacular display of light sen in the night sky at
high altitudes, occurring most frequently near the
earth's magnetic poles. The displays of aurora appear
as gaint curtains high up in the atmosphere
aurora is caused when the charged particles of the
solar wind get attractedbythe magnetic poles of the
earthand theretheyionise the atmospheric atoms or
molecules. The aurora in the northern hemisphere is
calledaurora borealisornorthern lightsand the
aurora in southern hemisphere is called aurora
australis orsouthern lights.
5.21TEMPORAL VARIATIONS IN THE
EARTH'S MAGNETIC FIELD'
27. Describe thevariations of earth's magnetic field
that have occurred with the passage of time.
Temporal variations in the earth's magnetic field.
The earth's magnetic field changes both in magnitude
and direction as time passes. These changes are of two
types:
(i)Short term changes.The position of the magnetic
poles of the earth keps shifting slowly with the
passage of time years, from 1580 to
1820, the magnetic declination at London has changed
by 35°. The magnetic south pole in the northern Arctic
region of Canada has ben found to shift in the
north-west direction at the rate of10 km per year in
recent times.
5.23
(ii)Long term changes.The changes in earth's
magnetic field over long term or glogical time scales
are interesting. The studies of basalt reveal that earth's
magnetic field reverses its direction every million
years or so. This means that once in a million years or
so, the currents in earth's core cool down, come to a
halt and then pick up sped in the opposite direction.
Basalt which contains iron, is emitted during
volcanic activity on the ocean floor. As it cools, it
solidifies and provides a picture of earth's magnetic
field. Its age can be determined by other means.
5.22NEUTRAL POINT
28. Define neutral point. How will you find the
magnetic moment of a bar magnet by locating its neutral
points, when the magnet is placed with its north pole
towards(i)north pole of the earth and(ii)south pole of
the earth?
Neutral point. Itisthe point where the magnetic field
dueto a magnet isequal andopposite to the horizontal
component of earth's magnetic field. The resultant magnetic
field at the neutral point iszero.If a compass nedle is
placedatsuch a point, it can stay in any position.
(i)Magnet placed in the magnetic meridian with its
north pole pointing north.Fig. 5.34 shows the
magnetic lines of force ofabar magnet placed in the
magnetic meridian with its north-pole pointing
towards theggraphic north of the earth. The fields
due to the magnet and the earth are in same directions
at points on the axial line and are in opposite directions
at pointson the equatorial line
stronger at axial points and weaker at equatorial
points. The two neutral points P and Q lie on the
equatorial line.
Let
South
Fig. 5.34 Field lines of a bar magnet with its
N-pole towards north.···3wx£n~m}r°n3lxv

447. 5.24
Then megnetic field strength at each neutral
point is
B = !-Io m
equa 41t .(r
2+[2l/2
For a short magnet, [«r,therefore,
_!-Io m
Bequa -41t.?
At the neutral point, the field of the magnet is
balanced by the horizontal componentBHof the earth's
magnetic field so that
BrI=!-I0.!!.:
41t'r3
Knowing randBrI,the value of the magnetic dipole
momentmcan be determined.
(ii)Magnet placed in the magnetic meridian with
its south-pole pointing north.Fig. 5.35 shows the
magnetic lines of force of a bar magnet placed in the
magnetic meridian with its south-pole pointing
towards the ggraphic north of the earth. Here the
fields due to the magnet and the earth are in the same
direction at points on the equatorial line and are in
opposite directions at points on the axial line of the
magnet. So the resultant field is weaker at axial points
and is stronger at equatorial points. In this case the two
neutral pointsPandQlie on the axial line near the
ends of the magnet.
N
W+E
Earth's S
field
South
Fig. 5.35 Field lines of a bar magnet with
S-pole towards north.
Suppose r be the distance of each neutral point
from the centre of the magnet. Let21be the length of
the magnet. Then magnitude of the magnetic field at
either of the neutral points will be
B.=!-Io 2mr
axial41t . (,1 _z2)2
PHYSICS-XII
For a short magnet,1««r,therefore
B.=!-Io 2m
axial 41t .r3
Again, at the neutral point, the field of the magnet
is balanced by the horizontal componentBrIof the
earth's magnetic field, so we have
B =!-Io 2m
H 41t' r3
Knowing the values of r andBrI,the magnetic
dipole momentmof the magnet can be determined.
Fo nu e Used
1.Declination(a)=Angle betwen ggraphic
meridian and magnetic meridian.
2.Relations betwen elements of earth's magnetic
field are
BH=Beas8and
~=tan8and
BH
3.For a magnet placed with its N-pole pointing
north, neutral points lie at its equatorial line
Bv=Bsin 8
B= ~B~+l\i
B=!-I0 m
H 471:'(r2+[2)3/2
-!-Iom
-471: .r3
[fori.lhort magnet]
4.For a magnet placed with its N-pole pointing
south, neutral points lie on its axialline
B=!-I0 2mr
H 471:'(r2_[2)2
Ilo2m
-- - [forashort magnet]
-471: .r3
Urn s Used
Magnetic fieldsB, BHandBvare intesla,
distancesrand [ in metre, magnetic momentmin
JT-1or Am2,angleaand8are in degres.
Coso sed
Ilo=471:x1O-7TmA -1.
Example 24The declination at a placeis15°west of north.
In which direction should a ship be steered so that it reaches a
place due east?
Solution. As the ship is to reach a place due east
i.e.,along OP (Fig. 5.36), so it should be stered at angle
of 15° +90°=105°with the direction of the compass
nedl.www4°±trsp€vvr4o±~

448. MAGNETISM
MNGN
_
-ME
P
GW--------.-~~----_.---GE
GSMS
Fig.5.36
Example25.Ashipisto reach a place 10° south of west. In
which directionshould it besteered ifthedeclination at the
place is18°west of north?
Solution. As the ship is to reach a place 10° south of
westi.e.,alongOP(Fig.5.37), so itshould be stered
west of magnetic north at angle of 90-18+10 = 820.
MNGN
P
MW---
_-ME
GW-------c~~----------GE
GSMS
Fig. 5.37
Example26.In the magnetic meridian of a certain place,
thehorizontalcomponent of the earth's magnetic field is0.26G
and the dip angle is60°.Whatisthemagnetic field of the
earth in this location? [NCERT]
Solution. Here BH =0.26 G,8=60°
As ~=Bcas8
B=~ = 0.26 G = 0.26 G =0.52 G.
cos8cos60° 0.5
Example 27.A compass needle whose magnetic moment is
60An?pointing geographical north at acertainplacewhere
thehorizontal component of earth's magnetic fieldis
40 IlWb /niexperiences a torque of1.2 x10-3Nm.Whatis
thedeclination of theplace? [Roorke 82]
Solution.In stable equilibrium, a compass nedle
points along magnetic north and experiences no torque
5.25
When it is turned through declinationa,it points along
ggraphic north and experiences torque,
't= mBsina
. r 1.2 x10-3 1
sma=-=-------,-
mB 60x40x10-6 2
a=30°.or
Example28.Thehorizontal andvertical components of
earth'sfield at aplace are0.22gauss and 0.38 gauss respec-
tively. Calculate the angle of dip andresultant intensity of
earth'sfield. [Haryana 93; Himachal 94]
Solution. Here BH = 0.22 G,By=0.38G
Now tan8= ~= 0.38 =1.7272
BH0.22
:. Angle of dip, 8 =59°56'
Resultant magnetic field of the earth is
B=~BH2+By2=~0.222+0.382 =0.427 G.
Example 29. Ifthehorizontalcomponent of earth's magnetic
field at aplace where theangleof dipis60°is0.4X1O-4T,
calculate thevertical component andtheresultant magnetic
fieldat that place. [CBSE 00 97C;Haryana01]
Solution. Here8=60°,BH =0.4x10-4T
By=BHtan8= 0.4x10-4tan60°
=0.4x10-4x.J3= 0.69x10-4 T.
Resultant magnetic field,
B 04 10-4
B=_H_= .x =0.8x 10-4 T.
cos 8cas 60°
Example30.If81and 82 be theangles of dip observedin
twovertical planes at right angles to each other and 8isthe
true angle of dip, prove that cot2 81+cot282=cot28.
[Punjab 96]
Solution. Let ~ and By bethe horizontal and
~
vertical components of earth's magnetic field B .Since 8
is the true angle of dip,therefore
tan 8= By
BH
B
or cot 8 = -1i. ...(1)
By
As shown in Fig. 5.38, suppose planes 1 and 2 are two
mutually perpendicular planes and respectively make
angles 8 and90°-8 with the magnetic meridian. The
vertical components of earth's magnetic field remain
same in the two planes but the effective horizontal
components in the two planes will be
Br= BH cos 8 and ~ = BH sin 8···3wx£n~m}r°n3lxv

449. 5.26
Pla
ne 2
Magnetic
meridian
"--'----- ..•• '----- Plane 1
Fig.5.38
Theangles of dip01 and 02in the two planes are
given by
or
tano=Bv= Bv
11\ BHcos8
BH cas 8
cot°1=--'-'--Bv--
tans=~ = Bv
2 ~ ~sin8
B sin 8
cot 02=----'-'H__
Bv
or
From equations (2) and (3), we have
2 2
cot2°+cot2°= BH (cos2 8+sin28) = BH
1 2 Hi Hi
or cot2 01+cot202= cot2 0.[Using equation (1)]
Example 31.A dipcircleshows an apparent dip of 60°at a
place where thetruedip is45°.If the dipcircle isrotated
through90°,what apparent dip will it show?
Solution. Here 01=60° and 0=45°.
As cot2 s= cot2 01+cot202
cot245°= cot2 60°+cot202
cot202= (1)2-(.1r= ~
cot 02 = 0.816
02=51°.
Example 32.Truevalue of dip at a place is45°.Theplaneor
of the dipcircleisturned through 60°from the magnetic
meridian. Find the apparent value of dip.
Solution. Here0= 45°, 8 = 60°, 0' =?
tan 0'=Bv= BH tan °
~ ~ cas 8
=tan° =tan 45° =2
cas 8 cas 60°
or
or
:. Apparent dip,
PHYSICS-XII
Example 33.Abar magnet of length 10emisplaced in the
magnetic meridian with its north pole pointing towards the
geographical north. A neutral pointisobtained at a distance
of12em from the centre of the magnet. Find the magnetic
moment of the magnet,if ~=0.34 G.
Solution. Here, 21= 10 em, 1=5em=5x1O-2m
r=12em=12x10-2m,~ =0.34 G=0.34 x10-4T
In this case, the neutral points lie on the equatorial
line of the magnet so that at any neutral point,
or~o m =B
411:.(r2+12)3/2 H
~qua =~
...(2)
:.Magnetic moment
m= ~. 411:.(1+12)3/2
~o
= 0.34 x10-4 x_1_[52+122]3/2(10-4)3/2
10-7
=0.747rr'.
Example 34.Themagneticmoment of a short bar magnet
is1.6A~.Itisplaced inthemagnetic meridian with north
polepointing south. Theneutralpointisobtained at distance
of20em from the centre of the magnet. Findthehorizontal
component ofearth's magnetic field. If the magnet be
reversed, i.e.,north pole pointing north, findtheposition of
neutral point.
Solution. Here m=1.6 Am2,r=20 em=0.20 m
...(3)
When N-poleof the magnet pointssouth,
neutral points lie on the axial line of the magnet.
Hence at the neutral point,
B - - ~o2m
axial - ~ -411:·r3
~=10-7x2x1.6=4x10-5T
(0.20l
the
or
When the magnet is reversed, its north pole points
north. The neutral points will lie on the equatorial line
of the magnet. Hence
_ _~o m
Bequa - ~ --. 1
411:r:
r3=~o. ~ =10-7x1.6=4.0x10-3m3
411: BH 4x10-5
r=(4.0x10-3)1/3
=1.6x10-1m=16 em.
Example 35.A magnet placed in the magnetic meridian
with its north pole pointing north of the earth produces a
neutral point at a distance of 0.15 m from eitherpole.Itis
then broken into twoequal pieces andone such piece is
placed ina similar position. Findtheposition of the neutral
point.···3wx£n~m}r°n3lxv

450. MAGNETISM
Soluti
on.Here the neutral points lie on the equatorial
line of the magnet at distancexfrom each of the two
poles.
_ _ 110 m
BH- Bequa -411:.x3
When the magnet is broken into two parts, its pole
strength remains unchanged.
Original magnetic moment,
m= qlllx21
Magnetic momentof each part,
21m
m=«;X2=2
B=110 m/2
H411:'x,3
110m _110 m
411:.x3-411:.2x,3
2x,3= x3
Hence
or
,x
x=21/3
0.15 f h I
x'=--=0.119 m, rom eac po e
1.26
Example36.Themagnetic field at a point on the magnetic
equator is3.1x10-5T. Taking theradius of the earth equal to
6400km,findthe magnetic momentof theassumed dipole at
~~~~~~ ~
Solution.Anypoint on the
magnetic equator lies in the
broad side on position of the
assumed magnetic dipole.
Hence
or
or
or
B - 110!!!...-
equa -411:.R3
411: 3Fig.5.39
m=-.Bequa R
~lO
= 107x3.1x 10-5x(6400x103)3
=8.1x1022Am2.
GS
Example37.The earth's magnetic field at the equator is
approximately 0.4G.Estimate the earth's dipole moment.
Radius of the earth = 6400km. [NCERT]
Solution. Here ~ = B = 110~
equa 411:r
0.4x10-4= 10-7 xm
(6.4x106)3
m=1.04x 1023Am2.
or
or
Example38.Ashort bar magnet isplaced in a horizontal
plane withitsaxisin themagnetic meridian. Null points are
found onitsequatorial line(i.e.,itsnormal bisector) at12.5em
5.27
from the centre of the magnet. The earth's magnetic field at
theplaceis0.38Gand the angleof dipiszero.
(i)Whatisthetotal magnetic field at points on the axis
of themagnet located at the same distance 02.5em)
asthenull-points from the centre?
(ii)Locate the null points whenthemagnet isturned
around by 180°.
Assume that the length of themagnet isnegligible as
compared to the distance of the nul/-points from the
centre ofthe magnet.
Solution. (a)At the neutral point on the equatorial
line of ashort magnet, we have
= 110.!!!.=B
Bequa 411:.r3 H
Magnetic field of the magnet on its axial line at the
same distance will be
1102m
Baxial=-.3=2BH=2x0.38 =0.76 G
411:r
At any point on the axial line,BHandBaxial are in
the same direction. So total magnetic field,
B=Baxial+BH= 0.76+0.38 =1.14G.
(b)When the magnet is turned through 180°, the
neutral points lie on the axial line.
B.=110 2m=B
axial 411:.x3 H
But ~ =110m 110m_110 2m
411:.r3 . .411:'r3-411:. ~
or x3=2r3
or x=(2)1/3r= 1.26 x12.5 em =15.75 cm.
cproblems For Practice
1.A ship is sailing due west accordingto Mariner's
compass. If the declination of the place is 15° eastof
north, what is true direction of the ship?
(Ans. 75°west of north)
2.Aship is sailing due east according to Mariner's
compass. If the declination of the place is 18° eastof
north, what is the true direction ofthe ship?
(Ans. 18°south of east)
3.The horizontal component ofearth's magnetic field
is0.2 G and total magneticfield is 0.4 G. Findangle
of dip. [Haryana96] (Ans.600)
4.Calculate earth's magnetic field ataplace, where
the angle ofdip is 60°andvertical component of
earth's field is 0.40 G. (Ans. 0.462 G)
5.A magnetic nedle fre to rotate in avertical plane
parallel to the magnetic meridian has its north tip
down 60° with the horizontal. The horizontal···3wx£n~m}r°n3lxv

451. 5.28
component of the
earth's magnetic field at the place
known to be 0.4 G. Determine the magnitude of the
earth's magnetic field at the place[CBSE F 111
(Ans. 0.8 G)
6.The vertical and horizontal components of earth's
magnetic field at a place are 0.2 G and 0.3464 G
respectively. Calculate the angle of dip and earth's
magnetic field at that place. (Ans. 30°,0.4 G)
7.A vertical wire in which current is flowing pro-
duces a neutral point with the earth's horizontal
field at a distance of 5 em from the wire in air. What
is current,ifBH = 0.18x10-4T ? (Ans. 4.5 A)
8.A compass nedle whose magnetic moment is 60 Am2
pointing ggraphical north at a certain place, where
the horizontal component of earth's magnetic field is
40IlT, experiences a torque of 1.2 x 10-3 Nm.What
is the declination at that place? (Ans. 30°)
9.A magnetic nedle fre to rotate about the vertical
direction (compass) points 3.5° west of the g-
graphic north. Another magnetic nedle fre to
rotate in a vertical plane parallel to the magnetic
meridian has its north tip pointing down at 18° with
the horizontal. The magnitude of the horizontal
component of the earth's magnetic field at the place
is known to be 0.40 G. What is the direction and
magnitude of the earth's magnetic field at the place?
(Ans.0.42 G, directed at an angle of 18° with horizontal
in the magnetic meridian towards the ground)
10.The true dip at a place is 30°. What is the apparent
dip when the dip circle is turned 60° out of the
magnetic meridian? (Ans. 49°71
11.The values of the apparent angles of dip in two planes
at right angles to each other are 30° and 45°.
Calculate the true value of the angle of dip at the
plac. (Ans. 26.6°)
12.A dip circle lying initially in the magnetic meridian
is rotated through an angleein the horizontal
plane
increased in the ratio sece:1
13.A short bar magnet of magnetic moment 0.5 JT-1is
placed with its magnetic axis in the magnetic meri-
dian, with its north pole pointing ggraphical north.
A neutral point is obtained at a distance of 0.1 mfrom
the centre of the magnet. Find the horizontal comfo-
nent of the earth's magnetic field. (Ans. 10-T)
14.A bar magnet 30 cm long is placed in the magnetic
meridian with its north pole pointing ggraphical
south. The neutral point is found at a distance of
30 cm from its centre
the magnet. Given BH = 0.34 G. (Ans.8.61 Am)
15.A neutral point is found on the axis of a bar magnet
at a distance of 10 cm from its one end. If the length
of the magnet be 10 cm and BH = 0.3 G, find the
magnetic moment of the magnet. (Ans. 0.012 Am2)
PHYSICS-XII
16.A magnet placed in the north pointing north
position, balances the earth's magnetic field at a
point, which is 27 cm from either pole
into thre pieces and one such piece is similarly
placed, find the position of the neutral point.
(Ans. 18.73 x 10-2 m, from either pole)
HINTS
4... B=~=~= 0.40 = 0.462 G.
sin0sin 60° 0.866
5.Here,0= 60°, BH = 0.4 G, B=?
B= BH =~= 0.4 =0.8 G.
cos0cos 60° 0.5
11I
7.As-o- =BH
2nr
1- 2r.:rBH _ 2r.:x5xlO-2 xO.18xl0-4 _
.. ---- 7 -4.5A.
110 4r.:x10-
8.1"=111BHsina
.. 1.2 x10-3= 60 x40x10-6xsin a
or sina=-1 .. a=30°.
9.Here 3.5° is the magnetic declination and 18° is the
angle of dip.
As BH=Bcoso
B= BHse0= 0.40 sec 18°
= OAOx1.0514 = 0.42G.
11.cot20= cot2 ~+ cot2 O2
= cot2 30°+ cot2 45°= 3 + 1 = 4
cot 0 = 2 :. 8 = 26.6° .
12.The true angle of dip0is given by tan0=Bv
BH
When the dip circle is rotated through angleS,the
apparent angle of dip0'is given by
tan 0'
:. -- =secO :1.
tan0
tan0/=Bv= tan o. see
BHcos0
~om
13.Use BH=-.3"'
4r.:r
14.Here the neutral points lie on the axial line.
110 2mr
..Baxia1=BHor 41t.(?_12)2 = BH
41t B(r2_12)2
or m=_.-,-,H,-,-_~_
~o 2r
1 0.34 x 10-4 (0.302 -0.152)2 2
=--7 . =2.582 Am
10- 2x0.30
m2582
Pole strength,qm=-=--= 8.61 Am.
21 0.30
~o2m .
15.Use BH =-. -3 .Herer=10+ 5= 15cm.
41tr
16.Proced as in Example 35 on page 5.26.···3wx£n~m}r°n3lxv

452. MAGNETISM
5.23SOME IMPORTANT TERMS USED TO
DESCRIBE MAGNETIC PROPERTIES
OF MATERIALS
29.Define the terms
magnetising field, magnetic
induction, intensiiuof magnetisation, magnetising field
intensity, magnetic permeability, relative permeability
and magnetic susceptibilitu. Write the various relations
among these quantities.
1.Magnetising field. When a magnetic material is
placed in a magnetic field, a magnetism is induced in
it.Themagnetic field that existsinvacuum and induces
magnetismiscalled magnetising field.For example,
consider a toroidal solenoid carrying currentIand
placed in vacuum. If the solenoid hasnturns per unit
length, then the magnetic field set up in the solenoid is
given by
Bo=llonI
This field is called themagnetising fieldcaused by
the so called fre current in the solenoid.
2.Magnetic induction. As shown in Fig. 5.40,
suppose the toroidal solenoid is wound round a ring of
-+
magnetic material. Under the influence of fieldfb,the
magnetic moments of the atomic current loops of the
magnetic material tend to align themselves with or
against the magnetising field%.This gives rise to a net
current on the surface of the material and is called
magnetisation surface current1Mas shown in Fig.5.41.
Spinning
currents
Fig.5.40Magnetic field in a
magnetic material.
Fig.5.41Magnetising
surface current.
-+
This current induces magnetic fieldBMinside the
material which is given by
BM=110n1M
Thetotal magnetic field inside a magnetic materialisthe
sum of the external magnetising field and the additional
magnetic field produced due to magnetisation of the material
-+
andiscalledmagnetic induction B.Themagnetic induction
may also be defined as the total number of magnetic lines of
force crossing per unit area normally through a material. or
Thus the51unit of magnetic induction istesla(T) or
webermetre=?(Wbm-2) which is equivalent to
Nm-1A-1 or JA-1m-2.
5.29
3. Magnetising field intensity. The ability of
magnetising field to magnetise a material medium is
-+
expressed by a vector H , calledmagnetising field inten-
sityormagnetic intensity. Its magnitude may be defined as
the number of ampere-turns (nI ) flowing round the unit
length of the solenoid required to produce the given
magnetising field.Thus
H=nI
%=llonI=lloH orH=%
110
The dimensions of magnetic intensity are[L-1A].Its
51unit isampere metre-1(Am-1)which is equivalent to
Nm-2T-1 or Jm-1Wb-1.
4.Intensityof magnetisation. When a magnetic
material is placed in a magnetising field, it gets magn-
tised.The magnetic moment developed per unit volume of a
material when placedina magnetising fieldiscalled intensity
of magnetisaiion or simply magnetisation.Thus
-+
-+m
M=-
V
If1Mis the surface magnetisation current set up in a
solenoid of cross-sectional areaAand havingnturns
per unit length, then magnetic moment developed per
unit length of the solenoid isnIMA.Therefore,
magnetic moment developed per unit volume or the
-+
magnetisation M is given by
mnIA
M---~M~-nI
-V- A - M
HenceBM=110n 1M=110M
Again, consider a bar of magnetic material having
cross-sectional areaaand length21.Its volume is
V=ax21
Suppose the bar develops pole strengthqrnwhen
placed in a magnetising field, then its magnetic moment,
m=qrnx21
M=m=qrnx21=!1JJL
Vax 2l a
Henceintensity of magnetisation may also be defined as the
polestrength deoelopedper unit cross-sectionalareaoja materia/.
As the total magnetic field or the magnetic induction
-+
Binside a magnetic material is the resultant of the
-+ -+
magnetising field%and the fieldBMproduced due to
the magnetisation of the material, therefore,
B=%+BM =lloH+l-loM
B=llo(H+M)
Clearly, both Hand Mhave the same units, namely
Am-1.···3wx£n~m}r°n3lxv

453. 5.30
5. Ma
gnetic Permeability. Permeability is the
measure of the extent to which a material can be pen-
trated or permeated by a magnetic field. The magnetic
permeabilityofamaterial may be defined asthe ratio of its
magnetic induction Btothemagnetic intensity H.
B
1-1=-
H
Clearly, 51 unit of1-1
tesla
ampere metre-1
=tesla metre ampere-lor TmA-1
Dimensions of1-1=[MLr2A-2].
6.Relative permeability. Permeability of various
magnetic substances can be compared with one another
in terms of relative permeability I-Ir.Itisdefined as the
ratio of the permeability of the medium to the permeability of
free space. Thus,
I-I=..e...
r1-10
Forvacuum1-1r=I,forair itis1.0000004 and for
iron, the value of1-1rmayexceed 1000.
7. Magnetic susceptibility. Magnetic susceptibility
measuresthe ability of a substance to take up magn-
tisation when placed in a magnetic field. Itisdefined as
theratio of the intensity of magnetisation Mtothe
magnetisingfield intensity H.1t is denoted byXm.Thus,
M
X/II=H
As magneticsusceptibility isthe ratio of two
quantities having the same units (Am-1),so it has no
units.
8.Relation betwen magnetic permeability and
magnetic susceptibility. Ifa linear magnetic material,
subjected to the action of a magnetising field intensity
H, develops magnetisation M and magnetic induction
B;then
But
B= l-Io(H +M)
B=I-I H
1-1H =l-Io(H +M)
1-1=1-10(1+ ~) oror
or
5.24CLASSIFICATION OF
MAGNETIC MATERIALS
30. How arematerialsclassified on the basis of their
behaviour in a magnetic field ?Giveexamples of each
type.
PHYSICS-XII
Classification of magnetic materials.On the basis of
their behaviour in external magnetic fields, Faraday
classified thevarious substances intothreecategories:
1.Diamagnetic substances. Diamagnetic substances
arethose which develop feeble magnetisation inthe opposite
direction of the magnetising field. Such substances arefeebly
repelled by magnets and tend to movefromstrongerto
weaker parts of amagnetic field.
Examples.Bismuth, copper, lead, zinc, tin, gold,
silicon, nitrogen (at 5TP), water, sodium chloride, etc.
2.Paramagnetic substances. Paramagnetic sub-
stances are those which develop feeble magnetisation inthe
direction of themagnetisingfield. Such substances are feebly
attracted by magnets and tend to move from weaker to
stronger parts ofamagnetic field.
Examples.Manganese, aluminium, chromium,
platinum, sodium, copper chloride, oxygen (at 5TP), etc.
3.Ferromagnetic substances. Ferromagnetic sub-
stancesarethose which develop strong magnetisation inthe
direction of the magnetising field. They are strongly attracted
by magnets and tend to move from weakerto stronger parts
ofamagnetic field.
Examples.Iron, cobalt, nickel, gadolinium and
alloys like alnico.
5.25ORIGIN OF DIAMAGNETISM
31. Explain the origin of diamagnetism. Why are the
diamagnetic substances repelled by magnets ?
Origin of diamagnetism. In atoms of some
materials like Bi, Cu, Pb, the magnetic moments dueto
different electrons cancelout.Insuch atoms, electrons
occur in pairs with one of them revolving clockwise
111
v
111
(a) (b)
t11l-/!.m
B v+/!.v
(d)(c)
Fig. 5.42An electron orbiting in an atom
produces a moment.www3«°tssr•wvs3q°~

454. MAGNETISM
and other
anticlockwise around the nucleus. Net
magnetic moment of an atom is zero, as shown in
Fig.5.42(a)and(b).
~
When such anatom is placed in a magnetic fieldB,
the speed of revolution of one electron increases and
that of other decreases. The magnetic moment of the
~ ~
former electron increases toIn+t'lmand that of the
~ ~
latter electron decreases tom -t'lIn.So each electron
~
pair gains a net magnetic moment 2t'lInwhich is
~
proportional to thefieldBbut points in its opposite
direction as shown in Figs. 5.42(c) and(d).A sufficient
magnetic moment is induced in the diamagnetic
~
sample in the opposite direction ofB.This sample
~
moves from stronger to the weaker parts of the field B,
i.e.,a diamagnetic substance is repelled by a magnet.
The behaviour of diamagnetic materials is
or
independent of temperature.
5.26ORIGIN OF PARAMAGNETISM
32. Explain the origin of paramagnetism. State
Curie'slawof magnetism.
Origin of paramagnetism. According toLangevin,
theatoms or molecules of a paramagnetic material
possess apermanent magnetic moment either due to the
presence of some unpaired electron or due to the
non-cancellation of the spins of two electrons because
ofsome special reason. In the absence of an external
magnetic field, the atomic dipoles are randomly
oriented due to their ceaseless random motion, as
shown in Fig.5.43(a). There is no net magnetisation.
Bo=O Bo~
-e- -e- -e--e-
-e--e- -e--e-
-e- -e- -e--e-
(b)(a)
Fig. 5.43 (a)Randomly distributed atomic dipoles in a para-
magnetic material in the absence of magnetic field.
(b)Alignment of dipoles in the presence of magnetic field.
~
When astrong enough fieldBaisapplied and the
~
temperature is low enough, the field Batends to align
the atomic dipoles in its own direction, producing a
~
weak magnetic moment in the direction ofBo'The
material tends to move from a weak field region to a
strong field region. This is paramagnetism.
5.31
At very high magnetic fields or at very low tempe-
ratures, the magnetisation approaches its maximum
value when all the atomic dipole moments get aligned.
This is called thesaturation magnetization valueMs.
Curie's law. From experiments, it isfound that the
intensity of magneiisaiion (M) of a paramagnetic
material is
(i)directly propertional to the magneiising field
intensity H,because the latter tends to align the
atomic dipole moments.
(ii) inversely proportional to the absolute temperature T,
because the latter tends to oppose the alignment
of the atomic dipole moments.
Therefore at lowH /Tvalues, we have
H
Moc-
T
M=C. H
T
or
M C C
HT orXm=T
Here C is curie constant andXmis the susceptibility
of the material. The above relation is calledCurie'slaw.
This lawstates that far away fromsaturation, the suscepti-
bilityof a paramagnetic material is inversely proportional to
theabsolute temperature.
Figure5.44shows the variation of intensity of
magnetisation M as a function of H /T.Beyond the
saturation valueMs'Curie law is not valid.
region
Curie law region
HIT~
Fig. 5.44MagnetisationMas a function ofHIT.
5.27ORIGIN OF FERROMAGNETISM
DOMAIN THEORY
33. Describe ferromagnetism on the basis of domain
theory. How does Curie's law get modified for ferro-
magnetic substances ?
Origin of ferromagnetism. Weiss explained
ferromagnetism on the basis of his domain thry. In
materials like Fe, Ni, Co, the individual atoms are
associated with large magnetic moments. The magnetic
moments of neighbouring atoms interact with each
other and align themselves spontanusly in a
common direction over macroscopic regions called···3wx£n~m}r°n3lxv

455. 5.32
domains. Each
domain has a typical size of about 1 mm
and contains about1011 atoms. Soeach domain
possesses astrong magnetic moment. In the absence of
anyexternal magnetic field, thesedomains arerandomly
distributedsothat the net magnetic moment is zero.
Domains
Fig.45Randomly oriented domains in a ferromagnetic substance.
When a ferromagnetic material is placed ina
magnetic field, all the domains align themselves along
the direction of the field leading to the strong magne-
tisation of the material along the direction of the field.
That is why theferromagnetic substances are strongly
attracted by magnets. The alignment of domains may
occur in either of thefollowing two ways :
1.Bydisplacement of the boundaries of domains.
When the external field Bois weak, the domains aligned
in the direction ofBagrow insizewhile those oppositely
directed decrease insize,as shown in Fig. 5.46(b).
WeakBo~ Strong Bo ~
(a)Unmagnetised (b)Magnetisation by(c)Magnetisation by
sample growing of domains rotation ofdomains
Fig.5.46Magnetisation of a ferromagnetic sample.
2.Byrotation of domains.When the external field
Bais strong, the domains rotate till their magnetic
moments get alignedin the directionofBaas shown in
Fig.5.46(c).
ModifiedCurie's law for ferromagnetic substances.
When a ferromagnetic sample is heated, its magn-
tisation decreases due to the increase in the randomi-
sation of its domains. Ata sufficiently high temp-
rature, the domain structure disintegrates and the
ferromagnetic substance becomes paramagnetic. The
temperature at which aferromagnetic substance becomes
paramagnetic iscalledCurie temperatureorCurie pointTe.
PHYSICS-XII
Above the curie pointi.e.,in the paramagnetic
phase,thesusceptibility varies with temperature as
C'
Xm=T _I (T>Te)
e
whereC'isa constant. Thisismodified Curie's lawfor a
ferromagnetic material abovethe Curie temperatur. It
isalsoknown asCurie-Weiss law. Thislaw states that
the susceptibilih) of a ferromagnetic substance above its
Curie temperatureisinversely proportional to the excess of
temperature above the Curie temperature.
Table 5.2 Curie Temperatures of some
Ferromagnetic Materials
Material Tc(K)
Cobalt 1394
Iron 1043
Fep3 893
Nickel 631
Gadolinium 317
5.28PROPERTIES OF DIAMAGNETIC
SUBSTANCES
34. Describe some of the important properties of
diamagnetic substances.
Properties of diamagnetic substances :
1.When placed inanexternal magnetic field, a
diamagnetic substance develops feeble magnisation in
the opposite direction ofthe applied field.
2.When a rod of a diamagnetic material is placed in
a magnetic field, poles are inducedon it inadirection
opposite to that of the inducing field. So the lines of
force prefer to passthrough the surrounding air than
to pass through the material itself i.e.,thelines of
force get expelledorrepelled, as shown in Fig. 5.47.
Consequently, the magnetic induction Binside the
material becomes lessthan the magnetising field,
Ba=J.10H.The reduction is very small, about 1part
in105.
Fig.5.47Reduction of lines of force in a diamagnetic rod.···3wx£n~m}r°n3lxv

456. MAGNETISM
3.When plac
edina non-uniform magnetic field, a
diamagnetic substance moves from stronger to the weaker
parts of the field.
When a watch glass containing a diamagnetic
liquid is placed over twoclosely lying (3-4 mm apart)
pole pieces of a magnet, the liquid is found to move
towards the poles causing a depression in the middle
[Fig5.4~(a)].Thisindicates that the field is strongerin
the middle than that near the poles. Now if the poles
are moved apart sufficiently, the magnetic field at the
middle becomes weaker than that near the poles.
Consequently, the liquid accumulates in the middle
and thins out near the poles [Fig. S.48(b)].
rDiamagnetic liquid\
ntf~~
Fig.5.48Effect of non-uniform magnetic field on a diamagnetic
liquid when(a)poles are quite close to each other,(b)poles are
sufficiently apart.
4.When a rod of a diamagnetic material is sus-
pended frely in a uniform magnetic field, it aligns
itself perpendicular to the magnetising field (Fig. 5.49).
Fig.5.49Afrely suspended diamagnetic rod
in a uniform field.
5. As a diamagneticsubstance dev ....lops a weak
magnetisation in the opposite direc.ion of the
magnetising field, the susceptibility(Xnz=M/ H) of
diamagnetic materials is small and negative
bismuth, Xm= -0.00015.
6. The relative permeabilityIlr(=1+X",) is positive
but less than 1 for a diamagnetic material.
7. The susceptibility
ofdiamagnetic sub- X'"
stances is independent
ofthe magnetising field 0 T
and the temperature, as
shown in Fig. 5.50. Xdi3~-----
8.The magnetisation
of a diamagnetic sub-
stance lasts so long as Fig.5.50Xm-Tgraph for dia-
the magnetising field ismagnetic material.
applied.
5.33
35.Briefly describe diamagnetism insuperconducting
metals.
Diamagnetism in superconducting metals.When a
metal is cooled to a temperature below its critical
temperature in a magnetic field, it attains both
superconductivity and perfect diamagnetism. The
magnetic lines of force get completely expelled from it
and it repels a magnet. For this material, X=-1and
Ilr=O.Thisphenomenon of diamagnetism in super-
conductorsiscalledMeissner effect. This effect forms the
basis for running magnetically levitatedsuperfast trains.
5.29PROPERTIES OF PARAMAGNETIC
SUBSTANCES
36. Describe some of the important properties oj.
paramagnetic substances.
Properties of paramagnetic substances:
1.When placed in an external magnetic field, a
paramagneticsubstance develops feeble magn-
tisation in the direction of the applied field.
2. When a rod of paramagnetic material is placed
in a magnetic field, the lines of force prefer to
pass through it than through the surrounding
airi.e.,the lines of force get slightly more
concentrated inside the material, as shown in
Fig. 5.51. The magnetic induction B becomes
slightly greater than the magnetising field,
Ba=lloH.The increase is very small, about
1 part in 105.
Fig.5.51Slightly cocentrated lines of force
in a paramagnetic rod.
3. When placed ina non-uniform magnetic field, a
paramagnetic substance movesfromweaker tothe
stronger parts of thefield.
When a watch glass containing a paramagnetic
liquid is placed over two closely lying pole
pieces of a magnet, the liquid accumulates and
elevates in the middle and thins out near the
poles [Fig.5.52(a)]. Thisis because the field in
the centre is the strongest. When the poles are
moved apart, the field at the poles becomes
stronger than that at the centre and the liquid
movestowards the poles [Fig.5.52(b)].···3wx£n~m}r°n3lxv

457. 5.34
Fig. 5
.52Effect of magnetic field on a paramagnetic
liquid when(a)poles are quite dose to
each other,(b)poles are farther apart.
4.When a rod of paramagnetic material is sus-
pended frely in a uniform magnetic field, it aligns
itself parallel to the magnetising field (Fig. 5.53).
Fig. 5.53A frely suspended paramagnetic rod in
a uniform magnetic field.
5.A paramagnetic material develops small magn-
tisation in the direction of the magnetising field,
so its susceptibility has small but positive value
For aluminium,X=1.8x10-6.
6. The relative permeability(ur=1+Xm)for a
paramagnetic material has a value slightly greater
than1.
7. The magnetic susceptibility of a paramagnetic
material varies inversely as the absolute temp-
rature,i.e.
1
Xex:-
mT
or
C
Xm=y'
where C is a constant called theCurieconstant
and this equation is known asCurie's law.
T~
Fig. 5.54Xm -Tgraph for a paramagnetic material.
8. For a given temperature, the intensity of magn-
tisation is proportional to the magnetising field,
PHYSICS-XII
so the susceptibility and permeability do not
~
show any variation with the field Ea .
9. As soon as the magnetising field is removed, a
paramagnetic substance loses its magnetism.
5.30PROPERTIES OF FERROMAGNETIC
SUBSTANCES
37. Describe some of the important properties of
ferromagnetic substances.
Properties of ferromagnetic substances.Ferro-
magnetic substances exhibit properties similar to those
of paramagnetic substances but in a highly dominant
manner. These are as follows:
1.When placed in an external magnetic field, a
ferromagnetic material develops strong magn-
tisation in the direction of the applied field.
2. When a ferromagnetic substance is placed in a
magnetic field, the lines of force concentrate
greatly into the material so that the magnetic
inductionBbecomes much more than the
magnetising field Ea.
Fig. 5.55Highly concentrated lines of force in a ferromagnetic rod.
3. When a ferromagnetic substance is placed in
non-uniform magnetic field, it moves from
weaker to the stronger parts of the field.
4. When a rod of a ferromagnetic material is sus-
pended frely in a uniform magnetic field, it
quickly aligns itself parallel to the magnetic field.
5. The intensity of magnetisation M is propor-
tional to the magnetising field intensity H for its
smaller values. For moderate values of H, M
increases rapidly and then finally attains constant
value for large H. This indicates the attainment
of the saturation stage of magnetisation.
6.The susceptibility of a ferromagnetic material
has a large positive value
M
XIII=H
andM» H for a ferromagnetic material. It is of
the order of several thousands.
7.The relative permeability(f..lr=1+XIII)of a
ferromagnetic material has a large positive
value
iron,f..lr=1000.www8notesdrive8com

458. MAGNETISM
8. The susceptibility of ferromagnetic material
decreases with temperature in accordance with
Curi-Weiss law:
C'
XI1l=T-
'[ (T>Tc)
c
9.At a certain temperature called theCurie point,
the susceptibility suddenly falls and the ferro-
magnetic substance becomes paramagnetic.
10. The magnetisation developed depends not only
on the value of magnetising field but also on the
past magnetic and mechanical history of the
material.
11.A ferromagnetic substance retains magnetism
even after the magnetising field is removed.
5.35
For Your Knowledge
~In the presence of anexternal magnetic field, magnetic
moments are induced in all materials. Hence
diamagnetism is universal. But paramagnetism and
ferromagnetism are much stronger than
diamagnetism, so it is difficult to detect diamag-
netism in para- and ferro-magnetic substances.
~Magnetic materials are broadly classified as
diamagnetic, paramagnetic and ferromagnetic.
However, there exist some other types of magnetic
materials with mysterious properties. These include
ferrimagnetic, anti-ferromagnetic, spin glass, etc.
~A very small variation in the value ofXIIImay lead to an
altogether different magnetic behaviour: diamagnetic
vs.paramagnetic. For diamagnetic materials, XIIIz:-10-5
whereXIII=+10-5for paramagnetic materials.
Table 5.3 Comparative study of the properties of dic-, para- and ferromaetic substances
.....
1.Effect oj
ma nets
2.Inexternal
magnetic field
3.In anon-uniform
magnetic field
4.Inauniform
magnetic field
5.Susceptibility
value(XIII)
6.Relative
permeability
value
7.Permeability
value
8.Effect oj
temperature
I. .•. •. Paramagnetic substances
A frely suspended ferro-
magnetic rod aligns itself
arallel to the field.
Tend to move slowly from Tend to move slowly from
stronger to weaker parts of weaker to stronger parts of
the field. the field.
They are febly repelled by They are febly attracted by They are strongly attracted
ma ets. ma nets. b ma nets.
Ferromagneticsubstances
A frely suspended diamag- A frely suspended
netic rod aligns itself paramagnetic rod aligns
er endicular to the field. itself arallel to the field.
Susceptibility is small and Susceptibility is small and
negativ.-1:::;Xm<0 positive. 0<Xm<s , wheree
is a small number
Slightly less than 1
O:::;l1r<l
Susceptibility is indepen-
dent of temperature.
Magnetisation lasts as
long as the magnetising
field is a lied.
M changes linearly with H.
Slightly greater than 1
l<l1r<l+s
Susceptibility varies
inversely as temferature :
Xm<x'y'
As soon as the magnetising
field is removed, magneti-
sation is lost.
M changes linearly with H
and attains saturation at
low temperature and in
ver stron fields.
Acquire strong magnetisa-
tion in the direction of the
ma etisin field.
Tend to move quickly from
weaker to stronger parts of
the field.
Susceptibility isvery large
and positiveXm>1000
11.
12. Physical state oj
the material
Of the order of thousands
Ilr>1000
11»110
Susceptibility decreases
with temperature in a
complex manner.
1
Xm<X.T _'[ (T> Tc)
c
9. Removal oj
magnetising field
Magnetisation is retained
even after the magnetising
field is removed.
10. Variation ojM
withH
M changes with H non-
linearly and ultimately
attains saturation.
Solid, liquid or gas. Normally solids only.Solid, liquid or gas.
Bvector shows h steresis.
Fe,Ni,Co,Gd,Fep3,Alnico.13. Examples
Bi,Cu,Pb,Si, Nz(atSTP),
Hp,NaCI
AI,Na,Ca,Oz(atSTP),CuClz···3wx£n~m}r°n3lxv

459. 5.36
5.31HYSTERESIS
38. Explain
the phenomenon of hysteresis in magnetic
materials. Whatisthe significance of the area of
hysteresis loop?State thepractical importance of
hysteresis loops.
Hysteresis.Whenaferromagnetic sampleis placed
inamagnetising field, thesample gets magnetised by
induction. As the magnetising field intensityHvaries,
the magnetic inductionBdoes not vary linearly with
H,i.e.,the permeability Il(=BIH)isnot constant but
varies with H.Infact,it also depends on the past
history of the sampl.
Figure 5.56shows thevariation of magnetic
inductionBwith magnetising field intensityH.Point0
represents the initial unmagnetised state of a
ferromagnetic sample
intensity Hincreases, the magnetic induction Bfirst
gradually increases and then attains a constant value.
In other words, the magnetic induction Bsaturates at a
certainvalue +Hmax'
B
Initial
buildup
-Hmax
-----r----+-~--r-----------------~H
+Hmax,
,
,
,
,
,
D'
Saturation
DB=Retentivity
DC=Coercivity
Fig.5.56Hysteresis loop for a ferromagnetic sample
Now if the magnetising field intensity His
gradually decreased to zero, Bdecreases but along a
n pathAB.Itis found that the magnetic inductionB
does not become zero even when the magnetising field
His zero,i.e.,the sample is not demagnetised even
when the magnetising field has ben removed. The
magnetic induction(=0 B)left behind in the sample after the
magnetising field has been removed iscalled residual
magnetism or retentivity orremanence.
To reduce the magnetism to zero, the field H is
gradually increased in the reversedirection, the
induction B decreases and becomes zero at a value of
H=OC.The value ofreverse magnetising field intensity H
required for the residual magnetism of a sample to become
zeroiscalled coercivity of the sample.
PHYSICS-XII
On further increasing H in the reverse direction to a
value -Hmax' we reach the saturation point Dlocated
symmetrically to pointA.Now ifHis decreased
gradually, the point Ais reached after going through
the pathDEFA.
The closed curve ABCDEFA which represents a
cycleof magnetisation of a ferromagnetic sample is
called itshysteresis loop. Throughout the cycle, the
magnetic fieldBlags behind the magnetising field
intensityH,i.e.,the value ofBwhenHis decreasing is
always morethan whenHis increasing. The pheno-
menon of thelagging of magnetic induction behind themagne-
tisingfield iscalledhysteresis. In fact, the word hysteresis
originates from a Greek word meaning' delayed'.
Significance of the area of hysteresis loop.The
productBH=B(~J=.».,has the dimensions of
11 Ilollr
energy per unit volume. Hence the areawithin the B-H
loop represents the energy dissipated perunit volume in
the material when it is carried through a cycle of
magnetisation. Thesourceisthesource of emf used in
magnetising the material and the sinkis the hysteretic
heat loss in the magnetic material.
Practical importance ofhysteresisloops.A study of
hysteresis loop provides us information about
retentivity, coercivity and hysteresis lossofamagnetic
material. Thishelps in proper selection of materials for
designing cores oftransformers and electromagnets
and in making permanent magnets.
39.Distinguish between soft and hardferromagnetic
materials. Draw theirhysteresis loops. Give examples of
each type.
Types of ferromagnetic materials.Ferromagnetic
materials can be divided into twocategories:
1.Soft ferromagnetic materials or soft ferro-
magnets. These are the ferromagnetic materials in which the
magnetisation disappears on the removal of the external
magnetizing field. Such materials have narrow hyst-
resis loop, asshown in Fig.5.57(a). Consequently, they
B B
--+i-r-----...H --f--++---..H
(a) (b)
Fig. 5.57 Magnetic hysteresis loop for
(a)soft,(b)hard ferromagnetic material.···3wx£n~m}r°n3lxv

460. MAGNETISM
have low retentivity, low coercivity,and low hyst-
resis
loss. But they have high relative magnetic per-
meability. They are used as cores of solenoids and trans-
formers.Examples.Soft iron, mu metal, stalloy, etc.
2.Hard ferromagnetic materials or hard ferro-
magnets.Theseare theferromagnetic materials which retain
magnetisation even after the removal of the external magne-
tisingfield.Such materials have wide hysteresis loop, as
shown in Fig.5.57(b).Consequently, they have high
retentivity, high coercivity and large hysteresis loss.
They are used for making permanent magnets.
Examples.Stel, alnico, lodestone, ticonal, etc.
Formulae Used
m
1.Intensity of magnetisation,M=V
B
3.Il=~2.Il=-
H r Il0
M
5.
C
[Curie's law]4.
Xm=fj Xm=T
6.B=llo(H+M) 7.Ilr= 1+Xm·
Units Used
Magnetising field intensity His in Am-1,field Bin
tesla, magnetisationMin Am-l,permeabilityIlin
TmA-1or Hm-l,susceptibilityXmand relative
permeabilityIl rhave no units.
Example 39.A magnet of magnetic moment2.5Arrt
weighs66g.If thedensity of the material of the magnetis
7500 kgm-3, find the intensity of magnetisation.
Solution. Volume,
66x10-3kg66x10-5 3
----"-7= m
7500kgm-3 75
V=Mass
Density
Magnetisation,
M=m 2.5
V66x10-5
--
75
=2.84x105Am-t.
2.5x75x105
66
Example 40. Obtain the earth's magnetisation. Assume
that the earth's field can be approximated by a gaint bar
magnet of magnetic moment8.0x1022Ant-. Theearth's
radiusis6400 km. [NCERT)
Solution. Here magnetic moment,
m=8.0x1022Am2
Radius of the earth, R = 6400krn= 6.4x106m
5.37
Magnetisation,
m m 8.0x1022x3
M =V=in?= 4x3.14x(6.4x106)3
3
=72.9Am-t.
Example 41. A domain inferromagnetic ironis intheform
of a cube of side length l)lm. Estimate the number of iron
atoms in the domain and the maximum possible dipole
moment and magnetisation of the domain.Themolecular
mass of iron is55g I mole and its densituis7.9g I crrf.
Assume that each iron atom has a dipole moment of
9.27x10-24AI;. [NCERT)
Solution. Each side of cubic domain,
1=Lurn=10-6 m
Volume of the domain,
V=13=(10-6 m)3=10-18m3 =10-12 cm'
Mass of domain=Volumexdensity
=10-12em:'x7.9gern-3
=7.9xl0-12g
Number of atoms in55g iron
= 1mole= 6.023x1023
:. Number of atoms in7.9x10-12g iron
6.023x1023x7.9x10-12
55
N=8.65x1.otOatoms.
Dipole moment of each iron atom,
m=9.27x1O-24Arn2
The dipole moment of the domain will be maxi-
mum when all its atomic dipoles get perfectly aligned.
Its value will be
m =mN=9.27 x10-24x8.65x1010
max
=8.0x10-13Am2.
The maximum possible magnetisation of the domain,
80x10-13Am2
M-mmax _. _--.;;--.".--_
- V 1O-18m3
=8.0x105Am-t.
Example 42. A magnetising field of1500AI m produces
a magnetic flux of2.4x10-5weber in a bar of iron of cross-
section0.5crrf. Calculate permeability and susceptibility of
the iron-bar used. [CBSE OD08)
Solution. Here H = 1500Am-I, <p==2.4x10-5Wb,
A=0.5x10-4m2
Magnetic induction,
_ <p _2.4x10-5 _0 48Wb -2
B--- - . m
A0.5x10--4
or···3wx£n~m}r°n3lxv

461. 5.38
Permea
bility,
Il=~ = 0.48 =3.2x10-4 TmA-1
H 1500
As Il=Il0(1+XIII)
.,Susceptibility,
=~-1=3.2x10-4 -1
XIIIIlo 4x3.14xlO-7
=254.77 -1 = 253.77.
Example43.Assume thateach iron atom has a permanent
magnetic moment equal to2Bohrmagnetons (1Bohr
magneton =9.27x10-24AI;). The number density of
atoms in iron is8.52x1028m-3.(i)Findthemaximum
magnetisation Min a long iron bar. (ii)Find the maximum
magnetic induction Binthebar.
Solution. (i)Number of atoms per unit volume,
n=8.52x1028m-3
Magnetic moment of each iron atom
=2IlB=2x9.27x 10-24 Am2
As magnetisation Misthe magnetic moment per
unitvolume, sothe maximumvalue of magnetisation is
~ax=nx21lB
(when all the dipoles getaligned)
=8.52x1028x2x9.27x10-24
=1.58x106Am -1.
(ii)Magnetic induction, B=llo (H+M)
As no magnetising field is applied, so H=O.Hence
B=IloM = 411: X10-7x1.58 x106=1.985T.
Example44.A solenoid of 500turns / miscarrying a
current of3AItscoreismade of iron which has a relative
permeability of5000. Determine themagnitudes of the
magnetic intensity, magnetisation and the magnetic field
inside the core. [NCERT]
Solution. Here n= 500 turns/m, I=3A Ilr=5000
Magnetic intensity,
H=nI= 500 m-1x3 A =1500Am -1.
As Ilr=l+XI/l
Xm=Ilr-1 = 5000-1=4999=-5000
Also, Ilr= ~ = 5000 or Il=5000Il0
Ilo
Magnetisation,
M=XmH=5000x1500
=7.5x106Am -1.
Magnetic field insidethecore,
B=IlH=5000Il0H
=5000x411:x10-7x1500=311:=-9.4 T.
PHYSICS-XII
Example45.The core ofa toroidhaving 3000turnshas
inner andouter radiiof11em and12cm respectively. The
magnetic field inthe corefor acurrent of0.70Ais2.5T.
Whatistherelative permeability of the core?
Solution. The magnetic field in the emptyspace
enclosed by a toroid is given by
B=llonI
wherenis the number of turns per unit length and Iis
thecurrent. If thespace is filled by a coreofperm-
ability u,then
B=IlnI
HereB=2.5 T,I=0.70 A
Mean radius,
r=11+12 em=11.5 em=11.5x10-2m
2
3000 3000
n----
- 211:r-2x3.14x11.5x10-2
Hence
B2.5x2x 3.14x11.5x10-2
Il=-;;r= 3000 x0.70
=8.6x10-4TmA-1
Relative permeability,
Ilr= ~ =8.6x10-4 =684.4.
Ilo 411:x10-7
Example46.An iron rod of volume 10-4m3and relative
permeability1000 is placed inside a long solenoid wound
with5turns per em.If a current of 0.5Aispassed through
the solenoid, find the magnetic moment of the rod.
Solution. The relationbetween the magnetic
inductionB,magnetising field intensity Hand the
magnetisation Mis given by
B=llo(H+M)
B IlH
M=--H=--H [.: B=1l H]
Ilo Ilo
=IlrH-H=(Ilr-l)H
But for a long solenoid, we have
H=nI
wherenis the number of turns per metre.
M=(Ilr-1)nI
HereJlr=1000, I=0.5 A
.5
n=-turns /m=500 turns / m
0.01
M= (1000 -1)x500x0.5=2.5x105Am-1
Magnetic moment,
m=MxV=2.5x105x10-4Am2 =25 Am2.www4°±trsp€vvr4o±~

462. MAGNETISM
Example47.The hystere
sislossfor aspecimen of iron
weighing 12kgisequivalent to300Im-3cycle-I. Find the
lossof energy per hour at50cycle s-l. Density ofironis
7500 kg m-3.
Solution.LetQbe the energy dissipated per unit
volume per hysteresis cycle in the given sample. Then
thetotal energy lost by the volumeVof the samplein
timetwill be
W=QxVXyxt
where yisthe number of hysteresis cycles persecond.
HereQ=300Jm-3 cycle-l, v=50cycle s-l,
t=1h=3600s
VolumeV=Mass=~ m3
, Density7500
.'. Hysteresis loss,
12
W=300x~-x50x3600J= 86400 J.
7500
Example48.Thecoercivity of a certain permanent magnet is
4.0x104Am-I. Thismagnetisplaced insideasolenoid15em
long andhaving 600 turns and acurrent ispassed in the
solenoid to demangneiise itcompletely. Find the current.
Solution. Thecoercivity of4x104Am-1of the
permanent magnet implies thatamagneticintensity
H=4x104Am-1isrequired to be applied in opposite
direction to demagnetise the magnet.
600 600
Heren=~~ = ?=4000turns/m
15ern15x10--m
AsH=nI
C I= H=4x104= 10A
.'.urrent,
n4000
jOrOblems For Practice
1.Abar magnet made of stel has a magnetic moment
of2.5Am2and a mass of6.6g.If the density of stel
is7.9x103kg m-3,find the intensity of magne-
tisation of the magnet. (Ans.3.0x106Am-1)
2.Themaximum value of permeability ofu-metal
(77% Ni,16%Fe,5%Cu,2%Cr)is0.126TmA-1.
Find the maximum relative permeability and -
susceptibility. (Ans.flr= 1.0 x105,X=-1.0x105)
3.Find the percent increasein the magnetic fieldB
when the space withinacurrent-carrying toroid is
filled with aluminium. Thesusceptibility of
aluminium is 2.1x10-5. (Ans.2.1x10-3)
4.The susceptibility of magnesium at 300Kis
1.2x10-5. At what temperature will the
susceptibility increase to 1.8 x 10-5. (Ans.200 K)
5.39
5.An iron rod of0.2em2cross-sectional area is
subjected to a magnetising field of1200Am-1.The
susceptibility of iron is599.Find the permeability
and the magnetic flux produced.
(Ans. 7.536x10-4Tm A -1,1.81x10-5Wb)
6.An iron rod0.2m long, 10mm in diameter and of
permeability 1000isplacedinside a long solenoid
woundwith300turns per metre. If a current of
0.5ampere is passed through the solenoid, find the
magnetic moment of the rod.(Ans.0.2325Am-1 )
7.An iron ring of mean circumferential length 30em
and cross-section 1em2is wound uniformly with
300turns of wire. When a current of0.032A flows
in the windings; the flux in the ring is2x10-6Wb.
Find the flux density in the ring, magnetising field
intensity and relative permeability of iron.
(Ans.2x10-2Wb m-2,32A turns m-1,500 )
8.Aniron ring having 500turns of wire and a mean
diameter of 12em carries a current of0.3A.The
relative permeability of iron is 600.What is the
magnetic flux density in the core ? What is the
magnetisation field intensity? What part of the flux
density is due to the electronic loop currents in the
core? (Ans. 0.3Wbm",397.9A turns m-I,
0.2995Wbm-2 )
HINTS
1.Proced as in Example 39on page 5.37.
2.Maximum relative permeability,
-~- 0.126 -10 105
flr-fl0-41tX107- •x
Maximum susceptibility,
X= flr-1=-1.0x105•
3.In the absence of aluminium,
Bo = floH
In the presence of aluminium
B = flH = fl0(1+X)H
Increase in field= B-Bo = fl0XH
Percent increase= B-Bo x100 = fl0XHx100
Bo floH
=Xx100=2.1x10-3•
..12=Xl.'Ii=1.2x10-5x:OO =200K.
X2 1.8xl0
5.HereA= 0.2em2= 0.2 X10-4m2,H= 1200Am-1,
Xm=599
Permeability,fl =flo (1+Xm)
=41tx10-7x( 1+ 599) = 7.536 x10-4 TmA-1···3wx£n~m}r°n3lxv

463. 5.40
Magnetic induction
,
B=~H=7.536x10-4x1200=0.904 T
Magnetic flux,
<p=BA=0.904x0.2x10-4 =1.81x10-5Wh.
6.Proced as in Example 46 on page 5.38.
7.HereI=30cm=0.30 m, A= lcm2 = 1O-4m2,
N = 300, I=0.032 A, <p= 2x10-6Wb,
n=N= 300 = 1000 m-I
I0.30
Magnetic flux density,
B=.!=2x~-6 =2x10-2Whm-2
A 10
Magnetising field intensity,
H=n1=1000x0.032 = 32 A turns m-1
Permeability,
_ ~ _2x10-2 _625104T A-I
~- - -. x m.
H 32
116.25x10-4
Relative permeability, ~l=-= 7=500.
r11041txl0-
8.HereN= 500, D = 12cm=0.12 rn, 1=0.3A,
~r= 600
Magnetising field intensity,
N N
H=n1=-.I=-.1
I 1tD
500x0.3 -1
= =397.9 A turns m
1tx0.12
Magnetic flux density, B=I1H=~r110H
=600x41t x10-7x397.9 =0.3 Wbm-1
Flux density dueto electroniccurrent loop is
~oM=.B -11oH=0.3 - 41tx10-7x397.9
= 0.3-0.0005 = 0.2995 Whm-2 •
5.32PERMANENT MAGNETS AND
ELECTROMAGNETS
40. Give a comparison ofthe magnetic properties of
soft iron andsteel.
Comparison of the magnetic properties of soft iron
and stel. Fig. 5.58 shows the hysteresis loops for soft
iron and stel.
Astudy of these B - H loops reveals the following
information:
1.Permeability.For a given H,Bis more forsoft
iron than stel. Sosoft iron has a greater permeability
(J..l=B/ H) than stel.
2.Susceptibility.As permeability ofsoft iron is
greater than stel, so soft iron has a greater suscep-
tibility(Xm=u,-1)than stel.
PHYSICS-XII
B
---~L-H-:+-+----.H
Fig.5.58Hysteresis loops of soft iron and stel.
3. Retentioitq.The retentivity of softiron(Ob') is
greater than the retentivity(Ob)of steel.
4.Coercivity.The coercivity of soft iron(Oc )is less
than the coercivity (Oc)of stel.
5. Hysteresis loss.Asthe area of the hysteresis loop
of soft iron is much smaller than that of stel, so the
hysteresis loss per unit volume per cycle is lessfor soft
iron than forstel.
We can summarise the above properties asfollows:
1.Permeability
2. Susceptibility
3.Retentivity
4. Coercivity
5. Hysteresis loss
1are greater for soft iron
than forstel
]
are less for soft iron
thanforstel
41.Howwill youselect materials formaking perma-
nent magnets, electromagnets and cores of transformers ?
Selection of magnetic materials. The choiceof
magnetic materials for making permanent magnets,
electromagnets and cores of transformers is decided
from the hysteresis loop of the material.
A.Permanent magnets. The material used for
making permanent magnets must have the following
characteristics:
1.High retentivity so that it produces a strong
magnetic field.
2. High coercivity so that its magnetisation is not
destroyed by stray magnetic fields, temperature
variations or minor mechanical damage.
3. High permeability.
Inspite of its slightly smaller retentivity thansoft
iron,stel is favoured for making permanent magnets.
Stel has much higher coercivity thansoft iron. The
magnetisationof stel is not easily destroyed by stray
fields. Once magnetised under a strong field, it retainswww4·»ttssryvt4r»«

464. MAGNETISM
magnetisation for a long duration.Other suitable
materials for
making permanent magnets are:
Cobalt stel 52%Fe, 36% Co, 7% W,
3.5% Cr,0.5% Mn,0.7% C
98%Fe,0.86% C,0.9% Mn
55%Fe,10%Al,17%Ni,
12%Co, 6% Cu
42%Co,26.5 Fe, 14%Ni, 8% AL
6.5 Ti, 3%Cu
Carbon stel
Alnico
Ticonal
B.Electromagnets.The material used for making
cores of electromagnets must have the following
characteristics:
1.Highinitial permeability so that magnetisation is
large even fora small magnetising field.
2. Low retentivity sothat the magnetisation is lost as
the magnetising current is switched off.
So soft iron is more suitable than stel for cores of
electromagnets.
C.Transformer cores.The material used for making
cores of transformers must have the following charac-
teristics:
1.High initial permeability so that the magnetic flux
islarge even for low magnetising fields.
2.Low hysteresis loss as the materials are subjected to
alternating magnetising fields of high frequency.
3.Low resistivity to reduce losses due to eddy
currents.
Soft iron is preferred for making transformer cores
and telephone diaphragms.
42.Mention three methods formaking permanent
magnets.
Methods for making permanent magnets.A hard
ferromagnetic material like stel can be converted into
a permanent magnet by anyof the following methods:
1.By holding the stel rod in north-south direction
and hammering it repeatedly.
2. Hold astel rod and stroke it with one end of a
bar magnet anumber of times, always in the
same sense to make a permanent magnet.
3. The most efficient way of making a permanent
magnet is to place a stel rod in a solenoid and
passa strong current. The rod gets magnetised
due to the magnetic field of the solenoid.
43.Brieflyexplain how anelectromagnetisformed.
Statesome uses ofelectromagnets.
Electromagnet. Asshown in Fig. 5.59, take a soft
iron rod and wind a large number of turns of insulated
5.41
copper wire over it. When we pass a current through the
solenoid, a magnetic field is set up in the space within
the solenoid. The high permeability of soft iron increases
the field one thousand times. The end of the solenoid at
which the current in the solenoid sems to flow
anticlockwise acts as -pole and other one as S-pole
When the current in the solenoid is switched off, the
soft iron rod loses its magnetism almost completely
due to its low retentivity.
I
+~I- (.
Battery Key
Fig. 5.59An electromagnet.
Uses of electromagnets:
1.Electromagnets are used in electric bells, loud-
speakers and telephone diaphragms.
2. Large electromagnets are used in cranes to lift
heavy machinery, and bulk quantities of iron
and stel.
3.In hospitals, electromagnets are used to remove
iron or stel bullets from the human body.
5.33TANGENT GALVANOMETER·
44.State tangent law of magnetism.
Tangent law.This law states that if a frely
suspended small magnet is acted upon by two uniform
mutually perpendicular magnetic fields Brand ~
simultanusly, then the magnet comes to rest in such
B2---------------------
5
Fig. 5.60Tangent law.
a position that the tangent of the angleethat the
magnet makes withBris equal to the ratio ~ /Brof the
two fields. That is,
tane= ~
Bl
or ~ =Brtane···3wx£n~m}r°n3lxv

465. 5.42
45. Describe
the principle, construction, theory and
working of a tangent galvanometer.
Tangent galvanometer. It is a device used to
measure very small currents. It is a moving magnet type
galvanometer. Its working is based on tangent law.
Construction.It consists of a circular frame of
non-magnetic material mounted on a horizontal turn
table. Thre coils having 2, 50 and 500 turns of
insulated copper wire are wound over it. The ends of
the coils are connected to thre base terminals.
A compass box of non-magnetic material is fitted at
the centre of circular frame
nedle pivoted at its centre with a long thin aluminium
pointer attached perpendicular to it. Both are fre to
move over a horizontal circular scale graduated in
degres and divided into four quadrants of 0° - 90°
each. A plane mirror is fixed at the base below the
pointer. This removes error due to parallax while
reading the position of the pointer.
Short
magnetic
nedle
Aluminium
pointer
Levelling
scrs
Fig.5.61Tangent galvanometer.
Adjustments The tangent galvanometer is levelled
with the help of levelling scrs and the coil is so
rotated that its plane becomes parallel to the length of
the magnetic nedle
coil lies along the magnetic meridian. The compass box
is rotated, so that the pointer comes along 0° - 0°lin.
Thry and working. The working of tangent
galvanometer is based on tangent law. When no
current is passed through the coil, the magnetic nedle
is influenced only by ~ of earth's magnetic field.
When a currentIis passed, there is a magnetic field B
along the axis of coil perpendicular to ~ , as shown in
Fig. 5.62. The magnetic nedle is influenced by two
perpendicular magnetic fields and it comes to rest at an
angle 8 with BHsuch that
PHYSICS-XII
B= ~ tan 8
This relation is known asTangent law.
Coil in
vertical
plane, N-Tums,
radiusR
B
BR(Resultant)
5
Fig. 5.62
Now magnetic field at centre of coil is
B=110NI
2R
where N and R are number of turns and radius of the
coil.
11NI
_0__ = ~tan 8
2R
I=2 R BHtane=Ktan 8
1l0N
2R~
whereK=--- is a constant for the tangent galvano-
1l0N
meter and is called itsreduction factor.
5.34OSCILLATIONS OF A FREELY
SUSPENDED MAGNET
46. Show that the oscillations of a freely suspended
magnet in a uniform magnetic field are simple harmonic.
Hence deduce an expression for its time period.
Oscillations of a frely suspended magnet in a
magnetic field. In the position of equilibrium, the
--+
magnetic dipole lies along B. When it is slightly
rotated from this position and released, it begins to
vibrate about the field direction under the restoring
torque,
t=-mBsin 8
The negative sign indicates that the direction of
toquetis such so as to decrease 8.
-->
m
-->
-----~~~~----~B
Fig. 5.63www4°±trsp€vvr4o±~

466. MAGNETISM
For small angular displacement 9, sin 9'"
9
r=-mB9
IfIis the moment of inertia of the magnet, then the
deflecting torque on the magnet is
d29
't= la=l-
dt2
In the equilibrium condition,
Deflecting torque =Restoring torque
d29
1-=-I11B9
dt2
d29=_mB9=_0)29
dt2 I
i.e.,angular accelerationd2~ex;angular displacement9.
dt
or
Hencethe oscillations of a freely suspended magnetic
dipoleina uniform magnetic field are simple harmonic. The
time period of oscillation is given by
T=2n=2n0.
0)f;;:;B
Example49.In Fig. 5.64,a magnetic needle isfree
tooscillate inauniform magnetic field. Themagnetic needle
has magnetic moment 6.7A~and moment of inertia
1=7.5 x10-6km~. It performs 10complete oscillations in
6.70s.Whatisthe magnitude field? [NCERT]
-- - - - ----- - - - - - - --~
----~--->/.e-l'!.-----~ --+
------ ---------~B
,-
,-
- - - -- -- ---- -- - --~
5
-- - - - ------ --- - - --~
Fig. 5.64
. 6.70 s 2
Solution.HereT=--= 0.67 s,m= 6.7 Am,
10
I =7.5 x10-6kg m2
As T =2n0
v-;;:;B
T2=4n2_1_
mB
The magnitude of the magnetic field is
B=4n21= 4x9.87x7.5xlO-6
mT2 6.7x(0.67)2
=9.8x10-5T.
or
5.43
5.35VIBRATION MAGNETOMETER *
47. Describe the principle, construction and working
of avibration magnetometer. Mention its uses.
Vibration magnetometer.It isan instrument used to
compare themagnetic moments of two magnets or to determine
thehorizontalcomponent of earth's magnetic field at a place.
Principle. When a magnet suspended frely in a
uniform magnetic field (like the one due to the earth),
is displaced from its equilibrium position, it begins to
vibrate simple harmonically about the direction of the
field. Theperiod of vibration is given by
T=2n~ I
mBH
where,m=magnetic moment of the magnet,
BH=horizontal component of earth's
magnetic field,
I=the moment of inertia of the magnet about
anaxisof rotation through its centre of mass
12+b2
and I=Massx ---
12
HereIis the length andbthe breadth of the magnet.
Construction.It consists of a short magnet enclosed
in a wooden box provided with glass windows. The
box has a narrow tube fixed on its top at the middle
The magnet is suspended horizontally in a lightbrass
stirrup by a silk thread which passes centrally down
the tube and is provided with a torsion-head at the top
of the tube
current and its top has two slits through which
vibrations of the magnet can be observed. A plane
mirror strip with a reference line on it is placed
lengthwise at the base of the instrument just below the
slits.
---Torsion head
Glasstube
Silkthread
Fig.5.65Vibration magnetometer.···3wx£n~m}r°n3lxv

467. 5.44
Adjustments
(i)Placea
compass needle onthe reference line
Rotate the box till the linebecomes parallel to
the compass nedJehis sets the magneto-
meter in the north-south direction.
(ii)To ensure that there is notwist in the thread,
place a brass bar of the same size as that of
the magnet inthe stirrupand allow the stirrup
to come to rest. Adjust the torsion head so that
thebrassbar is in the north-south direction.
(iii)Replace the brass bar by a small magnet with
its N-pole pointing ggraphic north. Bring a
powerful magnet near the box and remove it.
The suspended magnet starts oscillating. Note
its period of vibration.
48.Explain the various uses of a vibration magneto-
meter.
Uses of a vibration magnetometer:
1.Measurement of magnetic moment of a magnet.
Set the vibration magnetometerinthe north-south
direction. Place thebar magnet in its stirrup. Measure
thetime period of the bar magnet.
As T=2n~m~
or
The moment of inertia Ican be determined from the
gmetry of the magnet. Knowing Bwmagnetic
momentmcan be determined.
2.Comparison of magnetic moments of two
magnets of samesize and same mass.For the two bar
magnets of same size and mass, the moments of inertia
areequal. Withthe help of vibration magnetometer,
wemeasure thetime periods of vibration TlandT2of
the two magnets at a particular place
Tl =2n~~~ and T2 =2n~~~
or
3.Comparison of magnetic moments of two
unequal sizes and masses (sum and difference method).
Sum position.Placethetwo magnets in the stirrup of
vibration magnetometer so that theirnorth poles point
in the ggraphical north.
Moment of inertia of the combination =II+12
Magnetic moment ofthe combination = ~+ ~
PHYSICS-XII
Determine the period of vibration of the combi-
nation. Let it beT1.Then
...(1)
I=II+12
m='11-~ ~
5 NN 5
(a) (b)
Fig. 5.66(a)Sum position.(b)Difference position.
Difference position.Now placethetwo magnets in
the stirrup of the vibration magnetometer with their
opposite poles in the same direction.
Moment of inertia of the combination
=II+12
Magnetic moment of the combination
='11-111:2
Let period of vibration of the combination
=T2
Then ...(2)
or
('11-~)~
Dividing(1)by(2),we get
Tl_~'11-~
T2'11+~
T12_'11-~
Ti-'11+ ~
Bycomponendo anddividendo, we get
4.Comparison of horizontal components of
earth's magnetic field. With the help of a vibration
magnetometer, we measure the periods of vibration
TandT'of the same magnet at the two given places.
Let ~ and s;,be the horizontal components of earth's
magnetic field at these places.
Then T=2n ~ m~and
.. ~ =~ or
T'=2n~m~www4°±trsp€vvr4o±~

468. GUIDELINES TONCERT EXERCISES
5.1. Answer the following questions regarding earth's
magnetism:
(a)A vector needs
three quantities for its specification.
Name the three independent quantities conventionally used to
specify the earth's magnetic field.
(b) The angle of dip at a locationinsthern Indiaisabt
18°. Wld y expect a greater or smaller dip angleinBritain?
[CBSE Oi) 95C]
(c) If y made a map of magnetic field lines at Melbrne
in Australia, wld the lines seem to go into the grnd or come
t of the grnd?
(d) In which direction wld a compass free to move in the
vertical plane point to, if located right on the geomagnetic north
or sth pole? [CBSE 0 95C]
(e) The earth's field, it isclaimed, rghly approximates the
field due to a dipole of magnetic moment8x1022IT-1located
at itscentre. Check the order of magnitude of this numberin
someway.
(j)Geologistsclaim that besides the main magnetic N-S
poles, there are several local poles on the earth's surface oriented
in different directions. Howissuch a thing possible at all?
Ans.(a)The three independent quantities used to
specify the earth's magnetic field are
(i)magnetic declination, (ii)angle of dip, and
(iii)rizontal component of earth's magnetic field.
(b)Britain is closer to the magnetic north pole. So the
angle of dip is greater in Britain than that in India. It is
about 70° in Britain.
(c)Magnetic lines of force of earth's magnetism will
seem to come out of the ground at Melbourne in Australia
because this region lies in the southern hemisphere of the
earth where the earth's magnetic north pole lies.
(d)Earth's magnetic field is exactly vertical at the
poles and so the rizontal component of earth's field is
zero which makes the compass needle point in any
direction at the geomagnetic north or south pole.
-+
(e)Magnetic field B at an equatorial point of the
earth's magnetic dipole is given by
B -110 ~
- 41t.r3wwwXnotesdriveXcom

469. 5.62
Now m=8x1022JT-1
,r=6.4x106m
8x1022
B=10-7 x T
(6.4x106l
=0.3xlO-4T =0.3G
which is of the same order of magnitude as that of the
observed field on the earth
if)The earth's field is onlyapproximately a dipole
field. Local N-S poles may arise due to the different
deposits of magnetised minerals.
5.2.Answer the following questions:
(a) The earth's magnetic fieldvariesfrom point to point in
space. Does it also change with time? If so,on what time scale
doesit change appreciably ?
(b) The earth's coreisknown to contain iron. Yet geologists
do not regard this as a srce of theearth'smagnetism. Why ?
(c)The charged currents in the terconducting regions of
the earth's corearethght to beresponsible forearth's
magnetism. What might be the 'battery' (i.e.,the srce of
energy) to sustain these currents?
(d)The earth may have even reversed the directionof its
field several times during itshistory of4to5billion years. How
cangeologists knowabt the earth's field insuch distant past?
(e)The earth's field departs from itsdipole shape
substantially at large distances (greater thanabout30,000 km).
What agencies mayberesponsible for this distortion ?
(fJInterstellar spacehasanextremely weak magnetic field of
theorderof10-12T.Cansuch a weak field be of any significallt
consequence? Explain.
Ans.(a)Yes,it does change with time. Time scale for
appreciable change is roughly a few hundredyears. But
even on a much smaller scale ofa fewyears, its variations
are not completely negligible.
(b)The temperature of earth's core is very high, so iron
exists as molten iron which, being at a temperature higher
than Curie point is not ferromagnetic.
(c)Radioactivity may be one of the possible sources
for the charged current in the outer conducting regions of
the earth's core which are tught to be responsible for
earth's magnetism.
(d)Earth's magnetic field gets recorded weakly in
certain rocks during their solidification. An analysis of
these rocks may reveal the history of earth's magnetism.
(e)At large distances, the field gets modified due to
the field of ions in motion (in the earth's ionosphere). The
field of these ions, in turn, is sensitive to extraterrestrial
disturbances such as the solar wind.
if)When a charged particle moves in a magnetic field,
it gets deflected along a circular path of radius,
R=mv
eB
[ mv2]
.: evB=R
PHYSICS-XII
A weak field of 10-12Tbends the charged particle in a
circle ofverylargeradius. Overasmall distance, we may
not notice the deflection but oververylarge interstellar
distance, the deflection is quite noticeable.
5.3.Ashortbar magnet placed with itsaxis at30°witha
uniform external magnetic field of0.25Texperiences a torque of
magnitude equal to 45x10-2J.What isthe magnetic momen t
of the magnet ?
Ans.Here6 =30°, B =0.25T,r = 4.5 x1O-2J, m=?
Asr=mBsin 6,
m=_'t_= 4.5x10-2 =0.36JT-1.
Bsin60.25xsin30°
5.4.Ashort bar magnet of magnetic moment m =0.32111
isplaced in a uniform external magnetic field of0.15T. If the
barisfreeto rotate in the plane of the field, which orientations
wld correspond to its(i)stable and(ii)unstable equilibrium?
Whatisthepotentialenergy of the magnet in each case ?
Ans.Herem=0.32jr'. B=0.15T
(i)The bar will beinstable equilibrium when its
-> ->
magnetic moment mis parallel toB (6=0°).Its
potential energy is then minimum and is given by
Umin=-mBcos0°= -0.32 x0.15x1
=-4.Bx10-2J.
(ii)The barwill be in unstable equilibrium whenits
magnetic moment;;isantiparallel toB(6=IBOO).Its
potential energy is then maximum and is given by
Umax = -mBcosIBO° =-0.32 x0.15x(-1)
=+4.Bx10-2 J.
5.5.Aclosely wnd solenoid of800turnsandarea of
cross-section 25x10-4m2carries a current of3.0A.Explain in
what sense doesthe solenoid act like a bar magnet. Whatisits
associated magnetic moment ?
Ans.HereN=BOO,A=2.5x10-4m 2,1=3.0A
m=NIA=BOOx3x2.5x10-4 = 0.60 Jrl
The magnetic field of a solenoid has the same pattern
as that of a bar magnet. It acts along the axis of the solenoid.
Its direction is determined by the sense of flow of current.
5.6.If thesolenoid in Exercise 5.5isfreetoturn abt the
vertical direction andauniform horizontal magnetic field of
0.25Tisapplied, whatisthe magnitude of thetorque on the
solenoid when its axismakes an angle of 30°with the direction
of the magnetic field ?
Ans.Herem= 0.60rr'. B = 0.25T,6 = 30°
l'=mBsin6 = 0.60 x0.25 xsin30°= 7.5 x10-2 J.
5.7.Abarmagnet of magnetic moment 1.5IT-1 lies
alignedwith the direction of a uniform magnetic fieldof0.22T.
(a)Whatisthe amnt of work required to turn the magnet
soasto align its magnetic moment (i)normal to thefield
direction (ii)opposite to thefield direction?
(b)Whatisthetorque on the magnet in cases, (i)and(ii)?www5notesdrive5com

470. MAGNETISM
Ans.H
erem= 1.5 JT -1, B= 0.22 T
(i)Givenf\=0°,8 = 90°
.. W=-mB(cos82-cosf\)
= -1.5 x0.22 (cos 90° - cos 0°)
= -0.33 x (0-1)=+0.33J
Torque, 1:=mBsin 90° = 1.5 x 0.22 x 1= 0.33 Nm.
(ii) Given81= 0°,82= 180°
W= -1.5 x0.22xcos (180° - cos 0°)
= - 0.33 x (- 1-1) =0.66J
Torque, 1:=mBsin 180° = 1.5 x 0.22 x 0 =O.
5.8.A closely wnd solenoid of 2000 turns and area of cross-
section1.6 x 10-4 ni,carrying a current of 4.0 A,issuspended
thrgh its centre allowing itto turn in a horizontal plane.
(a) Whatisthe magnetic moment associated with the
solenoid?
(b) What are the force and torque on the solenoid if a
uniform horizontal magnetic field of75x10-2Tis
set up at an angleof30°with the axisof the
solenoid? [CBSE ODlSC]
Ans.Here N= 2000, A= 1.6 x 10-4m 2, I= 4.0 A
(a)Magnetic moment of solenoid ofturnsN,area of
cross-sectionAand carrying current Iis
m=NIA= 2000 x 4.0 x 1.6 x10-4 Am2
=1.28Am2
This magnetic moment acts along the axis of the
solenoid in a direction related to the sense of current via
the right-hand screw rule.
(b)Netforce experienced bythe magnetic dipole in the
uniform magnetic field
=0
TIlemagnitude of the torque1:exerted by the magnetic
-->
field Bon the solenoid is given by
1:=mBsin8=1.28x7.5 x 10-2 xsin 30°
=0.048Nm
This torque tends to align the axis of the solenoid (i.e.,
its magnetic moment vector n!)along the fieldB.
5.9.Acircular coil of16 turns and radius 10 em carryinga
current of 0.75 A rests with its plane normal to an external field
ofmagnitude 5.0x10-2T. The coilisfree to turn abt an axis
initsplane perpendicular to the field direction. When the coilis
turned slightly and released, it oscillates abt its stable
equilibrium with a frequency of2.0s-1. What isthe moment of
inertiaofthe coil abt itsaxisof rotation ?
Ans.HereN = 16,r= 10 ern = 0.10 m, I= 0.75 A
B=5.0xlO-2T, v=2.0s-1
Magnetic momentof the coil is m=NIA=NI . 7tr2
Frequency of oscillation, v= _1_1mB
27tVT
5.63
..Moment ofinertia is
I= ~ =NI7tr2 •B
47t~2 47t~2
16 x0.75 x (0.1)2 x 5x 10-2
4 x 3.14 x4
=1.2x10-4 kgm2•
5.10.A magnetic needle free to rotate in a vertical plane
parallel to the magnetic meridian has its north tip pointing
down at 22°with the horizontal. The horizontal component of
the earth's magnetic field at the placeisknown to be 0.35G.
Determine magnitude of earth's magnetic field at the place.
Ans.Here8=22°,BH= 0.35G, B=?
B=l!JL= 0.35 G=0.35G =0.38G.
cos8cos 22° 0.9272
5.11.At a certain locationinAfrica, a compass points 12°
west of the geographic north. The north tip ofthe magnetic
needle of a dip circle placed in theplane of the magnetic
meridian points60°above the horizontal. The horizontal
component of the earth's fieldismeasuredto be 0.16G.Specify
the direction and magnitude of earth's field at thelocation.
Ans.Here BH= 0.16G,8= 60°
.. B=l!JL= ~ =0.16 =0.32 G
cos 8cos 60° 0.5
Thus the earth'smagneticfield has a magnitude of
0.32 G and lies in a vertical plane 12° west ofthe geographic
meridian making an angle of 60° (upwards) with the
rizontal (magnetic south to magnetic north) direction.
5.12.A short bar magnet has a magnetic moment of
0.48 IT-1.Give the direction and magnitude of themagnetic
field produced by the magnet at a distance of10emfrom the
centre of the magnet on(i)the axis(ii)the equatorial line of the
magnet.
Ans.Herem= 0.48Jr1,r= 10cm = 0.10 m
. ~o2m
(I)For a short magnet, B.I= -.3
axia47t r
47tx 10 -7x 2 x 0.48 = 0.96 x 10-4 T = 0.96G.
47t (0.10)3
Thisfield actsalongS-Ndirection.
(ii)For a srt magnet,B= ~0 . ~
equa47tr
47tx 10-7 0.48 -4
--4-7t- x-(0-.1-0-)3= 0.48 x 10 T = 0.48 G.
This field acts along N-Sdirection.
5.13.A short bar magnet placed in a horizontal plane hasits
axis aligned along the magnetic north-south direction. Null
points are fnd on the axis of the magnetat14em from the
centre of themagnet. The earth's magneticfield at the place is
0.36Gand the angle of dip iszero. What isthetotal magnetic
field on the normal bisector ofthe magnet at thesame distance as
the null-points (i.e., 14em) from the centre ofthemagnet?www3~°trsqrvvr3p°z

471. 5.64
Ans.As the
null points lie on the axis of the magnet,
therefore
B -1102m -B
axial - 41t.r3-H
Magnetic field of the magnet on its normal bisector at
the same distance will be or
B=~. m=BH=0.36=0.18 G
equa41tr3 2 2
:.Total magnetic field at the required point on the
normal bisector is
Bequa+BH=0.18+ 0.36=0.54G.
5.14.If thebar magnet in Exercise 5.13isturned arnd by
180°,where will the newnull points belocated?
Ans.When the magnet is turned around by 180°,its
south pole will lie in the geographical south direction. The
null points will now lie on the equatorial line of the
magnet, say at distancexfrom the centre of the magnet.
Then
But
_ 110 m_
Bequa -- . 3- BH
41t X
B=110 2m
H 41t'r3
1102m 110 m
41t""7=41t.x3
r3
x3=-
2
r 14cm
x=i/3=1.26=11.1em.
[From Exercise 5.13]
or
or
5.15. A shortbar magnet of magnetic moment
5.25x10-2IT-1 isplacedwith its axisperpendicular to the
earth's field direction. Atwhat distance from the centre of the
magnet on (i)itsnormal bisector (ii)itsaxis,istheresultant
field inclined at 45°with the earth's field. Magnitude of the
earth's field at the place isgiven to be 0.42 G.Ignore thelength
of magnet in comparison to the distance involved.
Ans.Herem=5.25x10-2JT-1,
Bo=0.42G=0.42x10-4T
(i)Figure5.95(a) sws a point P on the normal bisector
of a magnet where the resultant field is inclined at45°with
Fig. 5.95
(a)
PHYSICS-XII
the earth's fieldBo'As point P isanequatorial point,
therefore, the resultant field must be such that
B
equa =tan450
Bo
B =Btan45°=R
equa '"1J
But for a srt magnet
B =110.!!!.-
equa 41t .r3
~.!!!.--B
41t.r3- 0
r3=110 ..!!!..=10-7.5.25x10-2
41tBo 0.42x10-4
=125x10-6
Hence r=5x10-2m=5 em.
(ii) Figure 5.95(b)shows a point Q on the axis of a
magnet where the resultant field is inclined at45°with the
earth's field Bo'In this condition,
B.I
~=tan45°orBaxial=Bo
Bo
or
n~rzJ----,i!
, ,
, ,
, ,
I.uuo~uu-e ~-- Q ..•'__u_
N : 5 Baxial
,......--r---+
,
s
(b)
Fig. 5.95
But for a srt magnet
B.=~ 2m
axial 41t.r3
1102m
41t'~=Eo
31102m
or r=-.-
41tBo
=125x10-6 x2=250x10-6
Hence r=(250)1/3 x10-2m
=6.3x10-2 m=6.3em.
5.16. Answer the following questions:
(a)Why does aparamagnetic sample display greater
magnetisation(for the same magnetising field) when coled ?
[CBSE OD 91 ; Himachal 98C]
(b)Whyisdiamagnetism, incontrast, almost independent
of temperature ? [Himachal 96]
(c)If a toroid usesbismuth for its core, willthe field in the
corebeslightlygreater orslightly less than whenthe coreis
empty?
(d)Isthepermeabilityof a ferromagnetic material
independent of themagnetic field? Ifnot,isitmorefor lower or
higher fields ?www5notesdrive5com

472. MAGNETISM
(e)Magne
tic field lines are always nearly normal to the
surface of a ferromagnet at everypoint. Why?
if)Wld the maximum possible magnetisation of a
paramagnetic sample beof the same order of magnitude as the
magnetisationof a ferromagnet? [NCERT]
Ans.(a)The tendency to disrupt the alignment of
dipoles (with the magnetising field) arising from random
thermal motion is reduced at lower temperatures.
(b)The induced dipole moment in a diamagnetic
sample is always opposite to the magnetising field, no
matter what the internal motion of the atoms is.
(c)As bismuth is diamagnetic, so the field in the toroid
with bismuth core will be slightly less thanwhen the core
is empty.
(d)No,the permeability of a ferromagnetic material is
not independentofthe magnetic field. This is evident
from theB-Hcurve which has greater slope (hence greater
u)at lower fields.
(e)The proof of this important fact is based on
-+ -+
boundary conditions of magnetic fields ( Band H )at the
interface of two media. When one of the media has p»1,
the field lines meet this medium nearly normally.
(/) Yes. Apart from minor differences in the strength of
the individual atomic dipoles of two different materials, a
paramagnetic sample with saturated magnetisation will
have the same order of magnetisation. But saturation
requires impractically high magnetising fields.
5.17. Answer the following questions:
(a)Explain qualitatively on the basis ofdomainpicture the
irreversibility in the magnetisation curve of a ferromagnet.
(b)Thehysteresis loop of asoft iron piecehasa much
smaller area than that ofa carbonsteelpiece.If thematerial isto
gothrgh repeated cyclesof magnetization, which piece will
dissipate greater heatenergy ?
(c)Asystem displaying a hysteresis loop such as a
ferromagnet is a device for storing memory ?Explain the
meaning of thisstatement.
(d) What kindofferromagnetic materialisused for coating
magnetic tapes in a cassette player, or for building 'memory
stores'ina modern computer?
(e) Acertain region of space isto be shielded from magnetic
fields. Suggest a method.
Ans.(a)In a ferromagnetic substance, the atomic
dipoles are grouped together in domains. All the dipoles
of a domain are aligned in the same direction and have net
magnetic moment. In an unmagnetised substance these
domains are randomly distributed so that the resultant
magnetisation is zero. When the substance is placed in an
external magnetic field, these domains align themselves
in the direction of the field. Some energy is spent in the
process of alignment. When the external field is removed,
these domains do not come back into their original
random positions completely. The substance retains some
5.65
magnetisation. The energy spent in the process of magne-
tisation is not fully recovered. The balance of energy is
lost as heat. This is the basic cause for irreversibility of the
magnetisation curve of a ferromagnetic substance.
(b)Carbon steel piece, because heat lost per cycle is
proportional to the area of the hysteresis loop.
(c)Magnetisation of a ferromagnet is not a
single-valued function of the magnetising field. Its value
for a particular field depends both on the field and also on
the history of magnetisationi.e.w many cycles of
magnetisation it has gone through etc. In other words, the
value of magnetisation is a record or 'memory' of its cycles
of magnetisation. If information bits can be made to
correspond to these cycles, the system displaying such a
hysteresis loop can act as a device for storing information.
(d)Ceramics (specially treated barium iron oxides)
also called farrites.
(e)Surround the region by soft iron rings. Magnetic
field lines will be drawn into the rings, and the enclosed
space will be free of magnetic field. But this shielding is
only approximate, unlike the perfect electric shielding of a
cavity in a conductor placed in an external electric field.
5.18.Along straight horizontal cable carries acurrent of
2.5 A in the direction 10°southofwest to10°northofeast.The
magnetic meridian oftheplacehappens to be10°west of the
geographic meridian. Theearth's magnetic field at the locationis
0.33G,and the angleof dip is zero. Locate theline of neutral points.
Ng
Sin
,
I?:,
C) •..•
c<>'
6',
,
W----------~~~--~~----E
,
'/Magnetic
\ meridian
,
,
,
S N",
g
Fig. 5.96
Ans.Suppose the neutral point lies at a distancerfrom
the cable. Then at the neutral point,
IloI _B
21tr -H
Il 0I 41tx10-7x2.5 -2
orr=-~ = 4=1.5x10 m =1.5em
21tBH 21tx0.33x10-
As the direction of the magnetic field of the cable is
-+
opposite to that of BH at points above the cable, so the line
of neutral points lies parallel to and above the cable at a
distance of 1.5 em from it.www3~°trsqrvvr3p°z

473. 5.66
5.19.A t
elephone cable at a place has fr long straight
horizontal wires carrying a current of1.0 Ain thesame
directioneast to west. The earth's magnetic field at the placeis
0.39G,and the angle of dip is35°.The magnetic declination is
nearlyzero.What are theresultant magnetic fields at points
4.0embelow, and above the cable?
Ans.Earth's field B =0.39 G, 0=35°
BH=Bcos0 =0.39cos 35° =0.319G
By=B sin0 =0.39 sin 35° =0.224 G
For the cable, we have
I =1.0 A, N =4, r=4.0 cm=4x10-2m
:. Magnetic field produced by the cable wires is
, 11 0NI41tx10-7 x4x1.0
B=-- = -----..,.-
21tr 21tx4x10-2
=0.2xlO-4T=0.2 G
Resultant field below the cable. At points below the cable
the·field B' is in the opposite direction ofBwSo the
rizontal component of the resultant field is
RH=BH - B'=0.319-0.2=0.119G
The vertical component of the earth's field remains
unaffected.
Rv=By=0.224G
:.The magnitude of the resultant field is
R=~RH2+ R/=~(0.119)2 + (0.224)2=0.254G
The angle that R makeswith the rizontal is
0=tan-1 ~=tan-10.224=tan-118.8.:::62°
RH 0.119
Resultantfield above the cable.In this case the field of the
cable acts in the direction of Bw
RH=BH + B'=0.319 + 0.2=0.519G
Rv=0·.224G
R=~r-(0-.5-19-;)2;-+-(-0.-22-4~)2 =0.566 G
Angle of dip,0=tan-10.224=tan-10.4316.::: 23°.
0.519
5.20.A compass needle free to turn ina horizontal planeis
placed at thecentre of circular coil of30turns and radius12cm.
Thecoilisinavertical plane making an angle of45°with the
magnetic meridian. When the currentinthe coil is0.35A,the
needle points west to east.
(a) Determine the horizontal componentof the earth's
magnetic field at the location.
(b) The current in the coilisreversed, and the coil is
rotated abt its vertical axis by an angle of 90°in
theanticlockwise sense loking from above. Predict
thedirectionof needle. Take themagnetic declination
at the places to be zero.
Ans.(a)Magnetic field set up at the centre of the coil is
B=lloNI
2r
PHYSICS-XII
This field acts along the axis perpendicular to the
plane of the coil. As the coil is in a vertical plane making
an angle of 45° with the magnetic meridian and the needle
points in thewest-east direction, it is obvious from
Fig. 5.97, that the needle is oriented at an angle of 45° with
the field B. Using law ofsines for a triangle,we get
N
Circular
coil
Fig. 5.97 5
.s«. B
sin 45° sin 90°
11NI
BH=Bsin 45° =_0-.sin 45°
2r
41tx10-7x30x0.35
--1-2-x-l-0--2"-- x0.7071 T
=3.8876x10-5 T.:::0.39G.
(b)The needle will reverse itsoriginal direction i.e.,it
will point east to west.
5.21. Amagnetic dipole isunder the influence of two
magnetic fields. The anglebetween the field direction is60°and
one of the fields has a magnitude of 12x10-2T.Ifthedipole
comes tostable equilibrium at an angle of 15°withthis field,
whatisthemagnitude of the other field?
or
Fig. 5.98
N
5
Ans.Here ~=1.2x1O-2T , ~ =15°
82=60°-15°=45°
In equilibrium,'1:1='1:2orm~sin ~=mE,.sin 82
or
E,.= ~sin~
sin82
1.2x10-2sin 15°
sin 45°
=4.4xlO-3T.
1.2x10-2x0.2588
0.7071www3~°trsqrvvr3p°z

474. MAGNETISM
5.22.Amonoene
rgetic(18keY) electron beam initially in
thehorizontaldirection is subject to a horizontal magnetic field
of0.40 Gnormal to the initial direction. Estimate the up or
down deflection of the beam over adistance of30em.
(me= 9.11x10-31kg, e=1.60x10-19C).
Ans.Kinetic energy of an electron
=..!mv2=eV v=~2ev
2 m
The magnetic field provides the centripetal force to
make the electron move in a circular path of radius R
Therefore,
mv2
-- =evB or
R
R_mv_m~2ev _ .J2meV
---- ------
eBeB m eB
Herem=9.11x10-31 kg,e= 1.6x10-19C
V= 18 kV=18x103V, B=0.40e= 0.40x1O-4T
~2x9.11x10-31 x1.6x10-19 x18x103
R= m
1.6x10-19 x0.40x10-4
= 11.3 m.
It is obvious from Fig. 5.99
that when the beam covers a
distance x,its up or down
deflection is y.
Nowfrom right t.OCB
sin 9 =~=0.3=~
R 11.3 113
[.:x= 30cm = 0.30 m]
/
/
I
I
,
I
\
\
,
,
Fig. 5.99
Hence up or down deflection of the electron beam is
y=OA -OC=R - Rcase
= R [1- cos 9] = R [1- (1- sin2 9)1/2]
~ R[ 1-(1-~Sin2 9)]
1 1 (3)3
="2Rsin2e="2x11.3x 113 m
=3.98x1O-3m ~4 mm.
5.23.Asample of paramagnetic salt contains 2.0x1024
atomic dipoles each of dipole moment 1.5x10-23IT-I. The
sample isplaced under homogenes magnetic field of 0.84 T
and coled to thetemperature of 4.2K.Thedegree of magnetic
saturation achieved isequal to15%.Whatisthetotal dipole
moment of the sample for a magnetic field of 0.98 T anda
temperature of2.8K(assume Curie's law)?
Ans.Dipole moment of each atomic dipole,
m= 1.5x10-23JT-1
Total number of atomic dipoles, N= 2.0x1024
Initial total magnetic moment at temperature
11=4.2Kis
5.67
~ = 15%of mN
=~x1.5x10-23 x2.0x1024JT-1= 4.5 JT-1
100
According to Curie's law,
B M2 Bz11
M=Constantx- -=-x-
T M1 ~12
Now ~ = 0.84 T, 11= 4.2 K,Bz= 0.98 T,12= 2.8 K
Hence the final dipole moment at temperature
12=2.8Kis
M2 =M1 x ~ x11= 4.5 x0.98x 4.2 JT-1
~ 12 0.84 2.8
=7.9JT-t.
5.24.ARowland ring of mean radius 15em has 3500 turns
of wire wnd on a ferromagnetic core of relative permeability
800.Whatisthemagnetic field (B) in thecorefora magneiising
current of 1.2A?
Ans.A Rowland ring is a circular ring of a magnetic
material over which is wound a toroidal solenoid. The
magnitude of the magnetic field in the core is given by
B=~nI
wheren= 3500.Isthe number of turns per unit length.
211:r
Now relative permeability,
~
~r=-, so that ~ =~0~r
~o
Hence
3500
B=~O~r·--·l
211:r
-7 3500
= 411:x10x800x 2 x1.2 T
211:x15x10-
=4.48 T.
5.25.Themagnetic moment vectors ~ and ~ associated
-->
with the intrinsic spin angular momentum 5and orbital
-->
angular momentumI,respectively, of an electron are predicted
by quantum theory (and verified experimentally to a high
accuracy) to begiven by:
-4 -4-4 -+
~s=-(e/m)S, ~1=-(e/2m)1
Which of these relations isin accordance with the result
expected 'classically' ?Outline the derivation of the classical
result.
Ans.The relationiI= -(2~)Iis in accordance with
classical physics. Por its derivation, refer answer to Q. 15
on page 5.13.
In contrast to ~ I /I,the magnitude of ~s /5ise/ m i.e.,
twice the classically expected value. This latter result
(verified experimentally) is an outstanding success of
modern quantum theory and cannot be derived from
classical physics.www3~°trsqrvvr3p°z

475. 5.68 PHYSICS-XII
Text Based Exercises
'-"'YPEA
:VERY SHORT ANSWER QUESTIONS (1 mark each)
1.What a magnet?
2.What a natural magnet? [Punjab02]
3.What does the word lodestone mean ?
(J4.Are the two poles of a magnet equally strong?
5.What a magnetic dipole? Give an example.
6.Define the term magnetic dipole moment. Give its
51unit. [CBSE on 95;Haryana02]
7.Is magnetic moment a vector ? If yes, give its
directn.
8.Define a unit magnetic pole.
9.Write the 51 unit of (i)pole strength and
(ii)magnetic moment of a bar magnet.
[CBSEF03]
10.What the directn of magnetic dipole moment?
[Haryana 02]
11.What the torque experienced by a magnetic
dipole moment mplaced with its ax at angle 8with
a uniform magnetic field B? [rSCE 93]
12.In a uniform magnetic field, when the torque on a
magnet(i)maximum and(ii)minimum?
13.Give the positn of a magnetic dipole held in a
magnetic field, where its potential energy
minimum. [CBSE on 93]
14.In which positn, the potential energy of a magnet
in a uniform magnetic field zero ?
15.On which factors does the pole strength of a magnet
depend?
16.Write the formula for the magnetic moment of a
current loop. [CBSEF02]
17.If a magnet broken into pieces, which one of the
following remains unchanged in each part-mass,
moment of inertia, magnetatn ?
18.Two iron bars attract each other, no matter in
which combinatn their ends are brought near
each other. What can we say about their state of
magnetatn ?
19.An electron which a charged particle in motn
has a magnetic moment. Why then does a neutron,
which has no charge, have a magnetic moment?
20.Define Bohr magneton and write its value.
[Punjab99C]
21.How much the magnetic moment of an electron
revolving in the nth orbit of hydrogen atom?
22.Who first dcovered earth's magnetm ?
23.How much the approximate dtance upto which
the earth's magnetic field extends ?
24.Give an order of magnitude of earth's magnetic
field near its surface.
25.What magnetic ax of the earth ?
26.What geomagnetic equator?
27.What geomagnetic meridian? [CBSED92]
28.What geographic meridian?
29.Name the elements or parameters of earth's
magnetic field.
[CBSEoo94 ;Punjab99C;Haryana 02]
30.Define declinatn at a place.
[Haryana02, 03;CBSED93C]
31.Define angle of dip (or magnetic inclinatn) at a
place. [CBSED93C;Haryana02]
32.What will be the angle of dip at a place on the
equator? [CBSEF95 ;D95]
33.What the angle of dip at magnetic poles?
[CBSE F 91]
34.How does dip angle vary from equator to poles?
[CBSE D 91;F 92]
35.What will be the value of the horizontal component
of earth's magnetic field at earth's geomagnetic
pole? [CBSEF95]
36.Horizontal components of Earth's magnetic field at
a place .J3times the vertical component. What
the value of angle of dip at th place?
[CBSE D 97]
37.The vertical component of Earth's mangetic field at
a place .J3times the horizontal component. What
the value of angle of dip at th place?
[CBSE D 06]
38.The needle of the dip circle vertical at the
magnetic poles. The dip circle rotated about a
vertical ax through 90°.What will be the positn
of the needle on the vertical circular scale?
[CBSE on 96C]
39.Where on the surface of earth the angle of dip
(i)0° and(ii)90°? [CBSE on 11]
40.A magnetic needle, free to rotate in a vertical plane,
orients itself with its ax vertical at a certain place
on the earth. What are the values of :
(a)horizontal component of earth's magnetic
field?
(b)angle of dip, at th place? [CBSEF12]www5~°trsq•vvr5p°z

476. MAGNETISM
41.The horizontal component
of the earth's magnetic
field at a place Band angle of dip 60°.What
the value of vertical component of earth's magnetic
field at equator? [CBSE0 12]
42.Where on the earth's surface the vertical
component of the earth's magnetic field zero?
[CBSE00 11;0 13C]
43.What are ogonic lines?
44.What are oclinical lines?
45.What an aclinic line?
46.What are odynamic lines?
47.Define neutral point in the magnetic field of a bar
magnet. [Punjab 02]
48.A magnet placed with the north pole towards the
north of the earth. Predict the positn of the
neutral points.
49.What the difference between a compass needle
and dip needle?
50.At a certain locatn in Africa, a compass needle
points15°West of the geographic north. What the
angle of declinatn at that point? [Haryana 97]
51.Torques'1and'2are required for a magnetic
needle to remain perpendicular to the magnetic
fields1\and ~ at two different places. What the
rat of the magnetic fields at the two places?
52.Give the order of magnetic moment of an atom.
53.What induced magnetm?
54.What is a magneting field ?
55.Define magnetic intensity. Give its SI unit.
56.Define magnetic inductn. Give its SI unit.
57.Define magnetatn of a material. Give its SI unit.
58.Define magnetic permeability. State its SI unit.
[ISCE 98]
59.What relative permeability ?
60.Define magnetic susceptibility. [Haryana 01]
61.Relative permeability of a material, Ilr=0.5.
Identify the nature of the magnetic material and
write its relatn to magnetic susceptibility.
[CBSE0 14C]
62.What the one quantum of atomic dipole moment
called?
63.What a diamagnetic substance? [Punjab 97]
64.Which of the following substances are diamagnetic?
Bi,AI,Na,Cu,Ca and Ni [CBSE0 13]
65.What a paramagnetic substance? [Punjab 98C, 99]
66.Which of the following substances are para-
magnetic? Bi, AI, Cu, Ca, Pb,Ni [CBSE0 13]
67.What a ferromagnetic substance?
5.69
68.Relative permeability(Ilr)of a material has a value
lying 1<Ilr<1+E(whereE a small quantity).
Identify the nature of the magnetic materials.
[CBSE0 14C]
69.Relative permeability of a material Ilr=400.
Identify the nature of the magnetic material.
[CBSE0 14C]
70.The permeability of bmuth 0.9983. To which
class of magnetic materials, does bmuth belong?
[CBSE011]
71.How does the intensity of magnetatn of a
paramagnetic sample vary with temperature?
[CBSE00 2000]
72.Will the neon gas be diamagnetic or paramagnetic?
Give reason.
73.What curie point?
74.Give two essential charactertics of a material used
for preparing an electromagnet. [CBSEF04]
75.Define hysteres.
76.What does the area of hysteres loop indicate?
77.The hysteres loop of a soft iron piece has a much
smaller area than that of a carbon steel piece. Ifthe
material to go through repeated cycles of
magnetatn, which piece will dsipate greater
heat energy?
78.What the basic use of a hysteres curve?
79.Name two alloys commonly used for making cores
of transformers.
80.Name two magnetic materials commonly used for
making permanent magnets.
81.Suggest two methods to destroy the magnetm of a
magnet. [Punjab 02]
82.What are permanent magnets? Give one example.
[CBSE0 13]
83.How does the magnetic permeability Ilrdiffer
for dia, para and ferromagnetic materials?
[ISCE 03]
84.Mentn the two charactertic properties of the
material suitable for making core of a transformer.
[CBSE0012]
85.A (hypothetical) bar magnet(AB) cut into two
equal parts. One part now kept over the other, so
that poleCz aboveC;. If M the magnetic
moment of the original magnet, what would be the
magnetic moment of the combinatn so formed ?
[CBSESample Paper 08]
Fig.5.100www5~°trsq•vvr5p°z

477. 5.70
86.An electric current of0.25A flows in a loop of
radius0.2cm. Calculate the magnitude of the
magnetic dipole moment of the dipole formed.
[ISCE 02]
Answers
PHYSICS-XII
87.Two circular loops, of radirand 2r, have currents, I
and
1/2 flowing through them in clockwe and
anticlockwe sense respectively. If their equivalent
magnetic moments areMlandM2respectively,
state the relatn betweenMlandM2•
[CBSE Sample Paper 2011]
•
1.A magnet a piece of material that has both
attractive and directive properties. It attracts small
pieces of iron, nickel, cobalt etc.
2.Natural magnet an iron ore called magnetite.
Chemically, it a black iron oxide of formula
Fe304·
3.The word lodestone means a leading stone. It
represents the directive behavur of a magnet.
4.Yes.The two poles of a magnet are always equally
strong.
5.Refer to point12of Glimpses on page5.78.
6.Refer to point13of Glimpses on page5.78.
7.Yes, magnetic moment a vector. Its directn
from south pole to north pole of the magnet.
8.A magnetic pole which when placedin vacuum at a
dtance of1m from an identical pole repels it with
a force of 10-7newton called unit magnetic pole.
9.(i)The S1 unit of pole strength ampere metre
(Am).
(ii)The S1 unit of magnetic moment ampere
metre/ (Am2) or joule/teslaorI).
10.The directn of magnetic dipole moment from
S-pole to N-poleof the magnet.
11.Torque,1:=mBsine.
12.(i)Torque maximum when the magnet lies per-
pendicular to the directn of the magnetic field.
(i)Torque minimum (zero) when the magnet
lies along the directn of the magnetic field.
13.The P.E. of a magnetic dipole minimum when
its dipole momentn; parallel to the magnetic
~
field B.
14.The P.E. of a magnetic dipole zero when its
dipole momentn; perpendicular to the fieldB.
15.The pole strength of a magnet depends on(i)its
area of cross-sectn(ii)nature of its material and
(iii)its state of magnetatn.
16.Magnetic moment of a current loop,m=IA.
17.Magnetatn.
18.One rod a permanent magnet while the other
an unmagneted iron rod.
19.Like mass and charge, magnetic moment a basic
property, one may consider neutron as a particle
containing equal amounts of positive and negative
charges spinning in opposite directns.
20.Refer to point22of Glimpses on page5.79.
21. fl/=nflB=n(~J.
41tme
22.
23.
William Gilbert.
Earth's magnetic field extends upto a dtance of
32000km, which about five times the radius of
the earth.
It 10-4T or 1 gauss (G). '.
The straight line passing through the magnetic
north and south poles of the earth called
magnetic ax of the earth.
It the great circle on the earth perpendicular to
the magnetic ax.
Refer to point26of Glimpses on page5.79.
Refer to point27of Glimpses on page5.79.
The elements of earth's magnetic field are
(i)Declinatn (ii)Dip
(iii)Horizontal component of earth's magnetic
field.
Refer to point29of Glimpses on page5.79.
Refer to point30of Glimpses on page5.79.
Angle of dip at the equator,cS=0°.
Angle of dip at magnetic poles,cS=90°.
The dip angle increases from0°to90°as one moves
from magnetic equator to poles.
At poles,cS=90°
:. Horizontal component,
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
BH=BcascS=Bcas90°=o.
e; Bv 1
Here tans=BH=J3Bv=J3
:. Angle of dip, 8=30°.
37.Here tan 8= ~ =J3BH =J3
BH BH
36.www:notesdrive:com

478. MAGNETISM
38.
39.
90°.
(i)A
ngle of dip 0°at magnetic equator. (i)Angle
of dip 90°at magnetic poles of the earth.
(a)Horizontal component ofearth's magnetic
field is zero.
(b)Angle of dip at the given place 90°.
At the magnetic equator, Bv=O.
At the magnetic equator, Bv=O.
The lines joining the places of equal declinatn are
called ogonic lines.
Thelinesjoining the places of equal dip or incli-
natn are called oclinical lines.
The line ofzerodipcalled aclinic line or magnetic
equator.
The line joining the places having the same value of
the horizontal component of earth's magnetic field.
are called odynamic lines.
Neutral point that point where the magnetic field
due to a magnet equal and opposite tothe
horizontal component ofthe earth's magnetic field.
The resultant magnetic field at the neutral point
zero.
The neutral points lie on equatorial line on either
side of the magnet.
Acompass needle is freetorotateabouta vertical
ax in ahorizontal plane while adip needle is free
to rotate about ahorizontal ax in a vertical plane.
a=15°,west of the geographic north.
Torque onamagnetic needle,
,=mBsin 90°=mBi.e.,,ocB
~=~
~ '2
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.10-23Am2.
53.When a piece of magnetic material placed in a
magnetic field, the material gets magneted. The
magnetm so produced in the material called
induced magnetism.
54.Refer to point34of Glimpses on page5.80.
55.Refer to point35of Glimpses on page5.80.
56.Refer to point 37of Glimpses on page 5.80.
57.Refer to point36of Glimpses on page5.80.
58.Refer to point 38of Glimpses on page 5.80.
59.Refer to point39of Glimpses on page 5.80.
60.Refer to point40of Glimpses on page5.80.
61.As 0<~r<1,the material diamagnetic.
~r=1+Xm·
62.Bohr magneton.
63.Refer topoint42of Glimpses on page 5.80.
5.71
64.Bi and Cu.
65.Refer to point43of Glimpses on page5.80.
66.Al and Ca.
67.Refer to point44of Glimpses on page5.80.
68.Paramagnetic.
69.Ferromagnetic.
70.As the permeability of bismuth less than 1,so it is
diamagnetic.
1
71.Moc-.
T
72.As there are no unpaired electrons in neon atoms,
so neon gas diamagnetic.
73.Curie point the temperature above which a
ferromagnetic substance becomes paramagnetic.
74.The material used for making an electromagnet must
have(i)high permeability and(iI)low retentivity.
75.The phenomenon of lagging of magnetic inductn
behind the magneting field in a magnetic material
called hysteresis.
76.The area of the hysteres loop gives a measure of
the energy wasted in a sample when it taken
through a cycle of magnetatn.
77.The carbonsteel piece will dissipate greater heat
energy as its hysteresis loop has greater area.
78.Thestudy of hysteres curve ofa material gives its
hysteres loss, retentivity and coercivity. Th
knowledge helps in selecting materials for making
cores of transformers, permanent magnets and
electromagnets.
79.(i)Mu-mctal, an alloy of nickel, iron, copper and
chromium.
(ii)Peum metal, an alloy of iron and nickel.
80.(i)Cobalt steel,an alloyof cobalt,tungsten and carbon
(ii) Alnico,an alloy of iron, nickel, aluminium,
cobalt and copper.
81. (i)Byheating the magnet.
(ii)By applying magnetic field in the reverse
directn.
82.The substances which at room temperature retain
their ferromagnetm for a long time are called perma-
nent magnetse.g.,cobalt, steel, alnico and ticonal.
83.For diamagneticsubstances, ~r<1
For paramagnetic substances, ~r>1
For ferromagnetic substances, ~r»1
84.The core material of transformer must have
(i)high initial permeability and(ii)lower
hysteresis loss.www5~°trsq•vvr5p°z

479. 5.72
85.Magnetic moment of the combinatn will be zero.
As poles ~ andc;will be of opposite nature, the
magnetic moments of
the two parts get aligned in
opposite directn in the combinatn. Hence
(M/2)-(M/2)=0.
PHYSICS-XII
86.m=IA=Ixrrr2=0.25x3.14x(0.2x10-2)2
=3.14x10-6Arn2
YPEB:SHORT ANSWER QUESTIONS (2or 3marks each)
1.Define the terms magnet and magnetm. What
the origin of the word magnetm ?
2.What meant by the terms magnetic field and
magnetic field intensity? [Haryana94]
3.State four basic properties of magnets.[Punjab 02]
4.Define the terms magnetic poles, magnetic ax,
magnetic equator and magnetic length with
reference to a bar magnet.
5.State Coulomb's law of magnetic forces. Hence
define a unit magnetic pole.
6.What a magnetic dipole? Define magnetic dipole
moment. Give its SI unit.
7.What a uniform magnetic field? Give an example
and sketch lines of force of such a field.
8.What are magnetic lines of force ? Sketch the
magnetic field lines of a bar magnet.
[CBSEco90C]
9.What are magnetic lines of force ? Give their
important properties. [Haryana94]
10.What the name given to the curves the tangent to
which at any point gives the directn of the
magnetic field at that point? Can two such curves
intersect each other? Justify your answer.
[CBSE Sample Paper 97]
11.Derive an expressn for the torque experienced by
a magnetic dipole in a uniform magnetic field.
Hence obtain the expressn for the potential
energy ofthe dipole. [CBSEF 03]
12.Derive an expressn for the torque on a magnetic
dipole placed in a magnetic field and hence define
magnetic dipole moment.
[Haryana 96 ; Himachal 2000]
13.Derive an expressn for the potential energy of a
bar magnet when placed in a uniform magnetic
field. [Himachal01]
14.Derive an expressn for the intensity of the
magnetic field at an axial point of a short magnetic
dipole.
15.Derive an expressn for the magnetic field due to a
magnetic dipole in broad-side on positn at a
dtancerfrom its centre. The length of the dipole
21and its magnetic moment m. [ISCE 98]
16.A current loop behaves as a magnetic dipole.
Obtain an expressn for the magnetic dipole
moment of a circular current loop. State the rule
used to find the directn of the magnetic dipole
moment. [CBSEF95]
17.Show that the electron revolving around the
nucleus in an orbit of radiusrwith speedvhas
magnetic moment evr/2.Hence using Bohr's
postulate of angular momentum, obtain the
expressn for the magnetic moment of hydrogen
atom in its ground state. [CBSEaD13C]
18.Give an experimental evidence in support of the
extence of earth's magnetic field.
19.Compare the magnetic fields of a b'ar magnet and
current carrying solenoid. [Punjab01]
20.Explain with the help of diagram(1)magnetic decli-
natn and(i)magnetic of dip at a place. In what
directn will a compass needle point when kept at
the(i)poles and(i)equator?[CBSED 04,OS,lSC]
21.Namethree elements required to specify the earth's
magnetic field at a given place. Draw a labelled
diagram to define these elements. Explain briefly
how these elements are determined to find out the
magnetic field at a given place on the surface of
Earth. [CBSED04 C, 08C]
22.Name and define the magnetic elements of earth's
magnetic field at a place. Derive an expressn for
the angle of dip in terms of the horizontal
component and the resultant magnetic field of the
earth at a given place. [CBSE DOl, 02]
23.Define neutral point. Locate the positns of neutral
points, when a small bar magnet placed with its
north pole(i)towards the north of the earth and
(i)towards the south of the earth.[CBSED9SC]
24.Sketch the magnetic field pattern of a bar magnet
placed in the magnetic meridian with its N-pole
pointing north. Show the neutral points. How will
you determine the dipole moment of the magnet?
[Punjab 2000]
25.Define neutral point. Sketch the lines of magnetic
field when a small bar magnet placed with
N-pole towards the south of the earth. Indicate the
positn of neutral points. [CBSEaD02C]
26.What "Aurora Boreal" (or northern lights) ?
Can it be seen in India?www<notesdrive<com

480. MAGNETISM
27.What is relative
permeability of a magnetic
material ? How it related to the magnetic
susceptibility ? [ISCE 97)
28.State and explain Curie law of magnetism.
[CBSE OD 97;Punjab2000,01)
29.What are dia, para and ferromagnetic substances?
Give one example of each. [Haryana02)
30.Classify materials on the basis of their behavur in
a magnetic field. Under which category does
iron come?Howdoesthemagnetic property of
ironchange withincrease oftemperature?
[CBSE D 95)
31.Dtinguh between diamagnetic and ferromagnetic
substances inrespect of (i)intensity of magne-
tatn, (i)behavur in a non-uniform magnetic
field and(ii)susceptibility. [ISCE 95; CBSE D 03)
32.What are diamagneticsubstances ? Explain the
origin of diamagnetm on the bas of electron
theory. [Haryana01)
33.What are paramagneticsubstances ? Explain the
origin of paramagnetm.
34.What are ferromagnetic substances ? Explain
briefly domain theory to explain ferromagnetm.
[Haryana 02)
35.Explain the phenomenonof hysteres in magnetic
materials. Draw a hysteresis loop showing remanence
and coercive force. [ISCE 96; Haryana01]
36.Define the terms retentivity and coercivity.
[Himachal 97]
37.Write two charactertic properties to dtinguh
between diamagnetic and paramagnetic materials.
[CBSEOD 05]
38.Draw a hysteres loop with axes labelled and
showing remanence and coercive force. What does
the area of the loop represent? [ISCE 02)
39.Define the term magnetic dipole moment of a
current loop. Write the expressn for the magnetic
moment when an electron revolves at a speed 'd,
around an orbit of radius'r'in hydrogen atom.
[CBSE OD 08, 10]
Answers
5.73
40.Permanent magnets are made of special alloys
while the cores of temporary magnets are made of
soft iron. Why ? [ISCE 2000)
41.Depict the field-line pattern due to a current
carrying solenoid of finite length.
(i)In what way do these lines differ from those
due to an electric dipole?
(i)Why can't two magnetic field lines intersect
each other? [CBSEF09]
42.Derive an expressn for the magnetic moment (~)
of an electron revolving around the nucleus in terms
->
of itsangular momentum (I).What the directn
of the magnetic moment of the electron with respect
to its angular momentum? [CBSE OD 14C)
43.Out of the following, indentify the materials which
can be classified as:(I)paramagnetic, (il)diamagnetic:
(a)Aluminium(b)Bmuth(c)Copper(d)Sodium
Write one property to dtinguh between para-
magnetic and diamagnetic materials.[CBSE D 09C)
44.(i)How does angle of dip change as one goes from
magnetic pole to magnetic equator of the earth ?
(i)A uniform magnetic field gets modified as
shown below when two specimens X and Yare
placed in it. Identify whetherspecimens X and Y
are diamagnetic, paramagnetic or ferromagnetic.
Fig. 5.101
(ii)How the magnetic permeability of specimen
Xdifferent from that of specimen Y ?
[CBSE F 09;D 09C)
45.(a)How an electromagnet different from a
permanent magnet?(b)Write two properties of a
material which make it suitable for making
electromagnets. [CBSE OD 14C)
1.Refer to point 1 of Glimpses.
2.Magnetic field. The space around a magnet within
which its influence can be experienced called its
magnetic field.
Magneticfieldintensity. The magnetic field
intensity at any point in a magnetic field defined
as the force experienced by a unit north pole placed
at that point. It a vector quantity whose directn
is same as the directn in which unit north pole
would tend to move if free to do so.
3.Refer to point 4 of Glimpses.
4.Refer answer to Q.4 on page 5.2.
5.Refer answer toQ.5 on page 5.3.
6.Refer answers to Q. 6 on page 5.3 and Q. 7 on page 5.4.www5~°trsq•vvr5p°z

481. 5.74
7.Refer answer toQA
on page 5.3. As
shown
in Fig. 5.102,
the lines of force of a
uniform magnetic
field are parallel to
one another.
--~~-----------.~ B
Fig. 5.102
8.Refer answer to Q.8 on page 5.6. For the lines of
force of a bar magnet, see Fig 5.24 on page 5.18.
9.Refer answer toQ.8 on page 5.6.
!'.J.Such curves are called magnetic lines of force. No
two such lines of force can intersect. If they do so,
then therewill be two tangents and hence two
directns of the magnetic field at the point of
intersectn which impossible.
11.Refer answers to Q. 12 and Q. 13 on page 5.11.
12.Refer answer to Q. 12 on page 5.11.
13.Refer answer to Q. 13 on page 5.11.
14.Refer answer to Q. 10 on page 5.7.
15.Refer answer to Q. 11 on page 5.7.
16.Refer answer to Q. 14 on page 5.12.
17.Refer answer to Q. 15 on page 5.13.
•18.Refer answer to Q. 20 on page 5.20.
19.Figure 5.24 shows the lines of force of a bar magnet
while Fig. 5.25 shows the lines of force of a straight
solenoid. The two patterns have a striking resem-
blance. So a solenoid behaves like a bar magnet.
The two ends of both have north and south polarity.
20.Refer answer to Q. 23 on page 5.21. At the poles,
the compass needle will stay in any directn. At
the equator,it will come to rest in the magnetic
north-south directn.
21.Refer answer to Q. 23 on page 5.21.
22.Refer answer to Q. 23 on page 5.21.
23.Refer answer to Q. 28 on page 5.23.
24.Refer answer to Q. 28 on page 5.23.
25.See Fig. 5.35 on page 5.24.
26.Refer answer to Q. 26 on page 5.23.
Th phenomenon cannot be seen anywhere in
India as it occurs onlynear the magnetic poles.
27.Refer to points 39 and 40 of Glimpses.
28.Refer answer to Q. 32 on page 5.31.
29.Refer answer to Q. 30 on page 5.30.
30.Refer answer to Q. 30 on page 5.30.
Iron a ferromagnetic substance. As temperature
increases, its magnetatn decreases due to
randomation of its domains.
PHYSICS-XII
31.See Table 5.3 on page 5.35.
32.Refer answer to Q. 31 on page 5.30.
33.Refer answer toQ.32 on page 5.31.
34.Refer answer to Q. 33 on page 5.3l.
35.Refer answer toQ.38 on page 5.36.
36.Refer answer toQ.38 on page 5.36.
37.(i)In a uniform magnetic field, a rod of diamag-
netic material aligns itself perpendicular to the
field while that of paramagnetic material
aligns itself parallel to the field.
(ii)The magnetic susceptibility of a diamagnetic
material small and negative while that of
paramagnetic material, thevalue small and
positive.
38.Refer answer toQ.38 on page 5.36.
39.Refer answer toQ.14 on page 5.12 andQ.15 on
page 5.13.
40.Permanent magnets are made of special alloys like
steel because of their high coercivity. The cores of
temporary magnets are made of soft iron because of
its high permeability and low retentivity.
41.See Fig. 5.25 on page 5.18.
(i)The magnetic lines of force of a solenoid form
closed loops while the electric lines of force of
an electric dipole start from the positive charge
and end at the negative charge.
(i)Refer answer to Q.10 on page 5.74.
42.Refer answer to Q.15 on page 5.13.
43.(i)Paramagnetic: Aluminium and sodium.
(ii) Diagmagnetic : Bmuta and copper.
Refer answer to Q.37 above.
44.(i)The angle of dip decereases from 90° to 0°as
one goes from magnetic pole to magnetic
equator of the earth.
(i)X diamagnetic as it expels field lines.
Y isparamagnetic or ferromagnetic as it pulls
in field lines.
(iii)Magnetic permeability of X less than 1 and of
Y greater than l.
45.(a)An electromagnet consts of a core made of a
ferromagnetic material placed inside a solenoid.
It behaves like a strong magnet when current
flows through the solenoid and effectively loses
its magnetm whenthe current switched off.
The substances which at room temperature
retain their ferromagnetm, after being magne-
ted once, are called permanent magnets.
(b)The core of an electromagnet must have (/) high
permeability(ii)low retentivity and(iii)low
coercivity.www<notesdrive<com

482. MAGNETISM
"YPE C:lONG ANSWER QUESTIONS (5
marks each)
1.(i)What the relatnship between the current and
the magnetic moment of a current carrying circular
loop?Use the expressn to derive the relatn
between the magnetic moment of an electron moving
in acircle and its related angular momentum?
(ii)A muon aparticle that has the same charge as
an electron but 200 times heavier than it. It we
had an atom in which the muon revolves around a
proton instead of an electron, what would be the
magnetic moment of the muon in the ground state
ofsuch an atom? [CBSESamplePaperOS]
2.Explain the possible causes of earth's magnetic
field. [Punjab98C,01]
3.Dtinguh the magnetic properties of a dia-, para-
and ferromagnetic substances in terms of(i)suscep-
tibility, (i)magnetic permeability and(iz)coercivity.
Giveone example of each of these materials.
Draw the field lines duetoan external magnetic
field near a(i)diamagnetic, (i)paramagnetic
substance. [CBSE0007]
4.(a)How does a paramagnetic material behave in
the presence of an external magnetic field ?
Answers
5.75
a
Explain with the help of an appropriate
diagram.
(b)What happens when the temperature of a
paramagneticsample is lowered?
(c) To which of the two - a polar dielectric or a
non-polar dielectric-does a paramagnetic
material correspond? Justify your answer.
[CBSESamplePaper 2011)
Asmall compass needle of magnetic moment
'm'isfree to turn about an ax perpendicular
tothe directn of uniform magnetic field'B'.
The momentof inertia of the needle about the
ax is '1'.Theneedle slightly disturbed from
its stable positn and then released. Prove that
it executessimple harmonic motn. Hence
deduce the expressn for its time perd.
(b)A compass needle, free to turn in a vertical
plane orients itself with its ax vertical at a
certain place on the earth. Find out the values
of(i)horizontal components of earth's
magnetic field and(ii)angle of dip at the place.
[CBSE013)
5.(a)
"
1.(i)Magnetic moment of a circular current loop
=CurrentxArea of the loop orm=LA
For magnetic moment of an electron, refer answer
toQ.15on page 5.13.
(ii)Magnetic moment of anelectron in ground state,
eh
~I(electron) =--
41tme
As mass of a muon 200times the mass of an electron,
i.e;its magnetic momentin the ground state
eh 1.6x10-19 x6.6x10-34
~ I(muon)=41tx200me 41tx200x9.1x1031
=4.63xlO-26 Am2.
2.Refer answer toQ.23 on page 5.2l.
3.See Table5.3on page5.35and see Fig. 5.76on
page 5.54.
'-"YPE D:VALUE BASED QUESTIONS (4marks each)
1.Once Indu noticed that her aunt wassuffering from
severe joint pain and could not take any pain killer
being allergic to them. Inher quest to help her aunt,
Indu found the use of magnets. She read
Dr.Philpott's work on magnetic therapy. She found
that many people are negative magnetic field
deficient due to electromagnetic pollutn. When
the body issupplemented with negative field
4.(a)
(b)
Refer answer toQ.32 on page 5.3l.
Asthe temperature of a paramagnetic
substance lowered, its atomic dipoles tend to
get aligned with the magneting field. So its
magnetatn increases until it reaches a
saturatnvalue at a stage when all the dipoles
get perfectly alignedwith the field.
(c)Paramagnetic material a kind of polar
dielectric. Th because the atoms/molecules
of sucha materialhavenon-zeromagneticmoment
Refer answer to Q.46 on page 5.42.
(i)Horizontal component ofearth's magnetic
field zero. BH=Bcos90° =o.
(i)Asthe compass needle stands vertically,
angle of dip(8)=90°.
5.(a)
(b)
energy, it brings about quick healing. Indu took her
aunt to a magnetic therapy centreon regular basis.
Her aunt recovered at a great pace.
(a)What are the values being highlighted by Indu?
(b)What the magnitude of the equatorial and
axial fields due to a bar magnet of length 5 cm
at a dtance of50cm from the midpoint? The
magnetic moment of the bar magnet 0.40 Am 2.www5~°trsq•vvr5p°z

483. 5.76
2.Rohit
'smother put many clothes for washing in the
washing, machine. She could not startthe washing
machine as an indicator was showing that the lidwas
notclosed. On seeing hismother struggling, Rohit
tried his best to close the lid butsoon he realised that
theclosing mechanism was notamanual system but a
magnetic system. He went to the shop andpurchased
a magnetic closer. Hereplaced the older closer by the
new one. Themachinestarted working. Themother
was happy that her son helped her to save three
hundred rupees.
(a)Whatare the values being dplayed by Rohit ?
(b)What are the values being shown by Rohit's mother?
(c)A magneted needle in auniform magnetic field
experiences atorque but no netforce. An iron nail
near a bar magnet, however, experiences a force of
attractn in additn to a torque.Why?
3.Shubham, oncewatching Dcovery Channel, saw
thatcertain organisms have the ability to sense the
field linesof earth's magnetic field. By following
these lines, they can travel from oneplace toanother.
Shubham, in h quest to know more about earth's
magnetic field, brought amagnetic compass. By
knowing the direction in which the compass comes
to rest, he found the magnetic meridian. Then he
fixed the compass on a cardboard and placed it
vertically along the magnetic meridian. Bymeasuring
theinclinatn of thecompass with thehorizontal,
he found the angle of dip at the given place.
(a)What values did Shubham have?
(b)Inwhich directn would a compass free to
movein the vertical plane point to, if located
righton the geomagnetic north or south pole?
Answers
PHYSICS-XII
4.While watching a science channel onTV,Renu once
sawthata spectacular dplay of light like the one
during commonwealth games could be seen in the
nightskyat high altitudes near the polar regions.
She got surprised and consulted her Physics
teacher to know more about this phenomenon. She
was explained thatthisphenomenon called aurora
caused when the charged particles of the solar
windgetattracted by themagnetic poles of the
earth and there they ne the atmospheric atoms
or molecules emiting green and pink light -the
glow of the aurora. The aurora in the northern hemi-
sphere is called aurora borealis and that in the
southern hemphere is called aurora austral. Renu
sharedth knowledge with her classmates also.
(n)What values did Renu have?
(b)The earth's fielddeparts from its dipole shape
substantially at large dtances (greater than
about 30,000km).What agencies may be
responsible for this dtortn?
5.Twofriends Deepa and Shilpa were good dancers
and used to perform in the school functns by
using their cassette player. Oneday when they
wereperforming, the tape got stuck up and the
music stopped. But Deepa was determined not to
let down the performance, soshedecided to sing
the song instead of dancing and Shilpa completed
the dance.
(a)What were the values displayed by Deepa and
Shilpa?
(b)What kind of ferromagnetic material used
for coating magnetic tapes in a cassette player,
orbuilding 'memory stores' in a modem
computer?
•
1.(a)Love, sympathy, diligence, punctuality and
regularity, maturity and responsibility.
(b)Referto thesolutn of Example 7 on page 5.8.
2.(a)Sympathy, responsibility, helping &self-reliance.
(b)Appreciatn, thankfulness&economicalnature.
(c)In the first case, the magnetic field uniform, and
forces actingon the two ends of the needle are
equal and opposite. So the netforce is zero.
However, a torque acts on the needle.
In the case of iron nail, there is aninduced
magnetm. The induced(say) south pole in the
nail, being closer to the north poleofthebar
magnet, experiences a larger attractive force than
the induced north pole. Sothe nail experiences
both a net attractive force and a torque.
3.(a)Nature of appreciatn, cursity, diligence,
self-reliance andcreative skill.
(b)Earth's magnetic field exactly vertical at the
poles and sothehorizontal component of
earth's magnetic field zero which makes the
compass needle pointin any directn at the
geomagnetic north orsouthpole.
4.(n)Natureofappreciatn, diligence, research
mindedness and communication skills.
(b)At large distances, the fieldgetsmodified due
to thefieldof ions in motn(in the earth's
ionosphere). The fieldof these ns,in turn, is
sensitive to extra terrestrial dturbances such
as solar wind.
5.(a)Team spirit,confidence, determination and
courage.
(b)Ceramics (specially treatedbarium iron
oxides) also called ferrites.www<notesdrive<com

484. Magnetism
GLIMPSES
1.Magnets and
magnetism.Amagnet is a piece of
material that has bothattractive and directive
properties. Itattracts small pieces of iron, nickel,
cobalt, etc. Thispropertyof attraction is called
magnetism.
2.Natural magnets.Anaturalmagnet is an iron
ore called lodestone (leading stone) or
magnetite. It is a black iron oxide, Fe304.
Natural magnets were found as early as the
sixthcentury B.C.in the province of Magnesia
in ancient Greece, from which the word
magnetism derives its name.
3.Artificial magnets.Pieces of iron and other
magnetic materials can be made to acquire the
properties of natural magnets. Such magnets
are called artificial magnets.
4.Basic properties of magnets.These areas
follows:
(i)Attractive property. A magnet attracts small
pieces of iron,nickel, cobalt, etc.
(ii)Directive property. Afreely suspended
magnet aligns itself nearly in the
geographic north-south direction.
(iij)Like poles repel and unlike poles attract. Thisis
afundamental law of magnetic poles.
(iv) Magnetic poles exist in pairs. Isolated
magnetic poles do not exist. If we break a
magnet into twopieces, we get two smaller
dipole magnets.
5.Magnetic field.The space around a magnet
withinwhich its influence can be experienced is
called its magnetic field.
6.Uniform magnetic field. A magnetic field in a
region is said to be uniform if it has same
magnitude and direction at all points of that
region.
7.Magnetic poles.These are the regions of
aparently concentrated magnetic strength in a
magnet where the magnetic attraction is
maximum.
8.Magnetic axis.The line passing through the
poles of a magnet is called its magnetic axis.
9.Magnetic equator.The line passing through the
centre of the magnet and at right angles to the
magnetic axis is called the magnetic equator of
the magnet.
10.Magnetic length. The distance between the two
poles of a magnet is called its magnetic length. It
is slightly less than the geometrical length of the
magnet.
11.Coulomb's law of magnetic force. This law
states that the force of attraction or repulsion
between two magnetic poles is directly
proportional to the product of their pole
strengths and inversely proportional to the
squareof the distance between them. Ifqand
nil
qmare the pole strengths of two magnetic poles
2
separated by distancer,then the force of
attraction or repulsion between them is
F=Ilo qlll1 qnl2
41t.r2
where110is the permeability of free space and
itsvalue is41tx10-7henry/metre or TmA-1.
(5.77)wwwtnotesdrivetcom

485. 5.78
12.Magnetic dipole. Any arrangement of
two equal
andopposite magnetic poles separated by a
small distance is called a magnetic dipole. A bar
magnet and a current-carrying loop are
magnetic dipoles.
13.Magnetic dipole moment. Itis equal to the
product of the polestrength(ql11) and the
magnetic length (2/)of the magnet.
m=ql11x21
The 51 unit of magnetic dipole moment is Aro2
or JT-1 •
14.Magnetic lines of force.A magnetic line of force
may be defined as the curve the tangent to
which at any point gives the direction of the
magnetic field at that point. Itmay also be
defined as the path along which a unit north
pole would tend to move if free to do so.
15.Properties of lines of force. These are as follows:
(i)Magnetic lines of force are closed curves
which start in air from the -pole and end
at the 5-pole and then return to the N-pole
through the interior of the magnet.
(ii)Thelines of force never cross each other.
(iii)They start from, and end on the surface of
the magnet normally.
(iv)The lines of force have a tendency to
contract lengthwise and expand sidewise.
This explains attraction between unlike
poles and repulsion between like poles.
(v)The relative closeness of the lines of force is
a measure ofthe strength ofthe magnetic
field which is maximum at the poles.
16.Magnetic field of a bar magnet at an axial point
(end-on position).
. _ Ilo 2mr
(z)Baxial -47t . (,2 _/2)2'
where ris the distance of the point from the
centre of the magnet.
(ii)For a short magnet, 1< <r,
B =Ilo2m
axial 47t' r3
The magnetic field at any axial point of
magnetic dipole is in the samedirection as that
of its magnetic dipole moment.
PHYSICS-XII
17.Magnetic field of a bar magnet at an equatorial
point (Broadside-on position).
(i)B _ Ilo m
equa - 47t .(,2+[2)3/2
(ii)Forshort magnet, 1< <r,
B =Ilom
equa 47t .r3 .
The magnetic field at any equatorial point of a
magnetic dipole isin the direction oposite to
thatof its magnetic dipole moment.
18.Torque on a magnet in a magnetic field.Ifa
magnet of dipole moment misplaced in a
....
magnetic fieldBmaking an angle 8 with it, then
torque acting on the magnet is
T=mBsin 8
....
In vector notation, l'=irtxB
The effect of the torque is to align the dipole
....
parallel to the field B.
If 8=90°, thenT=mB
Hence the magnetic dipole may bedefined as
thetorque acting on a magnetic dipole placed
perpendicular to a uniform magnetic field of
unitstrength.
19.Potential energy of a magnetic dipole in a
magnetic field. When a magnetic dipole is
rotated in a magnetic field against the torque
from initial position 81 tofinal position 82, the
work done or the potential energy stored is
given by
W=U= -mB(cos82-cos 81)
........
P.E.is zero whenml.B.Hence P.E. of the
dipole in any orientation 8 is
U=-mBcos8=-riz.B
Special Cases
1.When8=0°,U= -mB.Thus the P.E. of a
.... ....
dipole isminimum when IIIis parallel to B.
This isthe position of stable equilibrium.
2.When 8=90°, U=0.
3.When 8=180°, U=+mB.
Thus the P.E. ofthedipole ismaximum when m
....
is antiparallel to B. This is the position of
unstable equilibrium.www6notysxrivy6wom

486. MAGNETISM (Competition Section)
20.Current loop as
a magnetic dipole.A planar
current loopof area Aand carrying current I
behaves as a magnetic dipole of dipole moment,
m=IA
~
In vector notation, m=IA
The direction ofinis given byright hand thumb
rule.If we curl the fingers of the right hand
along the direction of currentin the loop, then
the extended thumb gives the direction of the
magnetic moment associated with the loop.
21.Magnetic dipole moment of a revolving
electron.The orbital magnetic moment of an
electron revolving around a nucleus innth orbit
of radiusrwith speedvisgiven by
~1=~r= 2~e1=n(4:~neJ
whereIisthemagnitude of the angular
momentum of the electron revolving around
the nucleus.
22.Bohr magneton.It is the magnetic dipole
moment associated with an electron due to its
orbital motion in the first orbit of hydrogen
atom. It is the smallest value of ~1•
eh
~B=(~l)min=-4-
1tme
=9.27 x10-24Am2
23.Gauss's law in magnetism.This law states that
the net magnetic flux through any closed
surface is zero.
Mathematically,
~ ~
4>B=fB.d5=O.
5
This law indicates that isolated magnetic poles
(also called monopoles) do not exist.
24.Basic difference between electric and magnetic
lines of force.The magnetic lines of force are
continuous and formclosed loops. They do not
start or end at a point. In contrast, the electric
lines of force start from a positive charge and
end on a negative charge or they fade out at
infinity.
25.Earth's magnetic field.When a magnet is
suspended freely, it orients itself roughly in the
geographical north-south direction. This
suggests that the earth behaves as a huge
5.79
magnet. Its field can be aproximated to that
of a magnetic dipole with dipole moment
8.0x1022Am2 with its axis aligned at a small
angle with the rotation axis of the earth. The
magnitude of the field on the earth's surface is
typically about1O-4T.
26.Magnetic meridian.The vertical plane passing
through the magnetic axis of a freely suspended
small magnet is called magnetic meridian. The
earth's magnetic field acts in the direction of the
magnetic meridian.
27.Geographic meridian. The vertical plane
passing through thegeographic north and
south poles is called geographic meridian.
28.Elements of earth's magnetic field. The earth's
magnetic field at a place can be completely
described by three parameters which are called
elements ofearth's magnetic field. These are
declination, dip and horizontal component of
earth's magnetic field.
29.Magnetic declination(ex).It isthe angle between
the geographic meridian and the magnetic
meridian at the given place.
30.Angle of dip(B).It is the angle made by the
earth's total magnetic fieldwith the horizontal
direction.
31.Horizontal component of earth's magnetic field.
Itis the component of the earth's total magnetic
field B actingin the horizontal direction.
BH=BcosB
At the magnetic equator,B= 0°,
BH= B cos 0° = B
At the magnetic poles, B= 90°,
BH= Bcos 90° =0.
32.Relations between the elements of earth's
magnetic field.
BH=BcosB
and Bv= Bsin B
B
~=tanB
BH
and B = ~B~+BJ
33.Neutral point. It isthe point where the magnetic
field of a bar magnet is completely cancelled bywww9notesdrive9com

487. 5.80
the horizontal component of earth's magnetic
f
ield.
(i)For a magnet placed with its N-pole
pointing geographic north, the neutral
points lie at the equatorial line.
Ateachneutral point,
B=~o m
H 4rc' (,.1+/2)3/2
[For a short magnet]
(ii)For a magnet placed with its N-pole
pointing geographic south, the neutral
points lie at the axial line. At each neutral
point,
B=~o 2mr
H 4rc' (,.1 _/2)2
_ ~o2m
-- - [For a short magnet]
4rc'r3
34.Magnetising field.The magnetic field that exists
in vacuum and induces magnetism is called
magnetising field. The magnetising field set up
ina solenoid carrying currentIand placed in
vacuum,
Ba=~onI
51unit ofBais tesla (T).
35.Magnetising field intensity or magnetic
intensity. Itisthe number of ampere-turns (n!)
flowing round the unit length of the solenoid
required to produce a given magnetising field.
Thus H =nI
Also, Bo= ~onI=~H
B
H=.-Jl.
~o
51unitof H is Am -1andits dimensions are
[L-1A).
36.Magnetisation or intensity of magnetisation. It
isthe magnetic moment developed per unit
volume of a material when placed in a magne-
tising field. Itis a vector quantity.
M=m
V
Its SI unit is Am-lor N Wb -lor Nm -2r1.
37.Magnetic induction. It is the total number of
magnetic lines offorce crossing per unit area
PHYSICS-XII
through a magnetic material. Its SI unit is
tesla ( T).
B=~o (H+M)
38.Magnetic permeability. It is the ratio of the
magnetic induction to the magnetising field
intensity.
B
~=-
H
Its SI unit is Tm A-lor Wbm-1 A-1.
39.Relative permeability.It is the ratio of the per-
meability of the material to the permeability of
free space.
u,=£..
~o
40.Magnetic susceptibility.It is the ratio of the
intensity of magnetisation(M)induced to the
magnetising field intensity (H).'
M
XIII=H
It can be shown that
~ =~o (1+XIII)
and ~r=1+Xm·
41.Classification of magnetic materials.Magnetic
materials are broadly classified as diamagnetic,
paramagnetic and ferromagnetic.
42.Diamagnetic substances. These are the
substances which when placed in a magnetising
field get feebly magnetised in the oposite
direction of the aplied field. Such substances
are feebly repelled by magnets and tend to
move slowly from stronger to weaker parts of a
magnetic field.
Examples are Bi, Cu, Pb, Si,N2(at STP), ~O
and NaCl.
43.Paramagnetic substances. These are the
substances which when placed in a magnetising
field get feebly magnetised in the direction of
the magnetising field. Such substances are
feebly attracted by magnets and tend to move
slowly from weaker to stronger parts of a
magnetic field. Examples are AI, Na, Ca,02(at
STP) and CuCI2.
44.Ferromagnetic substances. These are the
substances which when placed in a magnetising
field get strongly magnetised in the direction ofwww8notesdrive8com

488. MAGNETISM(Competition Section)
45.
the magnetising field. Such substances tend
to
move quickly from weaker to stronger parts of a
field. Examples are Fe, Ni, Co,Cd,etc.
The magnetic susceptibility Xissmall and
negative for diamagnetic substances, small and
positive for paramagnetic substance and large
and positive for ferromagnetic substances. The
relative permeability (u,=1+X)is slightly less
than1for diamagnetic substances, slightly
greater than 1for paramagnetic substances and
of the orderofthousands for ferromagnetic
substances.
Curie's law.The magnetic susceptibility of a
paramagnetic substance varies inversely with
its absolute temperature.
1
x.,o:y
C
XIII=y'
or
46.
where C is curie constant.
Curie temperature. The temperature above
which a ferromagnetic substance becomes
paramagnetic is called Curie temperature(Te).
The modified Curie law for ferromagnetic
substances above the Curie temperature is
C'
XIII=T-T (T>Te)
e
This relation is alsocalledCurie-Weiss law.
Hysteresis. Thephenomenon of lagging of the
magnetic induction behind the magnetising
field in a ferromagnetic material is called
hysteresis. The area of the hysteresis (B-H) loop
gives the energy wasted in a sample when it is
taken through a cycle of magnetisation.
Retentivity or remanence. The magnetic
induction left behind in the sample after the
magnetising field has been removed is called
retentivity.
Coercivity.The value of the reverse
magnetising field required to make the residual
magnetism of a sample equal to zero is called
coercivity.
Tangent law. When a short magnet is
suspended freely under the combined action of
two uniform perpendicular magnetic fields B
47.
48.
49.
50.
5.81
andBH,the magnet comes to rest making an
angle 8 with the direction ofBHsuch that
B=BHtan8.
51.Tangent galvanometer. Its working is based on
tangent law and is usedtqmeasure very small
currents. It consists of a circular coil of radiusR
andNturns. If a currentIin the coil produces
deflection 8in the compass needle at the centre
of the coil, then
[2 R BJI=__ H_tan8=Ktan8
~oN
52.
2RB
whereK=__ H_,isthereduction factor of the
~oN
tangent galvanometer.
Vibration magnetometer. It isused to compare
magnetic moments of two bar magnets or to
determine the horizontal component of earth's
magnetic field.
Ifabar magnet of magnetic moment Inand
momentof inertiaIoscillates under the
horizontal component(BH) of earth's magnetic
field, then its period of vibration is
T=2n;~ I .
mBH
53.Uses of vibration magnetometer.
(i)Measurement of magnetic moment of a
magnet,
4n;2I
In=--
~T2
(ii)Comparison of magnetic moments of two
magnets of same size and same mass.
I1'lz _T12
"1-T22
(iii)Comparison of magnetic moments of two
magnets of unequal sizes and masses.
mT2+T2
---I_2 1
lrL-T2_T2
--L 2 1
(iv)Comparison of horizontal component of
earth's magnetic field.
B T'2
H_
B,-y:z
Hwww6notysxrivy6wom

489. C H A PT E R
ELECTROMAGNETIC
INDUCTION
6.1MAGNETIC FLUX
1.Explain the
concept of magnetic flux linked with a
closed surface. Give its units and dimensions. When is the
magnetic flux said to be (i) positive and (ii) negative?
Magnetic flux.The magnetic flux tough any surface
placedina magnetic fieldisthe total number of magnetic
lines offorce crossing this surface normally.It is measured
as the product of the component of the magnetic field
normal to the surface and the surface area.
Magnetic flux is a scalar quantity, denoted by4>or4>B.
-t
If a uniform magnetic field B passes normally
through a plane surface areaA,as shown in Fig. 6.1(a),
then the magnetic flux through this area is
4>=BA
-t
If the field B makes angle 8 with the normal drawn
to the areaA,as shown in Fig.6.1(b),then the
Area=A
--+ 1\
B n
---+-.---.
(a) (b)
Fig.6.1Magnetic flux through an area depends on
its orientation w.r. t. the magnetic field.
component of the field normal to this area will be
Bcos 8, so that
4>=Bcos8xA
or
-t-t
4>=BAcos8=B.A
-t
Here the direction of vectorAis the direction of the
outward drawn normal to the surface
-t
In general, the field B over an areaAmay not be
-t
uniform. However, over a small area elementdA ,the
-t
field B may be assumed to be uniform. As shown in
-t
Fig. 6.2, if 8 is the angle betwen B and the normal
-t -t
drawn to area elementd A,then the component of B
-t
normal tod Awill be B cos 8.
A
n
Fig. 6.2SurfaceAin a magnetic field.
(6.1)www5~§tr€q•wvr5p§}

490. 6.2
~
:.Flux
through area elementd Ais
d<jl=B-1dA=Bcos8dA
~~
= BdAcos8=B .dA
~ .
Then the flux of B through the whole areaAIS
f~ ~
<jl=B.dA
A
Dimensions of magnetic flux.As we know that
1P=BA
F
B=---
qvsin 8
But <jl=F.A
qvsin 8
Dimensions of flux,
MLr2
<jl- L2
- C.Lrl.1·
ML2T-2
A
[<jl]=[ML2A-1T-2].
SI unit of magnetic flux.The SI unitofmagnetic
fluxisweber(Wb). One weber istheflux produced when a
uniform magnetic field of one tesla acts normally over an
area of111,(-.
1 weber = 1 teslax1 metre2
[.: 1cr1 =1A]
or
or 1Wb=1 Tin2
CGSunit ofmagnetic flux.The CGS unit of
magnetic flux ismaxwell (Mx). One maxwell istheflux
produced whena uniform nwgnetic field of one gauss acts
normally over an area of1 em".
1 maxwell = 1 gaussx1 ern2
1Mx=lGcm2or
or
Relation betwen weber and maxwell
1 Wb = 1 Tx1 m2=104Gx104em2
1 Wb = 108 maxwell.
Positive and negative flux.A normal to a plane can
be drawn from either side
plane points out in the direction of the field, then 8=00
and the flux is taken aspositive. If the normal points in
the opposite direction of the field, then 8 = 1800 andthe
flux is taken asnegative.
(1-r----. -> A
B Ii
~ ~o~~
~ ~
e=180°,$=-BAe=0°,$=+BA
Fig. 6.3 (a)Positive flux and(b)Negativeflux.
PHYSICS-XII
6.2ELECTROMAGNETIC INDUCTION
AN INTRODUCTION
2.Whatiselectromagnetic induction ?
Electromagnetic induction. Electricity and magne-
tism are intimately connected. In the early part of the
ninetenth century, the experiments of Oersted, Ampere
and others established that moving charges (currents)
produce a magnetic field. The converse effect is also
truei.e.,moving magnets can produce electric currents.
In 1831, Michael Faraday in Englandand almostsimul-
tanusly Joseph Henryin the U.s.A. discovered that
currents are produced in a loop of wire if a magnet is
suddenly moved towards the loop or away from the
loop such that the magnetic flux across the loop changes.
The current in the loop lasts so long asthe flux is
changing. This phenomenon is calledelectromagnetic
inductionwhich meansinducing electricity by magnetism.
Thephenomenon of production of induced emf (and
hence induced current) due to achange of magnetic
flux linked withaclosedcircuit iscalled electromagnetic
induction. •
The phenomenon of electromagnetic induction is of
great practical importance in daily life.It forms the
basis of the present day generators and transformers.
Modem civilisation owes a great deal to the discovery
of electromagnetic induction.
6.3FARADAY'S EXPERIMENTS
3. Describe the various experiments performed by
Faraday and Henry which ultimately led to the discovery
of the phenomenon ofelectromagnetic induction.
Faraday's experiments. The phenomenon of electro-
magnetic induction was discovered and understood
on the basis of the following experiments performed
by Faraday and Henry.
EXPERIMENT1.Induced emf with a stationary coil
and moving magnet. As shown in Fig. 6.4, take a
circular coil of thick insulated copperwire connected
to a sensitive galvanometer.
(i)When the N-pole of astrong barmagnet is
moved towards thecoil, the galvanometer
shows a deflection, say totherightofthe zero
mark [Fig. 6.4(a)].
(ii)When the N-pole of the bar magnet is moved
away from the coil, the galvanometer shows a
deflection in the opposite direction [Fig. 6.4 (b)].
(iii)If the above experiments arerepeated by
bringing the S-pole of the magnet towards or
away from the coil, the direction of current in
the coil is opposite to that obtained in the case of
N-pole.www5~§tr€q•wvr5p§}

491. ELECTROMAGNETIC INDUCTION
(iv)When the magnet is held stationary anywhere
near or inside
the coil, the galvanometer does
not show any deflection [Fig. 6.4 (c)].
0',
...
(a)N-pole moved towards coil
o
(b)N-pole moved away from coil
o
(c)Magnet at rest.
Fig. 6.4Induced emf with a moving magnet
and stationary coil.
Explanation. When a bar magnet is placed near a
coil, a number of lines of force pass through it. As the
magnet is moved closer to the coil, the magnetic flux
(the total number of magnetic lines of force) linked
with the coil increases, an induced emf and hence an
induced current is set up in the coil in one direction. As
the magnet is moved away from the coil, the magnetic
flux linked with the coil decreases, an induced emf and
hence an induced current is set up in the coil in the
opposite direction. As soon as the relative motion
betwen the magnet and the coil ceases, the magnetic
flux linked with the coil stops changing and so the
induced current through the coil becomes zero.
EXPERIMENT2.Induced emf with a stationary
magnet and moving coil. Similar results as in experi-
ment 1 are obtained if the magnet is held stationary
and the coil is moved, as shown in Fig. 6.5.When the
relative motion betwen the coil and the magnet is fast,
the deflection in the galvanometer is large and when
the relative motion is slow, the galvanometer deflec-
tion is small.
6.3
Fig.6.5 Electromagnetic induction with a
stationary magnet and moving coil.
Faster the relativemotionbetween the magnet and
thecoil,greateristhe rate of change ofmagneticflux
linked with thecoiland largeristheinducedcurrent
setup intryecoil.
EXPERIMENT3. Induced emf by varying current in
the neighbouring coil. Fig. 6.6 shows two coilsPand 5
wound independently on a cylindrical support. The
coilP,calledprimary coil,is connected to a battery and a
rhstat through a tapping keyK.The coil 5, called
secondary coil,is connected to a sensitive galvanometer.
(i)When the tapping key is pressed, the galvano-
meter shows a momentary deflection in one
direction (Fig. 6.6). When the key is released, it
again shows a momentary deflection but in the
opposite direction.
Battery K Rh
Fig. 6.6Electromagnetic induction by varying
current in the neighbouring coil.
(ii)If the tapping key is kept pressed and steady
current flows through the primary coil, the
galvanometer does not show any deflection.
(iii)As the current in the primary coil is increased
with the help of the rhstat, the induced
current flows in the secondary in the same
direction as that at themake of the primary
circuit.
(iv)As the current in the primary coil is decreased,
the induced current flows in the same direction
as that at thebreakof the primary circuit.
(v)The deflections in the galvanometer become
larger if we use a cylindrical support made of
iron.www5~§tr€q•wvr5p§}

492. 6.4
Explan
ation. When a current flows through a coil,
a magnetic field gets associated with it. Asthe primary
circuit is closed, the current through it increases from
zero to a certain steady value
with the primary and hence with the secondary also
increases. This sets up an induced current in the
secondary coil in one direction. As the primary circuit is
broken, the current decreases from the steady value to
zero, the magnetic flux through the secondary coil
decreases. An induced current is set up in the
secondary coil but in the opposite direction. When a
steady current flows in the primary coil, the magnetic
flux linked with the primary coil does not change and
no current is induced in the secondary coil.
From these experiments, we may conclude that:
1.Whenever the magnetic flux linked with a closed
circuit changes, an induced emf and hence an
induced currentisset up in it.
2.Thehigher the rate of change of magnetic flux linked
with theclosed circuit, the greateris the induced emf
or current.
6.4 LAWS OF ELECTROMAGNETIC INDUCTION
4.State the laws of electromagnetic induction.
Express these laws mathematically.
Laws of electromagnetic induction.There are two
types of, laws which govern the phenomenon of
electromagnetic induction:
A.Faraday'S laws which give us the magnitude of
induced emf.
B.Lenz's law which gives us the direction of
induced emf.
A.Faraday's laws of electromagnetic induction:
Thesecan be stated as follows :
First law. Whenever themagnetic flux linked with a
closed circuit changes, an emf (and hence a current) is
induced in it whichlasts only solong as the change in flux is
taking place. This phenomenon is called electromagnetic
induction.
Second law. The magnitude of the induced emf is equal
to the rate of change of magnetic flux linked with the closed
circuit. Mathematically,
lel=d~
dt
B.Lenz's law: This law states that the direction of
induced currentissuch thatitopposes the cause which
produces it, i.e., it opposes the change in magnetic flux.
Mathematical form ofthe laws of electromagnetic
induction: Expression for induced emf. According to
the Faraday's flux rule,
PHYSICS-XII
Magnitude of induced emf
=Rate of change of magnetic flux
lei=d~
dt
Taking into account Lenz's rule for the direction of
induced emf, Faraday'S law takes the form:
e= _dljl
dt
or
The negative sign indicates that the direction of
induced emfis such that it opposes the change in
magnetic flux.
If the coil consists ofNtightly wound turns, then
the emfs developed in all these turns will be equal and
in the same direction and hence get added up. Total
induced emf will be
e= -Ndljl
dt
If the flux changes from41to ~ in timei,then the
average induced emf will be
e=-N~-41.
t
If ~ is in webers andtinseconds, thenewill be in
volts.
6.5EXPLANATION OFlENZ'S LAW
5.State and illustrateLenz's law.
Lenz's law.In 1833, German physicistHeinrich Lenz
gave a general law for determining the direction of
induced emf and hence that of induced current in a
circuit.
Lenz's Jaw statesthat the direction of induced current
in acircuitissuch thatitopposes thecause orthe
change which producesit.
Thus, if the magnetic flux linked with a closed
circuit increases, the induced current flows in such a
direction so as to create a magnetic flux in the opposite
direction of the original magnetic flux. If the magnetic
flux linked with the closed circuit decreases, the indu-
ced current flows in such a direction so as to create a
magnetic flux in the direction of the original flux.
Illustrations of Lenz's law:
(i)When the north pole of a bar magnet is moved
towards a closed coil, the induced current in the coil
flows in the anticlockwise direction, as sen from the
magnet side [Fig.6.7(a)]closely. The face of the coil
towards the magnet develops north polarity and thus,
it opposes the motion of the north pole of the magnet
towards the coil which is actually the cause of the
induced current in the coil.···3wx§n£m~r°n3lxv

493. ELECTROMAGNETIC INDUCTION
s
v
-
-----~
Fig. 6.7(a)Direction
of induced current when a
magnet moves towards a coil.
In other words, the motionof the magnetincreases
the flux through the coil. The induced current generates
flux in opposite direction, and hence opposes and
reduces this flux.
(ii)When the north poleof a magnet is taken away
from a closed coil, the induced current in the coil flows
clockwise, as sen from the magnet side
the coil towards the magnet develops south polarity
and attracts the north pole of the magnet,i.e.,the
motion of the magnet away from the coil is opposed
which is really the cause of the induced current
[Fig.6.7(b)].
N
vII
-
-----~
Fig. 6.7(b)Direction of induced current when a
magnet moves away from a coil.
In other words, the motion of the magnet decreases
the flux through the coil. The induced current gen-
rates flux in the same direction and hence opposes and
increases this flux.
6.Show that Lenz'slawisa consequence of the law of
conservation of energy.
Lenz's law and law of conservation of energy.
Whether a magnet is moved towards or away from a
closed coil, the induced current always opposes the
motion of the magnet, as predicted by Lenz's law. For
example, when the north pole of a magnet is brought
closer to a coil [Fig. 6.7(a)],its face towards the magnet
develops north polarity and thus repels north pole of
the magnet. Work has to be done in moving the
magnet closer to the coil against this force of repulsion.
Similarly, when the north pole of the magnet is moved
away from the coil [Fig.6.7(b)],its face towards the
magnet develops south polarity and thus attracts the
north pole of the magnet. Here work has to be done in
6.5
moving the magnet away from the coil against this
force of attraction. It is this work done against the force
of repulsion or attraction that appears as electric
energy in the form of induced current.
Suppose that the Lenz's law is not valid. Then the
induced current flows through the coil in a direction
opposite to one dictated by Lenz's law. The resulting
force on the magnet makes it move faster and faster,
i.e.,the magnet gains sped and hence kinetic energy
without expending an equivalent amount of energy.
This sets up a perpetual motion machine, violating the
law of conservation of energy. ThusLenz's lawisvalid
andisaconsequence of thelaw of conservation of energy.
Formulae Used
-> ->
1.Magnetic flux, <jl=BAcose=B.A
2. Induced emf, e= -N d<jl
dt
3.Average induced emf, e= -N «l2 -<1\
t
lei
4.Induced current,I= -.
R
Units Used
Magnetic fieldBis in tesla, flux <jlinWb, areaAin
m2,induced emfein volt., induced currentIin
amper.
Example 1. Arectangular loop of area20 emx30emis
placed in a magnetic field of0.3 Twith its plane(i)normal to
the field (ii) inclined 30°to the field and(iii)parallel to the
field. Find the flux linked with the coil in each case.
Solution.A=20 ernx30em=6x10-2m2,
B=0.3 T
Letebe the angle made by the fieldBwith the
normal to the plane of the coil.
(i)Here 9 = 90° - 90° = 0°
.. <jl=BAcos9=0.3x6x10-2xcosOo
=1.8xlO-2Wb.
(ii)Here 9 = 90° - 30° = 60°
<jl= 0.3x6x10-2 xcos 60°
=0.9xlO-2Wb.
(iii) Here 9 = 90°
4>= 0.3 x6x10-2 xcos 90°
=zero.www5~§tr€q•wvr5p§}

494. 6.6
Exa pie 2.A small piece of
metalwireisdragged across
the gap between the polepieces of a magnet in 0.5 s. The mag-
neticflux between the pole piecesisknown to be8x1O-4Wb.
Estimate the emf induced in the wire. [NCERT]
Solution. Heredt= 0.5 s,
d<jl= 8x10-4-0 = 8xlO-4Wb
The emf induced in the wire,
lE.I= d<jl= 8x10-4 = 1.6 x10-3V.
dt 0.5
Example 3.The magnetic flux tough a coil perpen-
dicular to the planeisvarying according to the relation:
<jI=(5t3+4t2+2t-5)Wb
Calculate the induced current tough the coil at t=2s,
if the resistance of the coilis5 O. [Punjab 98C]
Solution. The magnitude of induced emf set up at
any instant twillbe
lE.I= d<jl=~(5t3+4t2+2t-5)=15t2 +8t+2
dt dt
Att=2s,
IE.I= 15(2l+8 (2)+2 = 60+16+2 = 78V
Resistanceofthe coil, R= 5 0
Induced current,
I=L§J=78=15.6 A
R 5
Example 4.Asquare loop ofside10em and resistance of
0.700isplaced vertically in the east-west plane. Auniform
magnetic field of0.10Tisset up across the plane in the north-
eastdirection. The magnetic fieldisdecreased to zero in
0.70sat a steady rate. Determine the magnitudes of induced
emf and current during this time-interval. [NCERT]
Solution. Clearly, the angle made by the normal to
the plane (east-west plane) of the coil with the
magnetic field in north-east direction is
e= 45°
Maximum flux through the coil,
<lkax=BAcosB=0.1x(0.10x0.10) cos 45°
=1~:::.0.7x1O-3Wb
This flux is set up in 0.7 s. So the magnitude of
induced emf, .or
lE.I = d<jl = <lkax-0 = 0.7 x1O-3Wb
dt di 0.7s
=1O-3V =1 mY.
The magnitude of in~uced current,
1=ili=1mV=1.4mA
R 0.70
PHYSICS-XII
Example 5. A10O-resistance coil has1000turns and at a
time5.5x10-4Wb offlux passestough it.Ifthe flux falls
to0.5x10-4Wb in0.1second, find the emf generated in
voltsand the charge flown tough the coil incoulombs.
Solution. R = 100, N = 1000, <l1= 5.5 x10-4 Wb,
42=0.5 x10-4Wb,t=0.1 s
Induced emf,
E.=_N42-<l1
t
= -1000x0.5x10-4 - 5.5 x10-4 = 5 V
0.1
Current through 100 resistance coil is
1= ~ =~ =0.5 A
R10
:.The charge passing through the coil in 0.1 sis
q=It=0.5 x0.1 =0.05 C
Example 6.Acoilwith an average diameter of0.02mis
placed perpendicular to a magnetic field of6000 T(tesla). If
theinducedemfis11Vwhen the magnetic field ischanged
to1000 Tin4s,whatisthe number of turns in the coil ?
[CBSE F 94 C]
Solution. Radius of coil,r= 0.02 =0.01 m
2
~ =6000 T, ~ =1000 T,t =4s,E.=11V
IE.I= N42-<l1=NA ~ - ~
t t
=N.1t1. ~-~
t
11=N.22x(0.O1l6000 -1000
7 4
Now
Number of turns,
N= 11x7x4 =28.
22x(o.01l x5000
Example 7. Awire88emlongbent into a circular loop
isplaced perpendicular to the magnetic field offlux density
2.5 Wbm-2. Within0.5 s, the loopischanged into 22em
square and flux densityisincreased to3.0Wbm-2• Calculate
thevalue of the emf induced.
Solution. For circular loop, 21tr= 88 em
88x7
r=-- =14 cm =0.14 m
2x22
..Initial area of loop,
~ =1t1= 22x(0.14)2 =0.0616 m2
7
Final area of loop,
~ = (0.22)2 =0.0484 m2
~ =2.5 Wbm-2,~ =3.0 Wbm-2, t=0.5s···3wx§n£m~r°n3lxv

495. ELECTROMAGNETIC INDUCTION
In
duced emf,
e=-(~~~)=_(~~~~Al)
= _(3 x0.0484-2.5x0.0616)
0.5
= _(0.1452 -0.154) = 0.0088 =0.0176 V.
0.5· 0.5
Example8.Acoil of mean area500artandhaving 1000
turnsishe/d perpendicular to a uniform field of 0.4gauss.
The coilisturned tough180°in 1110second. Calculate the
average induced emf
Solution. Here A=500 cm2 = 500x1O-4m2,
N=1000, B=0.4G=0.4xl0-4T, t=I/10s
When the coil is held perpendicular to the field, the
normal to the plane of the coil makes an angle of 0°
with the field B.
:.Initial flux, ell.=BAcos 0° =BA
Final flux, ~ =BAcos 180° = -BA
Average induced emf,
e=-N( ~ ~~)=-N(-BAt - BA)
= 2NBA= 2x1000x0.4x10-4x500x 10-4 V
t 1/10
=0.04V.
Example9.Acircular coil of radius10em,500turns and
resistance2nisplaced with its plane perpendicular to the
horizontal component of the earth's magnetic field. Itis
rotated about itsvertical diametertough180°in0.25s.
Estimate the magnitudes of the emf and current induced in
the coil. Horizontal component of the earth's magnetic field
at the place is3.0x10-5T. [NCERT ;CBSE F 15]
Solution. Here A=1t(0.10)2 =1tX 1O-2m2,
R=2n, N=500, t=0.25s, B=3.0xl0-5T
Initial fluxthrough each turn of the coil,
~=BAcos 91 =3.0x10-5x(1tx10-2) cos 0°
=31tx10-7Wb
Final flux through eachturn ofthe coil,
~ =BAcos 92 = 3.0x 10-5 x (1tx 10-2) cos 180°
= -31t x1O-7Wb
Estimated value of emf in the coil,
e=-N~-~
t
-31tx10-7-31tx10-7
= - 500x--------
0.25
500x61tx10-7=3.8x10-3V
0.25
6.7
Current induced in the coil,
I=.§..=3.8 x 10-3 V
R 2n
=1.9x10-3A.
Example 10.A coil of cross-sectional area A lies in a
uniform magnetic fieldBwith its plane perpendicular to the
field.In this position the normal to the coil makes an angle of
0°with the field. The coil rotates at a uniform rate to
complete one rotation in time T.Find the average induced
emfin the coil during the interval when the coil rotates:
(i) from0°to90°position
(ii)from90°to180°position
(iii)from180°to270°and
(iv) from270°to360°.
Solution. (i)For rotation from0°to 90°
~ =BAcos 0° =BA, ~=BAcos 90° = 0,t=T /4
:.Average induced emf,
e_ ~ - ~_0 -BA _4BA
---t---T74-T'
(ii)For rotation from 90°to 180°
~ =BAcos900=0, ~=BAcosI800=-BA,t=T/4
.. e=--BA-0=4BA.
T/4 T
(iii)For rotation from180°to 270°
~ =BAcosl80° =-BA, ~=BAcos270° =0,t=T/4
e_0+BA _ 4BA
.. --T74--y'
(iv) For rotation from270°to 360°.
~ =BAcos 270° = 0, ~=BAcos 360° =BA,t=T /4
e_BA-O _ 4BA
.. --T74--T'
Asthe sense of the induced emf in the second half
rotation is opposite tothat in the first half rotation, the
induced current will change itsdirection after first half
rotation.
Example11.Aconducting circular loopisplaced in a
uniform transverse magnetic field of 0.02 T.Somehow, the
radius of the loop begins to decrease at a constant rate of
1.0 mmls. Find the emf induced in the loop at the instant
when the radiusis2em.
Solution.Suppose risthe radius of the loop at time
t.Then magnetic flux linked with the loop is
cj>=n?B
IeI=dcj>= 21trBdr
dt dtwww5~§tr€q•wvr5p§}

496. 6.8
• dr
Her- =1.0 mms =
1.0 x 10-3 ms-I
dt
Whenr=2.0 em =2.0 x10-2m,the magnitude of
induced emf is
Ie!= 2x3.14x2.0x10-2x0.02x1.0x10-3
=2.5x10-6V=2.51lV
~rOblems For Practice
1.Find the magnetic flux linked with a rectangular coil
of size 6 emx8em placed at right angle to a mag-
netic field of 0.5 Wbm -2• (Ans.2.4x10-3Wb)
2.A square coil of 600 turns, each side 20 em, isplaced
with its plane inclined at 30° to a uniform magnetic
field of 4.5 xlO-4Wbm -2.Find the flux through the
coil. (Ans.5.4x10-3Wb)
3.The magnetic flux threading a coil changes from
12x10-3Wb to 6x10-3Wb in 0.01 s. Calculate the
induced emf. (Ans.0.6 V)
4.A magnetic field of flux density 1.0 Wbm -2acts
normal to a 80 turn coil of 0.01 m2.Find the emf
induced init,if this coil is removed from the field in
0.1 s. [Haryana02]
(Ans.8V)
5.A 70 turn coil with average diameter of 0.02 m is
placed perpendicular to magnetic field of 9000T. If the
magnetic field is changed to 6000 T in 3s,what is
the magnitude of the induced emf?[CBSE F 94C]
(Ans. 2.2 V)
6.A magnetic field of flux density 10 T acts normal to
a 50 turn coil of 100 em2area. Find the emf induced
in itifthe coil is removed from the field in 1/20 s.
(Ans.100 V)
7.A coil has 1000 turns and500 em2as its area. It is
placed at right angles to a magnetic field of 2x10-5
Wb m-2.The coil is rotated through 180° in 0.2 s.
Find the average emf induced in the coil.
(Ans.10 mY)
8.A coil of area 0.04 m 2 having 1000 turns is
suspended perpendicular to a magnetic field of
5.0x10-5Wbm-2. It is rotated through 90° in 0.2 s.
Calculate the average emf induced in it.
(Ans.0.01 V)
9.A wire 40 em long bent into a rectangular loop
15 em x 5 em isplaced perpendicular to the
magnetic field whose flux density is 0.8 Wbm -2.
Within 1.0 second, the loop is changed into a 10 em
square and flux density increases to 1.4 Wbm -2.
Calculate the value of induced emf.
(Ans. - 0.008 V)
PHYSICS-XII
10.An air-cored solenoid of length50 em and area of
cross-section 28 em 2has 200 turns and carries a
current of 5.0 A. On switching off, the current
decreases to zero within a time interval of 1.0 ms.
Find the average emf induced across the ends of the
open switchin the circuit. (Ans. 1.4V)
11.A closed coil consists of 500 turns on a rectangular
frame of area 4.0 em 2 and has a resistance of 50 n.
It iskept with its plane perpendicular to a uniform
magnetic field of 0.2 Wb m -2.Calculate theamount
ofcharge flowing through the coil when it isturned
over(rotated through 180°). Will this answer
depend on the sped with which the coil is rotated?
(Ans.1.6x10-3C, 0)
12.The magnetic flux through a coil perpendicular to
its plane and directed into paper isvarying according
to the relation<I>=(5t2+lOt+5) milliweber. Cal-
culate the emfinduced in the loop att=5s.
(Ans.0.06V)
HINTS
1<I>=BAcos 0° = 0.5x0.06x0.08xI =2.4x10-3 Wh.
.2.<I>=NBAcose
=600x4.5x10-4 x0.20x0.20xcas (90° - 30°)
=5.4x10-3 Wh.
3.e=_~ -<Il.=_6x10-3-12x10-3 =0.6 V.
t 0.01
4.e= -NABz.- 1\=-80x0.01x0 - 1.0 =8 V.
t 0.1
5e=_N.nr2Bz.-1\
t
= _7022(001)2 6000-9000 _
xx.. -2.2 V.
7 3
6.e=-NABz.-1\
1
=50x100x10-4 x10-0=100V.
1/20
7.Proced as in Example 8 on page 6.7.
8.e= _N BAcos 90~ -BAcas 0° = _NBAx0 - 1
t t
= -1000x5.0x10-5x0.04x0-1 =0.01V.
0.2
9.Here1\=0.8 Wbm -2,
~=15x5em2=75x10-4m2,
Bz.= 1.4 Wbm-2,
Az= 10xlOcm2 = 100x10-4m2, t=Is
e=_Bz.A2-1\~
t
1.4x100x10-4-0.8x75x10-4
1 =-0.008V.···3wx§n£m~r°n3lxv

497. ELECTROMAGNETIC I
NDUCTION
10.Proced as in Exercise 6.15on page 6.54.
11.6cjl=BAcos1800 -BAcos00=-2BA
E.= _ Ndcjl=_NM=2NBA
dt 6t 6t
q=16t=§...6t=2NBA
R R
2x500x0.2x4.0x10-4
50
=1.6x10-3 C.
12.Here q>=("St2+10t+5)mWb
=(St2+ lOt + 5)x1O-3Wb
I E I =d<l>= ~(5t2+lOt+5)x10-3
di dt: .
=(lOt-+-10)x10-3V
:.Att=5s,IEI=(lOx5+1O)xlO-3=O.06V.
6.6 MOTIONAL EMF FROM FARADAY'S LAWS
7. What is motional emf?Deduce anexpression for
the emf induced across theendsofaconductor moving in
a perpendicular magnetic field.
Motional emf from Faraday'slaw: Induced emfby
change of area of the coil linked with the magnetic
field.Theemfinduced across the ends of a conductor dueto
itsmotion in a magneticfieldiscalledmotional emf As
shown in Fig. 6.8, consider a conductor PQ of lengthI
fre to move on U-shaped conducting rails situated in
a uniform and time independent magnetic field B,
directed normally into the plane of paper. The con-
ductor PQ is moved inwards with a sped v.As the
conductor slides towards left, the area of the rectan-
gular loopPQRSdecreases. This decreases the
magnetic flux linked with theclosed loop. Hence an
emf is set up across the ends of conductor PQ because
of which an induced current flows in the circuit along
the pathPQRS. The direction of induced current can by
determined by using Fleming's right hand rule, stated
in the next section.
5
)(
X XXXIll1X
I •
X X X X X X
Px
X
X X X X X X X X
v,. ,
)i X A-X )X X
y XXXX X X
Q
X
•• I
XXXX·X X X X X
I
I
1
R
~I'-------x------~'I
Fig. 6.8 Induced current bychanging area of
the rectangular loop.
6.9
Suppose a length xof the loop lies inside the
magnetic field at any instant of time t.Then the
magnetic flux linked with the rectangular loopPQRSis
cjl=BA=Blx
According to Faraday's law of electromagnetic
induction, theinduced emf is
E=- dcjl.:«(Blx)=-BIdx
dt dt dt
or E=Blv
wheredx/dt=-o,because the velocity vis in the
decreasing direction of x.The induced emfBlvis called
motional emfbecause thisemfisinduced due to the
motionofa conductor inamagnetic field.
6.7FLEMING'S RIGHT HAND RULE
8. State a ruleto determine the direction of current
induced duetothemotion of a conductor in a perpen-
dicularmagnetic field.
Fleming's right hand rul. This rule gives the
direction of induced current set up in a conductor
moving perpendicular to a magnetic field and can be
stated as follows:
If we stretch the thumb and thefirsttwo fingers of our
right handinmutuallyperpendiculardirections and if
the forefinger points inthe direction of the magnetic
field, thumbinthe direction of motion of the con-
ductor;then the central finger. points inthedirection
of currentinduced inthe conductor.
Motion
up
".Motion
~
Induced
current
Fig. 6.9Fleming's right hand rule
This rule is also called thedynamoorgenerator rule
and has wide practical applications.
6.8MOTIONAL EMF FROM LORENTZ FORCE
AND ENERGY CONSIDERATION
9. Deduce an expression for the motional emf by
considering theLorentz force acting on the free charge
carriers of aconductor moving in a perpendicular
magnetic field. Also deduce expressions for the induced
current, forcenecessary to pull the conductor, power
delivered by theexternal source, and power dissipated as
Joule loss. Hence discuss theenergyconseruation.www5~§tr€q•wvr5p§}

498. 6.10
Motional emf
from Lorentz forceA conductor has
a large number of fre electrons. When it moves
through a magnetic field, a Lorentz force acting on the
fre electrons can set up a current. Fig.6.10shows a
rectangular conductor in which arm PQ is fre to
move. It is placed in a uniform magnetic field B,
directed normally into the plane of paper. As the arm
PQ is moved towards left with a sped v,the fre
..-
'( ( )(
1 ••
)
"
x
P
A-
XXXXX
)cX
XX'< X
v•
X X X
XX X X
Q
X
•
I
X X XXX XX
I.-
s.
i,
I)(
Fig. 6.10 Motional emf.
electrons of PQ also move with the same sped
towards left. The electrons experience a magnetic
Lorentz force,Fm=qvB.According to Fleming's left
hand rule, this force acts in the direction QP and hence
the fre electrons will move towards P. A negative
charge accumulates at P and a positive charge at Q. An
electric fieldEis set up in the conductor from Q to P.
This field exerts a force,Fe=qEon the fre electrons.
The accumulation of charges at the two ends continues
till these two forces balance each other,i.e.,
Fm=Fe
or qvB= qEorvB=E
The potential difference betwen the ends Q and Pis
V=El=vBl
Clearly, it is the magnetic force on the moving fre
electrons that maintains the potential difference and
produces the emf,
s,Blv
As this emf is produced due to the motion of a
conductor, so it is called amotional emf
Current induced in theloop. Let R be the resis-
tance of the movable arm PQ of the rectangular loop
PQRSshown in Fig.6.10.Suppose the total resistance
of the remaining armsQR,RS and SP is negligible
compared to R. Then the current in the loop will be
I=~ =Blv
R R
Force on the movable arm.The conductor PQ of
length 1and carrying current I experiences forceFin the
perpendicular magnetic field B.
PHYSICS-XII
The force is given by
F=IlBsin900=(B~V)LB= W~2v
This force (due to induced current)actsin the
outward direction opposite to the velocity of the arm
in accordance with Lenz's law. Hence to move the arm
with a constant velocityv,it should be pulled with a
constant forceF.
Power delivered by the external forceThe power
supplied by the external force to maintain the motion
of the movable arm is
W12v2
P=Fv=--
R
Power dissipated as Joule loss. The power
dissipated in the loop as Joule heating loss is
P=I2R=(BIV)2 R=WI2v2.
J R R
Clearly, PJ=P.Thus, the mechanical energy
expended to maintain the motion of the movable arm
is first convertedinto electrical energy (the induced
emf) and then to thermal 5 I P
energy. This is consistent
with the law of conser-
vation of energy. This
fact allows us to repr-
sent the electromagnetic
set up of Fig.6.10by
the equivalent electrical
circuit shown in
Fig.6.11.
Fig.6.11Equivalent electrical
circuit of the electromagnetic set
up of Fig. 6.10.
6.9RELATION BETWEENINDUCED CHARGE
AND CHANGE IN MAGNETIC FLUX
10. Prove that the induced charge does not depend on
the rate of change of magnetic flux.
Relation betwen induced charge and change in
magnetic flux. According to Faraday'S law, the magni-
tude of induced emf, IeI=~<P
M
If R is the total resistance of theclosed loop, then
the induced current will be
I=~ or
R
~q~<P1
M=M·R
Hence the charge induced in the circuit in timeM,
~<PNet change in magnetic flux
~q=-=----"'--------'''----
R Resistance
Clearly, the induced charge depends on the net
change in the magnetic flux and not on the time inter-
valMof the flux change
not depend on the rate of change of magnetic flux.www4z~tq•p·uvq4o~y

499. ELECTROMAGNETIC IND
UCTION
Exam /es Based on
...
Formulae Used
1.The emf induced in a conductor of length I
moving with velocity vperpendicular to field B,
€=Blv.
2. Induced emf developed betwen the two ends of
rodrotating at its one end in perpendicular
magnetic field, €=1BI2ro
2
Units Used
FieldBis in tesla or Wbm -2,length Iin metre,
velocity vin ms-1,angular spedcoin rad s-1and
emf€in volt.
Example 12. An aircraftwith a wing span of40m flies with
a speed of 1080 kmh-1 in the eastward direction at a cons-
tant altitudein the northern hemisphere, where the vertical
component of earth's magnetic field is 1.75x10-5T.Find the
emf that develops between the tips of the wings.[NCERT]
Fig.6.12
Solution. The metallic part betwen the wing- tips
canbe treated as a single conductor cutting flux- lines
due to vertical component of earth's magnetic field. So
emf is induced betwen the tips of its wings.
Here1=40m,Bv= 1.75x10-5T
v= 1080 kmh-I = 1080x1000 ms-1
3600
=300 ms-I
€=Bv Iv= 1.75x10-5x40x300 = 0.21 V.
Example 13. A jet plane is travelling west at450ms-1. If
the horizontal component of earth's magnetic field at that
placeis4x10-4tesla and the angle of dip is30°,find the emf
induced between the ends of wings having a span of30m.
[CBSEOD 08]
Solution.Here1=30m,v= 450ms ", ~ =30°,
~ =4x10-4 T
6.11
The wings of the airplane cut the flux-lines of the
vertical component of earth's magnetic field, which is
given by
Bv=BHtan 8 = 4x10-4tan 30° = 4x10-4x0.577 T.
The emf induced across the tips of the wings is
€=Bvlv=4x10-4x0.577x30x450 = 3.12 V.
Example 14.Arailway track running north-south has two
parallel rails 1.0m apari. Calculate the value of induced em!
betweentherails, when a train passes at a speed of90kmh" .
Thehorizontal component of earth's magnetic field at that
place is0.3x10-4Wbm-2 and angle of dip is 60°.
[Haryana 01)
Solution. Here I= 1.0 m, BH= 0.3x1O-4Wb m -2,
8=60°
Bv= ~ tan ~ = 0.3 x10-4tan 60°
= 0.3x10-4x1.732 = 0.52 x10-4 Wbm -2.
v= 90 kmh-I =90 x 1000=25 ms-I.
3600
€=Bv Iv=0.52x10-4 x1.0x25 =1.3x10-3V.
Example 15. A conductor of length 1.0 mfallsfreely under
gravity from a height of 10msothatitcutsthelinesofforce
of the horizontal component of earth's magnetic field of
3x10-5Wbm-2. Find the emf induced in the conductor.
Solution. The velocity vattained by the conductor
as it falls through a height of 10 m is given by
v2=u2+2gs=0+2x9.8x10=4x49
v=2x7=14 ms-I
Induced emf,
€= ~Iv=3x10-5x1.0x14 =4.2x10-4V.
Example 16. Twelve wires of equal lengths (each 10 em)
are connected intheform of a skeleton-cube. (i) If the cubeis
moving with a velocity of5ms-1 in the direction of a
magnetic field of0.05Wbm-2, find the emfinduced in each
arm of the cube.(ii)If the cube moves perpendicular to the
field,what will be the induced emf in each arm?
Solution. For the generation of motional emf :B,
Iandvmust be in mutually perpendicular directions.
If any two of these quantities are parallel, emf is not
induced.
v
Af--+--::-(
v
I)--+--}G
F
--+-
B
(a)
F
--+-
B
(b)
Fig. 6.13www5~§tr€q•wvr5p§}

500. 6.12
(i) In Fig.6.13(a),the velocity
of any conductor is
parallel tothe field B, so no emf is induced in any
conductor.
(ii)InFig.6.13(b),the armsAE, BF, CG and DH are
parallelto the velocity v,no emf is induced in these
arms.AJso,the arms AB, DC, EFand HG are parallel to
the field B, so no emf is induced in these arms.
The armsAD, BC, EHandFGare perpendicular to
bothBand v.Hence emf is induced in each of these
arms and is givenby
.E=Blv=0.05xlOx10-2x5 =2.Sx10-2V.
Example 17. Fig.6.14shows aconducting rodPQin
contact with metal railsRPandSQwhich are25em apart in
a uniform magnetic field of flux density0.4T acting per-
pendicular to the plane of thepaper. EndsRandSare con-
nected tough a50resistance. Whatisthe emf when the
rod moves to the right with a velocity of5ms-1?Whatisthe
magnitude and direction of the current tough5 0resis-
tance?If the rod moves to the left with the same speed, what
will be the new current and its direction? [rSCE 95]
x x
r-----+--
xX X X p
TR
XXX XX
S
~X 50XXX
N
1\
X X X X
XXXXQ
Fig.6.14
X
-1
v=5ms
X X
X X
Solution.Here B =0.4 T, v= 5ms ",
I=25 em =0.25 m
Induced emf, E=Blv= 0.4x0.25x5 =0.5 V
Current, I=!= 0.5V=0.1 A
R50
Applying Fleming right hand rule, the induced
current flows from Q to P,i.e., from the endRtoS
through the 50resistanc.
If the rod moves to the left with the same sped,
then the current of O.lA will flow through 50
resistance fromthe endStoR.
Example 18. A metallic rod of lengthLis rotated at an
angular speedronormal to a uniform magnetic fieldR
Derive expressions for the (i) emf induced in the rod
(ii) current induced and (iii)heat dissipation, if the
resistance ofthe rod isR. [CBSE D 08]
Solution. Suppose the rod completes one revo-
lution in timeT.Then change in flux
= Bx Area swept=Bx1t13
Ind d f Change in flux
uce em =--=----
Time
PHYSICS-XII
Fig. 6.15
xX X X
X
X
X
X
X
XX
E=Bx1tJ3=B1tJ3f
T
[.: T=7]
X
X
X
X
XX X X X
or
ro
Asf=- ,therefore
21t
~ 2co1 2
c,=B1t1:.. -=-B1:.r.u
21t 2
Induced current,
I=!=!BJ3ro
R 2 R
Heat dissipation in time t,
E2t 1 B2L4ro2t
Q=R='4 R .
Example 19. A metal disc of radius Rrotates with an
angular velocity coaboutan axis perpendicular to its plane
passing tough its centre in a magnetic fieldBacting
perpendicular to the plane of thedisc. Calculate the induced
emf between the rim and the axis of the disc.
Solution. Consider a disc of radius Rrotating in a
transverse magnetic field B with frequency f.In time
periodT,the disc completes one revolution.
.'.Change in flux = Bx Areaswept = Bx 1t R2
I d d f change in flux
n uce em =----"'-----
Time
E= B ~R2= B1tR2f[.:7=T]
ro
Asf=-
21t'
Example 20. A wheel with 10metallicspokes each 0.5m
longisrotated with a speed of120rev / minin a plane
normal to the horizontal component of earth's magnetic field
~ at a place. If ~=0.4Gat theplace,whatis the induced
emf between the axle and the rim ofthewheel ?Note that
1G= 10-4 T. [NCERT]
Solution. Here L=0.50 m, B = 0.40 G= 0.40x10-4T
f=120 rev = 120 rev =2 rps
min 60 se
Induced emf,
E= B 1t13f=0.40x10-4 x3.14x (0.50lx2
=6.28x10-5V.···3wx§n£m~r°n3lxv

501. ELECTROMAGNETIC INDUCTION
As all the ten spokes are connected with their one
end at the axle and the other end at the rim, so they are
connected in parallel and hence emf across each spoke
is same .
Example21.When a
wheel with metal spokes 1.0 m long
rotates in a magnetic field offlux density2xlO-4Tnormal
to the plane of the wheel, an emf of nx 10-2 Visinduced
between the rim and the axle of the wheel. Find the rate of
revolution of the wheel.
Solution. HereL= 1.0 m,B= 2x10-4T,
e=nx10-2V,f=?
As e=Bnr3f
e n x10-2
f=-- = = 50 rps.
BnL22x10-4 x rtx(1.0)2
Example 22.Ametallic rod of1 mlengthisrotated with a
frequency of50rev /s,with one end hinged at the centre
and the other end at the circumference of acircular metallic
ring of radius1m,about an axis passing tough the centre
and perpendicular to the plane of the ring.Aconstant and
uniform magnetic field of1Tand parallel to the axis is
present everywhere. Whatisthe emf between the centre and
the metallic ring? [Fig.6.15] [NCERT]
Solution. HereL= 1 m,f= 50 rps, B= 1.0 Wbm-2
.. e=Bnr3f
= 1.0 x3.14x(1)2x50 = 157V.
Example23.Acircular copper disc10cm in radius rotates
at 20n rad/s about an axis tough its centre and
perpendicular to the disc.Auniform magnetic field of0.2 T
acts perpendicular to the disc.(i)Calculate the potential
difference developed between the axis of the disc and the rim.
(ii) Whatisthe induced current, if the resistance of the disc
is2ohm? [CBSE OD 01]
Solution. Here R =10 em =0.10m,m=20nrad s-l,
B=0.2T
(i) P.D. developed betwen the axis and the rim
e=.!.BR2co=.!.x0.2x(0.10)220n
2 2
=0.0628V
(ii)Induced current,
I= ~ = 0.0628 = 0.0314 A.
R 2
Example24.A0.5m long metal rodPQcompletes the
circuit as shown in Fig.6.16.The area of the circuitis
perpendicular to the magnetic field offlux density0.15T.If
the resistance of the total circuitis3Q,calculate the force
needed to move the rodinthe direction as indicated with a
constant speed of2rns-1. [CBSE OD06]
6.13
x x x
xQ
x
x x x x x x x
>
x>x
>
xI---~vx x
x x x x x x x
x X·x x
p
Fig. 6.16
Solution. Heree=BlvandI= ~ =Blv
R R
F=liBsin900=Blv.lB=W/2v
R R
ButI=O.5rn, B=0.15T, R=3Q, .v=2ms-1
F= (0.15)2x(0.5)2x2 = 0.00375N.
3
<problems For Practice
1.A straight conductor1metre long moves at right
angles to both, its length and a uniform magnetic
field. If the sped of the conductor is 2.0 ms-1
and the strength of the magnetic field is 104 gauss,
find the value of induced emf in volt. [Punjab 96)
(Ans.2V)
2.If a 10m long metallicbar moves in a direction at right
angle to a magnetic field with a sped of 5.0 ms-1,
25Vemf is induced in it. Find the value of the
magnetic field intensity. [Punjab 99](Ans. 0.5 T)
3.A horizontal telephone wire 1 km long is lying
east-west in earth's magnetic field. It falls frely to
the ground from a height of 10m. Calculate the emf
induced in the wire on striking the ground. Given
BH= 0.32G. (Ans.0.448V)
4.A horizontal wire 24 ern long falls in the field of flux
density 0.8 T. Calculate the emf induced in it at the
end of 3s,after it was dropped from rest. Suppose
the wire moves perpendicular to its length as well
as to magnetic field.Takeg= 9.8 ms-2. (Ans.5.6V)
5.The two rails of a railway track insulated from each
other and the ground are connected to a milli-
voltmeter. What is the reading of the voltmeter
when a train travels at a sped of 180 kmh-1along
the track, given that the vertical component of the
earth's magnetic field is 0.2x10-4Wb m -2and the
rails are separated by 1 m? [lIT]
(Ans.1 mY)
6.A jet plane is moving at a sped of 1000 km h-1.
What is the potential difference across the ends of
its wings 20 m long. Given total intensity of earth's
magnetic field is 3.5 x10-4 tesla and angle of dip at
the place is 30°. (Ans. 0.97 volt)www5~§tr€q•wvr5p§}

502. 6.14
7.A straight rod 2 m long is placed in an aeroplane in
the east-west direction. The aeroplane lifts itself in
the upward direction at a sped of 36 km h-1.Find
the potential
difference betwen the two ends of the
rod if the vertical component of earth's magnetic
field is ~ gauss and angle of dip = 30°.
(Ans.5x10-4V)
8.Figure shows a rectangular conducting loopPQRS
in which the arm PQ is fre to move
magnetic field acts in the direction perpendicular to
the plane of the loop. Arm PQ is moved with a
velocityvtowards the arm R5. Assuming that the
armsQR,RSand SPhave negligible resistances and
the moving arm PQ
has the resistance R,
obtain the expression
for(i)the current in the
loop(ii)the force and
(iii)the power required
to move the armPQ.
[CBSE D 131
9.A rectangular loopPQMN with movable arm PQ of
.length 10emand resistance 2nis placed in a uniform
magnetic field of 0.1 T acting perpendicular to the
plane of the loop as is shown in the figure. The
x xx x x px
x x x x
x xv- X
x x
x x
x~xxxx
x x x x x QX
Fig. 6.17
XNX XXpXx 4.
x x x x x x
xxx x v 5.
xxxx x x
Fig. 6.18
XMX xxQx x
resistances of the arms MN, NPand MQ are
negligible. Calculate the(i)emf induced in the arm
PQ and(ii)current induced in the loop when arm
PQ is moved with velocity 20 ms-1. [CBSE D 14CI
(Ans.0.2V,0.1 A)
10.A whel with 8 metallic spokes each 50 ern long is
rotated with a sped of 120 rev/min in a plane
normal to the horizontal component of the Earth's
magnetic field. The Earth's magnetic field at the
place is 0.4 Gand the angle of dip is 60°. Calculate
the emf induced betwen the axle and the rim of the
whel. How will the value of emf be affected if the
number of spokes were increased? [CBSE OD 131
(Ans.3.14x10-5V)
11.A fan blade of length2arotates with frequencyf
cycles per second perpendicular to a magnetic field
B.Find the p.d. betwen the centre and the end of
the blade (Ans.-1tBa2f )
12.In a ceiling fan, each blade rotates in a circle of
radius 0.5 m. If the fan makes 20 revolutions per
PHYSICS-XII
second and if the vertical component of earth's field
is 8x10-5 Wb m-2,calculate the p.d. developed
betwen the ends of each bladeAns. 0.001 T)
13.A metal disc of radius 200emis rotated at a cons-
tant angular sped of 60 rad s-1 in a plane at right
angles to an external field of magnetic induction
0.05 Wbm -2.Find the emf induced betwen the
centre and a point on the rim. [Punjab 911
(Ans. 6V)
14.A copper disc of radius 10emplaced with its plane
normal to a uniform magnetic field completes 1200
rotations per minute. If induced emf betwen the
centre and the edge of the disc is 6.284 m V,find the
intensity of the magnetic field. Take1t= 3.142
(Ans. 10-21')
HINTS
1.Here 1= 1m,v= 2.0ms",B = 104 G= 1 T
. .E=Blv= 1x1x2 =2V.
E 25
2.B= - = -- '=0.5 T.
Iv10x5
3.Herev2=if+2gs= 0+2x9.8x10 = 4x49
or v=2x7=14ms-1
E= BHIv= 0.32x10-4 x1x103x14
=0.448V.
Herev=u+at=0+9.8x3 = 29.4 ms-1
.. E=Blu= 0.8x0.24x29.4 =5.6 V.
HereBv= 0.2 x10-4Wbm-2,I= 1 m
v=180kmh-1 = 180x1000 ms-1 =50ms-1
3600
:.Induced emf,
E=Bvlv= 0.2x10-4 x1x50
=1O-3V =1 mY.
Thus the reading of the milli-voltrneter is 1 mY.
6.Proced as in Exercise 6.10 on page 6.52.
7.Here1=2m, v=36kmh-1 =lOms-1, 8=30°
1 10-4
Bv=WG= 4.J3T
B =~ = 10-4 10-4 =.!x10-4T
Htan 84.J3tan 30° •t:1 4
'hI3 ..J3
.. E=BHlv=.!x10-4x2x10=5x10-4 V.
4
8.(i)I=§.=Blv
R R
(ii)F= llBsin 900=(B~V) lB=B2~2v
B2122
(iii)P=Fv= __ v .
R···3wx§n£m~r°n3lxv

503. ELECTROMAGNETIC INDUCTION
9.Here
B=0.1 T,1=lOcm=0.
10 m,v=20ms-1, R=2n
(i)e=Blv=0.1x0.10x20=0.2 V
(ii)I= ~ =0.2=0.1A
R 2
10.HereI=50 em=0.50 m,
f=120 rev=120 rev=2rps
min 60 se
BH=Bcos0=0.4 cas 60° =0.2G=0.2x10-4T
Induced emf,
e=BHnl2f=0.2x10-4x3.14x(0.50)2x2
=3.14x10-5V.
11.e= _d~=_!£(BA)= _BdA
dt dt dt
In time period T, the area swept by the blades isn a2.
. na2 na2 2
.. e= -B. -= -B. -= -rtBaf
T 1/f
12.Usee=Bxnr2xf.
~ 1 21 2
13.c=-Brro=-x 0.05 x 2x60=6 V.
2 2
14.Herer=1Dcm=0.10 m,f=1200/ 60=20s-\
e=6.284x10-3V=B xnr2xf
e 6.284x10-3 2
B=--= =10- T.
nr2f3.142x(0.10)2x20
6.10METHODS OF GENERATING
INDUCED EMF
11. Discuss the various methods of generating
induced emf
Methods of generating induced emf. An induced
emf can be produced by changing the magnetic flux
linked with a circuit. The magnetic flux,
~=BAcos 9
can be changed by one of the following methods :
1.Changing the magnetic field B,
2. Changing the areaAof the coil, and
3. Changing the relative orientation 9 of BandA.
1. Induced emf by changing the magnetic field B.
We have already learnt in section 6.3 how an induced
emf is set up in a coil on changing the magnetic flux
through it by(i)moving a magnet towards a stationary
coil,(ii)moving a coil towards a stationary magnet and
(iii)varying current in·the neighbouring coil.
2. Induced emf by changing the area of the coil.
Consider a conductor CD of length I moving with a
velocityvtowards right on U-shaped conducting rails
6.15
situated in a magnetic field B, as shown in Fig 6.19. The
field is uniform and points into the plane of the paper.
As the conductor slides, the area of the circuit changes
fromABCDtoABC [Yin timedt.
io-vdt..j
A xxxDxxD'
xxxx x x x
x xx xx
)(x x y--v x
x x x x xx
x x x xx
xx vx r
c
x
x
x
x
x
x
Cx
Fig.6.19Induced emf by changing area of the loop.
The increase in flux,
d<l>=Bxchange in area
=BxareaCOD'C'=B.I. vdt
This sets up induced emf in the loop of magnitude,
Ie1=d<l>=Blv
dt
According to Fleming's right hand rule, the
induced current flows in the anticlockwise direction.
3. Induced emf by changing relative orientation
of the coil and the magnetic field : Thry ofAC
generator. Consider a coilPQRSfre to rotate in a
---+
uniform magnetic field B. The axis of rotation of the
---+
coil is perpendicular to the field B. The flux through
the coil, when its normal makes an angle 9 with the
field, is given by
<l>=BAcos 9, whereAis the face area of the coil.
Axis of
Rotation
5
Fig. 6.20(a)Rotating coil in a magnetic field.
If the coil rotates with an angular velocity wand
turns through an angle 9 in timei,then
9=wt <l>=BAcoswt
As the coil rotates, the magnetic flux linked with it
changes. An induced emf is set up in the coil which is
given by
e= -d<l>= - ~(BAcoswt)=BAwsinwt
dt dt
If the coil hasNturns, then the total induced emf
will be
e=NBAcosinwtwww5~§tr€q•wvr5p§}

504. 6.16
Thus the induced emf varies sinusoidally with
timet.The value of
induced emf is maximum when
sinrot=1orrot= 90° ,i.e.,when the plane of the coil is
~
parallelto the field B. Denoting this maximum value by
Eo,we have
Eo=NBAro
E=Eosinrot=Eosin 2 rtIt
wherefis the frequency of rotation of the coil.
Uniform magnetic
field
,
R=N
,
5
P
,
5,
S~
(1)!(2) (3) (4) (5)
"U
QJW
u•.•..
..gEO
21t..sQJ
Fig. 6.20(b)Induced emf in a rotating coil.
Figure6.20(b)shows how the induced emf E
betwen the two terminals of the coil varies with time
We consider the following special cases:
1.Whenrot= 0°, the plane of the coil is
~
perpendicular to B,
.sinrot= sin 0° =0, sothatE=O
2. Whenrot=~, the plane of the coil is parallel to
2 .
fieldB,
sinrot= sin ~ = 1, so thatE= Eo
2
3. Whenrot= 1t, the plane of the coil
perpendicular toa
sinrot= sin rt =0, so that E =0.
3rt
4. Whenrot= - the plane of the coil
2
----
isagain
----
isagain
~
parallel to B, II
. . 31t 1 he e'
smrot= sm-=-, so t at ~ = - ~o
2
5. Wheprot= 27t, the plane of the coil again b-
~
comes perpendicular to B after completing one rotation,
sinrot=sin2rt =0, so thatE=0
As the coil continues to rotate in the same sense, the
same cycle of changes repeats again and again. As
PHYSICS-XII
shown in Fig.6.18(b),the graph betwen emf E and
timetis a sine curve sinusoidalor
alternating emfBoth the magnitude and direction of
this emf change regularly with time
The fact that an induced emf is set up in a coil when
rotated in a magnetic field forms the basic principle of
a dynamo or a generator.
For Your Knowledge
>The magnetic flux linked with a surface is maximum
when it is held perpendicular to the direction of the mag-
netic field and the flux linked is zero when the surface
is held parallel to the direction of the magnetic field.
>Induced emf is set up whenever the magnetic flux
linked with a circuit changes even if the circuit is open.
However, the induced current flows only when the
circuit is closed.
>No emf is induced when a coil and a magnet move
with.the ,same velocity in the same direction.
>No emf is induced when a magnet is rotated about its
own axis. However, emf is induced when a magnet is
rotated about an axis perpendicular to its length.
>No emf is induced when a closed loop moves totally
inside a uniform magnetic field.
>Just as a changing magnetic field produces an electric
field,a changing electric field also sets up a magnetic field.
>The electric fields created by stationary charges have
vanishing and path independent loop integrals. Such
fields are calledconservative fields.
ff.dt=·O
>The electric fields created by tim-varying magnetic
fields have non-vanishing loop integrals and are called
non-conservative fields. Their loop integrals are path
dependent.
!E.dt=dcp
:r dt
Electric potential is meaningful only for electric fields
produced by stationary charges. It has no meaning for
electric fields set up by magnetic induction.
The heart beating induces a.c. in the' surrounding
tissues. The detection and study of these currents is
calledelectrocardiographywhich provides valuable
information regarding the pathology of the heart.
>Migration of birds. Every winter birds from Siberia
fly unerringly to water spots in the Indian
subcontinent. It is believed that migratory birds make
use of earth's magnetic field to determine their
direction. As birds contain no ferromagnetic material,
so electromagnetic induction appears to be the only
mechanism to determine direction. However, very
small emfs induced across the bodies of these birds
create a doubt about the validity of this hypothesis. So
the migration pattern of birds is still a mystery ..···3wx§n£m~r°n3lxv

505. ELECTROMAGNETIC INDUCTION
Formulae
Used
1.e=sinwt
2.=NBAw,whereco=2rrf
e
3.Maximum induced current, 10= ~
R
Units Used
The induced emf eand maximum induced emf
are in volt, fieldBin tesla, areaAin m2, angular
frequencycoinrads ",current10in ampere,
resistanceRin ohm.
Example 25. A circular coil of area300crriand25turns
rotates about its vertical diameter with an angular speed of
40s-1in a uniform horizontal magnetic field of magnitude
0.05T.Obtain the maximum voltage induced in the coil.
rCERT]
Solution.HereA= 300em2=300x10--4m2,N=25,
ro=40s-l,B=0.05T
The maximum voltage induced in the coil is
eo=NBAw= 25 x 0.05 x300x10-4 x40=1.5v.
Example 26. A flat coil of 500 turns, each of area
5 x 10-3rrt-,rotates in a uniform magnetic field of0.14Tat
an angular speed of150rad s-l. The coil has a resistance of
5n.The induced emfisapplied to an external resistance of
10n.Calculate the peak current tough the resistance.
Solution. Here A=5x10-3m2, B=0.14T,
ro=150rad s-l
Total resistance, R= 5+10=15n
eo=NBAro
= 500x0.14x5 x 10-3 x 150 = 52.5 V
52.5
Peak current, 10=R=15=3.5 A
Example 27. An athlete peddles a stationary tricycle
whose pedals are attached to a coil having 100 turns each of
area 0.1nil.The coil, lying in the X-Y planeisrotated, in
this plane, at the rate of50rpm, about the Y-axis, in a region
-+ "
where a uniform magnetic field, B=(0.01) ktesla,ispresent.
Find the (i) maximum emftii)average emf, generated in the
coil over one complete revolution. [CBSE Sample Paper 13]
Solution. Here N= 100,A= 0.1 m2,
. 50 5
f=50rpm= -rps= -rps
60 6
-+ "
B = (0.01)ktesla~B=0.01T
6.17
(i)Maximum emf generated in the coil,
=NBAro= NBAx2 rrf
5 1t
= 100xO.01xO.1x2 x1tX- =- V:::.0.52 V.
6 6
(ii) As the generated emf varies sinusoidally with
time, so average emf generated in the coil over one
complete cycle=o.
Example 28.A rectangular coil of wire has dimensions
0.2mx0.1mThe coil has 2000 turns. The coil rotates in a
magnetic field about an axis parallel to its length and
perpendicular to the magnetic field of0.02Wb m-2. The
speed of rotation of the coilis4200 rpm. Calculate (i) the
maximum value of the induced emf in the coil (ii) the
instantaneous value of induced emf when the plane of the coil
has rotated tough an angle of30°from the initial position.
Solution. HereA=0.2x0.1m2=2x10-2m2,
N=2000, B=;,0.02Wbm-2
4200
f= 4200rpm= --rps= 70rps
60
ro=21tf=21tx 70 =140 1trad s-l
(i)The maximum value of the induced emf in the
coil is
=NBAw
= 2000x0.02x2x10-2 x140x2:; V=3520V.
(ii) The instantanus value of induced emf when
the coil has turned through 30°is
e=sinwt
=sin30°=3520 x ~V=1760 V.
Example 29.A rectangular coil of 200 turns of wire,
15cmx40em makes50revolutions/second about an axis
perpendicular to the magnetic field of 0.08 weber /rrt-.What
isthe instantaneous value of induced emf when the plane of
the coil makes an angle with the magnetic lines of (i)0°(ii)
60° and(iii)90°?
Solution. Here N=200,
A= 15emx40em=15x40 x 10--4 IP2=6x10-2m2
B=0.08Wb m-2, W=21tx50=100 1trad s-l
Induced emf at any instant is given by
e=sinrot=NBAwsinrot
Ifthe plane of the coil makes an angle awith the
magnetic lines of force, then the angle betwen the normal
tothe plane ofthe coil and the magnetic field will be
rot=90° -a
(i)Whena=0°, rot =90° _0° =90°
e=NBArosin90°
=200 x 0.08 x 6 x 10-2 x100 1tx1 V =301.6 V.www5~§tr€q•wvr5p§}

506. 6.18
(ii)When (l=60°
,rot=90° -60° =30°
e=NBArosin30°=301.6 x~=150.8 V.
(iii)When (l=90°, rot=90° -90° =0°
e=NBArosin0°=O.
~rOblems For Practice
1.A closelywound rectangular coilof200turns and size
0.3mx0.1m is rotating in a magnetic field of induc-
tion0.005Wb m-2with a frequency of revolution
1800rpm about an axis normal to the field. Calculate
the maximum value of induced emf. (Ans.5.65V)
2.A rectangular coil of dimensions0.1mx0.5m
consisting of2000turns rotates about an axis
parallel to its longer side, making2100revolutions
per minute in a field of0.1T.What is the maximum
emf induced in the coil?Also find the instan-
tanus emf, when the coil is60°to the field.
(Ans, 2200 V, 1100 V)
3.The armature coil of a generator has20turns and its
area is0.127m2.How fast should it be rotated in a
magnetic field of0.2Wbm-2,so that the peak value
of induced emf is160 V ? (Ans. 50rps)
4.A50turn coil of area500em2is rotating at a rate of
50rounds per second perpendicular to a magnetic
field of0.5Wbm-2.Calculate the maximum value
of induced emf. (Ans.392.5 V)
5.Calculate the maximum emf induced in a coil of100
turns and0.01m2area rotating at the rate of50rps
about an axis perpendicular to a uniform magnetic
field of0.05T.If the resistance of the coil is300,
what is the maximum power generated by it?
(Ans. 15.7 V, 8.23 W)
HINTS
1.=NBAx21tf
1800
=200 x 0.005 x 0.3x0.1x2x3.14x60=5.65V.
2.=NBAx21tf
22 2100
=2000 x 0.1 x 0.1 x 0.5 x2x7"x60=2200 V.
Hererot=90°- 60°=30°
..e=sin30°=2200xsin30°=1100V.
3.Frequency,f
e 160
_-,0,-- =50rps.
21tNBA 2x3.14x20x0.2x0.127
4.Use=NBAx21tf.
5.=NBAro=100x0.05x0.01x21tx50=15.7 V
[.: to=21tf1
Maximum current~ 1o=15.7=0.524A
30
Power generat~d =eol0=15.7x0.524=8.23 W.
PHYSICS-XII
6.11EDDY CURRENTS
12. What are eddy currents? Give some experiments
to demonstrate theirexistence. Whatistheeffect of eddy
currents in electrical appliances, where iron core isused?
How are eddy currents minimised?Describe some of the
important applications of eddy currents.
Eddy currents.Currents can be induced, not only in
conducting coils, but also in conducting shets or
blocks. Whenever the magnetic flux linked with a metal
shet or block changes, an emf is induced in it. The
induced currents flow in closed paths in planes perpen-
dicular to the lines of force throughout the body of the
metal. These currents look likeeddiesorwhirl- poolsin
water and so they are known aseddy currents.As these
currents were first discovered byFocaultin1895, so
eddy currents are also known asFocault currents.
Eddy currents are the currents inducedinsolid
metallic masses when the magnetic flux teading
tough them changes.
Eddy currents also oppose the change in magnetic
flux, so their direction is given by Lenz's law.
Experiments to demonstrate eddy currents:
Experiment1.Take a pendulum having its bob in
the form of a flat copper plate. 6.21(a),
it is fre to oscillate betwen the pole pieces of an
electromagnet. Inthe absence of any magnetic field,
the pendulum swings frely. As the electromagnet is
switched on, the oscillations of the pendulum get
highly damped and soon it comes to rest. This is
because as the copper plate moves in betwen the pole
pieces of the magnet, magnetic flux threading through
it changes. So eddy currents are set up in it which
according to Lenz's law, oppose the motion of the
copper plate in the magnetic field. Eddy currents flow
anti clockwise as the plate swings into the field and
clockwise as the plate swings out of the field.
Copper
plate
'---::::::::;;;;"Ic::::::-:-:-~..--- Eddy
currents
F· 6 21(a)Eddy currents damp the oscillations of19. .
a copper plateina magnetic field.···3wx§n£m~r°n3lxv

507. ELECTROMAGNETIC INDU
CTION
Experiment 2. Now take the pendulum of a flat
copper platewith narrow slots cut across it, as shown
.in Fig6.21(b). Asthe electromagnet is switched on,
eddy currents are set up in the plate. But this plate
swings for longer duration than the plate without slots.
This is because the loop has much larger paths for the
electrons to travel. Larger paths offer more resistance
to electrons andso the eddy currents are sufficiently
reduced. As a result, the opposition to the oscillations
becomes very small.
Fig. 6.21(b)Reduced eddy currentsin a slotted copper plate
Experiment 3. Take a cylindrical electromagnet fed
by an A.c. sourceand place a small metal disc over its
top.Asthe current is switched on, the magnetic field at
the disc rises from zero to a finite value, setting up
eddy currents which effectively convert it intoasmall
magnet. If initially the top end of the electromagnet
acquiresN-polarity, then by Lenz's law,the lower face
of the small magnetic disc will also have N-polarity,
resulting in a repulsive force. The disc is thus sen to be
thrown up as the current in the electromagnet is
switched on.
Light __©
metallic disc
Coil
Fig. 6.22A light metal disc on top of an electro-magnet is
thrown up as the current is switched on.
Undesirable effects Ofeddy currents.Eddy currents
are produced inside the iron cores of the rotating
armaturesofelectric motors and dynamos, and also in
the cores of transformers, which experience flux
6.19
changes, when they are in use. Eddy currents cause
unnecessary heating and wastage of power. The heat
produced by eddy currents may even damage the
insulation of coils.
Minimisation. The eddy currents can be reduced by
usinglaminated corewhich instead of asingle solid
mass consists of thin shets of metal, insulated from
each other by athin layer of varnish, asshown in
Fig. 6.23. The planes of theshets are placed
perpendicular tothe direction ofthe currents that
would beset up by the emfinduced in the material.
The insulation betwen the shets then offers high
resistance to the induced emf and the eddy currents
aresubstantially reduced.
Insulating
layers
(a) (b)
Fig. 6.23(a)Solid core,(b)Laminated core.
Applications of eddy currents. Although eddy
currents areundesirable, still they findapplications in
thefollowing devices:
1. Induction furnaces placed
in a rapidly changing magnetic field (produced by
high frequency a.c.), very large eddy currents are set
up. The heat produced is sufficient to even melt the
metal. Thisprocess is used in the extraction of some
metals from their ores.
2. Electromagnetic damping. When a current is
passed through.1galvanometer, itscoil suffers f
oscillations before coming to rest in the final position. As
the coil moves in the magnetic field, induced current is
set in the coil which opposes its motion. The oscillations
of the coil are damped. This is calledelectromagnetic
damping. The electromagnetic damping can be further
increased by winding the coil on a light copper or
aluminium frame. As the frame moves in the magnetic
field, eddy currents are set up in thefrarnewhich resist
the motion of the coil. This is how a galvanometer is
rendereddead beat, i.e., the coil does notoscillate - it
deflects and stays in the final position immediately.
3. Electric brakes. Astrong magnetic fieldis
applied to the rotating drum attached to the whel.
Eddy currents set up in the drum exert atorque on the
drum so as to stop the train.
4. Spedometers. In a spedometer, a magnet
rotates with thesped of thevehicle. The magnet is
placedinside an aluminium drum which is carefully
pivotedand held in position by a hairspring. As thewww5~§tr€q•wvr5p§}

508. 6.20
magnet rotates,
eddy currents are set up in the drum
which oppose the motion of the magnet. A torque is
exerted onthe drum in the opposite direction which
deflects the drum through an angle depending on the
sped of the vehicle.
5. Induction motor. In an a.c. induction motor, a
rotating magnetic field is produced by two single
phase alternating currents having a phase difference of
90°.A metallic rotor is placed in the magnetic field. The
eddy currents set up in the rotor tend to oppose the
relative motion betwen the rotating magnetic field
and the rotor. As a result, the rotor also starts rotating
about its axis.
6.Electromagnetic shielding. Eddy currents may
be used for electromagnetic shielding. As shown in
Fig.6.24,when a magnetic fieldB,directed towards a
metallic shet is suddenly switched on, large eddy
currents are produced in the shet. The change in the
magnetic field is only partially detected at points (such
asP)on the other side of the shet. The higher the
conductivity of the shet, the better the shielding of the
transient magnetic field.
-->
B
~
Conducting
!shet
P
Fig. 6.24Electromagnetic shielding.
7.Inductothermy. Eddy currents can be used to
heat localised tissues ofthe human body. This branch
is called inductothermy.
8.Energy meters. In energy meters used for
measuring electric energy, the eddy currents induced
inan aluminium disc are made use of.
For Your Knowledge
~Eddy currents are basically the induced currents set
up inside the body of conductor whenever the
magnetic flux linked with it changes.
~Eddy currents tend to follow the path of least resis-
tance inside a conductor. So they form irregularly
shaped loops. However, their directions are not
random, but guided by Lenz's law.
~Eddy currentshaveboth undesirable effects and
practically useful applications.
~Eddy currents can be induced in biological tissues.
For example, the cavity of the eye is filled with a
conducting fluid. A large transient magnetic field of
1 T alternating at a frequency of 60 Hz then induces
such a large current in the retina that it produces a
sensation of intense brightness.
PHYSICS-XII
6.12SELF-INDUCTION
13. What ismeant by self-induction ?Define
self-inductance. Give itsunits and dimensions.
Self-induction.When a current flows in a coil, it
gives rise to a magnetic flux through the coil itself. As
the strength of current changes, the linked magnetic
flux changes and an opposing emf is induced in the
coil.This emf is calledself-induced emforback emf
and the phenomenon is known as self-induction.
Self-inductionisthe phenomenon of production of
induced emfin acoil whenachanging current passes
toughit.
Figure6.2S(a)shows a battery and a tappingswitch
connected in series to a coil. As the switch is closed, the
current increases and hence the magnetic flux through
the coil increases from zero to a maximum value and
the induced current flows in the opposite direction of
the battery current. In Fig.6.2S(b),as the tapping switch
is opened, the current and hence the magnetic flux
through the coil decreases from a maximum value to
zero and the induced current flows in the same
direction as that of the battery current.
Induced
current
Induced
current
~~
Battery Opening
switch
c
~
a
t:'
~
~ ~~
Battery Closing
switch
(a) (b)
Fig. 6.25Induced current in a coil when the circuit
is(a)closed and(b)opened.
Coefficient of self-induction.At any instant, the
magnetic flux ~linked with a coil is proportional to the
currentIthrough it, i.e.,
~ex:I
or ~=LI ...(1)
where L is a constant for the given coil and is called
self-inductance or,more often, simplyinductance. It is
also calledcoefficient of self-inductionof the coil. Any
change in current sets up an induced emf in the coil
given by
e= _d~= _LdI (2)
dt dt ...
If in equation (I), 1=1, then ~=L
Thusthe self-inductance of a coilisnumerically equal to
the magnetic flux linked with the coil when a unitcurrent
flows toughit.www4z~tq•p·uvq4o~y

509. ELECTROMAGNETIC INDUCTION
Again from equation (2), ifdI= 1,thene= - L
dt
Thus
theself-inductance of a coil may be defined as the
induced emf set up in the coil due to aunit rate of change of
current tough it.
Units of self-inductanc. From equation(2),
L=_e_
dI/ dt
•.SI unit of L=~1 =1VsA-1=1henry (H)
1As-
Theself-inductance of a coilissaid to be one henryifan
induced emf of one volt is set up initwhen the current init
changes at therateof one ampere persecond.
From equation (1),one may note that self-
inductance is the ratio of magnetic flux and current. So
its SI unit isweber per ampere. Hence
1henry (H)=1VsA-1=1WbA-1
Dimensions of self-inductanceWe know that
L=1=BA= F.A [...F=qvBsinS]
I I qvsinSI
.'.Dimensions ofL
MLr2.L2 ML2T-2
___ =[ML2T-2A-2].
C.Lr1.A A.A [lCT-1=lA]
14.Describe an experiment to demonstrate the
phenomenon of self-induction.
Experiment to demonstrate self-induction. Take a
solenoid having a large number of turns of insulated
wire wound over a soft iron core. Such a solenoid is
called a choke coil. Connect the solenoid in series with
a battery, a rhstat and a tapping key. Connect a 6 V
bulb in parallel with the solenoid. Press the tapping
key and adjust the current with the help of rhstat so
that the bulb just glows faintly. As the tapping key is
released, the bulb glows brightly for a moment and
then goes out. This is because as the circuit is broken
suddenly, the magnetic flux linked with the coil
suddenly vanishes,i.e.,the rate of change of magnetic
flux linked with the coil is very large
induced emf and current are produced in the coil
which make the bulb glow brightly for a moment.
Coil
I-----{< ••~----i
~K
Battery Rh
Fig. 6.26 Demonstration of self-induction.
6.21
6.13SELF-INDUCTANCE OF A LONG SOLENOID
15.Derive anexpression for the self-inductance of a
longsolenoid. State the factors on which the self-
inductance of a coil depends.
Self-inductance of a long solenoid. Consider a long
solenoid of lengthIand radiusrwithr«Iand having
nturns per unit length. If a currentIflows through the
coil, then the magnetic field inside the coil is almost
constant and is given by
B=1l0nI
Magnetic flux linked with each turn
= BA=1l0nIA
whereA=rt,2= the cross-sectional area of the solenoid.
.'.Magnetic flux linked with the entire solenoid is
~ = Flux linked with each turnxtotal number of turns
=llonIAx nl =llon2lAI
But~=LI
.',Self-inductance of the long solenoid is
L=llon2 IA
IfNis the total number of turns in the solenoid,
thenn=N/land so
L=110N2A
I
If the coil is wound over a material of high relative
magnetic permeability11r(e.g.,soft iron), then
2 11 11N2A
L=1111on IA=r0
r I
Factors on which self-inductance depends. Obviously,
the self-inductance of a solenoid depends on its gmetry
and magnetic permeability of the core material.
1.Number of turns. Larger the number of turns in
the solenoid, larger is its self-inductance
Lex: N2
2.Area of cross-section.Larger the area of cross-
section of the solenoid, larger is its self-inductance
Lex:A
3.Permeability of the core material.The self
inductance of a solenoid increasesIlrtimes if it is
wound over an iron core of relative permeabilityIlr.
6.14 PHENOMENA ASSOCIATED WITH
SELF-INDUCTION
16.Describe some phenomena associated with
self-induction.
Phenomena associated with self-induction.
1.Sparking.The break of a circuit is very sudden.
When the circuit is switched off, a large self induced
emf is set up in the circuit in the same direction as the
original emf. This causes a big spark across the switch.www5~§tr€q•wvr5p§}

510. 6.22
2;Non-inductive winding.In re
sistance boxesand
post office boxes, different resistance coils havetobe
used. Here the wire is first r::::r==?
doubled over itself and then _U_
wound in the form of a coil over
a bobbin. Due to this, the
currents in the two halves ofthe
wire flow in opposite directions
as shown in Fig. 6.27.The
inductive effects of the two
halves ofthe wire, being in
opposite directions, cancel each
other. The net self-inductance
ofthe coil is minimum. Such a
winding of coils iscallednon-
inductive winding. The resis-
tance coils having no self- Fig.6.27Non-inductive
inductance are called non-windingof resistance
inductive resistances. coil.
3.Electromagnetic damping. Refer to applications
of eddy currents in section6.11.
r--- For Your Knowledge'I
~Inductance isa measure of the ratioof induced flux ~
to the current I.Itisa scalar quantity having the
dimensions of magnetic flux divided by current. Its
dimensions in terms of the fundamental quantities are
[ML2r2 A-2].ItsSI unit is WbA-lorVsA-1which is
calledhenry(H).It is named in honourofJoseph Henry
who discovered electromagnetic induction in USA
independently ofFaradayin England.
~Inductance playstherole of electrical inertia. The
analogue of self-inductance in mechanics ismass.
~A solenoid made from a thick wire has a negligible
resistancebuta sufficiently large self-inductance. Such
an element is called idealinductor, denoted by ~.
~A wire itself cannot actasaninductor because the
magnetic flux linked with the wire of negligible cross-
sectional area is zero.Only a wire bent into the form
of a coil can act as an inductor. Morver,the self-
induced emf appears only during the time the current
through it ischanging.
~The inductanceof a coil depends on its gmetry and
the intrinsic properties of the material that fills upthe
space inside it. In this sense, it bears similarity to capa-
citance and resistanceitance of a parallel
plate capacitor depends on the plate area and plate
separation (gmetry) and the dielectric constantKof
the interposing medium (intrinsicmaterial property).
Similarly, the resistanceof a conductor depends on its
length and cross-sectional area (gmetry) and
resistivity (intrinsic material property).
~The capacitance, resistance, inductance and diode
(described in chapter 14)constitute the fourpassive
elements of an electrical circuit. In fact, these are the
fouralphabetsof electrical/electronicenginering.
PHYSICS-XII
6.15MUTUALINDUCTION
17.What is meant by mutual induction ?Define the
termmutual inductance. Giveitsunitsanddimensions.
Mutual induction.Mutualinduction isthephenomenon
of production of induced emf in one coil due toa change of
current intheneighbouring coil.
As shown in Fig. 6.28,considertwo coilsPand5
placed close to each other. The coilPis connected in
series to a battery Band a rhstat Rh through a
tapping key K.The coil 5is connected to a galvano-
meterC.When acurrent flows through coil P,it
produces a magnetic field which produces a magnetic
fluxthrough coil 5.If the currentin the coil Pis varied,
the magnetic flux linkedwith the coil5changes which
induces an emf and hence a current in it, asis sen
from the deflection in the galvanometer. ThecoilPis
called theprimarycoiland coil 5,thesecondary coil,
because it is the former which causes an inducedemf
in the latter.
p
K
Rh">0011--------'
Fig.6.28Mutualinduction.
Coefficient of mutual induction. At any instant,
Magnetic flux linked with the secondary coil
o:current in the primary coil
i.e. 4>ccI
or 4>=MI ...(1)
The proportionality constant Miscalledthe mutual
inductance orcoefficient of mutual induction of the two
coils. Any change inthe currentIsets up an induced
emf in the secondary coil which is given by
e=_d4>= _M.dl
dt dt
Ifin equation(I),I=I,then 4>=M
Thus the mutual inductance of twocoilsisnumerically
equal to themagneticfluxlinked with one coil when a unit
current passes through the other coil.
Again, from equation(2),if
dI=1thene=-M
dt'
...(2)
Themutual inductance of two coils may be defined as the
induced emf setup in one coil whenthe current in the
neighbouring coil changesat the unit rate.···3wx§n£m~r°n3lxv

511. ELECTR
OMAGNETIC INDUCTION
Unit of mutual inductanceFrom equation (2), we
have
e
M=dI
dt
:.51 unit of M= ~1 =I VsA-1=I henry(H)
IAs-
Themutual inductance of two coils issaid to be
onehenryifan induced emf of one volt issetup in one coil
when the current inthe neighbouring coilchanges at the rate
of1ampere per second.
6.16MUTUAL INDUCTANCE OF TWO
LONG SOLENOIDS
18. Derive anexpression forthe mutual inductance of
two long coaxial solenoids. 5tate thefactors onwhich
mutual inductance depends. What iscoefficient of
coupling?
Mutual inductance of two long solenoids. As
shown in Fig.6.29, consider two long co-axial sol-
noids 51and52'with52wound over 51'
11
Fig. 6.29Two long coaxial solenoids of same lengthL
LetI=length of each solenoid
r1,r2=radii of the two solenoids
2
A=7tr1
=area of cross-section of inner solenoid 51
Nl,N2=number of turns in the two solenoids
First we pass a time varying current 12through52'
The magnet field set up inside52due to12is
~ =f.l0~I2
wheren2=N2 /I=the number of turns per unit length
of52'
Total magnetic flux linked with the inner solenoid
51is
41=~ANI=f.l0~I2 .ANI
:. Mutual inductance of coil I with respect to coil 2 is
M=!!.=f.l n.AN =f.lONIN2A
12I O"L 1 I
2
6.23
We now consider the flux linked with the outer
solenoid52due to the currentIIin the inner solenoid
51'The field ~ due toIIis constant inside51but zero in
the annular region betwen the two solenoids. Hence
~ =f.lon1I1
where~=N1 /I=the number of turns per unit length
of51'
Total flux linked with the outer solenoid52is
'"=RAN =unIAN_ f.lON1N2Al1
'12'"'1 2 011' 2- I
:.Mutualinductance of coil 2 with respect to coil I is
~1= ~ =f.lON1N2A
II I
Clearly M12=~1=M(say)
M=f.loN1N2A=f.l nn.AI-f.l n.n.m;2[
I 0l"L - O--l"-L 1
Thus, the mutual inductance of two coils isthe
property of their combination. It does notmatter which
oneof them functions as the primary or the-secondary
coil.This fact is known as reciprocity theorem.
Factors on which mutual inductance depends.The
mutual inductance of twosolenoids depends on their
gmetry and the magnetic permeability of the core
material.
1.Number of turns.Larger the number of turns in the
two solenoids, larger will be their mutual inductance
MCt:. N1N2
2. Common cross-sectional area. Larger the
common cross-sectional area of two solenoids, larger
will be their mutual inductance.
3. Relative separation.Larger the distance betwen
two solenoids, smaller will be the magnetic flux linked
with the secondary coil due to currentin the primary
coil.Hence smaller will be the value of M.
4. Relative orientation of the two coils.Mis
maximum when the entire flux of the primary is linked
with the secondary, i.e.,when the primary coil
completely envelopes the secondary coil. Mis
minimum when the two coils areperpendicular to
each other, as shown in Fig. 6.30.
(a) (b)
Fig. 6.30(a)Mismaximum when primary envelopes
secondary,(b) Mis minimum when
primary is perpendicular to secondary.www5~§tr€q•wvr5p§}

512. 6.24
5.Pe
rmeability of the COTematerial. If the two coils
are wound over an iron core of relative permeability
I.lr'their mutual inductance increases I.lrtimes.
Coefficient of coupling.The coefficient of coupling of
two coilsgives a measure of the manner in which the two coils
are coupled together.If4and ~ are the self- inductances
oftwo coils and M is their mutual inductance, then
their coefficient of coupling is given by
M
K=P1~
The value of Kliesbetwen 0and1.
When the coupling is perfecti.e.,the entire flux of
primary is linked with the secondary, Mis maximum
andK=l
When there is no coupling, M=0andK=o.
Thus Kis maximum for the coupling shown in
Fig.6.30(a)andminimum for the coupling shown in
Fig.6.30(b).
For Your Knowledge
>When two coils are inductively coupled, in addition
to the emf produced due to mutual induction,
induced emf is set up in the two coils due to
self-induction also.
>The mutual inductance of two coils is a property of
their combination. The value of M remains
unchanged irrespective of the fact thatcurrent is
passedthrough one coil orthe other.
>While calculating the mutual inductance of two long
co-axial solenoids, the cross-sectional area of the inner
solenoid is to be considered.
:==:While calculating the mutual inductance of two
co-axial solenoids of different lengths, thelength of
the larger solenoid is to be considered.
6.17GROUPING OF INDUCTANCES*
19.Acircuitcontains two inductors in series with
self-inductances ~and~and mut~al inductanceM.
Obtain a formula for theequivalent inductanceinthe
circuit. [NCERT]
Inductances in series. (i)Let the series connection
besuch that the current flows in the same sense in the
two coils as shown in Fig.6.31(a).
~rDO~~r
r--e, 1e, -I
I- eeq 1
Fig.6.31(a)Inductance in series when fluxes get added.
PHYSICS-XII
LetLeqbethe equivalent inductance of the two
self-inductances4and ~ connected in series. For the
series combination, the emfs induced in the two coils
get added up. Thus
eeq=e1+e2
If the rate of change of current in the series circuit is
dl
-,then
dt
e= _TdI_MdI,
1 Jdt dt
e=_Tdl_Mdl
2 ~dt dt
dl
eeq= -4qdt
The negative sign throughout indicates that both
self and mutual induced emfs are opposing the applied
emf. Using the above equations, we have
eeq=e,+e2
dl dl
or -4q.dt=-(4+M+~+M)dt
or Leq=4+ ~ +2M.
(i/)Let the series combination be such that the
current flows in opposite senses in the two coils, as
shown in Fig. 6.31(b).
and
Fig.6.31(b)Inductances in series when fluxes get subtracted.
The emfs induced in the two coils will be
e= _TdI+Mdl
1 '-1dt dt'
e=_Tdl+MdI
2 ~ dt dt
Here the mutual emfs act in the direction of applied
emf and hence positive. For this series combination
also,the emfs induced in the two coils get added up.
dl
Henceeeq= e1+ e2=-[4-M+~-M]-
dt
But eeq=-LdI
eqdt
dI dI
-4qdt=-[4+ ~-2M] dt
Leq=4+~-2M.or···3wx§n£m~r°n3lxv

513. ELECTROMAGNET
IC INDUCTION
20. Two inductors of self-inductances Lrand ~are
connected in parallel. The inductors areso far apart that
their mutual inductanceisnegligible. What isthe
equivalent inductance of the combination? [NCERT]
Inductances in parallel. For the parallel
combination, the total current Idividesupthroughthe
twocoils as
I=II+12
dI dI1 dI2
-=-+-
dt dt dt
I- e 'I
Leq
EJI
Fig. 6.32 Inductances in parallel
Forparallel combination, induced emf across the
combination is equal to the induced emf across each
inductancehus
e=_~dII or
dt
e= _ ~dI2 or
dt
edI1
-=--
~ dt
~ =_dI2
~ dt
This is because the mutual inductance Misnegli-
gibl. IfLeqisthe equivalent inductance of the parallel
combination, then
e=_L.dI=_L[dI1+dI2]
eqdt•eqdt dt
=4q[- ~: - ~~ ]
~q= [~+ ~] or
L=~
eq ~ + ~
If there is any mutualinductance Mbetwen the
coils,then
or
1 1 1
-=-+-
Leq ~ ~
or
6.25
Examples basedon
.•.. .
Formulae Used
1.For self-induction, <I>=LI
dI
2.Selfinduced emf, e=-L-
dt
,
3.For Mutual induction, <I>=MI
4. Mutual induced emf,e= -MdI
dt
5.Self-inductance of long solenoid,
~ N2A N
L=0 =~orfAI,where n=-
I. I
6. Mutual inductance of two closely wound
solenoids,
~ NNA
M= 01 2 =~o"11zAl,
I
N N
where" = _1 ,11z= ~
I I
Units Used
Flux<I>is inweber, inductances Land Min henry,
emfein volt, current Iinampere, cross-sectional
areaAin m2,number of turns per unit length
n, "and11zare inm-1.
Example 30. Whatistheself-inductance of a coil, in which
magnetic flux of40milliweberisproduced when 2Acurrent
flows tough it? [CBSE F 02]
Solution. Here <I>= 40 m Wb=40x10-3Wb,I= 2A
Self-inductance,
L=~ = 40x10-3=2x10-2H.
I 2
Example 31.A200turn coil of self-inductance 20mH
carries a current of4mA.Find the magnetic flux linked with
each turn of the coil.
Solution. Let <I>be themagnetic fluxlinked with
each of the N turns ofthe coil. Then
N<I>cx: I or N<I>= LI
Ll20x10-3x4x10-3
<1>=N = 200
=4x10-7Wb.
Example 32.If a rate of change of current of4As-I
induces an emf of20mVin a solenoid, whatisthe
self-inductance of the solenoid ? [CBSED96]
Solution. Here dI=4AS-I,
dt
IeI= 20 m V = 20x10-3Vwww5~§tr€q•wvr5p§}

514. 6.26
As Ie1= LdI
dt
L=J
.§..L=20x10-3
dl/ dt 4
=5x10-3H=5 mlf,
Example 33. A12ViJattery connected to a60,10 Hcoil
through a switch drives a constant current tough the
circuit. The switchissuddenly opened. Ifittakes1ms to
open the switch, find the average emf induced across the coil.
Solution.Steady-state current = 12 V=2A
60
.,
Finalcurrent =0
dI=(0-2)A=-2x10-3As-1
dt 1ms
Induced emf,
e=-LdI=-10x(-2x103)=20,000 V
dt
Such ahighemf usually causes sparks across the
open switch.
Example 34. An inductor of 5 Hinductance carries a
steady current of2 AHow can a50V self-inducedemf be
made to appearinthe inductor? [Punjab 01]
Solution.Suppose the current reduces to zero in
timetsecond. Thus
L= 5HdI= 0-2 = - 2A,e= 50V
As e=-LdI
dt
-2
50= -5x -ort= 0.2 s
t
Hence aninduced emf of 50Vcanbe generated by
reducing the current to zero in 0.2 s.
Example 35. Whatisthe self-inductance of an aircore
solenoid 50emlong and2emradiusifithas500turns?
Solution. Here I=50cm=0.50m,
r =2em =2 x10-2m,N= 500,Ilo=4nx10-7TmA-1
Self-inductance of the solenoid is
IlN2AIlN2nr2
L=0=--,,0 _
I I
4nx10-7x(500)2xnx(2x10-2)2
0.50
= 7.89x10-4H.
Example 36. An air-cored solenoid with length30em,
areaof cross-section 25~and number of turns 500, carries
a current of2.5A.Thecurrentissuddenly switched offina
brief time of10-3s.How muchisthe average back emf
induced across the ends of the open switchinthecircuit?
[NCERT]
PHYSICS-XII
Solution. I= 30 em = 0.30 m,
A=25cm2=25x10-4m2, N=500, dt=10-3s,
dI=0 -2.5 =-2.5A
2
Back emf=_LdI =_lloN A dI
dt Idt
4nx10-7 x(500)2x2.5x10-4 x(-2.5)
0.30x10-3
=6.542 V.
Example 37.A large circular coil, of radius Rand asmall
circular coil, ofradius r,are putinvicinity of each other. If
the coefficient of mutual induction, for this pair, equals1mH,
whatwould be theflux linked with the larger coil when a
current of0.5A flows tough the smaller coil ?
When thecurrent inthesmaller coil falls tozero,what
would be itseffectinthelargercoil? [CBSE D 08C]
Solution. Here M =1 mIi. I=0.5 A
Flux, <I>=MI= 1O-3Hx 0.5 A =5x10-4 Wh.
When the current in the smaller coil falls to zero, an
induced emf is set up in the larger coil due to the
decrease in the linked flux.
Example 38.Whatisthe mutual inductance of a pair of
coilsif a currentchange of six ampere inone coil causes
the fluxinthe second coil of 2000 turns tochange by
12x10-4Wb per turn?
Solution. Here N =2000, I=6 A
Flux per turn = 12x10-4Wb
Total flux, <1>=2000 x12x10-4=2.4 Wb
As <I>=MI
M= ~ = 2.4 = 0.4H.
I 6
Example 39. An emf of0.5Visdeveloped inthe
secondary coil, when currentinprimary coil changes from
5.0 A to2.0Ain300millisecond. Calculate the mutual
inductance of the two coils. [ISCE93]
Solution.Heree= 0.5V,dI= 2 - 5 =:.3A,
dt=300 ms = 300x10-3 S
Ase= -MdI
dt
0.5=-Mx -3
300x10-3
or M =0.05 H.
Example40.If the currentinthe primary circuit of a pair
of coils changesfrom5A to1Ain0.02s,calculate (i)induced
emfinthe secondary coilifthemutual inductance between
the two coilsis 0.5Hand(ii)the change offlux per turn in
the secondary,ifithas 200 turns.···3wx§n£m~r°n3lxv

515. ELECTROMAGNET
IC INDUCTION
Solution.(i)e= _ MdI= _ 0.5 x(1- 5) =_2_
dt 0.02 0.02
=100V.
(ii)e=-N~
0.02
:. Change in flux per turn,
d4>=- 100x0.02 = _0.01Wb.
200
Thenegative sign shows a decrease of magnetic flux.
Example 41. Over a solenoid of50em length and2em
radius and having500turns, is wound another wire of 50
turns near the centre. Calculate the(i)mutualinductance of
thetwocoils(ii)induced emf in the second coilwhen the
current in the primary changes from 0to5A in0.02s.
Solution.Here N, =500, N2 =50,r=2 cm=0.02 m
1=50cm=0.50 m, f..lo=41tx 10-7TmA-1
(i)The mutual inductance of the two coils is
M=f..lON1N2A =f..lON1N2·1tr2
1 I
41tx10-7x500x50x1tx(0.02)2
M=-----------'---'-
0.5
= 78.96 x10-6H=78.96f..lH.
(ii)The emf induced in the second coil is
e= _ MdI=-78.96x10-6. (5 -0)
dt 0.02
= -19.74 x10-3 V = -19.74 mY.
Thenegative sign indicates a back emf.
Example42.A solenoidal coilhas50turns per centimeter
along its length and a cross-sectional area of4x10-4 ~.
200turns of another wire are wound round thefirst solenoid
coaxially. The two coils areelectrically insulated from each
other.Calculate the mutual inductance between the two
coils. Givenf..lo= 41tx10-7NA-2. [ISCE98]
Solution.Heren1=50 turns per cm = 5000turns
permetre,~l=200 turns,A= 4x10-4m2
M=f..lo~~Al=f..lon1(~I)A
= 41tx 10-7 x5000x200x4x10-4
=5.027x10-4H.
Example43.A solenoid of length50cmwith20turns per
cm and area of cross-section40~completely surrounds
another co-axial solenoid of the same length, area of
cross-section25 ~with25turns per em. Calculate the
mutual-inductance of the system. [NCERT]
Solution. Here 1=50 em = 0.50m
Total no. of turns in outer-solenoid,
N1=n11=20x50=1000
6.27
Area of cross-section of outer solenoid,
~=40 em2 =40xl0-4m2
Total no. ofturns in inner solenoid,
N2=n21=25x50 =1250
Area of cross-section of inner solenoid,
~=25 cm2 =25x10-4m2
To determine mutual inductance, we take area of
cross-section of inner solenoid.
M= f..lo N1N2~
I
41tx10-7x1000x1250x25x10-4
0.50
= 7.85x10-3H = 7.85mHo
Example 44.(a) A toroidal solenoid with an air-core has an
average radius of15emarea of cross-section12C/~and
1200turns.Obtain the self-inductance of the toroid. Ignore
fieldvariation across the cross-section of the toroid.
(b) A second coilof300turns iswound closely on the
toroid above. If thecurrent inthe primary coilisincreased
fromzero to2.0A in0.05s,obtain the induced emf in the
second coil. [NCERT]
Solution. (a)Theuniform magnetic field set up
inside a solenoid is given by
B=f..lonI=~~~.I [.: n=2~r]
.,Total flux linked with theNturns is
f..lNI f..lN2IA
4>=NBA=N._0_ .A=--,,-0--
21tr 21tr
:.Self-inductance of the toroid is
L=~=f..loN2A
I21tr
41tx10-7x(1200)2 x 12x10-4
21txO.15
= 2.304x10-3H =2.3mfl,
(b)HereN1=1200,N2=300,dt=0.05s,
dI=2.0 -0 =2.0 A
The emf induced in the second coilis
e= MdI=f..loN1 N2A dI
dt 1 dt
41tx10-7x1200x300x12x10-4 2.0
x--
2nx0.15 0.05
=0.023 V. r.1=21tr]www5~§tr€q•wvr5p§}

516. 6.28
Example45.Fig.6.33shows a
short solenoid of length
4 em radius2.0em and number of turns 100 lying inside on
the axis of a long solenoid,80em length and number of turns
1500. Whatistheflux tough the long solenoidifa current
of3.0A flows tough the short solenoid?Also obtain the
mutual inductance of the two solenoids. [CERT]
~BID
Fig.6.33
Solution. As the short solenoid produces a compli-
cated magnetic field, so it is difficult to calculate
mutual inductance and flux through the outer sole-
noid. For this purpose we make use of the principle of or
reciprocity of mutual inductancei.e.,
~2=Mzl
Suppose 51represents the long solenoid and 52' the
short solenoid. Then
11= 80 em=0.80 m, Nl = 1500
12= 4cm=0.04m,N2=100,
~ =2.0 cm=0.02 m,12=3.0 A
The uniform magnetic field inside the long
solenoid is given by
I\=1l0N1Il
11
Since the short solenoid lies completely inside the
long solenoid, the flux linked with it is given by
~ =N2AzI\ =N2Az·1l0NIIl
11
:. Flux through each turn of short solenoid
=jL=1l0Nl II.Az=1l0N1 II . nRi
N2 11 11
By definition, ~ =M21 II
2
1l0Nl II . n~_
N2· - MzIIl
11
or
From the symmetry of mutual inductance, we have
2
M_ _1l0n~.N1N2
12-Mzl- I
1
4nx10-7 xnx(0.02)2x1500x100
= H
0.80
=2.96x10-4 H
The total flux linked with the long solenoid is
N1<1\=M12I2=2.96x10-4x3.0 Wb
= 8.88x10-4Wb =-8.9x10-4Wb.
PHYSICS-XII
Example46.Tee inductances are connected as shown
Fig.6.34Find the equivalent inductance. [Punjab931
L2=0.5 H
L3=0.5 H
Fig. 6.34
Solution. The equivalent inductancetof ~ and4
is given by
1 1 1 1 1
-=-+-=-+-=4
t: ~40.5 0.5
t=.!=0.25H
4
Now ~ andt:form a series combination, their
equivalent inductance is given by
L= ~ +t:=0.75 +0.25 =1 H.
~rOblems For Practice
1.Magnetic flux of 5 microweber is linked with a coil,
when a current of 1 mA flows through it. What is
the self-inductance of the coil? [Haryana 991
(Ans.5 mH)
2.Calculate the induced emf in a coil of 10 H induc-
tance in which the current changes from 8 A to 3 A
in 0.2 s. (Ans. 250V)
3.A magnetic flux of8x10-4Wb is linked with each
turn of a 200-tum coil when there is an electric
current of 4 A in it. Calculate the self-inductance of
the coil. (Ans.4x10-2H)
4.The selfinductanceof an inductor coilhaving 100turns
is 20ml-I,Calculate the magnetic flux through the
cross-section of the coil corresponding to a current
of 4 mA. Also, find the total flux.[CBSE PMT20001
(Ans. 8x1O-7Wb, 8x1O-5Wb)
5.A coil of inductance 0.5 H is connected to a 18 V
battery. Calculate the rate of growth of current.
(Ans. 36As-1 )
6.An average induced emf of 0.4Vappears in a coil
when the current in it is changed from 10 A in one
direction to 10 A in opposite in 0.40 second. Find
the coefficient of self induction of the coil.
[CBSE Sample Paper 111 (Ans. 8 mH)
7.A coil has a self-inductance of 10 ml-I,What is the
maximum magnitude of the induced emf in the
inductor, when a currentI= 0.1sin 200tampere is
sent through it. (Ans.0.2 V)www5~§tr€q•wvr5p§}

517. ELECTROMAGNETIC INDUCTION
8.What is the
self-inductance of a solenoid of length
40 em, area of cross-section 20 em2and total number
of turns 800 ? (Ans.4.02 mH)
9.The current in a solenoid of 240 turns, having a
length of 12 cm and a radius of 2 ern, changes at the
rate of 0.8 As-I. Find the emf induced in it.
(Ans. 6x10-4V)
10.Calculate the mutual inductance betwen two coils
when a current of 2 A changes to 6 A in 2 s and
induces an emf of 20 mVin the secondary coil.
[Punjab 991
(Ans. 10 mH)
11.The mutual inductance betwen two coils is 2.5 H.
If the current in one coil is changed at the rate
2.0 As-I, what will be the emf induced in the other
coil? (Ans. 5.0V)
12.In a carspark coil, an emf of 40,000Vis induced in
the secondary when the primary current changes
from 4 A to zero in Iflus. Find the mutual
inductance betwen the primary and secondary
windings of this spark coil. (Ans.0.1 H)
13.If the current in the primary circuit of a pair of coils
changes from 10 A to 0 in 0.1 s, calculate
(i)the induced emf in the secondary if the mutual
inductance betwen the two coils is 2 H, and
(ii)the change of flux per turn in the secondary if it
has 500 turns. (AhS. 200V,-0.04 Wb)
14.A conducting wire of 100 turns is wound over 1 ern
near the centre of a solenoid of 100 ern length and
2 cm radius having 1000 turns. Calculate the
mutual inductance of the two coils.[Haryana 941
(Ans. 1.58x10-4 H)
15.A solenoid has 2000 turns wound over a length of
0.3 m. The area of cross-section is 1.2 x10-3m 2.
Around its central section a coil of 300 turns is
closely wound. If an initial current of 2 A is
reversed in 0.25 s, find the emf induced in the coil.
(Ans.48 mY)
16.Calculate the mutual inductance betwen two coils
if a current 10 A in the primary coil changes the flux
by 500 Wb per turn in the secondary coil of
200 turns. Also determine the induced emf across
the ends of the secondary coil in 0.5 s.
(Ans.104Ii,2x105V)
HINTS
1.Here ~=:5 11Wb=5 x 10-6 Wb
Self-inductance,
~ 5x10-6 -3
L=-=:--3- =:5x10 H =:5 mH
I 10-
I=1 mA=10-3A
6.29
dI I -I 3-8
2.€=-L-= - L. _2__1= - 10x-- = 250 V.
~ t n2
3.N~=LI
N~ 200x8x10-4 2
..L=-= = 4x10-H.
I 4
4.Total magnetic flux,
N~=:LI=:20x10-3x4x10-3= 8 x 10-5 Wb
Magnetic flux through the cross-section of the coil,
~= 8x10-5 = 8x10-5 =8xl0-7 Wh.
N 100
5.dI=§.=~=36As-l.
dtL0.5
6.€=_LI2-II
t
0.4 = -L(-10 -10) orL=:0.4 = 0.008 H = 8 mHo
0.40 50
7.HereL= 10 mH = 10-2 H,I= 0.1sin 200t
€=:LdI= 10-2 x~ (0.1 sin 200 t)
dt dt
= 10-2 x0.1x200 cos 200t=:0.2 cas 200t
Clearly, €will be maximum when cas 200t=1
Therefore,
€max =:0.2x1=0.2V.
8.HereI= 40 cm=0.40 m,
A= 20 cm 2 = 20 x 10-4m 2,
N= 800,110= 41tx 10-7 TmA-1
Self-inductance of the solenoid is
L= 110N2A41tx10-7x( 800)2x20x10-4
1 0.40
= 4.02x10-3H = 4.02mrl,
9.I€1=LdI=110N2A dI
dt I dt
41tx 10-7 x(240)2 X1tx(0.02)2
-----'---'----'---'-- x0.8
0.12
=6 xlO-4V.
10.As€=: -MdI
dt
. . _ 20x10-3= - M 6-2 or M = 10 mH
2
11.I€I=MdI=:2.5x2.0 = 5.0V.
dt
12.€=-MdI
dt
0-4
.'.40,000 = -Mx 6
10 x 10-
or M =O.IHwww5~§tr€q•wvr5p§}

518. 6.30
13. (i)E=
-M dI = _2x0 - 10 =200 V.
dt 0.1
(ii)AsE= _Ndq, dq, =MdI
dt dt dt
ordq,=M . dI=2x (0 -10) =_0.04Wh.
N 500
14.HereNl= 1000, 1=100cm= 1m,
A=1tr2=1t(2x10-2)2 m2, N2 = 100
M=1l0Nl N2A
1
41tx10-7x1000x100x1tx4x10-4
1
= 161t2 x10-6.= 16 x9.87x10-6
=1.58x10-4 V.
PHYSICS-XII
15.M=110 Nl N2A
1
41tx10-7x2000x300x1.2x10-3
0.3
=3x10-3H
E=-MdI= _3x10-3(-2 - 2J
di 0.25
=48x10-3V
=48mV.
16.AsNq,=MI
:. 200x500 =Mx10
or M=104 H
Also,E=Ndq,= 200x500=2x10s V.
dt 0.5
VERY SHORT ANSWER CONCEPTUAL PROBLEMS
Problem 1. What is the basic cause of induced emf?
[Punjab01]
Solution. The basic cause of induced emf is the change
ofmagnetic flux linked witha circuit.
Problem 2. Will an induced emf develop in a
conductor, when moved in a direction parallel to the
magnetic field ?
Solution. No,because themagnetic fluxlinked with
the conductor does notchange when itismoved parallel
tothemagnetic field. Morver, the magnetic Lorentz
force on the fre electrons of the conductor is zero, so no
emf is induced acrossthe ends of the conductor.
Problem 3. A train is moving with uniform sped
from north to south.(i)Will any induced emf appear
across the ends of its axle ? (ii)Will the answer be
affected if the train moves from east to west?
Solution.(i)Yes, emf will appearbecause the axle is
intercepting the vertical component ofthe earth's magnetic
field. (ii)No,here also the axle intercepts the vertical
component of the earth's magnetic field, so emf is induced
acrossthe ends of the axle.
Problem 4.Does the change in magnetic flux induce
emf or current?
Solution.The change in magnetic flux always induces
emf. However, the current is induced only when the
circuit is closed.
Problem 5.Induced emf is also called back emf.
Why?
Solution. Itisbecause induced emf always opposes
any change in applied emf."
Problem 6. A wire kept along the north-south
direction is allowed to fall frely. Will an emf be
induced in the wire?
Solution. No, because neither horizontal nor vertical
component of earth's magnetic field will be intercepted by
the falling wire.
Problem 7. A metallic rod held horizontally along
east-west direction, isallowed to fall under gravity. Will
there beanemf induced at its ends ? Justify your
answer. [CBSED13]
Solution. Yes,because thehorizontal component of
earth's magnetic field is intercepted by the rod.
Problem 8. A cylindrical bar magnet is kept along the
axis of a circular coil. Will there be a current induced in the
coil if the magnet isrotated aboutitsaxis?Givereason.
Solution. 0,becausethe magnetic flux linked with
the circular coil does not change when the magnet is
rotated about itsaxis.
Problem 9. A vertical metallic pole falls down
through the plane of the magnetic meridian. Will any
emf be produced betwen its ends ? Give reason for
your answer. [CBSED95C]
Solution.No emf will be induced because the metallic
pole neither intercepts the horizontal component BHnor
the vertical componentBvof earth's magnetic field.
Problem10.The electric current flowing in a wire in
the directionBtoAis decreasing. Whatisthe direction
of induced current in the metallic loop kept above the
wire as shown in Fig.6.35. [CBSEOD 14]
o
• •• •
A B
Fig. 6.35www4z~tq•p·uvq4o~y

519. GUIDELINES TONCERT EXERCISES
6.1.Predict the direction of
induced current in the situations described by the following Figs.6.99(a) to(j).
p
Tapping key just closed
(c)
\
I
r-'~~""'"
Tapping key just released
(e)
Fig. 6.99
---G __NO
(a)+--
Oy Common
axis
z x
x
r
y
p q x y
(b)
z
---0,----
Rheostat setting
being changed
(d)
Current Idecreasing at
a steady ratewwwSnotesdriveScom

520. ELECTROMAGNETIC I
NDUCTION
Ans.(a)AstheS-pole of the magnet approaches the
coil, byLenz's law,the induced current in the coil
develops S-pole at the endq.For this theinduced current
should flow clockwe when seen from themagnet side.
Hence theinduced current flows along qrpq.
(b)Asthemagnet moved, its S-pole moves towards
pqcoilanditsN-pole moves away from xycoil.ByLenz's
lawthe induced current should develop S-pole at the end
qandalso aS-pole at the end x.Forththe current in the
two coils should flow clockwe when seen from the
magnet side. Hence the induced current flows along prqin
one coil and along yzxin the other coil.
(c)Asthecircuitof the left loop completed, current
flows in the directn of arrows shown. This current
develops south polarity on the side of right loop. The
induced current should flow clockwise (when seen from
left) in the right loop so as to oppose thegrowth of current
in left loop. Hence theinduced current flows alongyzx.
(d)Astherheostat is adjusted, theresistance decreases
and current increases. This increases the flux through the
neighbouring coil.ByLenz's rule, increase of flux should
be opposed. The induced current should produce fluxin
the opposite directn oforiginal flux.Hence the induced
current flows along zyx.
(e)Asthe circuit breaks, the flux decreases. The
induced current should flow along xryto supplement the
flux.
if)The circular field lines set up around the current
carrying wire liein the plane of theloop. The flux
threading the coil in the perpendicular direction zero.
Anychange incurrent will notchange thisflux.Hence no
induced current set up in the coil.
6.2.UseLenz's law to determine the direction of induced
current inthe situations described by Fig. 6.100.
----
· -+.
...~...
·.~..
b' c' d'
"....-.b-;- -;--.- -;-;"d-••~.
'-------~---------
· ·t·
x
x
o--....::.a_.r. • • •
x x x
Fig.6.100
(a)Awire of irregular shape turning into a circular
shape.
(b)Acircularloop being deformed into a narrow
straight wire.
Ans.(a)Astheloop changes from irregular to circular
shape, its area increases. Hence themagnetic fluxLinked
with it increases. According to Lenz's law,the induced
6.51
current should produce magnetic flux in the opposite
direction oforiginal flux. For th induced current should
flow in the anticlockwise directn, i.e.,along adcb.
(b)Asthecircular loop is transformed into a narrow
wire, itsareadecreases. Themagnetic fluxlinked with it
alsodecreases. ByLenz'slaw,theinduced current should
produce afluxin the directn of original flux. For this the
induced current should flow in the anticlockwise
directn,i.e.,along a'd'c'b'.
6.3.Along solenoid with15turns per em has a small loop of
area2.0cm2placed inside, normal to the axis of the solenoid. If
the current carried by the solenoid changes steadily from 2Ato
4 Ain0.1s,what istheinduced voltage in theloop while the
current ischanging?
Ans.Heren=15 turns/em =1500 turns/m,
A=2.0 cm2=2x10-4m2
IeI=d<j>=~(BA)=A.dB=A. ~(110nI)
dt dt dt dt
dI
=110nAdt
=4nx10-7 x1500x2x10-4 x(4-2J
0.1
=7.5x10-6V.
6.4.Arectangular wireloop of sides 8em and 2cmwith a
small cut ismoving out ofa region of uniform magnetic field of
magnitude0.3Tdirected normal to the loop.What is theemf
developed across the cut ifthe velocity of the loopis1ems-1in a
direction normal to the (i) longer side (ii)shorter side of theloop?
For how long doestheinduced voltage last ineachcase?
v=1em/s
xX X X X x
X X X X X XxT
2em
X X X X X X
Xl
XX X X X XX
I---8em---I
B=0.3T
XX X X X X X
X X X X X X X
-+--. v=1 cm/s
X X X X X X X
xX X X X X X
1---8cm--l
Fig. 6.101
Ans. HereB=0.3T,v=1cms-1 =0.01 ms-1
I=8cm=0.08m,b=2 cm=0.02 m
(/) Whentheloop moves normal to the longerside:
Inducedemf,
e=Blv=0.3x0.08 x 0.01 =2.4x10-4 Vwww5notesdrive5com

521. 6.52
:. Time for which emf lasts
=Time in which shorter side
moves out
of the field
=!:=0.02=2s.
v 0.01
(i)When the loop moves normal to the shorter side:
Induced emf,
e=Bbv=0.3x0.02x0.01 =0.6x10-4 V
:.Time for which emf lasts
= Time in whichthe longer side moves
out of the field
=.!..=0.08=8s.
v 0.01
6.5.A 1.0 m long metallic rodrotated with an angular
frequency of400rads-1about an axis normal to the rod passing
through its one end. The other end of the rodin contact with a
circular metallic ring. Aconstant and uniform field of 0.5T
parallel to the axis exists everywhere. Calculate the emf
developed between the centre and the ring.
Ans.Here L = 1.0 m,00= 400 rad s-l, B =0.5 T
The emf developed between the centre and the ring
e =-.!BL200=-.!x0.5x(1.0)2x400 =100 V.
2 2
6.6.Acircular coil of radiusB.Ocm and20turns rotates
about itsvertical diameter with an angularspeed of50rad S-1 in
a uniform horizontal magnetic field of magnitude 3.0x10-2T.
or
Obtain the maximum and the average emf induced in the coil. If
the coil forms aclosed loop of resistance 10n,calculate the
maximum value of current in the coil. Calculate the average
power loss due to Joule heating. Where does this power come
from? [CBSE SamplePaper 08]
Ans.Herer= 8 em = 0.08 m, N= 20,
00=50s-1, B=3.0xlO-2T
At any instant, emf induced in the coil given by
e =NBAoosinoot=eo sin 00t
:.Maximum induced emf in the coil
eo=NBAoo=NB1t r200
= 20x3.0 x 10-2 x3.14x(0.08)2<,x,50V
--,~
=0.603 V
Sincethe average value of sin ootover a cycle zero,
therefore,eav=0 .
M· .dd Ieo 0.603 3 A
axirnum in uce current, 0·=-'----=--=0.060
R 10
:.Power dsipated as heat
e2 2
=-.!.eI=-.! ~=-.!x(0.603)W=a.018W
2002 R 2 10'
The source of th power the external agent which
keepsthe coil rotating.
PHYSICS-XII
6.7.A horizontal straight wire10m long extending from
east to westfalling with a speed of5.0 m s-1at right angles to
the horizontal component of the earth's magnetic field,
0.30x10-4Wbm-2.
(a) Whatistheinstantaneous value of the emf induced
in the wire?
(b) What is thedirection of the emf?
(c)Which end of the wireisat the higher electrical
potential?
Ans.(a)Here, I= 10m,v=5.0ms ",
BH = 0.30x10-4Wbm-2
e = BHIv= 0.30x10-4 x10x5.0=1.5x10-3 V.
(b)According to Fleming's right hand rule, the
directn of emf fromwesttoeast.
(c)Western end of the wire at the higher electrical
potential.
6.8.Current in a circuit falls from5.0Ato0.0Ain0.1s.If
an average emf of200 Vinduced, giveanestimate of the
self-inductance of the circuit.
Ans.Here, II= 5.0 A, 12=0.0 A, t= 0.1s,e = 200V
As e= _LdI= _L12 -II
dt t
0.0- 5.0
200=-L =+50L
0.1
L=4H.
6.9.Apair of adjacent coils has a mutual inductance of
1.5 H. If the current in one coil changes fromato20Ain0.5s,
what is the change of flux linkage with the other coil ?
Ans.Here,M= 1.5 H,II= 0 A, 12= 20 A,t= 0.5 s
dlj> dl
As e=--=-L-
dt dt
.. dlj>=Ldl= 1.5x(20-0) =30 Wh.
6.10.A jet plane is travelling west at the. speed of
1800 km h-1.Whatthe voltage difference developed between
the ends of the wing25m long, if the earth's magnetic field at
the location has a magnitude of5.0x10-4T and the-dip angle
300? [CBSEOD09,15C).
Ans.v=1800 km h-1 '
= 1800x1000 ms-1 = 500 ms-1
3600
I= 25 m, B = 5.0 x10-4 T,0= 30°
Bv = Bsin0= 5.0'x'lO-4 xsin30°
= 2.5x10-4 T
->
Only the flux-lines of vertical component offieldB
are cutting across the horizontally moving jet plane.
..Induced emf,
e = Bv Iv= 2.5x10-4x25x500::::3.125V.www5notesdrive5com

522. ELECTR
OMAGNETIC INDUCTION
6.11.SupposetheloopinExercise 6.4 isstationary but the
current feeding the electromagnet that produces themagnetic
fieldisgradually reduced sothat"thefield decreases from its
initial value of0.3T at the rateof0.02Ts-1.If the cut isjoined
and the loop has a resistance of 1.6n,howmuch poweris
dissipated by the loopasheat?Whatisthe source of this
.power?
Ans.A=8x2 = 16 crn2 = 16x10--4m2,
dB= 0.02 Ts-1
dt
e=d~=A dB= 16x10-4x0.02
dt dt
=3.2x10-5V
R=1.60
Induced current,
e3.2x10-5 5
1=-= =2xlO-A
R 1.6
Power dsipated as heat,
P=12R= (2x10-5)2 x1.6 = 6.4x10-10 W.
The source of th power the external agency
responsible for changing the magnetic field with time.
6.12.A square loopofside12cm with itssides parallel to x-
and y-axes moves with a velocity of8cms-1 inthepositive
x-direction in anenvironment containing a magnetic field in
thepositive z-direction. The field isneither uniform inspacenor
constant in time. It has agradjent of 10-3Tem-Ialong the
negative x-direction (i.e.,itincreases by10-3T per emas one
movesin the-vex-direciion), andit isdecreasing in time at the or
rate of10-3Ts-1.Determine the direction and magnitude of the
induced current in theloopif itsresistance is4.5mO.
Ans.HereA=a2= (0.12 m)2 = 144xlO--4m 2
v= 8cm s-1 = 0.08 ms-1
Rate ofchange of magnetic field B with dtance
dB 10-3T
_= 10-3 Tcm-1 = =10-1Tm-1
.dx 10-2m
Rate of change of magnetic field B with time t
dB= 10-3 Ts-1
dt
Induced emf due to change in field B with positnx
given by
e=d~=.!!.-(BA)=A.dB=A.dx . dB
1dt dt dt dt dx
dB
=A.v.-
dx
= 144x10-4x0.08 x 10-1 V = 115.2 ~10-6V
Induced emf due tochange in field B with time t
e=d~=.!!.-(BA).; A dB
2dt dt dt
=144x10-4x10-3v = 14.4 x10-6V
6.53
..Totalinduced emf,
e=e1+e2= (115.2 +14.4) x 10-6 V
= 129.6 x10-6V
AsR= 4.5 mO =4.5x10-30
.', Induced current,
e129.6 x10-6 _ -2
I=R= 4.5 x10-3 A - 2.9x10 A
The two effects havebeen added up because both
cause a decrease in flux in-ve.z-direction. Thedirection of
induced current issuch as to increase thefluxthrough the loop
along +vez-directn. If for the observer the loop moves to
the right, the current will beseen to be anticlockwe.
6.13.Itdesired to measure themagnitude offield between
thepolesof a powe~lloudspeaker magnet. Asmallflatsearch
coilof area 2.0cmwith25closelywound turns ispositioned
normal to the field direction and then quickly snatched out of the
field region. (Equivalently, one can giveit a quick90°turn to
bring its plane parallel to the field direction). The totalcharge
flown inthe coil (measured bya ballistic galvanometer
connected to the coil) is 7.5mCTheresistance of the coil and
thegalvanometer 0.50n.Estimate the field strength of the
magnet.
Ans.Here<l\=BAand42= 0
Induced emf,
e=_N ~.\ =_N.0-BA=NBA
t t t
or!I.R=NBA
t t
or
IR=NBA
t
B=~
NA
q= 7.5 mC = 7.5 x1O-3C,
A= 2.0 cm2 = 2.0xlO--4m 2,
But
N= 25, R= 0.50n
7.5x10-3x0.50 Wb 2
B= =0.75 m"
25x2.0x10--4
6.14.Figure 6.102shows a metal rod PQresting on the rails
AB and positioned between the polesof a permanent magnet.
Therails, the rod and the magnetic field are in three mutually
perpendicular directions. A galvanometer Gconnects therails
through a switch K.Length oftherod=15em,B=0.50T,
Fig. 6.102www5notesdrive5com

523. 6.5-1
res
istance of theclosed loop containing therod=9.0 mn.
Assume. the field to be uniform.
(a) Suppose Kopen and therod moves with a speed of
12cms-1 inthe direction shown. Givethe polarity
andmagnitude of the induced emf
(b)Isthere an excess charge built lip atthe ends of the
rods when K open?What ifKisclosed?
(c)WithKopen andtherod moving uniformly, thereis
nonetforce on the electrons illtherod PQeven
thollgh thelJ do experience magnetic force duetothe
motion oftherod.Explain.
(d) Whatisthe retarding force on the rod when K
closed?
(e) How muchpower isrequired (by an external agent)
tokeeptherod moving at the same speed
(=12ems-1)when K isclosed?Howmuch power
isrequired when K isopen?
(f)How much power dissipated asheat in the closed
circuit? Whatthesource of the power?
(g) Whatisthe induced emf in the moving rod if the
magnetic field isparallel to therailsinstead of being
perpendicular?
Ars.(a)The magnitude of the induced emf is given by
€=Blvsine
--->
where Bis theangle made bythe rodPQwith the field B.
In th case B=90°so that
€=Blv=0.50x0.15x0.12V
=9.0x10-3V=9.0 mV
Astheconductor PQmovesin the directn shown,
the free electrons in it experience magnetic Lorentz force.
By Fleming's left hand rule, the electrons move from the
endPtowards the end Q. Deficiency of electrons makes
the endPpositive while the excess ofelectrons makes the
end Q negative.
(b)Yes,the excess charge built up at the ends of the
rod because induced emf set up across the ends as it
moves in the field even when the keyKisopen.
When the keyK closed, the excess charge
maintained by the continuous flow ofcurrent.
(c)The magnetic Lorentz force[~II=-e(i!xB)]is
---> --->
cancelled by theelectric force[~=eE]exerted by the
electricfieldset up by the opposite charges at its ends.
(d)Restance, R=9.0 mn=9.0x10-3n
€9.0x10-3 V
:.Induced current, l=-= 3 = 1.0 A
R9.0x10-o
Retarding force on the rod
F=IIB=1.0x0.15x0.50 =75x10-3N.
PHYSICS-XII
(e)Power expended by anexternal agent against the
retarding forceFto keep the rod moving uniformly at
12cms-1 given by
P=Forcexvelocity =Fv
=75x10-3x12x10-2W=9.0x10-3 W.
(f)Power dissipated as heat
=12R=(1.0)2x9.0x10-3 W=9.0x10-3 W
Thesource ofthispowertheexternal agent which
keeps the rod moving against the retarding force F.
(g)In th case, theangleBmade bytherodwith the
--->
fieldB zero.
€=Blvsin 0° =0
Th is because the motn of the loop does not cut
acrossthe field lines. There no change in magnetic flux.
So the induced emf is zero.
6.15.An air-cored solenoid with length 30em, area of
cross-section 25cm2and number of turns 500carriesacurrent
of2.5A.Thecurrent issuddenly switched off in abrieftime of
10-3s.Howmuch istheaverage backemfinduced across the
endsofthe open switch in the circuit? Ignore the variation in
magnetic field nearthe ends of the solenoid.
ns.The magnetic field inside a solenoid ofNturns,
length Iand carrying current I
B=lloNI
I
FluxlinkedwithNturns of the solenoid is
11NI 11AN2]
$=NBA=N_0_ .A=:......><...0--
I I
HereI=30cm=0.30m,A=25cm2 =25x10-4m2
I=2.5A,N=500, 110=47tx10-7 A-2
Initial flux linked with the solenoid is
47tx10-7 x25x10-4x(500)2 x2.5
~= O~ M
=6.54x1O-3M
Final flux linked with the solenoid (when the
current switched off)
$[=0
Average back emf
Totalchange in flux
€av
Total time
6.54x10-3 -0
-----,,3-- =6.54 V.
10-
6.16.(a)Obtain an expression for the mutual inductance
between alongstraight wireanda square loopof side aas shown
in Fig. 6.103.
Fig.6.103
awww9notesdrive9com

524. ELECTROMAGNE
TIC INDUCTION
(b) Now assume that the straight wire carries a current of
50 A and the loop is moved to the right with a constant uelociiv,
v=10mls.Calculate the induced emf in the loop at the instant
when x=0.2m.Take a=0.1m and assume that theloop has a
large resistance.
Ans.(a)As shown in
Fig.6.104, consider a
rectangular strip of small
widthdrof the loop at
dtance rfromthe wire.
dr
--+ -
Magnetic field at the
locatn ofthe strip
B=floI
2rrr
Th field pointsnormally into the plane of the loop.
Area ofthe strip, A=adr
Magnetic fluxlinked with the strip,
d<j>=BA =floI. adr
2rrr
a
r-----.j
Fig. 6.104
Total magnetic flux linked with the square loop,
<j>=fd<j>=
r=x+a r=x+a
_fl_o_I adr=_fl_o_l_a
2rer 2re
1
-dr
rf f
r=X r=x
=fl0Ia[Inr]x+a=fl0Ia[In(x+a)-Inx]
2re x 2re
(b)The square loop moving in a non-uniform
magnetic field. Themagnetic flux linked with the loopat
anyinstant is
<j>=fl 0IaIn(1+ ~)
2re x
Induced emf set upin the loop,
€=_d<j> = _ d<j> .dx=_vd<j>
dt dx dt dx
=-v~[flOIaIn(1+ ~)]
dx 2re x
=_vfl 0Ia__1_ (_~) =!:.Q. a2vI
. 2re '(1+:)' x22rex(x+a)'
4rex10-7 (0.1)2 x10
----x x50
2re 0.2 (0.2 + 0.1)
=1.67x10-sV =1.7x10-5V.
6.17.Aline charge Aper unit length is lodged uniformly
onto the wheel of massMand radius R.Thewheel has light
non-conducting spokes and free to rotate without friction
about its axis (Fig. 6.105). A uniform magnetic field extends
over a circular region within the rim.
6.55
I
I
I
I
I
I
I
I
I I
I-a-- I
I_R-----+4
Fig. 6.105
Itisgiven by
-> A
B=-Bok[r::;a, a<R]
=0 (otherwe)
What is the angular velocity of the wheel after the field is
suddenly switched off ?
Ans.Accordingto Faraday's law of electromagnetic
inductn, the induced emf
€=_ d<j>
dt
->
Th implies the presence of an electric field Etan-
gentto all points of the boundary of the circular region of
radiusa.If we move a test chargeqonce round th boun-
dary,a workq€will be done. The electric force on the
charge qEand work doneby th force round th boun-
dary qEx2rea.Equating the two works done, we get
q€=qEx2rea orE=~
2rea
€= _ d<j>
dtr
1 d<j> 1 d 2 adB
E=--.- =--.-(rea B)=---
2rra dt 2rea dt 2dt
In the given problem, totalcharge on the rim =Ax2rea.
Therefore, the force on th charge
adB
F=A.2rea.E=-A .2rea.-.-
2dt
[: F=ma=m ~~ ]
As
so
or
dv a dB
m. -= -A.2rca.--
dt 2dt
d 2dB
m.-(Rw)=-Area -
dt dt
da: 2dB
mR-= -Area-
dt dt
di»= _Area2 dB
mR
or
or
or
Integrating both sides,
Arca2B
W=---
mR
-> Brca2A A
rn=---k
mR
Invector notatn,
Thenegative sign indicates that the vector r;; inthe
negativez-directn.www5notesdrive5com

525. 6.56 PHYSICS-XI
I
Text Based Exercises
"YPE A:VE'RY SHORT ANSWER QUESTIONS (1 markeach)
1.When is themagneticfluxcrossing a given surface
areaheldin a magnetic field maximum ?
2.When is thefluxlinked witha closed coil held ina
magnetic field zero ?
3.Name the SI units of (i)magnetic flux and
(ii)magnetic induction (ormagnetic fluxdensity).
[Himachal 93]
4.Writethedimensional formula of magnetic flux.
[Punjab 99C]
5.Define weber. Howisitrelated toMaxwell ?
6.State Faraday's law of electromagnetic induction.
[CBSE OD09;F 09]
7.A coil of metal wireisheldstationary inanon-
uniform magnetic field.Is any emf induced in the coil?
8.Onwhat factors doesthemagnitude of the emf
induced in the circuit dueto magnetic flux
depend? [CBSEF 13]
9.Awire cuts across a flux of 0.2x1O-2Wb in 0.12s.
What istheemfinduced inthewire? [lCSE96]
10.A metallic wire 1minlength ismoving normally
across afield of0.1Twitha speed of 5ms-1.Find
the emf between the ends of the wire. [lSCE97]
11.How would you detect the pre~ence of magnetic
fieldon an unknown planet?
12.When current flowing in an inductive circuit is
switched off, will the induced current be in the
direction ofmaincurrent or inopposite direction?
13.Givethe expression forthe emf induced between
theends of a metalconductor moving perpen-
dicular to a uniform magnetic field.
[CBSE D93C]
14.A glass rodof length Imoves with velocity v
perpendicular to a uniform magnetic field B.What
isthe induced emf in therod?
15.A coil of areaAiskept perpendicular inauniform
magnetic field B.Ifthe coil isrotated by 180°,what
will bethe change in flux?
16.Abicycle generator creates a 3.0 V, when the bicycle
istravelling at a speed of 9.0krnh-1. How much
emf is generated when the bicycle is travelling at a
15krnh-1 ?
17.Statethelawthat.gives the polarity of the induced
~E ~~OO~
Or
State Lenz's law. [CBSE D13]
18.State the rule usedto determine the direction of
current induced ina conductor moving in a
perpendicular magnetic field.
19.A magnet is moving towards a coil with a uniform
speed vasshown in Fig. 6.106.Statethe direction of
the induced current in the resistor R.
[CBSESample Paper 13]
~-----N~I
Fig.6.106
20.A conducting loop is heldbelow a current carrying
wirePQas shown in Fig. 6.107. Predictthe
direction of the induced current in the loop when
the current in the wire is constantly increasing.
[CBSEOD14]
P Q
-----t---_--•• ••
o
Fi. 6.107
21.Predict the direction of induced current in metal
rings 1 and 2 when current 1inthe wire is steadily
decreasin. (Fi. 6.108) [CBSED12]
Fig.6.108
22.Predict the direction of induced current in ametal
ring when the ring is moved towards a straight
conductor with constant speed v.The conductor is
carrying current I in the direction shown in
Fig.6.109 [CBSED12]
!O
v [
••
Fig.6.109
23.Define the unit ofself-inductance.
[Haryana 98C ;Punjab 2000]www3notesdrive3com

526. ELECTROMAGNETIC INDUCTION
24.If the
number of turnsand the lengthof the solenoid
are doubled keeping the area of cross-section same,
how is the inductance affected? [CBSE.PMT 93]
25.Two coils of self inductance ~andLzare connected
in series, so that current flows in the same sense
and they have a mutualinductanceM.What is their
equivalent inductance?
26.If the numberofturns of a solenoid is doubled,
keeping the other factors constant, how does the self-
inductance of the solenoid change? [CBSE02000]
27.A coil of wire of certain radius has 600 turns and a
self-inductance of 108 ml-I,Whatwill be the seIf-
inductance of anothersimilar coil with500 turns?
28.Write three factors on which the self-inductance of
coil depends. [CBSE0095]
29.What is meant by mutual induction? [Himachal 98]
30.Definecoefficient of mutual inductance for apair of
coils. [rSCE 98]
31.What is theunit of mutual inductance. [rSCE 96]
32.Define mutual inductance of one henry.
33.Twocircular loops areplaced with their centres
separated by afixed distance. Howwould you
orient the loops to have (i)the largest mutual
inductance and(ii)thesmallest mutual inductance?
Answers
6.57
34.
35.
Find the dimensions of mutual inductance.
What will be the dimensions ofL/ R, if L is
inductance andRis resistance?
Show that the SI unit ofinductance, henry is equal
to volt second per ampere. usee 03]
When current in a coilchanges with time, how is
theback emf induced in thecoil related to it?
[CBSEOO 08]
Mention anyone useful application of eddy
currents. [CBSE009]
Define self-inductance. Give its SI units. [CBSE F 09]
Define mutual inductance. Give its SI units.
[CBSE F 09]
Howdoes the mutual inductance of a pair of coils
change, when (i)distance between the coils is
increased (ii)number of turns in each coil is
increased ? [CBSE0013]
36.
37.
38.
39.
40.
41.
42.Figure 6.110 shows a
currentcarryingsole-
noid moving towards a
conducting loop. Find
thedirection of the
current induced in the
loop. [CBSE015C]
armr-O
Fig.6.110
1.The flux ismaximum when the area isheld
perpendicular to the direction ofthemagnetic field.
2.When the plane of the coil is parallel to the
magnetic field, flux linkedwith it is zero.
3.(i)SI unit of magnetic flux isweber (Wb).
(ii)SI unit of magnetic induction is tesla(T).
4.$=BA=~
qvsine
MLT-2 L2
..[$j= '. = [ML2r2A-1j.
AT.LT-1 .1
5.One weber is the flux produced when a uniform
magnetic field of one tesla actsnormally over an
area of 1m2.
1weber =108maxwell.
6.Refer to point 4 of Glimpses.
7.Noemf is induced because the magnetic flux linked
with the stationary coil is notchangin.
8.The magnitude. of induced emf is directly
proportional to the rateofchange of magnetic flux
linkedwith the circuit.
9.Ie1=d$=0.2x10-2=0.0167V.
dt 0.12
10.e=Blv=0.1x1x5= 0.5 V.
11.Takeclosed coil connected to a sensitive galvano-
meter and rotate the coil. If thegalvanometer shows
a deflection, a magnetic field is present on the
planet, otherwise not.
12.When the circuit is switched off, induced current is
in thesame direction as th-emain current.
13.e=Blv.
14.Zero, because glass rod is an insulator.
15.Change in flux
=<12-<ll=BAcos1800 -BAcos00=-2BA.
16.Ase=Blv
or
e,I1
e2=v2
e2=5.0V.
3.0 9.0
..r=15
2
17.Lenz's law states that the direction of induced
.current is such that it opposes the cause which
produces it.
18.Fleming's right hand rule can be used to find the
direction of current induced in a conductor.
Stretch the thumb and the first two fingers of the
orwww3notesdrive3com

527. 6.58
right hand in
mutually perpendicular directions.
If the first finger points in the direction of the
magnetic field, thumb in the direction of motion of
the conductor, then the central fingerwill point in
thedirection of current induced in the conductor.
19.Asper Lenz's law,theinduced current flows from
XtoYthrough the resistor R
20.Anticlockwise.
2'Clockwiseinring1andanticlockwise in ring 2.
22.Induced current flowsclockwise in the metal rin.
23.TheSIunitof elf-inductance ishenry (H).The self-
inductance of a coil is said to be onehenry if an
induced emfof one volt is set up in itwhen the
current initchanges at the rate of one ampere per
second.
24.Inductance getsdoubled, because
~N2A
L=_o--
I
25.Leq=I,.+Lz+2M.
AsL=~oN2.A
I
26.
so when N is doubled, the self- inductance becomes
four times the original one.
27.Ast.«N2
Lz_N;
"4-N12
Lz=(500)2 x 108=75ml-l.
(600)2
28.Theself-inductance ofa coil depends on
(i)itsnumber ofturns,
(ii)itsarea ofcross-section, and
(iii)thepermeability of the core material.
29.Thephenomenon of production of induced emf in
onecoil due toachange of current in the neigh-
bouring coil is called mutual induction.
"YPE B
PHYSICS-XII
30The mutual inductance of two coils isdefined as
the induced emf set up in one coil when the
current in the neighbouring coil changes at the
unit rate.
31.TheSI unit of mutual inductance ishenry.
32.Themutual inductance of two coilsis said to be one
henry if an induced emf of 1 volt is set up in one coil
when the current in the neighbouring coil changes
at the rate of 1 ampere persecond.
33.(i)When the planes of the twoloops are held
parallel to each other, mutual inductance is
largest.
(ii)When the planes of the loops are held perpen-
dicular to each other, mutual inductance is
smallest.
34.M=~= BA =_F_. A
I / qvsineI
.'. Dimensions of M
MLT-2.t2
CLT-1.A
35.MOLOT 1.
36.AsL=_e_
d// dt
IV
SI unit ofL=~~-1
lAs-
or 1 henry (H)=1VsA -1.
37.Backemf induced ccRateof change of current in
the coil
eo:dl
dt
or
3 .Eddy currents areused inelectric furnaces tomelt
metals.
39.Refer answertoQ.14on page 6.60.
40.Refer answer to Q.16on page 6.60.
41. (i)Mdecreases (ii)Mincreases.
42.Clockwise when seenfrom thesolenoidside.
SHORT ANSWER QU ESTIONS (2 or 3 marks each)
1.Define magnetic flux. Compute its dimensions.
2.When is the magnetic flux taken as (i)positive and
(ii)negative?
3.What is electromagnetic induction ?Give an
experiment which demonstrates thisphenomenon.
4.State Faraday's lawsof electro magnetic induction.
Express them mathematically. [CBSEOD95C, 99]
5.State Lenz's law.Give one example to illustrate this
law.The "Lenz's lawisa consequence of the law of
conservation 01ener". Justify this statement.
[CBSE D 09]www3notesdrive3com

528. ELECTROMAGNETIC INDUCTION
6.Describe a simple
experiment (or activity) to show
that the polarity of emf induced in a coil is always
such that it tends to produce a current which
opposes the change of magnetic flux that produces
it. [CBSE D 14]
7.State Lenz's law. Prove that the charge induced is
independent of time. [CBSE D 92C rPunjab 95]
8.Prove that the magnitude of the emf induced in a
conductor of lengthIwhen it moves at vm/s
perpendicular to a uniform magnetic fieldBisBlv.
[ISCE 03; CBSED06]
9.Derive an expression for the induced emf produced
by changing the area of a rectangular coil placed
perpendicular to a magnetic field. [CBSE D 92]
10.A rectangular coil ofNturns, areaAis held in a
uniform magnetic fieldB.If the coil is rotated at a
steady angular speed00,deduce an expression for
the induced emf in the coil at any instant of time.
Or rCBSE 9' ]
A coil of number of turnsN,areaAis rotated at a
constant angular speed00in a uniform magnetic
fieldB,and connected to a resistorR.Deduce
expressions for: (i)Maximum emf induced in the
coil(ii)Power dissipation in the coil.
[CBSE 06C, 08]
11.What are eddy currents ? How are these
minimised ? Mention two applications of eddy
currents. [CBSE OD 06 C, 09]
12.Define the term 'eddy currents'. State the main
undesirable effect of these currents and give the
method used to minimise this undesirable effect.
Write any two applications of eddy currents.
[CBSE OD 06 ; D 07C]
13.What is electromagnetic damping ? How is a
galvanometer made dead beat?
14.Define the term self-inductance. Write its51unit.
Give two factors on which self inductance of an
air core coil depends. [CBSE OD 15]
Answers
6.59
15.Define the term 'self-inductance'. Give its unit.
W rite an expression for the ener stored in an
inductor when a steady current '1'is passed
through it. Is this ener electric or magnetic?
[CBSE OD 04C]
Define mutual inductance. Write its51unit. Give
two factors on which the coefficient of mutual
inductance between a pair of coils depends.
[CBSEDII:OD 15C]
Derive an expression for the mutual inductance of
two long solenoids wound over one another, in
terms of their number of turnsNt,N2 ;common
cross-sectional areaAand common length /.
[CBSf. ODq ]
Derive the expression for the self-inductance of a long
solenoid of cross-sectional areaAand lengthI,
havingnturns per unit length.[CBSE D 12 rOD 13C
Define self-inductance and give its51unit. Derive
an expression for self-inductance of a long, air-
cored solenoid of lengthI,radiusrand havingN
number of turns. [CBS[ OD liS DO]
Or
Deduce an expression for the self-inductance of a
long solenoid ofNturns, having a core of relative
permeability).!r . CBSL D (jhC 0]
Define the coefficient of mutual induction.
A long solenoid, of length 1and radius1.,is
enclosed coaxially within another long solenoid of
lengthIand radiusr2(r2>1.andI>r2).Deduce the
expression for the mutual inductance of this pair of
solenoids. [C' S".Q ";D OgC]
Obtain the expression for the mutual inductance of
a pair of coaxial circular coils of radiirandR
(R»r)placed with their centres coincidin.
[CBSE D Ob]
(i)How are eddy currents reduced in a metallic
core?
16.
17.
18.
19.
20.
21.
22.
(ii)Give two uses of eddy currents. [CBSE F 09]
1.Refer answer to Q.1on page6.1.
2.Refer answer toQ.1on page6.1.
3.Refer answer to Q. 3 on page6.2.
4.Refer answer toQ.4on page6.4.
5.Refer answer toQ.6on page6.5.
6.See illustrations in the answer ofQ.5on page6.4.
7.Refer answer toQ.10 on page 6.10.
8.Refer answer to Q.11(2)on page6.15.
9.Refer answer toQ.11(2)on page6.15.
10.Refer answer toQ.11(3)on page6.15.
Instantaneous power,
C2 C 2
P=_=_0_sin2wt
R R
Average power dissipated per cycle,
C02 .2 C02 1
P=-<smwt>=-.-
avR R 2
N2B2 A2w2
2Rwww3notesdrive3com

529. 6.60
11.Eddy
currents are the currents induced in solid
metallic masses when changing magnetic flux
passes through them.
The eddy currents can·be reduced by using
laminated iron cores which consist of thin iron
sheets insulated from each other. The eddy currents
can also be reduced by using slotted iron blocks.
Eddy currents have following applications:
(i)they are used to make galvanometers dead beat.
(ii)In electric furnaces to melt metals.
12.Eddy currents are the currents induced in solid
metallic masses when changing magnetic flux
passes through them. Changing magnetic fields set
up current loops in the metallic masses. These loops
are irregularly shaped.
Eddy currents are considered undesirable in a
transformer because they dissipate ener in the
form of heat. To reduce the loss, the core is made of
a large number of thin layers separated by
insulating material. This increases the resistance of
the possible paths and hence reduces the eddy
currents. For applications of eddy currents, refer
answer to the above question.
13.Refer answer to Q. 12 on page 6.19.
14.The self-inductance of a coil is defined as the
induced emf set up in the coil when the current
through it changes at the unit rate.
PHYSICS-XII
SI unit of self-inductance is henry (H).
The self-inductance of an air-core depends on
(i)the number of turns in the coil, and
(ii)the area of cross-section of the coil.
15.Refer answer to above question.
When a currentIflows through an inductor of
inductanceLthe ener stored in it is
U=..!LI2
2
This ener is stored as magnetic ener.
16.Mutual inductance.The mutual inductance of two
coils is defined as the induced emf set up in one coil
when the current in the neighboring coil changes at
the unit rate.
SI unit of mutual inductance is henry (H).
The mutual inductance of two coils depends on
(i)the number of turns and the geometrical shape
of the two coils.
(ii)the relative orientation of the two coils.
17.Refer answer to Q. 18 on page 6.23.
18.Refer answer to Q. 15 on page 6.21.
19.Refer answer to Q. 15 on page 6.21.
20.Refer answer to Q. 18 on page 6.23.
21.Refer to the solution of Problem 29 on page 6.40.
22.Refer answer to Q.12 above.
"YPE C:LONG ANSWER QU ESTIONS (5markseach)
1.State Faraday's laws of electromagnetic induction
and explain three methods of producing induced
emf. [Punjab02, 03]
2.(a)State the law which relates to generation of
induced emf in'a conductor being moved in a
magnetic field.'
(b)A rod of lengthIis moved horizontally with a
uniform velocity'tr'in a direction perpen-
dicular to its length through a region in which
a uniform magnetic field is acting vertically
downward. Derive the expression for the emf
induced across the ends of the rod.
(c)How does one understand this motional emf
by invoking the Lorentz force acting on the free
charge carriers of the conductors? Explain.
[eBSEOD14;SP 15]
3.(a)StateFaraday's law ofelectromagneticinduction.
(b)A horizontal straight wire of length Lextending
from east to west is falling with speedvatright
angles to the horizontal component of Earth's
magnetic field B.
(i)Write the expression for the instantaneous
value of the emf induced in the wire.
(ii)What is the direction of the emf?
(iii)Which end of the wire is at the higher
potential? [eBSEOD11]
4.What are eddy currents? How are they produced?
Describe briefly three main useful applications of
eddy currents. [eBSEF 15]
5.Derive an expression for the induced emf set up in a
coil when it is rotated in a uniform magnetic field
with a uniform angular velocity. Explain how does
the emf vary when the coil turns through an angle
of21t? What is the instantaneous value of induced
emf when the plane of the coil makes an angle of 60°
with the magnetic lines. [eBSEOD95C]www6notesdrive6com

530. ELECTROMAGNETIC INDUCTION
6.Distinguish between self-induction and mutual
induction. Calculateself-inductanceof a long solenoid
of length I, number of turns N and radius r.
[eBSE0 92C]
7.(a)Define mutual
inductance. Deduce an
expression for the mutual inductance of two
long coaxial solenoids having different radii
and different number of turns.
(b)A coil is mechanically rotated with angular
speed<.0in a uniform magnetic field which is
perpendicular to the axis of rotation of the coil.
The plane of the coil is initially perpendicular
Answers
6.61
to the field. Plot a graph showing variation of
(i)magnetic fluxcpand(ii)the induced emf in
the coil as a function oftct.[eBSE0009,F13)
8.(a)Define mutual inductance and write its 51units.
(b)Derive an expression for the mutual
inductance of two long co-axial solenoids of
same length wound one over the other.
(c) In an experiment, two coils ~ and ~ are
placed close to each other. Find out the
expression for the emf induced in the coil ~
due to a change in the current through the coil
~. [eBSE015)
••
1.Refer answers toQ.3on page6.2&Q.4on page6.4.
2.(a)According to Faraday's flux rule, the magni-
tude of induced emf is equal to the rate of
change of magnetic flux linked with the closed
circuit.
e=_dcp
dt
(b)Refer answer to Q. 7 on page6.9.
(c) Refer answer toQ.9 on page6.10.
3.(a)Refer to point 4 of Glimpses
(b)(i)e=BHLv
(ii)Induced emf will be in the direction from
west to east.
(iii)West end of the wire will be at higher
potential.
4.Refer answer toQ.12on page6.18.
5.Refer answer toQ.ll(3)on page6.15.
Instantaneous value of induced emf,
e=eosin(90°- 60°)
=eosin30°=eo /2
6.Refer to solution of Problem 7 on page6.45.Also,
refer answer toQ.15on page6.21.
7.(a)Refer answer toQ.18on page6.23.
(b)The variation of magnetic flux(cp=BAcos<.Ot)
and induced emf(e=eosinrot)"asa function of
rotis shown below.
+BA
-BA
- eo
Fi. 6.111
8.(a)Refer answer toQ.17on page6.22.
(b)Refer answer toQ.18on page6.23.
(c) Flux linked with coil ~ocCurrent in coil ~
or<1\oc12 or <1\=M12
.. d<l\=MdI2
dt dt
EMF induced in coil ~,
e= _d<l\= _MdI2
dt dt
~YPE D:VALUE BASED QUESTIONS (4 marks each)
1.Vikas once observed that the large number of
electric generators used in areas where small
workshops existed, produced lot of pollution. He
decided to do something for controlling pollution.
He alongwith his some friends, made a survey and
concluded that like in society apartments, a
common generator could be set up for all such
workshops, so that noise and air pollution could be
reduced considerably. They had a tough time in
convincing the local bodies and they were going towww3notesdrive3com

531. 6.62
some NGOs and other financiers to
help them to set
up a big generator. Itwasadmirable tosee their
perseverance.
(a)What are the values being displayed by Vikas
and his friends?
Answer
PHYSICS-XII
(b)State the factors on which the induced emf in a
coil rotating in a uniform magnetic field
depends.
1.(a)Team spirit, patience, tolerance, magnani-
mity, determination, responsibility and
dutifulness.
(b)Induced emf,
E.=NBArosinrot
Clearly,induced emf setup in the coildepends on
(i)number of turns of the coil.
(ii)area of the coil.
(iii)angular speed of rotation of the coil and
(iv)strength of the magnetic field.www4yztpso·tvp4nzx

532. Electromagnetic Induction
GLIMPSE
S
1.Magnetic flux. The number ofmagnetic lines of
force crossing a surface normally iscalled
magnetic flux linked with the surface. If the
normal drawn to the surface area Amakesangle
~
ewith the field B, then the magnetic flux is
~ ~
$=BAcose=B.A
Magnetic flux is a scalar quantity. Its
dimensions are[ML2A-IT-2].
2.SI unit of magnetic flux is weber (Wb). It isthe
fluxproduced when auniform magnetic field of
1 T crosses normally anareaof 1 m2. The CGS
unit of magnetic flux is maxwell.
1 weber =108ma~well
3.Electromagnetic induction. It is the pheno-
menon of production of induced emf (and
hence induced current) due to a change of
magnetic flux linked with aclosed circuit. The
term electromagnetic induction means inducing
electricity bymagnetism.
4.Faraday's laws of electromagnetic induction.
First law.Whenever the magnetic flux linked
with aclosed circuit changes, an emf is induced
in it which lasts only so long as the change in flux
istaking place.
Second law.The magnitude of induced emf is
equal to the rate of change of magnetic flux
linked with the closed circuit. Mathematically,
lei=d$
dt
5.Lenz's law. Itstates that the direction of induced
current is such that it opposes the cause which
produces iti.e.,it opposes the change in flux.
6.Mathematical form of the laws of electro-
magnetic induction.For acoil of one turn,
inducedemf is given by
e= _d$=_<i2-<11
dt t
Fora coil of N turns, e=-Nd$= -N<i2 - <11.
dt t
The negative sign shows that induced emf
opposes thechange in flux. It isin accordance
with Lenz's law.
7.Motional emf. The emf induced across the ends
of a conductor due to its motion in a magnetic
field is called motional emf. It is produced due
to the magnetic Loren force acting on the free
electrons of the conductor. If a conductor of
lengthlmoves with velocity vin a magnetic
field Bperpendicular to both its length and the
direction of the magnetic field, then the emf
induced across itsends is given by e=Blv
eBlv
Induced current, I=-=-
R R
Force necessary to move the conductor,
Wl2V
-F=--
R
Power dissipated as Joule heating loss,
WPv2
P=Fv=--
R
8.Fleming's right hand rule. It gives the direction
of induced emf. If we stretch the thumb and the
first two fingers of our right hand mutually
perpendicular to each other and if the forefinger
points in the direction of the magnetic field,
thumb in the direction of motion of the
conductor, then the central finger points in the
direction of current induced in the conductor.
(6.63)wwwmnotesdrivemcom

533. 6.64
9.Relation between induced charge and change in
magnetic flux.The induced charge flowing
through a circuit depends on the net change in
magnetic flux linked with circuit and is inde-
pendent of the time interval of the flux change.
6<1>Net change in magnetic flux
6q=--=
------~----~------
R Resistance
10.Methods of generating induced .emf.As
<I>=BAcos 9, so the magnetic flux linked with a
loop can be changed and hence induced emf can
be produced by three methods:(i)bychanging
the magnetic field B,(ii)by changing the areaA
of the loop and(iii)by changing the relative
orientation 9 of the loop and the magnetic field.
11.Motional emf in a rotating coil. If an N turn coil
of areaAis rotated with uniform angular
velocityroin a uniform magnetic field B about
an axis perpendicular to the field B, then the
motional emf set up across the ends of the coil is
given by
e=NBArosinrot=eosinrot=eosin 21tft
whereeo=NBAco=peak value of induced emf.
Both the direction and magnitude of the induced
emf change regularly with time. Such a sinu-
soidally varying emf is called alternating emf.
12.Eddy currents.These are the currents induced
in solid metallic masses when the magnetic flux
threading through them changes. Such currents
flow in the form of irregularly shaped loops
throughout the body of the metal and their
direction is given by Lenz's law. Eddy currents
cause unnecessary heating and wastage of
power. They are reduced by using laminated soft
iron cores. They are useful in (i)electric brakes,
(ii)speedometers,(iii)induction furnaces and
(iv)electromagnetic shielding.
13.Self induction.It is phenomenon of production
of induced emf in a coil when a changing
current passes through it.
14.Self-inductance or coefficient of self induction
(L).When a currentIflows through a coil, flux
linked with it is<I>=LI
Induced emf,e= _d<l>=_LdI
dt dt
Thus self-inductance of a coil is the induced emf
set up in it when the current passing through it
changes at the unit rate. It is a measure of the
opposition of the coil to the flow of current
through it.
PHYSICS-XII
15.Self-inductance of a long solenoid.Theself-
inductance of a long solenoid of length I, area of
cross-section Aand having N turns is
~ N2A N
L=--o---=lIon2IA wheren=-
1 ..' 1
When the solenoid is wound over asoft iron
core of relative permeability~r' L=~r ~on21A
The value of self inductance depends on the
number of turns in the solenoid, itsarea of
cross-section and the relative permeability of its
core material.
16.Mutual induction.Itisthe phenomenon of
production of inducedemf in one coil when the
current through the neighbouring coil changes
at the unit rate.
17.Mutual-inductance or coefficient of mutual
induction(M).If a currentIflowing through
one coil generates flux<I>in the neighbouring
coil,then <I>=MI
d<l> dI
Induced emf, e=--=- M-
dt dt
Thus mutual-inductance of two coils may be
defined as the induced emf set up in one coil
when the current in the neighbouring coil
changes at the unit rate.
18.Mutual inductance of two long solenoids.The
mutual inductance of two long co-axial
solenoids wound over one another is
wherenI,nzare the number of turns per unit
length of thesolenoids, 1is their common length
andA=1t'12= cross-sectional area of the inner
solenoid.
19.Henry.It is SI unit for both self and mutual-
inductances. Inductance is one henry if an
induced emf of 1 volt is set up when the current
changes at the rate of 1 ampere per second.
1 henry (H)= 1 VsA-I =1 Wb A-I.
20.Coefficient of coupling. It gives a measure of the
manner in which two coils are coupled together.
If4andLzare the self-inductances of the two
coils and M is the mutualinductance, then the
coefficient of coupling
K=~4MLz
The value ofKlies between 0 and 1.www7notesdrive7~om

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621. C H A PT E R
ELECTROMAGNETIC
WAVES
8.1INTRODUCTIO
iefrve te electromagnetic"aues.We
ha ·elearnt that an electric rrent produces a mag-
.c eld. Also a magnetic field changing with time
u an electric field. Can an electric field
changing with time produce a magnetic field?James
Cleric,\fuxwel/ (1831-1879), argued that this was indeed
the case-an electric field changing with time produces a
magnetic field. Maxwell noticed that Ampere's
circuital law is inconsistent namely, makes non-unique
predictions for the magnetic field in situations where
electric rrent changes with time. He showed that
consisten requires an additional source of magnetic
field, this is calleddplacement current. This made the
laws of electrity and magnetism symmetrical.
Maxwell formulated a set offour equations, called
Maxwell's equatns. With the help of these equations,
he predicted that electric and magnetic fields
dependent on time and space propagate as transverse
waves, calledelectromagnetic waves.His discovery that
electromagnetic waves travel with the speed of light
led him to a remarkable conclusion that light is an
electromagnetic wave. Heinrich Hert, in1865,
sucessfully demonstrated the existence of electro-
magnetic waves. A fewyears later,Guglielmo Marconi
of Italy suceeded in transmitting electromagnetic
wavesover distances of several kilometres. His
experiments brought a revolution in communication
which we are witnessing even today.
8.2 MAXWELL'S DISPLACEMENT CURRENT
1. Discuss the inconstency inAmpere's circuital
law.What modificatn was made by Maxwell inth
law?Whatisdplacement current?Conductn and
displacement currents are individually dcontinuous,
but their sumiscontinuous. Comment.
Inconsistency of Ampere's rital law. Acording
to Ampere's rital law, the line integral of the
~
magnetic field Balong any closed loopC proportnal to
the current I passing through theclosed loop, i.e.,
fB.dt=~oI ...(1)
c
In 1864, Maxwell showed that the equation (1) is
logically inconsistent. Toprove this inconsistency, we
consider a parallel plate capator being charged by a
battery as shown inFig.8.1(a).As the charging
Capacitor plates
I ~ I
_Cj \) _
Key
(a) (b)
Fig.8.1A parallel plate capacitor being
charged by a battery.
(8.1)°°°1uv{lzkyp®l1jvt

622. 8.2
continues, a
rrentIflows through the connecting
wires, which of course changes with time. This rrent
produces a magnetic field around the capator.
Consider two planar loops C1 andC2,C1just left of the
capator andC2in between the capator plates, with
their planes parallel to these plates.
Now the rrentIflows across the area bounded by
loopC1because connecting wire passes through it.
Hence from Ampere's law, we have
fB.dl=~oI ... (2)
c1
But the area bounded byC2lies in the region
between the capator plates, so no rrent flows
across it.
fB.dl=o
c2
Imagine the loopsc;.andCztobe infinitesimally close to
each other, as shown in Fig.8.1(b). Thenwe must have
fB.dl=fs.s
C1 c2
This result is inconsistent with the equations(2)
and(3). Soa need for modifying Ampere's law was felt
by Maxwell.
Maxwell's modification of Ampere's law : Dis-
placement rrent. To modify Ampere's law, Maxwell
followed a symmetry consideration. By Faraday'S law,
a changing magnetic field induces an electric field,
hence a changing electric field must induce a magnetic or
field. As rrents are the usual sources of magnetic
fields, a changing electric field must be assoated with
a rrent. Maxwell called this rrent as the
dplacement currentto distinguish it from the usual
conduction rrent caused by the drift of electrons.
Displacement currentisthat current which comes
into existence, in addition tothecondudioncurrent,
whenever the electric field and hence the eledric flux
changes withtime.
To maintain the dimensional consisten, the
displacement rrent is given the form:
_d4t
Id -EO-
dt
where4t=electric fieldxarea=EA,is the electric flux
across the loop.
:.Total rrent across the closed loop
=Ie+Id=Ie+EOd4t
dt
Hence the modified form of the Ampere's law is
fB.dl=~O[Ie+EOd~] ...(5)
PHYSICS-XII
Unlike the conduction rrent, the displacement
rrent exists whenever the electric field and hence the
electric flux is changing with time. Thus acording to
Maxwell, the source of a magnetic field is not just the
conduction electric rrent due to flowingcharges, but
also the time-varying electric field. Hence the total
rrentIisthesum of the conduction rrentIeand
displacement rrentId
I=i,+Id= '.+EOd~
Consisten of modified Ampere's law. For loop
Cl,there is no electric flux(4t=0).Therefore, from
equation(5)we have
...(3)
fB.dl=~OI
c1
For loopC2,conduction rrentI=0butId*0,
because a time-varying electric field exists in the
region between the capator plates. Hence
fB.dl=~oEOd4t ...(7)
C dt
2
...(6)
...(4)
IfAbe the area of the capator plates andqbe the
charge on the plates at any instanttduring the
charging process, then the electric field in the gap will
be
E=-q-
EoA
EA=J...-
EO
or
or Flux
This agrees with the equation (6), proving the
consisten of the Ampere's modified law (5).
Property of continuity.The sum(Ie+Id)has the
important property of continuity along any closed
path even when individually IeandIdmay not be
continuous. In Fig. 8.1,for example, a rrentIeenters
one plate and leaves the other plate of the capacitor.
The conduction rrent
I=dq
edt
is not continuous across the capator gap asno
charge is transported across this gap. The displacementwww2z~tpsoruvp2n~y

623. ELECTROMAGNETIC WAVES
rrent1.1is zero
outside the capator plates and in the
gap, it has the value
EOdclt=EU ~(EA)=EU ~(!L)=dq ...(8)
dt dt dt EUdt
which is exactly the value of the conduction rrent in
the lead wires. Thus the displacement rrent satisfies
the basic condition that the rrent is continuous.
The sumIe+EOdclthas the same value along the
dt
entire path (both inside and outside the capator
plates), although individually the two rrents are
discontinuous. Clearly, outside the capator plates,
we have only conduction rrentI.=I,and there is no
displacement rrent(Id=O~ While inside the capa-
tor plates, there is only displacement rrent1.1=I,
and there is no conduction rrent(I.:=0). But in any
general medium, bothIeand1.1are present. However,
Ieis larger than1.1in a conducting medium while1.1is
larger thanIein an insulating medium.
2. Is a dplacement current associated with a
magnetic fidd?Or, can a changing electric flux induce a
magnetic field?Explain it with the help of an example.
Induced magnetic field.A displacement rrent
produces the same physical effects as the conduction
rrent. Like a conduction rrent, a displacement
rrent is also assoated with a magnetic field.
Consider, for example, the charging of a parallel plate
capator by a constant rrentIin the connecting
wires [Fig.8.2(a)].This increases the charges on the
capator plates at a steady rate. Consequently, the
electric field between the plates also increases at a
steady rate. Between the capator plates, there exists a
displacement rrent due to time varying electric field.
In such a region, we expect a magnetic field though
there is no source of conduction rrent nearby.
~
Experiments have shown that a magnetic field B is
Q
--x-x--
"' ....x x~ .... x',
I ;'xl ;;:x x ;;'xx' \
I XI X X X xrX'xE'
IxI'x xx x x ~x\
IB \ \
,x x x x x x x x
Ix\x x x x xtx:
\x\' x x x x I ~I
\xvx rx.I
\ , I
\Xx....Xx x x....x I
'-,x~x-;; x...
•..B ~....._ ~ _x_-
-->
B
---+.. -
--'--'--+
+: :-
i(t)-.-f--.,--....--i.-+- --
+ .. -
~
~
~c
---"--
+ - -
(a) (b)
--> -->
Fig. 8.2(a)Electric and magnetic fieldsEandBat any pointQ
between the capator plates.(b)A cross-sectional view of
Fig.8.2(a)
8.3
indeed induced (say at a point Q)between the
capator plates and has same magnitude as that just
outside (say at pointPl.
In Fig.8.2(a),the direction ofEis from the positive
..... .
plate to the negative, whereas the direction of B at\.2is
perpendilar to the plant! of paper. Fig.8.2(b)shows a
cross-sectional loop parallel to the plant! of the plates.
The fieldEis directed normally into the plane paper,
~
as shown by crosses. The induced field B is clockwise
along the tangents un a rcle in this cross-sectional
plane.
3. Statetheimportant consequences vfdplacement
current.
Consequences of displacement rrent. The
concept of displacement rrent has made the lawsvi
electrity and magnetism symmetrical. Acording tv
Faraday'S law of electromagnetic induction, the
magnitude of induced emfisequal to the rate ofchange
magnetic flux.But the emf between two points is the
work done in taking a unit charge from one point tu
another against the electrostatic forces. 'This implies
the existence of an electric field in the region. So
Faraday'S law simply states that a time varying
magnetic field gives rise to an electric field.
By symmetry, a time varying electric field should
give rise to a magnetic field. This is an importance
consequence of displacement rrent which is a source
of magnetic field.
Another very important consequence of the
symmetry of electrity and magnetism is the existence
of electromagnetic waves, so important for modern
communication.
4. State the important properties of dplacement
current.
Important properties of displacement rrent.
These are as follows :-
1. Displacement rrent exists whenever there is a
change of electric flux. Unlike conduction
rrent, it does not exist under steady conditions.
2. It is not a rrent. It only adds to rrent density
in Ampere's rital law. As it produces
magnetic field, so it is called a rrent.
3. The magnitude of displacement rrent is equal
to the rate of displacement of charge from one
capator plate to the other.
4. Together with the conduction rrent,
displacement rrent satisfies the property of
continuity.www2z~tpsoruvp2n~y

624. 8.4
....
...
Formulae Used
d
~E
1.Displacementcurrent, Ia=EOat
d dE
AlsoIa=Eodt (EA) =EoAdt
=A ~ (V)=EoA dV=CdV
Eodtd d dt dt
2.Modified Ampere's cirital law,
fss=!loUe+ Ia)
Units Used
Electric fieldEis in Vm-1,voltage Vin volt, area
Ain m2,distancedin m, current Iinampere and
fieldBin Wb m -2or tesla.
Constants Used
Permittivity constant, Eo=8.85x1O-2C2N-1m-2
Permeability constant, !l0=4nx10-7TmA-1.
Example 1.A parallel plate capacitor has circular plates,
each of radius5.0em Itisbeing charged so that electric field
in the gap between its plates res steadily at the rate of
1012Vm-1 s-1.Whatisthe dplacement current?
Solution. Herer=5em =5x10-2m,
dE=1012Vm-1s-1
dt
Displacement rrent,
_ d ~_ dE_ 2dE
Io -EO -- -EOA --EO •ttr: -
dt dt dt
= 8.85 x10-12 x1tx(5x10-2) x1012A =0.07 A.
Example2.The voltage between the plates of a
parallel-plate capacitor of capacitance1.0J.lF changing at
the rate of 5Vs-1.Whatisthe dplacement currentinthe
capacitor?
Solution. HereC=1.0J.lF=1.0x1O-6F,
dV=5Vs-1
dt
Displacement rrent,
Io=EOd~=EO ~(EA)=EO~(VA)
dt dt dt d
=EOA dV=CdV
d dt dt
= 1.0x10-6 x 5 A=5.0J.lA
PHYSICS-XII
Example3.A parallel plate capacitor of area50 ~and
plate separatn3.0mmischarged initially to80J.lc.Due to
radactive source nearby, the medium between the plates gets
slightly conducting and the plate loses charge initially at the
rate of1.5 x10-8Cs-1.Whatisthe magnitude and directn
of dplacement current? Whatisthe magnetic field between
the plates?
Solution. Due to leaking, there is a flow of+ve
charge from the+veplate to the -ve plate (or the flow
of-ve charge in the reverse direction). Thus the
'conduction rrent within the plates is from the+ve
plate to the -ve plate. Now the displacement
current is
_d~ _ dE
10 -EO - - EOA -
dt dt
d [ qJ 1dq
= EoA dtEOA=EOAEOA dt
10=dq=1.5:10-8 Cs-1
dt
or
The direction ofdisplacement rrent is opposite to
that of electric field Eand hence opposite to the
conduction rrent. But its magnitude is same as that
of the conduction rrent. The net rrent between the
plates is zero.
Using Ampere's law (with 10there replaced by
I=Ie+'»=0),
fB.dl =J.loI=J.l0(Ic+1o)=J.loxO=0
So the magnetic field within the plates is zero at all
points.
Example4.Refer to Fig.8.3.Use modified Ampere's
circuital law and the symmetry in the problem to calculate
magnetic field between the - - -- --
plates at a point (i) on the
ax(ii)6.5em from theaxis
(iii)15em from the ax.
(io)At what dtance from
the axisthe magnetic
field due to dplacement
current greatest?Obtain
the maximum value of the
field.
.._--_.....
Fig. 8.3
Solution. Consider a rlar loop of radius r
between the plates and co-axial with them,.e.,its
centre lies on the axis of the plates and its plane is
normal to the axis, as shown in Fig.8.3.By symmetry,
Bis tangential to the rcle at every point and equal in
magnitude over the rcle. Therefore,
fB.dl=B·f dl=Bx21tr=21trB
Circlewww2z~tpsoruvp2n~y

625. ELECTROMAGN
ETICWAVES
Astheconduction rrent, I=0in the region
between the plates, therefore, from modified Ampere's
law
fB.ell=21trB=1l0 ld
= 11OEOxRateofchange of electric
fluxthrough the area1t?
When r~R.Let<IEbe the flux through area1tR2.
Then flux through area1t?(which is less than1tR2)
1t? ?
=<IE1tR2=<IER2
•.Rateofchangeof flux through the area 1t?
d<IE?
=Tt·R2
Hence,
d<IEr2 ?
21trB= 110EOTt·R2=ld·R2
.',Forr~R,
B=1l0rId
21tR2
When r ~R.In this case the capator plates are
totally enclosed by the area1t?of the larger rcle, so
the total rrent through the area1t?isld•Thus
21trB=110ld
..Forr~R,
B=110ld
21tr
(i)From equation (a),the magnetic field on the axis
(r=O)
B=O
(ii)Herer=6.5em andR=12em,i.e.,r<R.Again
using equation(a),we get
B= 41tx 10-7x6.5x10-2x0.15
21tX(12x10-2)2
=1.35x10-7T [ld=0.15 A]
(iii)Bismaximum at r=R.From(a)or(b),we have
B =11Old= 41txlo-7xO.15
max21t R 21t x12x10-2
=2.5x10-7T.
Example5.Aparallel plate capacitor with circular plates
of radius1m has a capacitance of1nF.Att=0,itis
connected for charging in series with arestance R=1M!1
across 2Vbattery. Calculate the magnetic field at a pointP,in
betweenthe plates and halfway between the centre and the
periphery oftheplates,after 10-3 s. [NCERT]
8.5
Solution. Herer=1 m, C= 1nF=10-9F
R=1Mn=106!1, V=2V,t=10-3 s
Time constant of theRCrit is
"t =RC=106 x10-9 =10-3 s
QR=lm
PQ=lm
2
-..--
Q P
-Lv t----JV\I\r---'
IMn
...(a)
Fig.8.4
Charge on the capator plate at any instant
during charging is given by
q(t)=CV[ l-e-;C ]=10-9x2[I-e-1:-3]
The electric field in between the plates at timetis
E=q(t)=q(t;=q(t) ...(1)
EoA EOltl E01t
Now consider a rlar loop of radius 1/2m
parallel to the plates passing throughP.The magnetic
fieldBat all points on this loop is along the loop and of
the same value.
The flux<IEthrough this loop is
<IE=Exarea of the loop
(1)2 E
=Ex1tX2"= 1t4 = 4 :0 [Using(1)]
The displacement rrent is
t,= EOd~= EO:t(4!J
=!:.dq=!:.~ [2x1O-9(I_e-1:-3) ]
4 dt4dt
= _!:.x2x1O-9.e-t/1O-3. (__1_)
4 10-3
= 0.5 x10-6e:t/1O-3
Att=10-3s, ld=0.5 x10-6e-l
Applying Ampere's rital law to this loop, we get
B.21tx!:. = 110(Ie+ld)=110 (0+0.5x1O-6e-1)
2
B= 110 x0.5x10-6 41tX10-7 x0.5x10-6
1t e 1tX2.718
=0.74x10-13T.
... (b)www2z~tpsoruvp2n~y

626. 8.6
flroblems For Practice
1.How would
you establish a displacement rrent of
2.0A in the space between the two parallel plates of
1JlF capator?'
(Ans.By changing p.d. across the capator
plates at the rate of 2x10"Vs-1)
2.A capator consists of two rlar plates each of
radius10.0em and separated by2.0mm. The
capator is being charged by an external battery.
The charging rrent is constant and equal to0.5A.
Callate(a)the capatance,(/1)the rate of change
of potential difference across the plates and(c)the
displacement rrent.
(Ans. 13&94pF,3.nx109Vs-1, 0.5A)
3.A parallel plate capator has two metal plates of
size30emx15em and separated by2.0mm. The
capator is being charged so that the charging
rrent has a steady value of100mA. Callate the
rate of change of potential difference between the
capator plates. What is the displacement rrent
in the region between the capator p~ates ?
(Ans.5 x 108 Vs-1, 100mA)
4.A parallel plate capator of capatanceC=0.1JlF is
connected across an a.c. source of (angular) fre-
quen500rad s-1.The value of conduction rrent
is1mA.What is the rms value of the voltage from
the source ? What is the displacement rrent
across the capator plates? (Ans. 20V,1mA)
5.A parallel capator made of rlar plates of
radius10.0em has a capatance of200pF. The
capator is connected to a200a.c. supply with an
angular frequen of200rad s-l,
(i)What is the r.m.s. value of conduction rrent?
(ii)Is the conduction rrent equal to displace-
ment rrent ?
(iii)Find peak value of displacement rrent? •
(iv)Determine the amplitude of magnetic field at a
point2.0em from the axis between the plates.
[Ans.(i)8JlA(ii)Yes(iii)11.312JlA
(iv)4.525 x 10-12 T]
HINTS
1.As proved in Example 2,
1=CdV
D elt
dV1D 2.0
Tt=-C-= lO-n
=2 xlOn V,,-l.
2.Proceed as in ExerseKlon paReK32.
PHYSICS-XII
3.I=dq= ~(CV)=CdV=EoA . dV
dt dt dt d dt
dV Id
.. -=--
dt fllA
100x10-3x 2 x 10-3
=8.85 x 10-12 x 0.30 x 0.15
=5x108 Vs-1•
4.HereC=O.IJlF=10-7F,00=500rad s-l,
I,ms=ImA=10-3A
1
V,ms=Xc .I,ms= - .I,ms
ooC
17x 10-3 =20V.
500x10-
5.HereR=10em=0.10m,
C=200pF=2 x 10-10 F,V,ms=200V
00=200rad s-l,r=2.0 x 10-2 m
V V
(I)I = --11!&=-.-!.!!1L=00CV
,msXc 1/ec rms
=200x2xl0-10 x200
=8x10-6A=8JlA.
(ii)Yes, becauseID=I
(iii)10=.fiIrms=.fix 8 x 10-6
=11.312 x 10-6 A
=11.312JlA.
(iv)Consider a rlar loop of radius rbetween
the plates and co-axial with them. Area of the
loop,A'=1t?By symmetry,Bis tangential to
the rcle at every point and is equal in
magnitude over the rcle. Here only a part of
the displacement rrentIDthreads the loop of
areaA'.
:. Current through areaA'
ID 2r2
=--2X1tT =2·ID
1tR R
Using modified Ampere's rital law,
!-+ -+
:rB .dl=JloxCurrent through areaA'
?
B .21tT=Jl0R2ID
B_JloIDr
- 2n R2
=--------..-----www2z~tpsoruvp2n~y

627. ELECTROMAGN
ETICWAVES
8.3MAXWELL'S EQUATIONS
Maxwell found that all the basic prinples of electro-
magnetism can be formulated in terms of four funda-
mental equations called Maxwell's equations. Assuming
that no magnetic or dielectric material is present, the
four basic equations can bestated as follows:
1. Gauss law of electrostatics. This law states that
theelectric flux through a closed surfaceSis ~timesthe
EO
totalchargeqenclosed bythe surface S.
fE.JS=!L
s EO
Important consequences of this lawarethat(i)the
charge on an insulated conductor resides only on its
outer surface, and(ii)the electrostatic force between
two chargesis inversely proportional to the square of
the distance between them.
i.e.,
2. Gauss law of magnetism. Acording to this law,
the magnetic flux through any closed surface is always
zero.
i.e., fB.dS=0 ...(10)
s
This lawimplies that isolated magnetic poles or
magnetic charges do not exist, i.e.,it explains the
absenceof magnetic monopoles.
3. Faraday's law of electromagnetic induction.
Thislaw tells that achanging magnetic field induces
an electric field. Acording to thislaw,the induced emf
set upin a closed ritCis equal to the rate of change
of magnetic flux linked with the closed rit.
i.e.,
f
E.d!=_d<JB
c
dt
E=-~[f~~lB.dS ...(11)
dtc
or
4. Modified Ampere's law. This law states that the
line integral of the magnetic field around anyclosed
rit C is equal toJ.lotimes the total rrent (the sum
of conduction and displacement rrents) threading
theclosed rcuit.
fB.d!= J.lo[Ie +Id]=J.lo[Ie+EOd<lt_]
c dt
This law implies. thefact that not only a con-
duction current but a displacement current, asso-
atedwith achanging electricfield, also produces a
magnetic field.
i.e., ...(12)
8.7
...(9)
8.4 MAXWELL'S PREDICTION OF
ELECTROMAGNETIC WAVES
5. Explainclearly how Maxwell was led to predict the
existence ofelectromagnetic waves. How can these waves
be represented mathematically?
Maxwell's prediction of electromagnetic waves.In
1865, Maxwell theoretically predicted the existence of
electromagnetic waves. Acording to Faraday's law of
electromagnetic induction:
Atime-varying magnetic field isasourceofchanging
electricfield.
On the basis of his theoretical studies, Maxwell
argued that
Atime-varyingelectric field isasource of changing
magnetic field.
This means that the change in either field
(electric/magnetic) produces the other field. Maxwell
further showed that these variations in electric and
magnetic fields or in mutually perpendilar
directions and have wave like properties. He was thus
led to the idea that a wave of electric and magnetic
fields both varying with space and time should exist,
one providing the source of the other. Such a wave is
called anelectromagnetic waveand it indeed exists.
Anelectromagnetic wave isawave radiatedbyan
accelerated charge and which propagates through
spaceascoupled electric and magnetic fields,
oscillating perpendicular to each other and to the
directionof propagatn of the wave.
Mathematical representation of electromagnetic
waves.Figure 8.5 shows a plane electromagnetic wave
travelling along X-axis. The electric field Eosllates
~
along Y-axis while the magnetic fieldBosllates
along Z-axis.
z
Direction of propagation ~
Fig.8.5A plane electromagnetic wave
travelling along X-axis.
The values ofelectric and magnetic fields shown in
theabove figure depend only on xandt.The electricwww2z~tpsoruvp2n~y

628. 8.8
field vector can
be represented mathematically as
follows:
~" "
t:=Eyj=Eosin(kx -cor)j
=Eosin[2n(i-vt)]j
=EoSin[2n(i-~)]j...(1)
where k=2n /f...is the propagation constant ofthe
wave and angular frequen,ro=2 zrv,
Clearly, Ex=Ez=a
The magnetic field vector may be represented as
-;::t" "
Jj=Bzk= ~sin(kx - wt) k
= ~sin[2n (i-vt)]k
=~sin[2n(i-~)]k
Clearly, Bx=By=O.
HereEoand ~ aretheamplitudesof the electric field
Eand magnetic field 13,respectively.
Equations (1)and(2)show that thevariatnsin
electric and magnetic fields are insame phase, i.e.,both
attain their maxima and minima at the same instant
and at the same place (x).
The magnitudes of Eand13are related as
.!i=corEo=c
B ~
Maxwell also showed that the speed of an e.m.
wavedepends on the permeability and permittivity of
the medium through which it travels. The speed of an
e.m.wave in free space is given by
1
c=--
~J.lOEO
Permeability of free space,
J.lo=4nx10-7Ns2C-2
Permittivity of free space,
EO=8.85x10-12C2N-1m-2
1
c=-r====~========~
~47tx10-7x8.85x10-12
=3.0x108ms-l
which isthe speed of light in vaum. This fact led
Maxwell topredictthat light is an electromagnetic
PHYSICS-XII
wave. The emergence of the speed of light from purely
electromagnetic considerations is the crowning
achievement of Maxwell's electromagnetic theory.
The speed of an e.m. wave in any medium of
permeability11and permittivityEwill be
1 1
v=--=-r====
.fill~K EO J.l r110
C
~KJ.lr
whereKis the dielectric constant of the medium andJ.lr
is its relative permeability.
As the electric and magnetic fields in an e.m. wave
are always perpendilar to each other and also
perpendilar to the direction of wave propagation, so
e.m. waves are transverse in nature.
...(2)
8.5SOURCES OF ELECTROMAGNETIC
WAVES
6.Briefly explain howisan accelerating charge a
source of an electromagnetic wave.
An acelerating charge as a source of an
electromagnetic wave.A stationary charge produces
only an electrostatic field while a charge in uniform
motion produces a magnetic fieldthat does not change
withtime. Thus, neither stationary charges nor charges
in uniform motion (or steady rrents) can produce
electromagnetic waves. Acording to Maxwell, an
accelerating charge produces electromagnetic waves.
Consider a charge osllating harmonically with time.
This is an example of an acelerating charge. This
chargeproduces an osllating electric field in its
neighbourhood. This field, in turn, produces an
osllating magnetic in the neighbourhood. The
process continues because the osllating electric and
magnetic fields act as sources of each other. Hence an
electromagnetic wave originates from the osllating
charge. The frequency of the electromagnetic wave isequal
to the frequency of oscillatn of the charge. The energy
carried by the wave comes from the source which
makes the charge osllating.
Fromthe above disssion, we may note that in
order to generate an electromagnetic wave of
frequenv,we need to set up an a.c. rit in which
the rrent osllates at the frequen v. Hence it is
easier to generate low frequen e.m. waves, such as a
radiowave. However, it is not possible to experi-
mentally demonstrate the existence of high frequen
e.m. waves, such as visible light. For example, the
generation of yellow light requires an osllator of
frequen 6x1014Hz, while the modern osllators
have frequen hardly above1011Hz.
Inthe next section, we disss Hertz's experiment
for demonstrating the existence of low frequen
radiowaves.°°°1uv{lzkyp®l1jvt

629. ELECTROMAGNETIC WAVE
S
8.6HERTZ'S EXPERIMENT
7.Describe Hertz's experiment for producing and detec-
ting electromagnetic waves.How werethe various pro-
perties ofelectromagnetic wavesdemonstrated by Hertz?
Hertz's experiment. Maxwell predicted the
existence of electromagnetic waves in1865.This
prediction had to wait for about 22years before a
German physist, Heinrich Hertz, suceeded inexperi-
mentally confirming the existence of electromagnetic
waves in1887.
In the osllations of anLC-rit, we know that
the charge oscillates across the capacitor plates. Since
an osllating charge has non-zero aceleration, it will
continuously emit electromagnetic waves. As shown
in Fig. 8.6,Hertz used an oscillatory LC-rit for
produng electromagnetic waves.
Metal
plate
p
.~CS5 551L~®_5~1_
-eU--=- .:..j2~ ~ 5'2
.s 5
Detector
Metal
plate
p
Fig. 8.6Hertz's experimental set-up for produng
and detecting e.rn. waves.
Thetransmitterconsists of two large square metal
(brass) plates with sides oflength 40 . These are
placed in the same vertical plane with their centres
about 60 em apart. The plates are connected to two
thick wires ending in highly polished brass spheres
51and52'The distance between the two spheres is 2 to
3 . The two thick wires are connected to the
secondary terminals of an induction coil.
Every time the rrent in the primary rit of the
induction coil is interrupted, a large p.d. is set up across
51and52and the metal plates getcharged. The high p.d.
ionises the air in the gap and makes the gap conduc-
ting. The electrons and ions so produced osllate back
and forth across the gap 5152, An oscillatory discharge
of the plates ors through the conducting gap. The
process results in the production of e.m. waves.
The metal plates form a capator of low
capacitance C and connecting wires offer a low
inductance L The system generates e.m.waves of high
frequency (v) given by'
1
v=--,==
2rc.J[C
8.9
Thereceiverordetectorconsists of an almost closed
cirlar stout wire terminating at the two ends in two
small polished brass spheres ~ andS2'The electro-
magnetic waves reaching the gap of the detector are
associated with a suffiently strong electric field
which sets up a high p.d. across the gapSl S2' This
causes tiny sparks jumping across the gap, thereby
proving the existence of e.m. waves.
Hertz demonstrated the various properties of e.m.
waves as follows:
1. Hertz observed that maximum sparks are
produced across the detector gap when this gap is
parallel to the transmitter Bap. When these two gaps
are perpendilar to each other, no sparkes are
produced across the detector gapi.e.,no electro-
magnetic waves are detected. This means that electric
field assoated with the waves radiated from the
transmitter is parallel to the two gaps.e.,the direction
of the electric field is perpendilar to the direction of
propagation of the e.m. wave. This clearly demon-
strates that thee.m. waves are transverse in nature.
2. Hertz not only produced and detected electro-
magnetic waves, but also demonstrated their properties
of reflection, refraction and interference and so
established beyond doubt that thee.m.radiatn has a
wave nature.
3. Hertz allowed the e.m. waves to fall on a large
plane sheet of zinc. The reflected waves superimposed
on the indent waves, produced stationary e.m.
waves. The wavelength of these waves was deter-
mined by measuring the distance between two nodes.
The frequen of the wave was equal to that of the
oscillator, i.e.,
1
v=--,==
2rc.J[C
Hence the speed of the e.m. wave was determined
by using the formulav=v "- It was found thate.m.
waves travel with the same speed as the speed of light.
4. Electromagnetic waves can be polared.To test
this fact, take a portable AM radio provided with a
telescopic antenna. It responds to the electric
component of the e.m. signal from the broadcasting
station. When the antenna is turned horizontal, the
signal is greatly diminished. The portable radios
having horizontal antenna inside them are sensitive to
the magnetic component of e.m. wave. The signal is
best received when such a radio is held horizontal.
In Hertz set-up, the frequen of the e.m. waves
produced was 5x107Hz.So the wave length of the
e.m.waves produced is given by
c3x108
A,=- =-- =6m.
v5x107www2z~tpsoruvp2n~y

630. 8.10
8.7HISTORY OF
THE OBSERVATION
OF EM WAVES
8. Give a brief htory of the observatn of electro-
magnetic waves.
History of the observation of electromagnetic
waves. In 1865,Maxwellpredicted the existence of
electromagnetic waves purely from theoretical consi-
derations. He showed that an acelerating charge
produces electromagnetic waves. Since an osllating
charge is acelerated continuously, it would conti-
nuously produce e.m. waves of same frequen as that
of the osllating charge.
In 1887,Hertzsuceeded in experimentally confir-
ming the existence of e.m. waves. He used an osllatory
LC-rit for produng these waves. He was able to
produce and detect e.m. waves of wavelength around6m.
In 1885,SirJ.e.Bosesuceeded in produng e.m.
of much shorter wavelength(5mm to25mm) with the
help of a self-designed radiator. He was able to
transmit e.m. waves over a distance of about20m.
In 1896,Guglielmo Marconi, discovered that if one of
the spark gap terminals is connected to antenna and the
other terminal is earthed, then e.m. waves can be trans-
mitted over distances of several kilometers. He suceeded
in transmitting e.m. waves across the British Channel
in 1899 and across the Atlantic Ocean in 1901. His experi-
ments marked the beginning of radio communication.
8.8TRANSVERSE NATURE OF
ELECTROMAGNETIC WAVES
9.Prove mathematically that electromagnetic waves
are transverse in nature.
Transverse nature of electromagnetic waves.
Consider a plane electromagnetic wave travelling in
the X-direction. The assoated wavefront lies in the
YZ-plane (a wavefront is the los of continuous
points having same phase of vibration) andABCDis a
portion of it at any timet.The electric and magnetic
fields at timetwiJI be zero to the right ofABCD. To the
left ofABCD, they depend onxandt,but not onyand
z, as we are considering only a plane wave.
x
F~~ ~
I
" I
n---+-E'xl
I
).--------
"H dx
"-'+--~n
dy
D
E
A
z
Fig.8.7Elementary parallelepipedABCDEFGH
chosen as a closed surface.
PHYSICS-XII
We use Gauss's law to prove the transverse nature
of electromagnetic waves. Consider the closed surface
enclosed by the parallelopipedABCDEFGHof sidesdx,
dyanddz.The total electric flux through this closed
surface must be zero as it does not enclose any charge.
Hence
f
ABCDEFGH
~ ~
E.dS=0
or [fE.+fE.tiS]+[fE.dS+fE.is]
ABCD EFGH ABFE DCGH
+[fE.e•fE.ts]=O
AEHD BFGC
The paired integrals are the contributions from the
faces normal to X,Yand Z-axis, respectively. Since E
does not depend onyandz ,the contributions from the
faces normal toYand Z-axis cancel out in pairs, so the
above equation becomes
fE.dS+fE.ts=0
ABCD EFGH
LetExandE~be thexcomponents of the electric
field at the facesABCDandEFGH, respectively. The
outward normals on these faces are oppositely
directed (along X-axis), therefore,
...(1)
fE.dS=Ex .dydz
ABCD
fE.= -E~dydz
EFGH
Hence equation(1)becomes
Ex .dy dz -E~.dy dz=0orEx=E~
.e.,the component of the electric field along the
direction of propagation is constant. But a constant or
static field cannot produce a wave, so this constant
must be equal to zero,.e.,Ex=0
Similarly, we can prove thatBx=o.Thus the
electric or magnetic fields have no component along
the direction of propagation. Or,in an electromagnetic
wave both the electric and magnetic fields are perpendicular
to the directn of propagatn, .e., the electromagnetic
waves are transverseinnature.
8.9ENERGY DENSITY, INTENSITY AND
MOMENTUM OF ELECTROMAGNETIC
WAVES
Jo.nhtain expressn for the energy density of an
cicctromognctic 7Jlal1e. In an electromagnetic wave, show
that the arcrog« energy density of the Efield equals the
a71ernge energy dr.nsity oftheBfield.www4~«trsq•vvr4p«z

631. ELECTROMAGNETIC WAVES
Energy density of an
electromagnetic wave.Electro-
magnetic waves carry energy as they travel through
space and this energy is shared equally by the electric
and magnetic fields.Energy density of an e.m. waveis
the energyinunit volume of the space through which the
wave travels.
We knowtha+energy is stored in space wherever
electric and magnetic fields are present.
In free space, the energy density of a static fieldEis
_ 1 E2
uE-"2EO
Again in free space, the energy density of a static
magnetic field is
uB=_1_B'-
2110
The total energy density of the static electric and
magnetic fields will be
1 2 1....2
U="e+"e= -EOE+ --I:)
2 2110
But in an electromagnetic wave, bothEandBfields
vary sinusoidally in space and time. The average
energy densityuof an e.m. wave can be obtained by
replangEandBby their rms values in the above
equation. Thus
1 2 1....2
U=-EOErms+ --I:)rms
2 2110
oru-1EE2+ 1 02[..E - Eo B _ Bo]
- 40 0 4110 'iJ.rms-..fi' rms-..fi
Moreover,Eo=cIbandc?=_1_ , therefore
110 EO
1 21 2
uE=4EOEo=4EO(cIb)
=.!.EO . ~ =_1_ ~ =uB
4 110 EO 4110
Hence inan electromagnetic wave, the average energy of
theEfield equals the average energy density of theBfield.
It may be noted that
1212122
U=-EOEo+ -EOEo=-EOEo=EOErms
4 4 2
Also,
u=_l 02+_1 02=_1 o2=...!.-B'-
411 0'iJ411 0'iJ211 0'iJ110 rms
11.Define intensity of an electromagnetic wave.
Obtain an expressn for it.
Intensity of an electromagnetic wave. The energy
crossing per unit area per unit timeina directn perpen-
dicular to the directn of propagatniscalled intensity of
the wave.
8.11
Suppose a plane electromagnetic wave propa-
gates along X-axis with speed c. As shown in Fig.8.8,
consider a lindrical volume with area of cross-
sectionAand lengthcStalong the X-axis. The energy
~I'---------c~t-------~'I
,
,
\
\
I
I
,
,
,
-+-~c
Area=A
Fig. 8.8Callation of intensity.
contained in this linder crosses the areaAin time.M
as the wave propagates with speed c.The energy
contained is
U=Average energy densityxvolume
=uxc.MxA
Intensity of the wave,
1=Energy= ~ =uc
AreaxTimeA.M
or
1=_1_o2c=...!.- Ifc
2110'iJ110 rms
Thusthe intensity of an electromagnetic waveis
proportnal to the square of the electric/magnetic field.
Conversely, the size of the electric/magnetic field of an
electromagnetic wave is proportional to the square
root of its intensity.
12.Write an expressn for the momentum carried by
an e.m. wave.
Also,
Momentum of an e.m. wave.An electromagnetic
wave transports linear momentum as it travels
through space. If an electromagnetic wave transfers a
total energy U to a surface in timet,then total linear
momentum delivered to the surface is
U
p=-
c
[For complete absorption of energy U]
If the wave is totally reflected, the momentum
delivered will be 2 U / ,because the momentum of the
wave will change frompto -p.
13.Write an expressn for the pressure exerted by an
electromagnetic wave.
Pressure exerted by an e.m. wave.When an electro-
magneticuiaoefalls on a surface,itexertspressureon the surface.www2z~tpsoruvp2n~y

632. 8.12
Th pressureiscalled radiatn pressure. The radiation
pressure for
an electromagnetic wave of intensityIis
given by
It is because of the radiation pressure of the solar
radiation that the tails of comets point away from the sun.
8.10PROPERTIES OF ELECTROMAGNETIC
WAVES
14. Mentn the varus properties of electro-
magnetic waves.
Properties of electromagnetic waves.These are as
follows:
1.The electromagnetic waves are produced by
acelerated charges and do not require any
material medium for their propagation.
2.The directions of osllations ofEandBfields
are perpendilar to each other as well as
perpendilar to the direction of propagation of
the wave. So the electromagnetic waves are
transverse in nature.
3.The osllations ofEandBfields are in same
phase.
4.All electromagnetic waves travel in free space
with the same speed,
c=_l_ =-3x108ms-1
~flOEO
In a material medium, the electromagnetic
waves travel with the speed,
1 c c
V=--=--=-
~ ~IJrEr n
wherenis the refractive index of the medium.
5. The amplitude ratio of the electric and magnetic
fields isEo=c= ~
1b -yflOEO
6. The electromagnetic waves carry energy as they
travel through space and this energy is shared
equally by the electric and magnetic fields. The
average energy density of an e.m. wave is
1[ 2 ~]U=uE+uB=-Eo11)+ -
2 IJo
7. Electromagnetic waves transport linear
momentum as they travel through space :
U
p=-.
c
8. Electromagnetic waves are not deflected by
electric and magnetic fields.
PHYSICS-XII
9. Electromagnetic waves obey the prinple of
superposition. They show the properties of
reflection, refraction, interference, diffraction
and polarisation.
10.The electric field of an electromagnetic wave is
responsible for its optical effects, because
Eo»lb·
~------
For Your Knowledge
An acelerating or osllating charge is a source of
electromagnetic waves. An electric charge osllating
harmonically with frequen v,produces electro-
magnetic waves of frequen v. An osllating electric
chpole radiates electromagnetic waves.
All types of electromagnetic waves travel through
vaum with the same speed but they travel with
different speeds in any material medium.
__The frequen of an electromagnetic wave is its
inherent characteristic. When an electromagnetic
wave travels from one medium to another, its
wavelength changes but frequen remains unchanged.
The ratio0) /c gives the magnitude of the propagation
vectorkfor an electromagnetic wave,
k=21t=~
A.c
The direction of propagation of an electromagnetic
wave is same as that of the vectorExB.
The speed of an electromagnetic wave through any
medium depends on its permeability IJand
permittivity e.
1 c
v=--=---
~ [10;
Half of the intensity of an electromagnetic wave is
provided by its electric field and half by the magnetic
field. So the power delivered by the magnetic field of
an electromagnetic wave is equal to the power delivered
by its electric field, but the magnetic field strength is
much weaker than the electric field strength. In fact,
E
--.!l.=c
1b
The fact that electromagnetic waves can carry energy
from one place to another, is of great technical
importance. They transmit energy from radio and TV
stations to our homes. Light carries energy from the
sun to the earth, thus making life possible on the earth.
In 1903, the American sentistsNicolsandHullsucess-
fully measured the radiation pressure of visible light
and found it to be of the order of 7x10--6Nm -2.Thus,
on a surface area of 10 crn2, the force due to radiation
is only about 7x10-9N.°°°1uv{lzkyp®l1jvt

633. ELECTROMAGNETIC WAVES
•....-
1. Wave veloty, c =VA
hc
2. Energy of photon,E= hv =-
A
3.Speed of e.m. wave in vaum,c= ~
"J.loSo
4.Speed of e.m. wave in a
material medium,c=b
-yJ.ls
5.For a wave of frequen v, wavelength A.,
propagating alongx-direction, the equations for
electric and magnetic fields are
~ =SJsin(kx -rot)=SJsin [ 21t(i-~)]
Bz=Eosin(kx-rot)=Eosin [ 21t(i-~)]
6. Amplitude ratio of electric and magnetic fields,
SJ=c=_l_
Eo ~J.lOSo
7P· k21tto
. ropagation constant, = - = -
Ac
8. Average energy density of E-field,
1 21 2
UE="4Eo~="2 SoE;;".s
9. Average energy density of Bfield,
U__1_R2__1_ B2
B-4J.lO lJ-2J.lo rms
10. Average energy density of e.m. wave,
1 1 ~
U - -S-,:2+_B2 -1'_-,:2 _ rms
av -20';ems 2J.lOrms - -u =rms-J.lo
12121212
oruav="4So ~+4J.l0Bo ="2 So~ = 2 J.l 0Bo
11. Momentum delivered by an e.m. wave.
U
p=-
c
In. f Energy / time Power
12. tensity0a wave = = --
Area Area
or I=uavc=Soe:c.
Units Used
8.13
Example 6. Electromagnetic waves travel in a medium at a
speed of2.0xlOBms-1. The relative permeability of the
mediumis 1.0.Find the relative permittivity.
Solution. Speed of an e.m. wave in a medium is
given by
1 1
v---- r=====
-..flE-~Ilrllos,So
lIe
= ~lloSo~Ilrsr=~llrSr
2?
v=--
Ilrsr
Hence relative permittivity,
? (3xlOBl
€=--= =2.25
rIlrV2 1.0x(2xlOB)2 .
Example 7. A plane electromagnetic wave of frequency
25 MHz travels in free space along the x-directn. At a
particular point in space and time,E=6.3]Vm-1.Whatis
---+
Bat th point? [NCERT; CBSE D 06C]
Solution. Magnitude ofE,E=6.3 Vm-1
---+
Magnitude of B ,
B= ~ = 6.3 Vm-1 =2.18x10-8T
c 3xlOBms-1
We note thatEis along y-direction and the wave
propagates along x-direction. Now an e.m. wave
propagates in the direction of the vectorExIf.Sothis
vector must point along x-direction. As
(+])x(+k)=+i
---+
Clearly, vector B is along z-direction. We can write
If=2.18x10-BkT.
Example 8. The magnetic field in a plane electromagnetic
waveisgiven by
By =2x10-7sin(0.5x103x+1.5x1011t) T.
(a) Whatisthe wavelength and frequency of the wave?
(b) Write an expressn for the electricfield.[NCERT]
Solution. Given
By =2x10-7sin (0.5 x103x+1.5+1011t)tesla
On comparing with the standard equation,
Wave veloty s in ms ", wavelength Ain metre,
frequen v in Hz, fieldEin Vm-I,field Bin tesla,
energy densitiesupuBanduavare in [m-3,
intensityIin Wm-2. we get,www2z~tpsoruvp2n~y

634. 8.14
:. Wavelength,
2
1t 2 x 3.14
A=-------::-
0.5 x103 0.5x103
=1.26x1O-2m = 1.26 em.
Also, 21tV= 1.5 x1011
1.5x1011 1.5 x1011
v=----
21t 2 x3.14
= 23.9 x109Hz = 23.9 GHz.
(b)lb =2x10-7T
:. Eo =clb =3 xlOBx2x10-7=60 Vm-1
or
The electric field is perpendilar to the direction of
propagation(x-axis) and the direction of magnetic field
(y-axis). So the expression for electric field is
Ez=60 sin (0.5 x103x+1.5x1011t)Vm-1.
EXAMPLE 9.Light with flnenergy flux of18watts/cn1
fallson a non-reflecting surface at normal incidence. If the
surface has an area of20 cnl,find the average force exerted
on the surface during a30minute time span. [NCERT]
Solution. Energy flux
=18 W em-2 =18 Js-1em-2
Area =20 em2
Time= 3Umin = 1800 s
Total energy falling on thesurface
U= Energy fluxxtimexarea
= (18 Js-1em -2)x 1800s x20 em2
= 6.48x105J
The total momentum delivered to the surface,
U6.48x105J
p=-=
c3x10Bms-1
=2.16x1O-3kgms-1
The average force exerted on the surface,
F=E=2.16x10-3= 1.2x10-6N
t 1800
In case, the surface is a perfect reflector, the change
of momentum
and
=p-(-p)=2p
F=2x1.2x10-6= 2.4x10-6N.
Example 10. Calculate the electric and magnetic fields
produced by the radiatn coming from a 100watt bulb at a
dtance of3m Assume that theefficiency of the bulbis2.5%
and itisa pointsource. [NCERT]
Solution. The bulb, as a point source, radiates light
in all directions uniformly. At a distance of 3 m, the
surface area of the surrounding sphere is
A=4n?= 4rc(3)2 = 113 m2
PHYSICS-XII
The intensity at this distance is
I=Power = 2.5% of 100W
Area 113 m2
=0.022 Wm-2
Halfof this intensity is provided by the electric
field and half by the magnetic field.
But intensity of e.m.wave
=EOE;msc
or
1 1 2 1 -2
2"I=2"EO Errnsc=2" X0.002 Wm.
E2=0.022= 0.022 =8.286
rms EOC8.85x10-12x3xlOB
Erms= 2.878 ::. 2.9 Vm-1
Peakvalue,
Eo=.fi Errns= 1.414x2.9 = 4.1 Vm-1
Thestrength of the magneticfield,
B Errn' 2.9Vm-1 8
,=__s= =9.6x 10-T
rrns C 3x108ms-1
lb=.fiBrms= 1.414x9.6x10-8
= 1.4x10-7T.
It may be noted that although the power in the
magnetic field is equal to the powerin the electric field,
yet the magnetic field strength is very weak.
Example 11.A plane electromagnetic wave in the vible
regn moving along the z-direction. The frequency ofthe
wave 0.5x1015Hzand the electric field at any point is
varying sinusoidally with time with an amplitude of 1Vm-1.
Calculate theaverage values ofthedensities of the electric
andthe magnetic fields.
Solution. Average energy density of the electric
field is
Also,
1 21 -12 2
uE= -EOEo=- x8.85x10xl
4 4
= 2.21x10-12Jm-3
Average energy density of the magneticfieldis
U_lb2_! E02
B- -
41-l0 4 I-lo ~
1 12 1
=-x x--~~
4 41t X10-7(3 xlOB)2
= 2.21x10-12Jm-3.
Example 12.The electric field in ane.m. wave given
by
E=50 sin 21t (ct -x)Ne1
A
Findthe energy contained in a cylinder ofcross-sectn
10 em2andlength 50emalong the x-ax.°°°1uv{lzkyp®l1jvt

635. ELECTROMAGNETIC WAVES
Solution. Here Eo = 50NC1
.',
Average energy density of the e.m. wave is
u=.!..10E2=.!..x8.55x10-12x(50)2
av2aa2
=1.1xlO-BJm-3
Volume of the linder,
V=10 2 x50 em =500'=5x10-4m3
Energy contained in the linder is
U= Volumexenergy density
= 5x10-4 x1.1x10-8=5.5x10-12J.
Example13.A plane e.m. waveispropagating in the
x-directn has a wavelength of6.0 mm The electric field is
in the y-directn and its maximum magnitudeis33Vm-1.
Write suitable equatns for theelectric and magnetic fields
as afunctn of x and t.
Solution. Heref....= 6.0 mm = 6x10-3m,
Eo=33 Vm-1
21tc 21tx3xlOB 11 -1
co= 21tv =-= =1tx10 rad s
f....6x10-3
Ba=Eo =~ =1.1 x 1O-7T
c3xlOB
The equation for the electric field along y-axis can
be written as
E = Ey = Eo sinw(t - ~)
= 33 sin 1tx1011(t - ~) Vm-1.
The equation for the magnetic field along z-axis
can be written as
B= Bz =Basinw(t -~)
= 1.1x10-7sin 1tx1011(t-~)tesla.
Example14.A laser beam has intensity2.5x1014Wm-2.
Find the amplitudes of electricand magneticfields in the beam.
Solution. The intensity of a plane e.m wave,
1 2
I=u .c=-f:oEoc
av 2
:.Amplitude of electric field,
fIr 2x2.5x1014
Eo =VlOac=8.85x10-12x3x108
=4.3x108Vm-1
Amplitude of magnetic field,
B = Eo = 4.3 x 108 = 1.44 T.
ac3xlOB
8.15
Example15.A light beam travelling in the x-directnis
described by the electric field:Ey = 270 sinco (t - ~ ) .An
electronisconstrained to move along the y-directn with a
speed of2.0x107ms-1. Find the maximum electric force
and maximum magnetic force on the electron.
Solution. Maximum electric field,
Eo =270 Vm-1
Maximum magnetic field,
Ba= Eo =~=9x 1O-7T,
c3x108
directed along z-direction
Maximum electric force on the electron,
Fe=qEu= 1.6 x10-1'1x270 =4.32x 10-17 N
Maximum electric force on the electron
Fm=qvlb= 1.6 x10-1'1x2.0x107x9x10-7
=2.88x10-18N.
~rOblems For Practice
1.The electric field vector of a plane electromagnetic
wave osllates sinusoidally at a frequen of
4.5xuroHz. What is the wavelength?
[CBSE OD 91]
(Ans.6.67x10-3m)
2.The maximum electric field in a plane electro-
magnetic wave is 600 NC-1. The wave is going in
the x-direction and the electric field is in the
y-direction. Find the maximum magnetic field in
the wave and its direction.
(Ans.2 x 10-6T, z-direction)
3.The frequenes of radio waves in the AM
broadcast band range from 0.55 x106Hz to
1.6 x 106 Hz. What are the longest and the shortest
wavelengths in this band ?
(Ans.5.45x102m, 1.87 x102m)
4.A radio transmitter operates at a frequen of
880 kHz and a power of 10 kW. Find the number of
photons emitted per second. [CBSE OD 90]
(Ans. 1.71x1031)
5.The permittivity and permeability of free space are
lOa= 8.85x1O-12C2N-1m-2
andfl 0= 41tx10-7TmA-1,respectively.
Find the veloty of the electromagnetic wave.
(Ans.3x108ms ")
6.In a plane electromagnetic wave of frequen
1.0 x 1012 Hz, the amplitude of the magnetic field iswww2z~tpsoruvp2n~y

636. 8.16
5.0 x 1O-6T. (a)Callate
the amplitude of the
electric field.(b)What is the total average energy
density of the e.m. wave?
(Ans.1.5 x 103 Vm-1,1.0 x 10-SJm -3)
7.A plane electromagnetic wave is moving along
x-direction. The frequen of the wave is1015Hz
and the electric field at any point is varying sinu-
soidally with time with an amplitude of 2 Vm-1.
Callate the average densities of the electric and
magnetic fields.
(Ans.8.85x10-12Jm-3, 8.85 x 1O-3Jm-3)
8.The magnetic field in a plane e.m. wave is given by
B= (200IlT)sin (4.0xlO-5s-1) (t -~)
Find the maximum electric field and the average
energy density corresponding to the electric field.
(Ans. 6 x i04Vm -1,0.008 Jm-3)
9.A millimetre wave has a wavelength of 2.00 mm
and the osllating electric field assoated with it
has an amplitude of 20 Vm -1. Determine the
frequen of osllations of the electric and magnetic
fields of this electromagnetic wave. What is the
amplitude of the magnetic field osllations of this
wave? (Ans.1.5x1d1 Hz,6.67xlO-B T)
10.In a plane electromagnetic wave, the electric field
varies with time having an amplitude 1Vm-1. The
frequen of wave is 0.5 x 1015Hz.The wave is pro-
pagating along z-axis. What is the average
energy density of(i)electric field(it)magnetic field
(iii)total field, and(iv)what is amplitude of
magnetic field ?
[Ans.(i)2.21x10-12J m-3(ii)2.21x10-12J m-3
(iii)4.42 x 1O-12Jm -3(iv)3.33x10-91']
HINTS
c 3xlOB -3
1.A.= - -ao= 6.67 xl0m.
v4.5x hr
2R=EU=~=2xl0-6 T
.-n C3xlOB
-+ -+
As the directions ofE, Band direction of
-+
soBpropagation are mutually perpendilar,
'should be along the z-direction.
3.Proceed as in Exerse 8.5 on page 8.33.
4.Herev= 880 kHz = 880 x 103 Hz,
P= 10kW= 10 x 103 W
Number of photons emitted per second,
P 10 x103 31
n= - = 34 3= 1.71 x10 .
,hv6.6 x 10- x 880 x 10
PHYSICS-XII
1
5.Usec= ~
,,11OEO
6.(i)EU=cBo= 3xlOBx5.0x10-6= 1.5xl03 Vm-1.
(ij)U=..!EO1='2=..!x8.85x10-12x(1.5x103)2
av2 "1J 2
=1.0xl0-5 Jm-3•
7.Proceed as in Example 11 on page 8.14.
8.Here Eo= 200 IlT = 2x10-4T
EU= cEo=3x10B x2xlO-4 =6xl04 Vm-1
"t:=..!EoEJ=..!x8.85x10-12x(6x104)2
4 4
= = 0.008Jm-3
c3xlOB
9.Frequen,v=-= 3= 1.5x1011Hz
A2x10
EU20 8
Eo=-=--8 = 6.67 xl0 T.
c3x10
10.(i)"e=..!EoEJ=..!x8.85x10-12x12
4 4
= 2.21x10-12Jm-3
(it)uB=uE= 2.21x10-12Jm-3
(iit)Uav= ~+UB= 2x221x10-12
= 4.42x10-12Jm-3
(iv)Eo=EU=_1-8 =3.33x10-9T.
c3x10
8.11ELECTROMAGNETIC SPECTRUM
15. Whatiselectromagnetic spectrum?Name the
main parts of the electromagnetic spectrum giving their
frequency range and source of productn. Also give their
important properties and uses.
Electromagnetic spectrum. All the known
radiations form a big family of electromagnetic waves
which stretch over a large range of wavelengths.The
orderly dtributn of the electromagnetic waves in
accordance with their wavelength orfrequency into dtinct
groups having widely differing properties iscalled
electromagnetic spectrum.As shown in Fig.8.9,the main
parts of the e.m. spectrum are y-rays, X-rays,
ultraviolet rays, visible light, infrared rays microwaves
and radio waves in the order of increasing wavelength
from 10-2 Aor 10-12 m to 106 m.
The various regions of the e.m. ,spectrum do not
have sharply defined boundaries and they overlap.
The classification is based roughly on how the waves
are produced and/or detected.
We now describe the various regions of the
electromagnetic spectrum in the order of increasing
frequen.°°°1uv{lzkyp®l1jvt

637. ELECTROMAGN
ETIC WAVES
Frequen, Hz
-
r
-
I-
- Gamma rays
-
~x-,,~
I:
-
\..f-
<-
-
-
-
\.r-
- Ultraviolet
-:.
-
Visible
-
:.,.
- Infrared
-
-
"\ >- -
-
(
Microwaves
-
-
Short radio waves -
-
r--Television and FMradio--{
-
-
-
-
AM radio -
--<
-
-
-
-
Long radio waves -
-
-
-
-
-.-
\.
Fig. 8.9Electromagnetic spectrum.
1.Radiowaves.These arethe e.m.wavesof longest
wavelength and minimum frequency.
Wavelength range 600 m to 0.1 m
Frequency range 500 kHz to 1000 MHz
Source Aceleratedmotion of charges in con-
----
ducting wires or osllating rits.
Dcovered by Marconi in 1895
----
Properties Reflection, diffraction
Uses of radio waves:
(i)Inradio and television communication systems.
(ii)Inradioastronomy.
Table 8.1 Someimportant wireless
communication bands
Freauencv band Service
540 - 1600kHz Medium wave AM band
3-30MHz Shortwave AM band
88 -108 MHz FM broadcast
54-890MHz TVWaves
840- 935MHz Cellular Mobile radio
8.17
Wavelength
-14
Violet
10 400nrn
-13
10
-12
10
-11
10
450nrn
-10
10
10-~': 1 nrn
Blue
,.'8
W
-7
10 500nrn
-6
10 Lum Green
-5
lO
-4
550 nrn
10
'.-3
In
~2
1 em
Yellow
10'.
lO-('"
600nrn
1
'..1m
Orange
1
lO
2
10
650nrn
103 tkm-.
Red
4
10
105
700nrn
106
7
10
2. Microwaves. They are the e.m. waves having
wavelengths next smaller to radiowaves.
Wavelength range0.3 m to 10-3 m
Frequency range 109Hz to 1012 Hz
Source Oscillating rrents inspeal
vaum tubes like klystrons,
magnetrons and Cunn diodes.
Dcovered by Marconi in 1895
Properties Reflection, refraction, diffraction
and polarisation. Due to their
shorter wavelengths, they can
travel as a beam in a signal.
Uses of microwaves:
(i)In radarsystemsforaircraft navigation.
(ii)In long-distance communication systems via
geostationary satellites.
(iii)In microwave ovens.
3. Infrared waves. These radiations lie close to the
low-frequen or long-wavelength of the visible
spectrum. Infrared waves produce heating effect, sothey
are also known as heat wavesorthermal radiation.www2z~tpsoruvp2n~y

638. 8.18
The water moleles
(and alsoCO2, NI\ moleles)
present indifferent materials readily absorb infrared
waves, increase the thermal motions and hence heat up
the materials and their surroundings.
Wavelength range.2x10-3mto10-6m
Frequency rang!:, 1(y.1Hz to5x1(y.4Hz
--
Source Hotbodies and moleles.
...--
Discovered .!!Ji William Herschel in1800.
---
Properties Heating effect, reflection, refrac-
I
tion, diffraction and propagation
through fog.
Uses of Infrared waves:
(i)Inthe remote control of a TVor VCR, the keypad
of which contains a small infrared transmitter.
(ii)In green houses to keep the plants warm.
(ii)Inhaze photography because infrared waves
are less scattered than visible light by atmos-
pheric particles.
(iv)Infrared lamps in the treatment of muslar
complaints.
(v)In reading the secretwritings on the anentwalls.
(vi)In knowing the molelar structure.
4. Visible light. It is a very small part of the e.m.
spectrum towards which the human retina is sensitive.
The visible light emitted or reflected from bodies
around us gives information about the world.
Wav~ngth range 8x10-7m to 4 x10-7m.
Frequency range4 x1(y.4Hzto 7 x1014Hz
Source Radiated by exted atoms in
ionised gas and incandescent
bodies.
Properties Reflection, refraction, interference,
diffraction, polarisation, photo-
electric effect, photographic action,
sensation of sight.
Uses of visible light:
(i)It provides usthe information of the world
around us.
(ii)It can cause chemical reactions.
The approximate wavelength ranges for lights of
different colours are as follows:
Table 8.2 Visible Spectrum
Colour
Wavelength
Colour
Wavelength
ranee ranee
Violet, indigo 4000 - 4500AYellow5700 - 5900A
Blue 4500 - 5000AOrange5900-6200A
Green 5000 - 5700A Red 6200 - 7500A
PHYSICS-XII
5.Ultraviolet light. This region of the e.m.
spectrum has wavelengths just shorter than visible
light and can be detected just beyond the violet end of
the solar spectrum .
Wavelength range3.5 x10-7m to1.5x10-7m
Frequency rangel(j6Hz toio"Hz
Source High voltage gas discharge tubes,
--
arcs of iron and merry, the sun
Dcovered by Ritter in1800
Properties Effect on photographic plate, fluo-
rescence, ionisation, highly energeti,
tanning of the human skin.
Uses of ultraviolet light:
(i)In food preservation.
(i)In the study of invisible writings, forged
doments and finger prints.
(ii)In the study of molelar structure.
The ultraviolet light in large quantities has harmful
effects on human beings. But fortunately, most of the
ultraviolet light coming from the sun is absorbed by
the ozone layer in the atomosphere at an altitude of
about 40 -50km.
6. X-rays. These e.m. waves have wavelengths just
shorter than ultraviolet light. As X-rays can pass
through many forms of matter, so they have many
useful medical and industrial applications.
Wavelength range100Ato0.1A
Frequency range1018Hz to102°Hz '1
Source Sudden deceleration of fast moving
-----
electrons by a metaltarget.
1
Discovered by Rontgen in1895
Properties Effect on photographic plate,
ionisation of gases, photoelectric
effect, fluorescence, more energetic
than UV rays.
Uses of X-rays:
(i)In medical diagnosis because X-rays can pass
through flesh but not through bones.
(i)In the study of crystals structure because
X-rayscan be reflected and diffracted by crystals.
(ii)Inengineeringfor detecting faults, cracks, flaws
and holes in the finished metal products.
(iv)Indetective departments to detect explosives,
diamond, gold, etc. in the possession of
smugglers.
(v)In radiotherapy to re untracable skin
diseases and malignant growths.°°°1uv{lzkyp®l1jvt

639. ELECTROMAGNETIC WAVES
7.Gammarays. These are e.m. radiations of highest
frequen range and lowest wavelength range. These
are most penerating e.m. waves.
Waveleng!..h ran~1O-14m to1O-10m.
Preauencurange H
y8Hz to1022Hz.
Source Radioactive nucleiand nuclear
reactions. Co -60isa pure y-ray
--
source.
Dcovered bu HenrvBecqurelin1896
Properties Effecton photographic plate, fluore-
scence, ionisation, diffraction,
high penetrating power.
Table8.3 The electromagneticspectrum
8.19
Uses of y-rays :
(I)In radiotherapy for the treatment of malignant
turnours.
(il)In the manufacture of polyethylene from
ethylene.
(iii)To initiate some nuclear reactions.
(iv)To preserve food stuffs for a long time
because soft y-rays can kill micro-
organisms.
(v)To study the structure of atomic nuclei.
Name
Frequen
I
Wavelength
IProductionIDetectionI
Main properties
range(Hz) range and uses
Radiowaves 104to 108 >0.1 m 'Rapid aceleration Receivers Different wavelengths find
and deacelerationsaerials. spealised uses in radio
of electrons in communication.
aerials.
Microwaves 109to 1012 0.1 m to Klystron valve or Point contact (a)Radar communication.
Imm magnetron valve. diodes. (b)Analysis of fine details of
molelar and atomic
structure.
(c)SinceA.=3x10-2m,useful
for demonstration of all wave
properties on macroscopic
scale.
Infrared 1011to 1 mm to Vibration of atoms Thermopiles (a)Useful for eludating
5x1014 700 nm and moleles. Bolometer molelar structure.
Infrared
(b)Less scattered than visible
photographic light by atmospheric particles-
film.
useful for haze photography.
Visible light4x1014to 700 nm to Electrons in atomsHuman eye (a)Detectedby stimulating
7x1014 400nm emit light when they Photocells nerve endings of human
move from one Photographic retina.
energy level to a film.
(b)Can cause chemical reaction.
lower energy level.
Ultraviolet 1016to 1017 400 nm to Inner shell electronsPhotocells (a)Absorbed by glass
1 nm in atoms moving Photographic (b)Can cause many chemical
from one energy film. reactions,e.g.,the tanning of
level to a lower level. the human skin.
(c)Ionize atoms in atmosphere,
resulting in the ionosphere.
X-rays 1016to 1019 1 nm to X-ray tubes or innerPhotographic (a)Penetrate matter
1O-3nm shell electrons. film, Geiger (e.g.,radiography)
tubes, Ionization(b)Ionize gases
chamber.
(c)Cause fluorescence
(d)Cause photoelectric emission
from metals.
(e)Reflected and diffracted by
crystals enabling ionic lattice
spang and N A(or wave-
length) to be measured.
Gamma rays 1018to 1022 <10-3nm Radioactive decay of Photographic film, Similar to X-rays.
the nucleus. Geiger tubes,
Ionization chamber.www2z~tpsoruvp2n~y

640. 8.20
ForYour Knowledge
~Allelectromagnetic
waves travel through vaum
withthe same speed. They differ basically intheir
wavelengths or frequenes. Asa result, different e.m.
waves interact differently with matter.
~The e.m. waves interact with matter through their
electric and magnetic fields. These fields set into
oscillation the charges present in the matter. The mode
of interaction (absorption, scattering, etc.) depends on
the wavelength of the e.m. wave and the natureof the
atoms and moleles constitutingthematter.
~The wavelength of the e.m. wave radiated by any
charge system depends on the size of that system.
Atomic nuclei radiatey-rays of wavelengths 10-14 to
1O-15m. Heavy atoms emit X-rays. Electrons osllating
ina rit give rise to radiowaves. A transmitting
antenna radiates most effectively theradiowaves of
wavelength equal to the size of the antenna.
~The infrared waves indent on a substance set into
osllation all its electrons, atoms and moleles. This
increases the internal energy and hence the
temperature of the substance. That is why infrared
waves are also called heat waves.
~Our eyes are most sensitive tothe most intense wave-
lengths of the solar spectrum. That isthe centre of
sensitivity of oureyescoindes with the centre of the
wavelength distribution of thesolar radiation.
8.12EARTH'SATMOSPHERE
16. Give classification of earth's atmosphere into
different layers.
Earth's atmosphere.Thethickenvelope of airthat
surrounds the earth iscalled earth's atmosphere. It extends
to about 400 krn above the earth. As we go up, the air
pressure decreases gradually. Broadly, theearth's atmos-
phere can be divided into the following layers or zones.
1. Troposphere. This layer extends to a height of
12 krn from the earth's surface. Its upper boundary is
calledtropopause.As height increases, temperature
decreases from 290Kto 220K.This layer contains a
large amount of water vapour and cloudsareformed
in it. It is responsible for all the important wheather
phenomena that affect our environment.
2.Stratosphere. This layer extends from 12 krn to
50krnand its upper boundary iscalledstratopause.
The lower portion of this layer contains a large
concentration of ozone, resulting from the dissoation
of molelar oxygen by solar ultraviolet radiation in
the upper atmosphere. This layer iscalledozone layer
orozonosphereand extends from 15 krn to about
30km.Thetemperature of stratosphere rises from
220 K to 280 K.
PHYSICS-XII
Appleton layer Approx.
400km
Kennelly Heaviside layer
Thermosphere
Mesopause 80km
Mesosphere
Stratopause 50km
Stratosphere
30km
Ozone layer
15km
Tropopause 12km
Troposphere
Sea
level
Fig. 8.10Various layers of earth's atmosphere.
3. Mesosphere.This layer extends from50km to
80 km. Its upper boundary is calledmesopause. The
temperature of thisregion falls from 280 K to 180 K.
4.Ionosphere.This layer extends from 80kmto
400 km. Itstemperature increases with height from
180Kto700K.The ionsophere is mostly composed of
electrons and positive ions. This ionisation is caused by
ultraviolet radiation and X-rays corning from the sun.
The lower portion of the ionosphere extending from 80
km to95krn is calledthermosphere. The concentration
of electrons (i.e.,electron density) isfound to be very large
in a region beyond 110 krn from the surface of earth
which extends vertically for a few kilometres. This
layer of electrons is calledKennelly Heavide layer.
Beyond this layer, the electron density decreases consi-
derably untilat a height of about250krn, alayer of
electrons is again met. Thislayer is calledAppleton layer.
Except for the ionosphere, the rest of the
atmosphereiscomposed mostly of neutral moleles.
Table 8.4 Salient features of earth's atmosphere
Extent Fallinden-
Behaviour of
Name of in 10nsityinterms
temperature in
the layer aboveofground
thelayer
earthlevelvalue
Troposphere oto 12 From Falls uniformly
1to 10-1 from 290K to
220K
Stratosphere 12 to 50 From Risesuniformly
10-1 to 10-3 from 220Kto
280K
Mesosphere 50 to 80 From Falls uniformly
10-3to 10-5 from 280 K to
180K
Ionosphere 80 to From Risesuniformly
300 10-5to10-10 from 180 K to
700Kwww4~«trsq•vvr4p«z

641. ELECTROMAGNETIC WAVES
8.13EFFECT OF
EARTH'S ATMOSPHERE ON
ELECTROMAGNETIC RADIATION
Introduction
The sun is the main source of the electromagnetic
radiation that we receive on the earth. The atmosphere
is transparent to the visible radiation as we can see the
sun and the stars through it clear ly. However, the
other components such as infrared and ultraviolet
radiations from the sun are absorbed by different
layers of the atmosphere.
17.What Greenhouse effect for the atmosphere of
the earth and what its importance?
Greenhouse effect. This isthephenomenon which
keeps the earth's surface warm at night.
The radiation from the sun heats up the earth. Due
to its lower temperature, the earth re-radiates it mostly
in the infrared region. These infrared radiations cannot
pass through the lower atmosphere, they get reflected
back by gas moleles. Low lying clouds also reflect
them back to the earth. These radiations heat up the
objects on the earth's surface and so keep the earth's
surface warm at night.
8.21
Solar energy
heating earth
Atmosphere reflects
Infrared rays
J
Infrared waves
radiated by earth
Fig.8.11Greenhouse effect.
18.What the importance of ozone layer in the
atmosphere?
Importance of ozone layer. The solar radiation
consists of ultraviolet and some other lower wave-
length radiations which cause genetic damages to
living cells. The ozone layer absorbs these radiations
from the sun and prevent them from reaching the
earth's surface and causing damage to life. Moreover,
it also keeps the earth warm by trapping infrared
radiation.www2z~tpsoruvp2n~y

642. 8.32 PHYSICS-X
II
GIDELINESTONCERT EXERCISES
8.1.Figure 8.13shows acapacitor made of twocircular
plates each of radius 12cm and separated by 5.0m. The
capacitorisbeingcharged by an externalsource.The charging
current isconstant andequal to0.15A.
Fig.8.13
(a) Calculate the capacitance and the rate ofchange of
potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c)IsKirchhoffs first law (junction rule) valid at each
plateofthecapacitor?
Ans.Radius, R=12em=12x10-2m
Plate separation, d=5.0mm=5x10-3m
Plate area, A=1tR2=1tx(12x1O-2)2m2
(a)Capacitance,
C=EoA
d
8.85x10-12 x 1tx(12x10-12)2
5x10-3
=80.1x10-12 F=80.1pF
Now the charge at any instant on a capacitor
plate is
s=CV
1=dq=C.dV
dt dt
Thus the rate of change of potential difference
between the plate is
dV I
-=-
dtC
0.15
=80.1x10-12
=1.875xl09 Vs-1
(b)From the property of continuity,
Displacement current==Conduction current
ID=1=0.15A
(c)As the sum (I+ID)is continuous, so the Kirchoffs
first law is valid at capacitor plate if the current in the law
is the sum of conduction and displacement currents.
8.2.A parallel plate capacitor made of circular plates each of
radius R=6.0em has a capacitance C=100pF. The capacitor
isconnected to a230Va.c. supply with a (angular) frequency of
300 rads-1.
(a) What isthe r value of the conduction current?
(b)Isthe conduction current equal tothe displacement
current?
(c) Determine theamplitude ofBat a point3.0emfrom
theaxisbetween the plates.
230 V, 50 Hz
Fig.8.14
Ans.HereR=6.0em,C=100pF=10-10F
Vr=230V, 00=300rads-1
(a)Impedance of the capacitor is
1
Xc=-
wC
V
I=r
rms Xc
=Vrms' wC
=230x300x10-10 A
=6.9x10-6A=6.9IlA.
(b)Yes, the conduction current is equal to the
displacement current even if Iis oscillatingin time.
(c)The formula
B=llo...!:....I =llo...!:....I
21tR2 D21tR2
isvalid even if ID(and henceB)oscillates in time. The
formula shows thatBandIDare in phase.
Hence R -110.L:I
"1J-21tR2 0
whereIbandIoare the amplitudes of the oscillating
magnetic field and current, respectively.
Now 10=.,fiIr=-Iix6.911A=9.7611A
R=2x10-7x3x10-2 x9.76x10-6
"1J (6x102)2
=1.63xl0-11T.www4z~tqspruvq4o~y

643. ELECTROMAGNETIC WAVES
8.3.What physical quantityisthe same for X-rays of
wavelength 1O-
10m red light of wavelength6800Aand
radiowaves of wavelength500m?
Ans.The wave speed in vacuum is the same for
all radiations:c= 3x10Bms -1.
8.4.Aplane electromagnetic wave travels in vacuum along
Z-direction. What can you say about the directions of its electric
and magnetic field vectors?If the frequency of the waveis
30 MHz,whatisits wavelength?
~ ~
Ans.The electric and magnetic field vectorsEandB
of an e.m. wave must be perpendicular to each other and
alsotothe direction of propagation of the wave. Hence
~ ~
vectors EandBlie inx-yplane in mutually
perpendicular directions.
Frequency, v= 30 MHz =30x 106 Hz
c3 x lOB
.. Wavelength,A=-= 6= 10m.
v30 x 10
8.5.Aradio can tune to any station in the7.5MHz to
12 MHzband. What isthe corresponding wavelength band?
Ans.HerevI=7.5MHz =7.5x 106 Hz,
v2= 12 MHz = 12 x 106 Hz
~=~
vI
3 x lOB
--....,-=40m
7.5x106
C
A2=-
v2
3 x lOB
--....,-=25m
12x106
Thus the wavelength band is 40 m -25 m.
8.6.Acharged particle oscillates abouts its mean
equilibrium position with a frequency of109Hz.Whatisthe
frequency of the electromagnetic waves produced bythe
oscillator ?
Ans.According to Maxwell, a charged particle
oscillating with a frequency of 109 Hz,produces electro-
magnetic waves of the same frequency 109 Hz.
8.7.The amplitude of the magnetic field part of a harmonic
electromagnetic waveinvacuumis80=510nT. Whatisthe
amplitude of the electric field part of the wave?
Ans.Here80= 510 nT = 510 x 1O-9T
Amplitude of the electric field,
EO=cBo'
=3x10Bx 510 x 10-9
=153NC1.
8.33
8.8.Suppose that'theelectric field amplitude of an
electromagnetic wave isEO= 120NC1and thatits frequencyis
v =50.0 MHz.(a) Determine, 80,0>,kandA.(b) Find
~ ~
expressions for EandB.
Ans.Here EO = 120NC1,
v= 50.0 MHz = 50x106Hz
0>=27tV=2x3.14x50x106
=3.14xlOB rad s-l.
0>3.14xlOB -1
k= -= B= 1.05 m .
c3x10
c3xlOB
A=-= 6= 6.00m
v50x10
(b)If the wave is propagating along x-axis, then field
~ ~
Ewill be along !taxis and fieldBalongz-axis,
~ A
.. E=EOsin(kx-o>t)j
orE=120sin(1.05x -3.14xlOBt)1NC1
where xisin metre and t in second.
~ A
B=Bosin(kx-o>t)k
=4x10-7sin(1.05x-3.14x108t)jtesla.
8.9.The terminology of different parts of the
electromagnetic spectrum isgiven in the text. Use the formula
E=hv(forenergy ofaquantum of radiation :photon) and
obtain the photon energy in units ofeV for different parts of the
em spectrum. In what wayarethedifferent scales of photon
energies that you obtain related to the sources of electromagnetic
radiation?
Ans.Photon energy forA= 1m,is given by
hc
E=hv =-
A
6.63x10-34x3x10B J
1
6.63x10-34x3xlOB
----~~-eV
1.6x10-19
= 12.43xlO-7eV
= 1.24x10-6 eV
Photon energy for other wavelengths in the figure for
electromagnetic spectrum can be obtained by multiplying
appropriate power of ten, as indicated on next page.www8notesdrive8com

644. 8.34
Wavelength (in
m) Energy of photon (in eV)
1O-15m 1.24 x 109 eV
1O-12m 1.24 x 106 eV
1O-9m 1.24 x 103 eV
5xlO-7m 2.5eV
10-3m 1.24 x 1O-3eV
10m 1.24 x 10-7 eV'
103m 1.24 x 10-geV
Energy ofphoton thatasource produces indicates the
spacings of the relevant energy levels of the source. For
example,/...=1O-12m corresponds to photon energy
=1.24x106eV=1.24 MeV. This indicates that nuclear
energy levels (transitions between which causey-ray
emission) are typically spaced by 1 MeV or so. Similarly, a
visible wavelength /...=5 x10-7m corresponds to photon
energy=2.5eV. This implies that energy levels
(transitions between whichgive visible radiation) are
typically spaced by a few eV.
8.10.Ina plane e.m. wave, the electric field oscillates
sinusoidally at a frequency of 2.0x1010Hzand amplitude
48 Vm-1.
(a) Whatisthe wavelength of a wave?
(b) Whatisthe amplitude of the oscillating magnetic
field?
-4
(c)Show that the average energy density of the E field
-4
equals the average energy density of theBfield.
[CBSE OD 90]
Ans.(a)Wavelength,
c3 xlOB
/...=~=2.0xHYO
=1.5x10-2m.
(b) ~=fv=~
'1J C 3xlOB
=1.6xl0-7T.
-4
(c)Average energy density of Efield,
1 2
UE=-EOEQ
4
-4
Average energy density of Bfield,
1 2
UB=- BQ
4110
1
But fv =cBaand?= -
1l0Eo
PHYSICS-XII
121 2
uE=4EoEQ=4EO(cBa)
1 1 2 1 2
=-Eo.-- .BQ= --Bo=uB•
4 11oEo 4110
8.11.Suppose that the electric field part of an electro-
magnetic wave in vacuumis
-4 8 A
E=(3.1NIC)cos [(1.8 radlm)y+(5.4x10radls)tJi
(a) Whatisthe direction of propagation?
(b) Whatisthe wavelength /... ?
(c)Whatisthe frequencyv?
(d) What is the amplitude of the magnetic field part of
the wave?
(e) Write an expression for the magnetic field part of the
wave.
Ans.(a)The wave is propagating along negative
y-direction or its direction is -J.
(b)Comparing the given equation with the standard
equation,
we get,
E=fv cos [ 21t(t+vt)]
21t=1.8
x
:. Wavelength,
/...=21t=2x3.14 =3.5 m.
1.8 1.8
(c)Also,Znv=5.4x106
5.4x106
v=---
21t
5.4xlOB
--- =85.9x106Hz.
2x3.14
(d)
'" 86 MHz.
B _fv_3.1NC1
0-c-3x10Bms-1
=10.3 x 10-9 T=10.3 nT.
-4 A
B=Bacos(ky+wt)k
=(10.3 nT) cos [(1.8 rad / m) y
+(5.4 x lOB rad/ s)t]k.
8.12.About5%of the power of a100Wlight bulbis
converted to visible radiation. Whatisthe average intensity of
visible radiation:
(e)
(a) at a distance of1m from the bulb?
(b) at a distance of10m?
Ans.The bulb, asa point source, radiates light in all
directions. At a distance ofrm, the surface area of the
surrounding sphere,
A=41tr2www8notesdrive8com

645. ELECTROMAGNETIC WAVES
:. Average Intensity
_ Energy / time _ Power _ Power
- Area - Area - 4 n?
(a)Average intensity of
visible radiation at a
distance of1m
=5%of100 W=0.4Wm-2
411:(1m)2
(b)Average intensity of visible radiation at a
distance of10m
=5%of100 W=0.004Wm-2.
411:(10m)2
8.13.Use the[ormuluAmT=029emKto obtainthe
characteristic temperature rangesfordifferent parts.of the e.m.
spectrum. What do the numbers that YDUobtain tell YDU?
Ans.A body at temperature Tproduces a continuous
spectrum. For a black body, the wavelength corres-
ponding to maximum intensity of radiation is given by
Wien's law:
AmT=0.29emK=0.0029mK T=0.0029K
Am
From the above formula, the temperatures for
various wavelengths of the e.m. spectrum will be as
follows:
Part of the Wavelength Temperature
Spectrum (Amin m) (Tin K)
1.
y-rays 10-12 29x108
2.X-rays 10-10 29x106
3.Ultraviolet 3x10-7 104
4.Visible 5x10-7 6x103
5.Infrared 3x10-5 102
6.Microwaves 3x10-3 1
7.Radiowaves 1 29x10-4
Thesenumbers tell us thetemperature ranges
required to produce e.m. radiations of different wave-
lengths. To produce visible radiation of A=·5x 10-7 m, we
needto have source at a temperature of 6000K.A source
at lower temperature will also producethis wavelength
but not with maximum intensity.
8.14.Given below are some[amousnumbers associated
with electromagneiic radiation in different contextsin
physics. State the part.of the e.m. spectrum to which each
belongs :
(a)21em (wavelength emitted by atomic hydrDgen in
interstellar space).
(b)1057MHz (frequency o] radiation arisingfromitoo
close energy levels in hydrDgen;knoumasLamb
shift)·
8.35
(c)2.7K(temperature associated with the isotropic
radiation filling all space-thoughtto be a relic.ofthe
'big-bang' origin .of universe).
(d)5890A-5896A(doublet lines.ofsodium).
(e)14.4keV(energy.of a particulartransition)inFe57
nucleus associated with afamous high resDlutiDn
speciroscopic method (MDssbauer spectroscopv).
Ans.(a)RadiDwave(short wavelength end).
(b)Radiouiaue (high frequency or short wavelength
end).
(c)Here T=2.7K
By Wien's law,
AmT=0.29emK
A=0.29em K.:::0.11em
m 2.7K
This wavelength lies in the microuiaoe region of the
e.m. spectrum.
(d)Yellow part of thevisible regionof the e.m.
spectrum.
(e)Here
As
E=14.4keV=14.4 x103x1.6x10-19J
E=hv
_.!:._14.4x1.6xlO-16 -35 1018H
v- - 34 -.x z
h 6.6 xlO
This frequency lies in theX-rayorSDft y-rayregion of
the e.m. spectrum.
8.15.Answer the fDllowing questions :
(a) Long distance radio broadcasts useshort-uiaue
bands.Why? [CBSE D 05]
(b) Itisnecessary to use satellites jorlongdistance TV
transmission. Why ? [CBSE DOS]
(c) Optical and radioielescopes are built .on theground
but X-ray asironomu ispossible .onlyfromsatellites
orbiting theearth. Why? [CBSEOD09]
(d)The small DZDne layer.on tDP .of the stratosphere is
crucial forhumansurvival. Why?
[CBSED14;OD09]
(e)If the earth did not have an atmosphere, uiouldits
average surface temperature behigher Driotoerthan
what it is nDW? [CBSE D 14]
if)Some scientists havepredicted that a globa!
nuclear war .onthe earth uiouldbe[ollouied by a
severe 'nuclear winter' with a devastating effect on
life.onearth. What might be the basis .of this
prediction ? [CBSEOD95]
Ans.(a)It isbecause the radiowaves of shortwave
band are easily reflected back to the earth by the
ionosphere.www8notesdrive8com

646. 8.36
(b)TVsignals being
of high frequency are not
reflected by the ionosphere. Also, ground wave
transmission is possible only upto a limited range.
That is why satellites are used for long-distance TV
transmission.
(c)The earth's atmosphere is transparent to visible
light andradiowaves but it absorbs X-rays. X-ray
astronomy is possible only from satellites orbiting the
earth. These satellites orbit at a height of 36,000km, where
the atmosphere is very thin and X-rays are not absorbed.
(d)Ozone layer absorbs ultraviolet radiation from the
sun and prevents it from reaching the earth and causing
damageto life.
PHYSICS-XII
(e)The earth radiates infrared waves which are
reflected by the gases in the lower atmosphere. This
phenomenon, called Green house effect, keeps the earth
warm. So if the earth did not have atmosphere, its average
temperature would be low due to the absence of Green
house effect.
if>The clouds produced by a global nuclear war
would perhaps cover substantial parts of the sky
preventing solar light from reaching many parts of the
globe. This would cause a 'nuclear winter'.www4z~tqspruvq4o~y

647. Text Based Exercises
YPE A:VERY SHORT ANSWER QUESTION
S(1 mark each)
1.Name the scientt who first predicted the extence
of electromagnetic waves.
2.Name the Indian scientt who first produced the
electromagnetic waves.
3.What dplacement current?
4.Write the SIunit of dplacement current.
5.What modificatn was made by Maxwell in
Ampere circuital law ? [Haryana94]
6.Write anexpressn forthe dplacement current.
7.Does the dplacement current satfy the property
of continuity ?
8.The charging current for a capacitor 0.25A. What
the displacement current across its plates?
9.Dtinguh between conductn current and
dplacement current.
10.Namethe laws associated with the following
equations :
f~~ df~ ~
(ii)B.dS=J!oEo - E.dS .
dt
11.Canwe apply Maxwell's equatns to different
types of media, like dielectrics, conductors,
plasmas, etc?
12.Are the Maxwell's equatns true for arbitrary high
~ ~
and low values of EiB,q,I?
13.Write down Maxwell's equatns for steady electric
field. [Haryana93]
14.Write down Maxwell's equatns for steady
magnetic field.
15.Which of the four Maxwell's equatns shows that
electric lines of force cannot form closed loops?
16.Which of the four Maxwell's equatns shows that
magnetic lines of force cannot start from a point nor
end at a point?
17.If magnetic monopolies exted, which of the four
Maxwell's equatns be modified?
18.A capacitor has been charged by ade source. What
are the magnitudes of conductn and dplacement
currents, when it fully charged ?[CBSE D13]
19.A capacitor connected to an a.c. source. Is the
conductn current in connecting wires equal to the
dplacement current in the capacitor?
20.What are electromagnetic waves?
21.Write an expressn for the speed of e.m. waves in
free space. [Haryana98]
22.For an electromagnetic wave, write the relatnship
between amplitudes of electric and magnetic fields
in free space. [CBSE OD 94 ;D 95]
23.State two charactertics of an electromagnetic
wave. [CBSE D 98C;Punjab02)
24.Can we produce a pure electric or magnetic wave in
space?
25.Is the light emitted by an ordinary electric lamp an
electromagnetic wave?
26.Namethe basic source of electromagnetic waves.
27.A plane electromagnetic wavetravels in vacuum
along x-directn. What can you say about the
directns of electric and magnetic field vectors ?
[CBSE D11)wwwrnotesdrivercom

648. ELECTROMAGNETIC WAVES
28.What are the
directns of electric and magnetic
field vectors relative to each other and relative to
the directn of propagatn of electromagnetic
waves? [CBSEOD 12]
29.What was the range of wavelengths of e.m. waves
produced by ProfessorJ.e.Bose? [CBSED 93C]
30.What was the range of wavelengths of electro-
magnetic waves produced by Hertz?
31.Name the scientt who first set up the transmitter
and receiver of electromagnetic waves.
32.What the significance of theyear 1887 in the
htory of rad communicatn?
33.What electromagneticspectrum ?[Haryana97]
34.Name the electromagnetic waves that have
frequencies greater than those of ultravlet light
but less than those of gamma rays.[CBSED 04C]
35.What approximate wavelength range for vible
spectrum ? [Haryana99C]
36.Write the frequency limit of vible range of
electromagnetic spectrum in kHz. [CBSED 98]
37.Name the electromagnetic radiatn which has the
largest penetrating power. [CBSEOD99; D 2000]
38.Name the electromagnetic radiatn to which the
following wavelengths belong :
(a)10-2m (b)1A. [CBSED 06C]
39.What approximate wavelength of X-rays?
[CBSEOD90]
40.Name the part of electromagnetic spectrum whose
wavelength lies in the range of 10-10 m.Give its one
use. [CBSEOD 10]
41.Arrange the following in descending order of
wavelength:X-rays, Radwaves, Blue light,
Infrared light. [CBSEOD 10]
42.Which of the following has the shortest
wavelength : microwaves, ultra-vlet rays and
X-rays? [CBSEOD 10]
43.Arrange the following in the descending order of
wavelengths : y-rays, infrared rays, microwaves,
yellow light, rad waves. [CBSED 13C]
44.Arrange the following e.m. radiatns in the
increasing order of frequency: X-rays, radwaves,
ultra-vlet light, blue light, red light and infrared
light. [CBSED 98]
45.What common between different types of e.m.
radiatns?
46.Give the wavelength range and frequency range of
y-rays.
47.What the order of magnitude of the frequency of
vibratn of the longest and shortest waves in the
electromagnetic spectrum ?
8.37
48.Name the electromagnetic radiatns used for
viewing objects through haze and fog.
[CBSED 97, 04]
49.Name the part of the electromagnetic spectrum
which used in 'green houses' to keep plants
warm. [CBSEOD95C]
50.What 'greenhouse effect' ? [CBSEF 94C]
51.Which part of electromagnetic spectrum absorbed
from sunlight by ozone layer? [CBSED 10]
52.Name the part of the electromagnetic spectrum
suitable for radar systems used in aircraft
navigatn. [CBSED 04, 09,10;F 12]
53.What are microwaves? [Haryana96]
54.Name the different layers of the earth's atmosphere.
55.Which layer of the earth's atmosphere useful in
long dtance ra, transmsn ?
56.What the cut-off frequency beyond which the
nosphere does not reflect electromagnetic
radiatns? [CBSEOD94C]
57.What the range of wavelength of televn
broadcast. [CBSEF 95]
58.What type of waves are used in telecom-
municatn ?
59.What the nature of the waves used in radar ?
What their wavelength range?
[CBSESamplePaper 1997]
60.Name the part of the electromagnetic spectrum of
wavelength 102 m and mentn its one applicatn.
[CDSE008]
61.Name the part of the electromagnetic spectrum of
wavelengthla-2m and mentn its one applicatn.
[CBSED 08, 09]
62.Identify the part of the electromagnetic spectrum to
which the following wavelengths belong:
(i)1 mm(i)10-11 m. [CBSEOD08]
63.Identify the part of the electromagnetic spectrum to
which the following wavelengths belong :
(i) 1O-1m ; (i) 1O-12m [CBSEOD08]
64.From the following, identify the electromagnetic
waves having the(i)Maximum,(i)Minimum
frequency.
(i)Radwaves (i)Gamma-rays
(ii)Vible light (iv)Microwaves
(v)Ultravlet rays, and(vi)Infrared rays.
[CBSESamplePaper08]
65.Name the charactertic of electromagnetic waves
that:(i)increases;(i)remains constant in the
electromagnetic spectrum as one moves from
radwave regn towards ultravlet regn.
[CBSESamplePaper08]www8notesdrive8com

649. 8.38
66.Stat
e which ofthetwo rats, a and 13, defined
below; is greater thanor lessthan unity :
a=vX-ray; 13=A.vible
Vinfrared A.microwaves [CBSE D 07C]
67.The frequency values, Vjand v2' fortwospectral
lines of the e.m. spectrum, are foundto be 5 x1020
Hz and2.5xH11HZrespectively. Find the rat,
~ / 1..2 of their wavelengths. [CBSE OD 07C]
68.Thewavelength values, ~ and 1..2oftwospectral
lines, of the e.m. spectrum, are1800nmand0.12nm
respectively. Calculate the rat vI/ v 2oftheir
frequencies. [CBSE OD 07C]
69.The bombardment of a metal target, by high energy
electrons, can result inthe production of e.m.
waves. Name these waves. [CBSE D 08C ; F09]
70.Special devices, likethe.Klystron valve or the
magnetron valve, areused forproduction of
electromagnetic waves. Name thesewaves and also
writeone of their applications. [CBSE D 08C ; F09]
Answers
PHYSICS-XII
71.The frequency of oscillatn of the electric field
vector of a certain e.m. wave 5 x1014Hz. Whatis
the frequency of oscillatn of the corresponding
magnetic field vector and to which part of the
electromagnetic spectrum does itbelong?
[CBSEOD 08]
72.Write the following radiatns in ascending order
inrespect of their frequencies:
X-rays, microwaves, UV rays and radwaves.
[CBSE D 09]
73.Name the EM wavesused for studying crystal
structure of solids. What itsfrequency range?
[CBSEOD 09]
74.Name the electromagnetic waves which (i)maintain
the earth's warmth and(i)areused in treatment of
cancer tumours. [CBSE F 12]
75.Towhich part of the electromagnetic spectrum
does a wave of frequency (i)5 x](19Hz(i)
3x1013Hz (ii)5 x1011Hz belong? [CBSE OD 14]
1.James Clerk Maxwell.
2.j.c.Bose.
3.Dplacement current is that current which comes
into existence, inaddition to the conduction
current, whenever the electric field and hence the
electricflux changes with time.
4.The SIunitofdisplacement current ampere (A).
d~
5.A dplacement current, ID=EO_Ewasadded to
dt
the current term ofAmpere circuital lawtomakeit
logically constent so that themodified form of
Ampere's circuital law is
IB.dl=Jlo(IC+ID)
d~ dE
6.Dplacement current, I D=EO--X=EOA- .
dt dt
7.Yes,thesum of dplacement andconductn
currents remains constant along a closed path.
8.According to theproperty of continuity,
Dplacement current=Charging current =0.25A.
9.Conductncurrent is due to the flowof electrons
in a circuit. Itexts even if electrons flow at a
uniform rate. Dplacement current due to
time-varyingelectric field. It does not ext under
steady conditions.
10. (i)Gauss law of electrostatics.
(i)Faraday's lawof electromagnetic inductn.
•
(ii)Modified Ampere's law, the termontheright
hand side is Maxwell's dplacement current.
11.Yes,Maxwell's equatns are universally valid.
12.Yes.
13.(/)f
I--> -->
(i)j'E .dl=0
I--> -->
14.(i)j'B.dS=0
I--> -->
(i)rB.dl=0
15.Faraday's law of electromagnetic inductn i.e.,
I--> --> d~B
E.dl=--
dt
shows that electric lines of forcecannot form closed
loops.
I--> -->
16.Gauss'slaw ofmagnetostatics i.e., B.dS=0
showsthat magnetic lines of force always form
closed loops.
17.IB.dS=O.Th equatn isbased on the fact that
magnetic monopoles do not exist. If they exist, then
R.H.s. would containaterm representing pole
strength of the monopole, similar to Gauss law in
electrostatics.
I-->--> d~
Also,equatn E .dl=-_Bwould be modified.
dt
Anadditnal term1mrepresenting the current due
to flow of magnetic charges would have tobe
included onR.H.s.www5~£trsqrvvr5p£z

650. ELECT
ROMAGNETIC WAVES
18.When the capacitor gets fully charged,
Conductn current =Dplacement current =0
19.Yes.
20.Electromagnetic waves are the waves radiated by
accelerated charges and const of sinusoidally
varying electric and magnetic fields, the oscillatns
of the two fields are mutually perpendicular to each
other as well as to the directn of the propagatn
of the waves.
21.The speed of an electromagnetic wave in free space

1
c=---
~Il 0 EO .
22.The rat of the amplitudes of electric and magnetic
fields equals the speed of light.
FU=c=_l_
Bo ~lloEo
23.(a)The electromagnetic waves travel in free space
with a speed of3x108m / s.
(b)They are transverse in nature.
24.No, it not possible.
25.Yes, it an e.m. wave.
26.Anelectric dipole a basic source of electro-
magnetic waves.
27.Electric field vectors are alongy-/z-directn and
magnetic field vectors are alongz-/y-directn.
28.Thedirectns of electric and magnetic field
vectors areperpendicular to each other as well as
to the directn of propagatn of electromagnetic
waves.
29.[.C,Bose was able to produce e.m. waves of the
wavelength ranging from 25 mm to 5mm.
30.Hertz was able to produce e.m. waves of
wavelength around 6m.
31.Hertz.
32.Itwasinthe year 1887 that a German physict,
Heinrich Hertz, succeeded for the first time in
producing and detecting e.m. waves of wavelength
around 6 m.
33.Theorderly dtributn of electromagnetic radiations
ofalltypes according to their wavelength or fre-
quency into dtinct groups having widely differing
properties called electromagnetic spectrum.
34.X-rays.
35.The wavelength range for the vible part of the
electromagnetic spectrum 3900 Ato7600 A
36.The frequency range for the vible part of the e.m.
spectrum 4 x "kHzto 7.7 x1011kHz.
37.y-rayshavethe largest penetrating power.
8.39
38.(a)Microwaves (b)X-rays.
39.The approximate wavelength range for X-rays
0.lA-100 A.
40.X-rays. These areusedin medical diagnos.
41.Inthedescending order of wavelength,
Radwaves>Infrared light> Blue light> X-rays
42.X-rays.
43.Inthe descending order of wavelength, wehave
Radwaves, infrared rays, red light, yellow light,
y-rays.
44.Inthe increasing order of frequency, we have
Radwaves, infra-red light, red light, blue light,
ultra-vlet light, X-rays, y-rays.
45.All e.m.radiations havetransverse wave nature
and travel with the same speed of 3 x108ms-1in
free space.
46.Wavelength range of y-rays from0.01to1.4 Aand
frequency range from 1019to1024Hz.
47.Frequency of vibration of longest waves (rad-
waves) of the order of 105Hz. Frequency of
vibration ofshortest waves (y-rays)of the order
of1019to1024Hz.
48.Infra-red radiatns.
49.Infra-red radiatn.
50.Greenhouse effect the phenomenon which keeps
the earth's surface warm at night. Theearth reflects
back infrared part of solar radiatn. Infrared rays
are reflected back by lowlying clouds and lower
atmosphere and keep the earth's surface warm at
night.
51.Ozone layer absorbs ultravlet radiatn coming
fromthe sun.
52.Microwaves.
53.Microwaves are e.m. waves of wavelength range
1O-3m to0.3m.
54.Earth's atmosphere can be divided into four layers.
These are:
(i)Troposphere (i)Stratosphere
(ii)Mesosphere (iv)Ionosphere.
55.Ionosphere.
56.Ionosphere cannot reflect electromagnetic
radiations having frequency higher than40MHz.
57.The e.m. waves used in T.V. broadcast have
wavelength rangefrom1.36m to3m.
58.Microwaves.
59.Microwaves.These are electromagnetic waves of
the wavelength range 1O-3m to 0.3m.
60.Radiowaves, used in rad and televn
communicatn.
61.Microwaves, usedin microwave ovens.www8notesdrive8com

651. 8.40
62.(i
)Microwaves, (ii)Gamma rays.
63.(i)Radwaves, (ii)Gammarays.
64.(i)Gamma-rays havethemaximum frequency.
(ii)Radwaves have the minimum frequency.
65. (i)Thefrequency of the electromagnetic waves
increases.
(ii)The speed of theelectromagnetic waves remains
constant (c =3x108ms-I).
a>landp<1
A,=v2
A,2 VI
2.5x1011 10-10
-----,.,20" =5x .
5x10
66.
67.
0.12 1
--=--
1800 1500
PHYSICS-XII
69.X-rays,used in medical diagnos.
70.Microwaves, used in microwave ovens.
71.Frequency of oscillatn of the magnetic field vector
=5x1014Hz
Th frequency belongs to thevisible region of the
e.m. spectrum.
72.Radwaves<Microwaves<UV-rays<X-rays.
73.X-rays are used to study crystal structure. Their
frequency range from 1016 Hzto 1019 Hz.
74.(i)Infra-red radiatn
(ii)Gamma rays.
75.(i)X-rays/y-rays
(ii)infra-red radiatn
(ii)microwaves.
"YPE B :SHORT ANSWER QU ESTIONS (2 or 3 marks each)
1.What dplacement current? Why was th
concept introduced?
2.State and explain Maxwell's modification of
Ampere's law. [HPB1997]
3.Give four charactertics of dplacement current.
4.Explain how one' observes an inconstency' when
Ampere's circuital law applied to the process of
charging a capacitor.
How th 'contradictn' gets removed by
introducing the concept of an 'additnal current',
known as the 'dplacement current' ?
[CBSESP 15]
J.Write the generaled expression forthe Ampere's
circuital law in terms of the conduction current and
dplacement current. Mentn the situation when
there :(i)Only conduction current and no
dplacement current(ii)Dplacement current and
no conductn current. kBSE F 13]
6.(a)A capacitor connected inseries toan
ammeter across a d.c. source. Why does the
ammeter show a momentary deflectn during
the charging of the capacitor? Whatwould be
the deflectn when it is fully charged ?
(b)How the generalized form of Ampere's
critical law obtained to include the term due to
dplacement current? [CBSEOD 14C]
7.Conductn and dplacement currents are
individually dcontinuous, but theirsum is
continuous. Comment. [Haryana 93,94]
8.State Maxwell equatns. [CBSED 92C]
9.Briefly explain how Maxwell was led to predict the
extence of electromagnetic waves.
10.How can we express mathematically a plane
electromagnetic wave propagating along X-ax?
Alsorepresent it graphically.
11.(a)An e.m. wave travelling in a medium with a
velocityIf=vi.Draw a sketch showing the
propagatn of the e.m. wave, indicating the
directn of the oscillating electric and
magnetic fields.
(b)How are the magnitudes of the electric and
magnetic fields related to the velocity of the
e.m. wave? [CBSED 13]
12.Briefly explain, how does an accelerating charge act
as a source of an electromagnetic wave?
13.Briefly describe the work of Maxwell and Hertz in
thefield of electromagnetic waves. [CBSED 97C]
14.What were the contributns of J.e.Bose and
Guglielmo Marconi in the field of rad
communicatn ?
15.Give a brief htory ofthe observatn of
electromagnetic waves. [Punjab 91 ; Haryana 97]
16.Showthatelectromagnetic waves are transverse in
nature. [Haryana 01]
17.Obtain an expressn for the energy density of an
electromagnetic wave.
18.Inan electromagnetic wave, show thatthe energy
density of the E-field equals the energy density of
theBfield,www.notesdrive.com

652. ELECTROMAGNETIC
WAVES
19.Define intensity of an electromagnetic wave.
Obtain an expression for it.
20.Write expressions for(i)linearmomentum and
(ii)pressure, exerted by ane.m.wave on asurface.
21.State four basic properties of electromagnetic
waves. [CBSEOD95;Punjab97, 01]
22.Givetwocharactertics of electromagnetic waves.
Write the expressn for the velocity of electro-
magnetic waves in terms of permittivity and
magnetic permeability. [CBSED93C]
23.Whatmeant by electromagneticspectrum? Give
its four uses. [Punjab99C]
24.Write one property andoneuse each of infrared
rays,ultraviolet rays and radiowaves. [Punjab02]
25.Draw a sketch of electromagnetic spectrum,
showing the relative positions of UV, IR,X-rays
and microwaves with respect to vible light. State
approximate wavelength of any two. [ICSE 2000]
26.Briefly describe any four regions of the
electromagnetic spectrum, mentioning their special
properties/features. [ICSE 2002]
27.What are microwaves? Give their anyone use.
[Haryana9S]
28.What is meant by thetransverse nature of
electromagnetic waves? Draw a diagram showing
the propagatn of an electromagnetic wave along
the x-directn, indicatingclearly the directns of
the oscillating electric and magnetic fields
associated with it. [CBSEODOS]
29.Identify the following electromagnetic radiatns as
per the wavelengths given below. Write one
applicatn of each.
(a)1rom(b)1O-3nm (c)1O-8m [CBSEODOS]
30.Identify the following electromagnetic radiatns as
per the frequencies given below. Write one
application of each.
(a)1020Hz(b)109Hz(c)1011Hz.[CBSEODOS]
31.The following table gives the wavelength range of
some constituents of the electromagnetic spectrum:
S.Na. WavelenQth range
1.
1rom to700nm
2. 400nm to1nm
3. 1nm to10-3nm
4. <10-3 nm
Answers
8.41
Select the wavelength range and name the
electromagnetic waves that are:
(i)widely used in the remote switches of house-
hold electronic devices.
(i)produced in nuclear reactns.
[CBSE D OSC]
32.Name the electromagnetic radiatns having the
wavelength range from1nm to10-3nm.Give its
two important applicatns. [CBSEF09]
33.Name the electromagnetic radiatn having the
wavelength range from10-1m to10-3m. Give its
two important applicatns. [CBSED09]
34.Name the electromagnetic radiatns having the
wavelength range from1rom to700nm. Give its
two important applicatns. [CBSEF09]
35.Arrange the following electromagnetic radiatns
in ascending order of their frequencies:
(i)Microwave(i)Radwave
(iii)X-rays (iv)Gamma rays. [CBSED10]
36.Answer the following questns: [CBSED14]
(a)Name the e.m. waves which are produced
during radactive decay of a nucleus. Write
their frequency range.
(b)Welders wear special glass goggles while
working. Why ? Explain.
(c)Why are infrared waves often called as heat
waves? Give their one applicatn.
37.Answer the following:
(a)Name the e.m. waves which are used for the
treatment of certain forms of cancer. Write
their frequency range.
(b)Thin ozone layer on top of stratosphere
crucial for human survival. Why ?
(c)Why the amount of the momentum
transferred by the e.m. waves incident on the
surface so small? [CBSED14]
38.Answer the following questns:
(a)Show, by giving a simple example, how e.m.
waves carry energy and momentum.
(b)How are microwaves produced ? Why it
necessary in microwave ovens to select the
frequency of microwaves to match the
resonant frequency of water molecules?
[CBSE D14C]
"
1.Refer answers to Q.3andQ.5on page8.38.
2.Refer answertoQ.1on page 8.1.
3.Refer answer toQAon page8.3.
4.See property of continuity inQ.1on page 8.2.www8notesdrive8com

653. 8.42
5.Modified
Ampere's circuital law
fB.di=~o[Ie+Eod:tE]
(i)In the connecting wires,
$E=0, soId=0.
There only conductn current,
I_dq
d -dt
(i)In the regn between the capacitor plates,
Ie=0
andId=EOd$E= Eo!£(!L] =dQ.
di dt EOdt
6.Refer to the solutn of Problem 1 on page 8.24.
7.See property of continuity in Q.1 on page 8.2.
8.Refer to point 3 of Glimpses.
9.Refer to solutn of Problem 4 on page 8.24.
10.Refer to point 4 of Glimpses, see Fig. 8.5 on page 8.7.
11.(a)See Fig. 8.5 on page 8.7. (b)c=Eo.
Eo
12.Refer answer toQ.6 on page 8.8.
13.Refer answer toQ.8 on page 8.10.
14.Refer answer toQ.8 on page 8.10.
15.Refer answer to Q. 8 on page 8.10.
16.Refer answer toQ.9 on page 8.10.
17.,18.Refer answer to Q. 10 on page 8.11.
19.Refer answer toQ.11 on page 8.11.
20.Refer answers toQ.12 andQ.13 on page 8.11.
21.Refer to solutn of Problem 5 on page 8.24.
22.Refer answer to Q. 14 on page 8.12.
23.Refer answer toQ.15 on page 8.16.
24. (i)Infrared rays produce heating effect. They are
useful in haze photography.
(i)Ultravlet rays cause tanning of human skin.
They are used in the study of molecular
structure.
(ii)These are the e.m. waves of longest
wavelength. They are used in rad and TV
transmsns.
25.Refer answer toQ.15 on page 8.16.
26.Refer answer toQ.15 on page 8.16.
27.Refer answer toQ.15 on page 8.16.
-+ -+
28.The directns of oscillatns of E and B fields are
perpendicular to the directn of propagatn of
e.m. waves. So these waves are transverse in nature.
PHYSICS-XII
29.(a)Microwaves, used in radar systems for aircraft
navigatn.
(b)j-rays, used in the treatment of malignant
tumours.
(c)Ultravlet rays, used in food preservatn.
30.(a)X-rays, in the study of crystal structure.
(b)Radwaves, in rad and televn
communicatn systems.
(c)Microwaves, in microwave ovens.
31.(i)1runto 700run,infrared radiatns.
(i)Ie<10-3run,Gamma rays.
32.X-rays. They are used in
(i)medical diagnos and
(i)in the study of crystal structure.
33.Microwaves. They are used
(i)in radar systems and(i)in microwave ovens.
34.Infrared radiatns. They are used
(i)in remote control devices and
(i)in haze photography.
35.The radiatns in ascending order of frequencies
are:
Radwaves<Microwaves<X-rays<Gamma rays
36. (a)y-rays. Range:1019Hz to 1023Hz
(b)Welders wear special glass goggles to protect
their eyes from large amount of harmful UV
radiatn produced by welding arc.
(c)Infrared waves incident on a substance
increase the internal energy and hence the
temperature of the substance. That why they
are also called heat waves.
Infrared waves are used in remote control of
TV, in green houses, in the treatment of
muscular complaints, etc.
37. (a)X-rays/y-rays. Range:1018Hz to 1022 Hz.
(b)Ozone layer absorbs ultravlet radiatn from
the sun and prevents it from reaching the earth
and causing damage to life.
(c)If an e.m. wave transfers a total energy U to a
surface, then total momentum delivered to the
surface
U
p=-
c
[For complete absorptn of energy U]
As the speed of light c has a very large value,
so the value of momentum transferred very
small.
38.(a)Consider a plane perpendicular to the
directn of propagatn of the wave. An
electric charge, on the plane, will be set inwww8notesdrive8com

654. ELECTROMAGNETIC WA
VES
motn by the electric and magnetic fields of
-emwave, incident on this plane. This
illustrates thate.m. waves carry energy and
momentum.
(b)Microwaves are produced by special vacuum
tubes like the Klystron/Magnetron/Cunn
diode.
8.43
The frequency of microwaves selected to
match the resonant frequency ofwater
molecules, so that energy transferred
efficiently to the kinetic energy of the
molecules.
"VPE C :LONG ANSWER QUESTIONS (5 marks each)
1.Describe Hertz's experimentforproducing and
detecting electromagnetic waves. Howdid Hertz
experimentally establish that (i)e.m. waves are
transverse in nature, (ii)e.m.waves travel with the
speed of light, and(iii)e.m.waves can be polarised ?
2.Draw a labelled diagram of Hertz's experimental
set-up toproduce electromagnetic waves. Explain
Answers
the generatn of electromagnetic waves using th
set up. Explain how the electromagnetic waves are
detected. [CBSED 06; OD06C]
3.What electromagnetic spectrum? Name its main
parts, giving the frequency rangeandsource of
production ineach case. Also give one important
useof each part.
•
1.Refer answer to Q.7 onpage 8.9.
2.Refer answer toQ.7 on page 8.9.
3.Refer answer to Q.15 on page 8.16.
"VPE D :VALUE BASED QUESTIONS (4 marks each)
1.Two friends were passing through the market.
They saw two welders using welding machines.
One welderwas using the goggles and face masks
with window in order to protect h face. The other
one was weldingwith naked eyes. They went to the
welder who was not using face mask and explained
him the advantages ofgoggles and masks. Next
day,thewelder bought a set of goggles and began
to do hwork fearlessly.
(a)What values were displayed by twofriends?
(b)Why do welders wear glass goggles or face
masks with glass windows while carrying out
welding?
2.Many people like to watch CID programme on a TV
channel. In th programme, a murder mystery
shown. Th mystery solved by a team of CID
people. Each member of the team works with full
dedicatn. They collect informatn and evidences
from all possible sources and then tend to lead to
correct conclusn. Sometimes they also use
ultravlet rays in the forensic laboratory. Some
people get surpred to know the advantage of
ultravlet rays because they are only aware of the
fact that ultraviolet rays coming from the sun
produce harmful effects.
(a)What values were displayed by the members
of CID team ?
(b)What the use of ultravlet rays in a forensic
laboratory ?
3.Sonam's mother was heating food on a gas stove.
Her friend Payal came andsawher mother heating
food on the gas stove. Payal told Sonam's mother,
"Why don't you buy a microwave oven" ? Sonam's
mother replied at once that she doesn't like to use
microwave oven. Sonam and Payal convinced her
that microwave not harmful for cooking food.
Th aneasy,safe and efficient process. Sonam's
mother got convinced and immediately ordered for
a microwave oven.
(a)What were the values dplayed by Sonam and
Payal?
(b)Why does a microwave oven heat up a food
item containing water moleculesmost efficiently?www8notesdrive8com

655. 8.44
Answers
PHYSICS-XII
•
1.(a)Knowledge,
creating awareness and social
responsibility.
(b)Welders wear special glass goggles or face
masks with glass windows to protect their eyes
from large amount of harmful UV radiatn
produced by welding arc.
2.(a)Team spirit, sense of responsibility and
awareness.
(b)UV rays are used in the detectn of forged
documents, finger prints, etc.
3.(a)Caring and creating awareness.
(b)Th because the frequency of the
microwaves matches the resonant frequency of
water molecules.www8notesdrive8com

656. Electromagnetic Waves
GLIMPSES
1.Displacement current.It is that current which
comes into existence (in addition of conduction
current) whenever
the electric field and hence
the electric flux changes with time. It is equal to
EOtimes the rate of change of electric flux
through a given surface.
_ d<% _ dE
ID-EO--EO A-
dt dt
2.Modified Ampere circuital law.It states that the
line integral of the magnetic field13over a
closed path is equal to!-totimes the sum of
conduction currentIeand the displacement
current(ID)threading the closed path.
fB.dz=!-to(Ie+ID)
=!-t0(Ie+EOd~)
The sum of conduction and displacement currents
has an important property ofcontinuityi.e.,the
sum remains constant along any closed path.
3.Maxwell's equations. These are as follows:
(i) Gauss law of electrostatics:
~ ~q
fE.dS=-
EO
(ii) Gauss law of magnetism:
fB.dS=0
(iii) Faraday's law of electromagnetic induction:
(iv)Modified Ampere's circuital law :
4.Electromagnetic wave.An electromagnetic
wave is a wave radiated by an accelerated
charge and which propagates through space as
coupled electric and magnetic fields, oscillating
perpendicular to each other and to the direction
of propagation of the wave.
5.Principle of production of electromagnetic
waves.An accelerated charge produces electric
and magnetic fields, which vary both in space
and time. The two oscillating fields act as
sources of each other and hence sustain each
other. This results in the propagation of an
electromagnetic wave through space.
6.Source of an electromagnetic wave.An accele-
rating charge produces electromagnetic waves.
An electric charge oscillating harmonically
with frequency v, generates electromagnetic
waves of the same frequency . An electric
dipole is a basic source of electromagnetic
waves.
An LC-circuit containing inductance L and
capacitance C produces electromagnetic waves
of frequency,
1
v=------.==
2rc.J[C
7.Mathematical representation of electromagnetic
waves.For a plane electromagnetic wave of
frequency v,wavelengthA,propagating along
(8.45)wwwmnotesdrivemcom

657. 8.46
x-axis, the
electric and magnetic fields maybe
represented as follows:
= Eosin[2t:(i-vt ) ]j
=EOSin[2n(i-f)]j
and
~" "
B=Bzk= 1b sin (kx-rot)k
=1b sin[2n (i-vt ) ]k
=1bSin[2n(i-f)]k
B=B=0
x y
Here Eoand 1b are theamplitudes of the electric
and magnetic fields respectively.
Propagation constant,
k=2~=~.
A. C
8.Hertz's experiment. In1888, Hertz succeeded in
experimentally confirming theexistence of
electromagnetic waves. Byusing oscillatory
LC-circuits, he not only producedand detected
electromagnetic waves, but also demonstrated
their properties of reflection, refraction, inter-
ference and polarisation and established beyond
doubt that light radiation has wavenature. He
was able to produce e.m. waves of wavelength
around6m.
9.Contributions ofJ.e.Bose and Marconi. In1895,
SirJ.e.Bose succeeded in producing and
observing electromagneticwaves of much
shorter wavelength(25mm to5mm).
In1896,Guglielmo Marconisucceeded in trans-
mitting electromagnetic waves over distances of
many kilometres. His experiments marked the
beginning of radio communication.
10.Basic properties of electromagnetic waves.
These are
(1)The e.m. waves are produced by accele-
ratedcharges and do not require any
medium for their propagation.
PHYSICS-XII
~ ~
(ii)The oscillations ofE andBfields are
perpendicularto each other as well astothe
direction of propagation ofthe wave. Sothe
e.m. waves are transverse in nature.
~ ~
(iii)The oscillations of E andBfields are in
same phase.
(iv)All electromagnetic waves travel in free
space with same speedcgiven by
c=__l_ "'-3x108ms-1
~flo EO
In a material medium of refractive indexn,
the speed of an electromagnetic wave is
given by
1 c c
v=--=---=-
.J;f. ~flrErn
(v)Theratio oftheamplitudes of electric and
magnetic fields is
E 1
~=c=---
1b ~flOEO
(vi)The electromagnetic waves carry energy as
they travel through space and this energy is
shared equallybythe electric and magnetic
fields. The average energy of an electro-
magnetic wave is
_ _1[ 21b2l
U-UE+UB-2EOEo+ -;- .
11.Momentum of an electromagnetic wave.Ifan
electromagnetic wave transfers a totalenergy U
to asurface in timet,then total linear momen-
tum delivered to the surface is p=U
c
12.Intensity of an electromagnetic wave. The
energy crossing perunitarea per unittime ina
direction perpendicular to the direction of
propagation of thewaveiscalled intensity of
the wave.
In.Energy / time Power
tensity = =---
Area Area
For an e.m. wave of average energy density u,
I=uc
AI 1-1 E2 -E2
so, -2:EO 0 C-EO TIllSC
andwww=notesdrive=com

658. ELECTROMAGNETIC WAVES (Competition Section)
Electromagnetic spectrum.The
orderly distri-
bution of the electromagnetic waves in
accordance with their wavelength or frequency
into distinct groups having widely differing
properties is calledelectromagnetic spectrum.
The main parts of an electromagnetic spectrum
in the order of increasing wavelength from
8.47
10-2Aor 10-12 m to lO-6m are y-rays, X-rays,
ultraviolet rays, visible light, infrared rays,
microwaves and radiowaves, The different parts
of the e.m. spectrum differ in their methods of
production and detection.www=notesdrive=com

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761. RAY OPTICS AND OPTI
CAL INSTRUMENTS
GIDELINES TONCERT EXERCISES
9.1.Asmallcandle 2.5eminsizeisplaced27eminfront of
aconcave mirrorof radius of curvature 36em.Atwhatdistance
from the mirror should ascreen be placed inorderto receive a
sharpimage ?Describe thenature andsize of the image. If the
candle ismovedcloserto themirror, how would the screen have
tobemoved?
Ans.Here, f;= 2.5em, u= -27cm, R=-36em
R
f=-=-18cm
2
[.:Ris- ve for a concave mirror]
By mirror formula,
1 1 1 1 1 -3+2 1
-=---=-+ -=---=--
vfu-1827 54 54
or v= -54em
Thus the screen should be placed at 54em from the
mirror on thesame side as the object.
Magnification,
m=hz=_~ =_-54 = - 2
f; u -27
:. Size ofimage,
hz=-?'<2.5=-5 em
Negative sign shows that theimage isreal and
inverted.
If the candle is moved closer to the mirror, the image
moves away from mirror, so the screen would have to be
moved farther and farther from the mirror. Closer than
18 em from the mirror (when the focal point is crossed),
the image becomes virtual and cannot be taken on screen.
9.2.A4.5em needle isplaced 12em away from a convex
mirror of focal length 15em.Givethe location of the image and
the magnification. Describe what happens astheneedle is
moved farther from the mirror.
Ans.Here, hI=4.5 em, u=-12em,f= + 15 em
[.: fis+ve foraconvex mirror]
By mirror formula,
1 1 1 1 1 4+5 9 3
-=---=-+-=--=-=-
vfu15 12 60 60 20
20
v=+-=+6.67em
3
or
Asvis+ve,image is virtual and erect and is formed at
6.67 em behind the mirror.
m==hz=_~=_ 20 5
f;u 3x(-12)9
Magnification,
Size of image,
5 5
hz=-xf;=-x4.5 =2.5em
9 9
9.131
Asthe needle is moved farther from the mirror, the
image shifts towards the focus (but never beyondF)and
goes on decreasing in size.
9.3.A tankisfilled with water toa height of12.5em.The
apparent depth of a needle lying at the bottom of the tankis
measured by a microscope tobe9.4em.Whatisthe refractive
indexof water ?Ifwaterisreplaced by a liquid of refractive
index1.63up to the same height, bywhat distance would the
microscope havetobemoved to focus on the needle again?
[Himachal 2000;CBSE D 09C]
Ans.Tankfilled with water:
Real depth=12.5em
Apparent depth = 9.4em
Refractive index of water is
all=Real depth=12.5=1.33
WApparent depth 9.4
Tank filled with liquid:
Realdepth=12.5em
Real depth
Refractive index of liquid = --------'---
Apparent depth
1.63= 12.5
Apparent depth
:.Apparent depth with liquid
= 12.5em = 7.669em c:7.7em
1.63
or
Distance through which the microscope has to be
moved
=9.4-7.7=1.7em.
9.4.Figures.9.199(a) and (b) show refraction of an incident
rayinair at60°with the normal to a glass air and water air
interface, respective/yo Predict the angle of refraction of an
incident rayinwater at45°with the normal to a water glass
interface [Fig.9.199(c)}.
(a) (b) (c)
Fig.9.199
Ans.From Fig. 9.199(a),
a sinisin 60° 0.8660
11=-=--=--=1.51
gsinrsin 35° 0.5736www5notesdrive5com

762. 9.132
From Fig.9.199(b),
a sinisin 60° 0.8660
Il=
-=--=--=1.32
Wsinrsin 41° 0.6561
From Fig. 9.199(c),
a
w Ilg smI
Il=-=--
gall", sinr
or
1.51 sin 45° 0.7071
--=--=--
1.32 sinrsinr
. 1.32 x 0.7071
or sm r= = 0.6181
1.51
.. r -=38.2°.
9.5.A small bulb is placed at the bottom of a tank
containing watertoa depth of80cm. What is the area of the
surface of water through which light from the bulb can emerge
out? Refractive index of water is1.33.Consider the bulb to be a
point source.
Ans.The light rays from the small bulb5which are
incident at an anglei>icare totally internally reected
and carmot emerge out of water surface. The light from
the bulb5comes out through a circular patch of radiusror
given by
. OA r
tanz =-=-
c05h
orr=htanic
Fig.9.200
.. 1 1 3
smz =-=--=-
cIl1.33 4
cosI,=~l-(~r= ~
.3 4 3
tanz =-x-=-
c4J7 J7
Area of the patch,
=1t?=7th2tan2i
c
= 3.14x(0.80)2x~m2
7
= 2.58m 2c:2.6 m2.
9.6.A prism is made of glass of unknown refractive index. A
parallel beam of light is incident on a face of the prism. The
angle of minimum deviation is measuredtobe 40°. What is the
refractive index of the material of the prism. ?The refracting
angle of the prism is60°.If the prism is placed in water
(refractive index1.33), predict the new angle of minimum
deviation of a parallel beam of light.
PHYSICS-XII
Ans.When the prism isplaced in air:
8m= 40°, A=60°
:. Refractive index of the prism material is
A+8 60°+40°
sin msin---
2
all
g
2
· A
sm--
2
. 60°
sm-
2
sin 50° 0.7660
=--=--=1.532
sin 30° 0.5000
Whentheprismis placed in water :
A+8'
sin m
2
·A
sm-
2
or
60°+8'
sin m
2
·60°
sm-
2
60°+8'
sin m
2
1.532
1.33 sin 30°
or
60°+8' 1532
sin m=_.- x0.5 = 0.5759
2 1.33
8'
30°+-1!1.= sin-1(0.5759)= 35°10'
2
or 8~= 10°20'.
9.7.Double-convex lenses are tobe manufactured from a
glass of refractiveindex1.55,with both faces of the same radius
of curvature. What is the radius of curvature required ifthe
focal length of the lens istobe20em? [CBS£:0014C]
<\1S.HereIl= 1.55, f= 20em
If ~ =R,then ~ = -R
As7=(Il -1) [~-~]
or
1 [ 1 1]-= (1.55 - 1) - +-
20 R R
1 2
-= 0.55 x-
20 R
or R = 0.55x2x20 em = 22.0 em.
9.8.A beam of light convergestoa pointP.A lens isplaced
in the path of the convergent beam12cm fromP.At what point
does the beam convergeifthe lens is (a) a convex lens of focal
length20em, (b) a concave lens of focal length16em?
[CBSEOO 06]
Ans.Here the point P on the right of the lens acts as a
virtual object but the imageIis real, as shown in
Figs. 9.201(a)and(b).www3xy°o£n~s•o3myw

763. RAY OPTICS AND OPTICAL INSTRUMENTS
I+--u=12em--
o{
)- - - --p
(a)
I+--u=12em---+t
(b)
Fig. 9.201
(a) For convex lens: u= + 12ern,f= + 20em
Now
1 1 1 1 1 1
---=-
v12 20v uf
1 1 1 3+5 8
-=-+-=--=-
v20 12 60 60
15
v=-=7.5em
2
Thus the beam converges at a point 7.5 em to the right
of the lens.
(b) For concave lens:U=+ 12em,f=-16em
1 1 1 1 1 -3 +4 1
Now -=-+-=-+-=---=-
vfu-16 12 48 48
or
or
.. v=4Bem
Thus the beam converges at point 48 cm to the right of
the lens.
9.9.An object of size 3.0cmisplaced14eminfrontof a
concave lens of focal length21em.Describe the image produced
by the lens. What happensifthe objectismovedfurther away
from the lens ?
Ans.Here'1=3em,U=-14em, f=-21em,v=?
1 1 1
For a lens, - - -= -
v uf
1 1 1 1 1 -2 - 3 -5
-=-+-=-+-=--=-
vfu-21 -14 42 42
or v=-B.4em
Negativevindicates that the image is virtual, erect
and is formed at 8.4 em from the lens on the same side as
the object.
As
9.133
:.Size of image,
v -8.4
~ =-x'1= -- x 3 em =1.Bem
u -14
i.e.,the image is diminished in size.
As the object is moved away from the lens, the virtual
image moves towards the focus of the lens (but never
beyond it) and progressively diminishes in size.
9.10.Whatisthe focal length of a combination of a convex
lens of focal length30cm and a concave lens of focal length
20em?Is the system a converging or a diverging lens? Ignore
thickness of the lenses.
Ans.Here,It= + 30em (convex lens)
f2=-20 (concave lens)
Focal length of the combination is given by
1 1 1 1 1 1
-=-+-=-+--=--
fItf230-20 60
or f=-60em
The negative value offindicates that the combination
behaves as a diverging lens.
9.11.A compound microscope consists of an objective lens
of focal length2.0em and an eyepiece of focal length625em
separated by a distance of15em.How far from the objective
should an object be placed in ordertoobtain the final image at
(i) the least distance of distinct vision(25em),(ii)infinity?
Whatisthe magnifying power of the microscope in each case?
[CBSE OD 08]
Ans.Herefo= 2.0em, fe= 6.25em,Uo=?
(i)When the final imageisobtained at the least distance of
distinct vision:
ve= -25em
As
11 1
-
veuefe
1 1 1 1 1
.. ---=-----
»:vet.-256.25
-1-4 -5 1
=--=-=--
25 25 5
or ue= - 5em
Now distance between objective and eyepiece
=15 em
:. Distance of the image from objective is
Vo= 15 - 5 = 10em
or
1 1 1 1 1 1-5 2
---=---=--=
UoVo fo10 2 10 5
5
Uo= -'2= -2.5em
.. Distance of object from objective = 2.5 emwww5notesdrive5com

764. 9.134
Magnifyin
g power,
m= ~xme=~[1+ ~)
= ~(1+ ~) =20.
2.5 6.25
(ii) When thefinal image is formed at infinity:
Hereve=00, fe= 6.25 em
1 1 1 1 1 1
As ..---=-
veuefe 00uet.
or ue=-fe=-6.25 cm
Distance between objective and eyepiece = 15 cm
:.Distance of the objective from the image formed by
itself,
or
Va=15-6.25 = 8.75 ern
fa=+2.0 em
1 1 1 1 1 2 - 8.75 - 6.75
-=---=----=---=--
"'0va fa8.75 2 17.5 17.5
"'0=-17.5= - 2.59 em
6.75
:.Thedistance of the object fromobjective = 2.59 em
Magnifying power,
V 25
m=~xm=-.!!.x--
e"'06.25
=27x4 = 13.46 = 13.5.
8
Also
9.12.Aperson with a normal near point(25em)using a
compound microscope with an objective of focallength8.0mm
and eyepiece of focal length 25em can bring an object placed
9.0mmfrom the objective in sharp focus. What isthe separation
between the two lenses? Howmuchisthe magnifying power of
the microscope ?
Ans.Herefa= 0.8 em, "'0=-0.9 em,va=?
1 1 1
As---=-
va"'0fa
1 1 1 1 1
-=-+-=---
va fa "'00.8 0.9
0.9-0.8 0.1
0.9x0.80.8 x 0.9
0.8x0.9
va= = 7.2em
0.1
or
Now for the eyepiece, we have
fe= 2.5 em, ve= -D=-25em, ue=?
1=~_~=_~_~= -1-10 = -11
ueveI,252.5 25 25
or
25
u=--=-2.27em
e11
PHYSICS-XII
Hencethe separation betweenthe two lenses
=va+IueI= 7.2 + 2.27 = 9.47 em
Magnifying power,
m= ~xme=I ~ I [1 +~)
= 7.2(1+25)=88.
0.9 2.5
9.13.Asmall telescope has an objective lens of focal length
144em and aneyepiece of focal length 6.0emWhatisthe
magnifying power of the telescope ?What isthe separation
between the objective and the eyepiece?
Ans.Here fa= 144 em,fe= 6 em
For the small telescope set in normal adjustment, the
magnifying power is
m=fa= 144 =24
I,6
Separationbetween the objective and the eyepiece
=fa+fe= 144+6 =150em.
9.14.(i)Agiant refracting te/escope at an observatory has
an objective lens of focal length 15m.If an eyepiece of focal
length1.0cmisused,whatisangular magnification of the
telescope?
(ii)Ifthis te/escope is used toview the moon, what isthe
diameter of the image of the moon formed by the objective lens?
The diameter ofthe moonis 3.48x106m;and the radius of
lunar orbitis3.8x108m. [CBSE0008, 11 ;015]
Ans.Herefa= 15 m, fe=1.0 em=0.01 m
(i)Angular magnification,
m=fa= ~=1500.
fe0.01
(ii)Letdbe the diameter of the image in metres. Then
angle subtended by the moon will be
Diameter of moon 3.48 x106
a= =-----,n
Radius of lunar orbit 3.8x·108
Angle subtended by the image formed by the
objective will also be equal to aand is given by
Diameter of image of moond
a= =-
fa 15
d3.48x106
15=3.8x108
Diameter of image of moon,
d_3.48 x 106 x15 3.48x15x10-2-13 73
- 8 -•em.
3.8xlO 3.8
9.15.Usethemirrorequation to deduce that:
(a) an object placed between f and 2fof aconcavemirror
produces a real image beyond2f [CBSE015]www3xy°o£n~s•o3myw

765. RAY OPTICS A
ND OPTICAL INSTRUMENTS
(b) a convex mirror always produces a virtual image
independent of the location oftheobject.
(c)thevirtual image produced by aconvexmirroris
always diminished insizeandislocated between the
focus and the pole.
(d) an object placed between the pole and focus ofa
concave mirror produces avirtual andenlarged
image. [CBSE OD 11]
Ans.(a)From mirror formula, ~ = 2-~
vfu
Now for a concave mirror, f<0 and foran object on
the left, u<0
2f<u< f
11 1
or->->-
2f uf
1 1 1
--<--<--
2f u f
111111 11
-- - < ---<---or-< - <0
f 2f f u f f 2f v
This implies thatv<0so that image is formed on left.
Also the above inequality implies
2f>v
or 12fl<lvl [.,' 2fandvarenegative]
i.e.,the real image is formed beyond 2f
(b)For aconvexmirror, f>0 andfor an object on left,
u<O.From mirror formula,
1 1 1
vfu
or
or
This implies that~>0 orv>0
v
Thisshows thatwhatever be the value of u,.a convex
mirrorforms avirtual image on the right.
(c)Forconvex mirror, f>0 and for an object on the
leftu<0,somirror formula,
1 1 1
vfu
1 1
implies that->-
vf
[1. .]
.,'-; ISa+vequantity
or v<f
This shows that the image is located between the pole
and the focus of the mirror.
(d)From mirror formula,
111
vfu
For a concave mirror, f<0 and for an object located
between the pole and focus ofaconcave mirror,
f<u<O
1 1
- >-.or
fu
1 1
--->0
fu
or
1
->0
v
i.e.,avirtual image isformed on the right.
9.135
Also
1 1
-<-
VIul
orv>1ul
v
Iml=->l
Iul
i.e.,image is enlarged.
9.16.A smallpin fixed on atabletopisviewed from above
from a distance of50cm.By what distance would the pinappear
toberaisedifitisviewed from the same point through a 15cm
thick glass slab held parallel to the table?Refractive indexof
glass=1.5.Doestheanswer depend on the location of the slab?
Ans.The distance through which the pin appears to be
raised is
d= Real thickness of slab
- Apparent thickness of slab
R 1thickn fIb Real thickness of slab
= ea 1ess0sa---------
11
Heret= 15 em, 11= 1.5
(1)(1.5 -1)
:. d=15 1-- =15 --
1.5 1.5
=5cm
The answerdoesnot depend on the location of the slab.
9.17.Figure9.202 shows a cross-section ofa 'light- pipe'
madeof a glass fibre of refractive index 1.68.The outer covering
of thepipeismade of amaterial of refractive index 1.44.Whatis
the range of the anglesof the incident rays with the axisofthe
pipefor which total reflections inside the pipetakeplaceas
shown.
Fig. 9.202
Ans.Given 112=1.68, III=1.44, 11=112= _1_
IIIsini/
:.Critical anglei~is given by
sini~= ~ =1.44=0.8571 =>i~=59°
112 1.68
Total internal reection will occur if theanglei'>i~,
i.e.,ifi'>59°or whenr<rmax,wherermax= 90° - 59°=31°.
Using Snell's law,
sinimax= 1.68
sinrmax
or sin imax= 1.68 xsinrmax
= 1.68 xsin31°= 1.68 x0.5150=0.8662
Thus all incident rays which make anglesin the range
0<i<60°with the axis of the pipe will sufer total
internal reections in the pipe.www5notesdrive5com

766. 9.136
(a)
9.18.Answer the following ques
tions:
(a) You have learnt that plane and convex mirrors
produce virtual images of objects. Can they produce
real image under somecircumstances ?Explain.
(b) A virtual image, we always say, cannot be caught
on a screen. Yet when we 'see'a virtual image, we
are obviously bringing it on to the'screen' (i.e.,the
retina) of our eye. Is there a contradiction ?
(c) A diver under water, looks obliquely at a fisherman
standing on the bank of a lake. Would the fisherman
look taller or shorter to the diver than what he
actuallyis?
(d) Does the apparent depth of a tank of water change if
viewed obliquely?If so, does the apparent depth
increase or decrease ?
(e) The refractive index of diamondismuch greater
than that of ordinary glass. Is this fact ofsome use
to a diamond cutter?
Ans.(a)Yes,a plane or convex mirror can produce a
real image if the object is virtual. As shown in Figs. 9.203(a)
and(b),if a plane or a convex mirror is placed in the path
of rays converging to a point, the rays get reected to a
point in front of the mirror. Real image can be obtained on
a screen.
image
,'
<>;
Vi::'.J°
object
Real
Iimage
Virtual
I.object
_".0
....---:,."
, •••. .,.;1'
(b)
Fig. 9.203
(b)When the reected and refracted rays are diver-
gent,the image is virtual. These rays are converged by the
eyelens to form a real image on the retina. The virtual image
serves as a virtual object. Also the screen is not located at
the position of virtual image. So there is no contradiction.
(c) The man looks taller to a diver under water. As the
fisherman is in air, the light rays travel from rarer to
denser medium. They bend towards the normal and
hence appear to corne from a larger distance, as shown in
PHYSICS-XII
Fig.9.204. It may be noted that the points P andQare,in
fact,so close that the rays through these points can enter
the small aperture of the eye of the fish. Here
AB=Real height of the man,
AB'=Apparent height of the man.
Fig. 9.204
Fig.9.205
(e)Yes. Refractive index of diamond ishigh
( J..l=2.42), so its critical angle is small (ic=24°). A
diamond cutter makes use of this large range of angle of
incidence(24°to 90°) to ensure that light entering
diamond sufers total internal reection several times.
When light emerges out, it produces sparkling efect.
9.19.Theimage of asmallelectric bulbfixed on the wall toa
roomistobe obtained on the opposite wall3m away bymeans
of alarge convex lens.What is themaximum possible focal
length of the lensrequired for the purpose ?
Ans.The minimum distance (as proved in Problem 14
on page 9.124) between an object and its real image is4f.
D 3rn
4fmax =Dorfmax =- =-- = 0.75 m.
4 4
9.20.A screen is placed 90em from an object, theimageof
theobject on the screen isformed by a convex lens at two
different locations separated by 20em.Determine the focal
lengthofthelens.www3xy°o£n~s•o3myw

767. RAY OPTICS AND OPTICAL INSTRUMENTS
Ans. As shown in
Fig. 9.206, let 0andIbe the
positions of object and image respectively and ~ andLz
be the two conjugate positions of the lens.
Ll L2
Fig. 9.206
Obviously,x+ 20 +x= 90cm orx= 35cm
When the lens is in position ~, we have
u= -x= - 35ern,v= 20 +x= 20 + 35 = 55cm
1 1 1 1 1 7 + 11 18
.. -=---=-+-=--=-
fv u55 35 385 385
385
orf=-= 21.4cm.
18
9.21.(a) Determine the 'effective focal length' of the
combination of the two lenses in Exercise 9.10,ifthey are placed
8.0em apart with their principal axes coincident. Does the
answer depend on which side a beam of parallel light is incident?
Isthe notion of effective focal length of this system useful at all?
(b) An object15em in size is placed on the side of the convex
lens in the above arrangements. The distance between the object
and the convex lens is40em.Determine the magnification
produced by the two-lens system, and the size of the image.
Ans.(a)(i)Let a parallel beam of light be incident from
the left on the convex lens first. Then
.Ii= 30cm
~=-oo
1 1 1
---=-
11.~.Ii
1 1 1 1 1 1
-=-+-=---=-
11..Ii ~30 0030
or 11.= + 30cm
This image becomes avirtualobjectfor the second lens
so that
Now
f2= -20cm,
~ =+(30 -8)=+22cm
1 1 1
-=-+-
v2f2 ~
1 1 -11 + 10 -1
=--+-=---
20 22 220 220
v2=-220 emor
9.137
The parallel incident beam appears to diverge from a
point 220- 4 = 216cm from the centre of the two-lens
system.
(ii)Let the parallel beam be incident from the left on
the concave lens first. Then
.Ii=-20cm, ~= -00
1 1 1
As
11.~.Ii
1 1 1 1 1 1
-=-+-=--+-=--
11..Ii ~ -20 - 00 20
11.= - 20cm
This image becomes arealobjectfor the second lens so
that
or
Now
f2= -30cm
~ =-(20+ 8)=-28cm
1 1 1 1 1 14- 15 - 1
-=-+-=---=---=-
v2f2 ~ 30 28 420 420
v2=-420em
Thus the parallel incident beam appears to diverge
from a point 420- 4 = 416em on the left of the centre of
the two-lens system.
Clearly, the answer depends on which side of the lens
system the parallel beam is incident. The notion of
efective focal length, therefore, does not seem to be
meaningful for this system.
(b)Here ~ = - 4Ocm,.Ii= 30cm
1 1 1
---=-
11.~.Ii
111
-+-=-
11.40 30
or
As
1 1 1 4-3 1
or -=---=--=-
11.30 40 120 120
or 11.= 120cm
Magnitude of magnification due to the first (convex)
lens is
This image becomes a virtual object for the second
lens so that
Now
~ = + (120-8)= + 112cm
f2= -20cm
111
-=-+-
v2f2 ~
1 1 - 112+ 20 - 92
=--+-=----
20 112 112x 20 112x 20
112x20
v2=-92 cm = - 24.9cmorwww5notesdrive5com

768. 9.138
Magnitud
e of magnification due to the second
(concave) lens is
IV2I112x20 20
~=-;;;= 92x112 = 92
Net magnitude of magnification due to the two-lens
system is
m=n;x~
3x20
=--=0.652
92
Size of image,
~ =mh;= 0.652 x1.5=0.98 cm.
9.22.Atwhat angle should a ray of light be incident on the
faceof a prism of refracting angle 600sothatit just suffers total
or
internal reflection at the other face? Therefractive indexof the
prismis 1.524.
Ans.Therefracted rayQRwill just sufer total internal
reflection if itis incident at thecritical angle ic.
Thus r2=ic
Now sini=.-! = _1_ = 0.6542
c~1.524
ic= sin-\0.6542) c:410
Fig. 9.207
But'i+r2=A
'i=A -r2=A-ic= 600 -410=190
From Snell's law,
sin~
.f.!=-.-
sm'i
sin~ =Iisin'i= 1.524xsin 190
= 1.524x0.3256= 0.4962
Hence ~ =sin-\0.4962)-"'30°.
9.23.You are given prisms made of crown glass and flint
glass with a variety of angles. Suggesta combination of prisms
which will (a) deviate a pencil of white lightwithout much
dispersion, (b) disperse (and displace) a pencil of white light
without much deviation.
Ans.Two identical prisms made ofthe same material
placed with their base on 'opposite sides (or the incident
white light) and facestouching (or parallel) will neither
deviate nor disperse, but will merely produce a parallel
displacement of the beam.
PHYSICS-XII
Now, angular dispersion produced by crown glass
prism is
°b-o'=(~b-~,)A
Mean deviation produced by crown glassprismis
Oy=(~y-1)A
Angular dispersion produced by int glass prism is
0;'-0;= (~~ -1)A'
o~= (~~ -1)A
(a)To deviate a beam without dispersion, the net
angular dispersion produced by thecombination must be
zeroi.e.,
(~b- ~,)A+(~;,- ~;)A'=0
A'=-(~b-~,) A
(~;,-~;)
Negative sign shows that the two prisms must be placed
with their bases on opposite sides. As ~;, -~;)for int
glass ismore than ~b- ~ ,)for crown glass, therefore, a int
glass prism of smaller refracting angle should be combined
with a crown glass prism so that the dispersion due to the
first is nullified by the second as shown in Fig. 9.208(a).
(a)
(b)
Fig.9.208
(b)To produce dispersionwithout deviation, the net
mean deviation should be zero, i.e.,
(~y-1)A+ (~ -l)A'= 0
~ -1
or A,=--y-A
~~-1
We take a crown glass prism of certain angle and go on
increasing the angle of int glass prism till the deviations
duetothe two prisms are equal and opposite. However,
the int glass prism angle willstill be smaller than that of
crown glass because int glass has higher refractive index
than that of crown glass as shown in Fig. 9.208(b).
Due to the adjustments involved for many colours, the
above combinations are not very accurate arrangements
for the purposes required.
9.24.For anormaleye,thefar pointisat infinity andthe
near point of distinct visionisabout 25eminfrontof the eye.
The cornea of the eyeprovides a converging power of aboutwww3xy°o£n~s•o3myw

769. RAY OPTICS AND OPTI
CALINSTRUMENTS
40 dioptres, and the least converging power of the eyelens
behind the cornea is about 20 diopters. From this rough
dataestimate therangeof accommodation (i.e.,rangeof
converging power ofits eyelens) of a normaleye.
Ans.To see objects at infinity, the eye uses its least
converging power
= 40+20 = 60dioptres
.',Approximate distance between the retina and the
cornea eyelens
= focal lengthof the eyelens
100 100 5
=-=-=-em
P 603
To focus an object at the near point on the retina, we
have
5
u=-25em, v=-em
3
..Focal lengthfshould be given by
1 1 1
fv u
31 15+1 16
=-+-=--=-
5252525
f=25 em
16
..Corresponding converging power
= 64dioptres
Power of the eyelens = 64-40 = 24dioptres
Thus the range of accommodation of the eyelens is
roughly 20 to 24 dioptres.
9.25.Does short-sightedness (myopia) or long-sightedness
(hypermetropia) imply necessarily that the eye has partially lost
its ability ofaccommodation ?If not, what might causethese
defects ofvision?
Ans.No,it does not imply necessarily that the eye has
lost its ability of accommodation. A person may have
normal ability of accommodation of the eye lens andyet
maybemyopic or hyperopic. Myopia arises when the eye
ball from front to back gets too elongated, hypermetropia
ariseswhenitgetstooshortened. In practice, in addition
the eye lens may also lose some of its ability of
accommodation. When the eye ball has normal length but
the eyelens loses partially its ability of accommodation (as
happens with increasing age for normal eye), the defectis
calledpresbyopia and is corrected in the same manner as
hypermetropia.
9.26.A myopic person has been using spectacles of power
-1.0dioptrefor distantvision. During old age he also needs to
use separate reading glass of power+2.0dioptres.Explain what
may have happened.
Ans.Here, P= - 1.0dioptre
.. f = 100 = 100 =_100em
P-1
9.139
Thus, the far point of the person is 100em, on the other
hand, his near point may have been normal (about 25 em).
The objects at infinity producevirtual images at
100 cm (using spectacles).
Toseecloser objects, i.e.,those which are (or whose
images using the spectacles are) between 100 cm and
25 em, the person uses the ability of accommodation of his
eyelens. Thisability usually gets partially lost in old age
(presbyopia). The near point of the person recedes to
50 cm.
Soto view the objects at 25 emclearly, we have
u=-25em, v=-50 em
1 1 1
fvu
1 1 -1+2 1
=--+-=---=-
5025 50 50
f= 50cm
100 100 .
P=-=- =+2 dioptres
f 50
Thus the person needs a converging lens of power +2
dioptres.
9.27.Aperson lookingat aperson wearing a shirtwitha
pattern comprising vertical and horizontal lines isabletoseethe
vertical lines more distinctly than the horizontalones.Whatis
thisdefect due to?Howissuch a defect ofvision corrected ?
Ans.This defect is calledastigmatism. It arises because
the curvature of the cornea plus eyelens refracting system
is not the same in diferent planes. The eyelens is usually
spherical, i.e.,has the same curvature in diferent planes
but the cornea is not spherical in case of an astigmatic eye.
In the present case, the curvature in the vertical plane is
enough, sosharp images ofvertical lines can be formed on
the retina. But the curvature is insuficient in the
horizontal plane, so horizontal lines appear blurred. The
defect can be corrected by using a cylindrical lens with its
axisalong thevertical. Clearly, parallel rays in the vertical
plane will sufer noextra refraction, but those in the
horizontal plane can get the required extra convergence
due to refraction by the curved surface of the cylindrical
lensif the curvature of the cylindrical surface is chosen
appropriately.
9.28.A man with normal near point(25ern)reads a book
withsmall print using a magnifying glass: a thin convex lens of
focal length 5em.
(a) What is theclosest and the farthest distance at which he
can read thl! book whenviewing through the magnifying glass?
(b) Whatisthe maximum andthe minimum angular
magnification (Magnifying power) possible using the above
simple microscope ?
Ans.For theclosest distance:
or
Hence,
v= - 25em, f= 5 em, u=?www5notesdrive5com

770. 9.140
As
1 1 1
vuf
1 1 1 1 1 -1-5 -6
.. -=---=----=--=-
uvf
-255
25 25
25
or u=--em=-4.2em
.6
This is the closest distance at which the man can read
the book.
For the farthest image:
v=oo, f=5em, u=?
111
u vf
=-.!.-~=O-~=-~
005 5 5
U=-5em
This is the farthest distance at which the man can read
the book.
(b)Maximum angular magnification is
~=~=6
umin 25/6
Minimum angular magnification is
~= 25=5.
umax 5
9.29.A card sheet divided into squares each of size1mm2is
being viewed at a distance of9 em through a magnifying glass
(a converging lens of focal length10em) held close to the eye.
(a) Whatisthe magnification produced by the lens?
How muchisthe area of each square in the virtualor
image?
(b) Whatisthe angular magnification (magnifying
power) of the lens?
(c) Is the magnification in (a) equal to the magnifying
power in (b)?Explain.
Ans. (a)Here, area of each square (or object)
=1mm2
u= -9 em,f=+10 em
As
1 1 1
---=-
vuf
1 1 1
.. -=-+-
Vf
u
1 1 9-10 1
=---=--=--
10 9 90 90
or v= -90 em
Magnitude of magnification is
v90
m=-=-=10
Iul9'
Area of each square in the virtual image
=(10)2xl =100 mm2 =1em2
PHYSICS-XII
D 25
(b)Magnifying power,M =-=-=2.8.
I ul 9
(c)No.Magnification of an image by a lens and
angular magnification (or magnifying power) of an
optical instrument are two separate things. The latter is
the ratio of the angular size of the object (which is equal to
the angular size of the image even if the image is
magnified) to the angular size of the object if placed at the
near point (25 em). Thus magnification magnitude isI~I
dgnifyin .25
an ma g power 1S - .
Iul
Only when the image is located at the near point
IvI=25 em, the two quantities are equal as will be seen in
the next exercise.
9.30.(a) At what distance should the lens be held from the
figure in Exercise9.29in order to view the squares distinctly
with the maximum possible magnifying power?
(b) Whatisthe magnification in this case?
(c) Is the magnification equaltothe magnifying power in
this case?Explain.
Ans.(a)Maximum magnifying power is obtained
when the image is at the near point (25 em). Thus
v= -25 em,f=+10 em,u=?
1 1 1
As
v uf
1 1 1 1 1 -2-5 -7
-=---=----=--=-
uvf 2510 5050
50
u=--=-7.14em
7
So lens should be held 7.14em away from the figure.
(b)Magnitude of magnification is
v 25
m=- =--=3.5.
Iul50/7
( ) Mifyi D 25
cagm mg power=- =-- =3.5
lu150/7
Yes, the magnifying power is equal to the magnitude
of magnification because image is formed at the least
distance of distinct vision.
9.31.What should be the distance between the object in
Exercise 9.30 and the magnifying glass ifthe virtual image of
each square in thefigureisto have an area of625mm2?Would
you be abletosee the squares distinctly with your eyes very close
to the magnifier?
Ans.Here, the magnification in area
=6.25 ~2=6.25
1mm
:.Linear magnification,m=../6.25=2.5
As m=~ .. v=mu=2.5u
uwww4°»ttssrxvt4r»~

771. RAY OPTICS AND OPTICAL INSTRUMENTS
or
Now
1 1 1
-
vUf
1 1 1
---- -
2.5U U10
1-2.51
-
2.5U10
or 2
.5U=-1.5 x 10
2.5U=-1.5x10
1.5x10
U=---- =-6em
2.5
or
or
Hencev= 2.5 U= 2.5x ( -6)=-15em
As the virtual image is closer than the normal near
point (25em),it cannot be seen by the eye distinctly.
9.32.Answer the following questions:
(a) The angle subtended at the eye by an objectisequaltothe
angle sub tended at the eye by the virtual image produced by a
magnifying glass. In what sense then does a magnifying glass
provide angular magnification?
(b) In viewing through a magnifying glass, one usually
positions one's eyes very close to the lens. Does angular
magnification changeifthe eyeismoved back?
(c) Magnifying power of a simple microscopeisinversely
proportional to the focal length of the lens. What then stops us
from using a convex lens of smaller and smaller focal length and
achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a
compound microscope have short focal lengths?[CBSE 0010]
(e) When viewing through a compound microscope, our
eyes should be positioned not on the eye-piece but a short
distance away from it for best viewing. Why?How much
should be that short distance between the eye and eyepiece?
Ans.(a)It is true that the angle subtended at the eye by
an object is equal to the angle subtended at the eye by the
virtual image produced by a magnifying glass. When a
magnifying glass is not used, an object has to be placed at
a distance of 25 cm. But the use of a magnifying glass
allows us to place the object much closer to the eye than at
25 cm. The closer object has larger angular size than the
same object at 25em.It is in this sense that a magnifying
lens produces angular magnification.
(b)Yes, the angular magnification decreases slightly if
the eye is moved back. This is because angle subtended at
the eye would be slightly less than the angle subtended at or
the lens. The efect is negligible when image is at much
larger distance.
(c)First, grinding lenses of very small focal lengths is
not easy. More important, if we decrease focal length,
both spherical and chromatic abberations become large.
So in practice we cannot get a magnifying power of more
than3or so with a simple convex lens. However using an
aberration corrected lens system, one can increase this
limit by a factor of 10 or so._
9.141
(d)The magnifying power of a compound microscope
is given by
m=mxm=vax[1+DJ
o eUo Ie
Angular magnification(ma)of objective will be large
whenUois slightly greater thanfa.Since microscope is
used for viewing very close objects, soUois small.
Consequentlyfahas to be small.
Moreover, the angular magnification(me)of the
eyepiece will be large iffeis small.
(e)Refer to the solution of Problem30(a)on page 9.129.
9.33.An angular magnification (magnifying power) of
30 Xisdesired using an objective offocal length125em and an
eyepiece offocal length5em.How will you set up the compound
microscope?
Ans.We assume the microscope in common usage,
i.e.,the final image is formed at the least distance of
distinct vision,
D= 25ern,t.= 5ern
:.Angular magnification of the eyepiece is
D 25
m=1+-=1+-=6
e t, 5
As total magnification,m=mex%
:. Angular magnification of the objective is
m 30
%=-=-=5
me6
As real image is formed by the objective, therefore,
v
% = ~ = - 5 or va=-5Uo
Uo
fa= 1.25cm
1 1 1
---=-
vaUofa
1 1 1
----=--
- 5Uo Uo 1.25
- 6 1
--=--
5Uo1.25
6x1.25
Uo=--5- =-1.5em
Thus the object should be held at 1.5 cm in front of the
objective lens.
Now
or
or
Also va=-5Uo= - 5x ( -1.5) = 7.5 ern
1 1 1
---=-
ve ue t,
Aswww5notesdrive5com

772. 9.142
1 1 1 1 1
=---=----
vefe
-25 5
[ve=-D=-25cm] or
-1-5 6
=--=--
25 25
- 25
u= -- =-4.17em
e6
:. Separation between the objective and the eyepiece
=I -, I+IVoI
= 4.17+7.5 =11.67em.
9.34. Asmall telescope has an objective lens of focal length
140 cm and an eyepiece of focal length 5.0 em. What is the
magnifying power of the telescope for viewing distant objects
when
(a) the telescope is in normal adjustment(i.e.,when the
final image is at infinity),
(b) the final image is formed at the least distance of
distinct vision(25em)? [CBSE OD 13C)
Ans.Herefo= 140em,fe= 5.0 em
(a) In normal adjustment:
Magnifying power,
m=fo= 140 =28
I,5
(b) When the final image is formed at the least distance of
distinct vision(25em) :
m=j:(1+ ~)
= 140 (1+ ~)= 28x1.2 =33.6.
5 25
9.35. (a) For the telescope described in Exercise 9.34(a),
whatisthe separation between the objective lens and the
eyepiece? [CBSE OD 13C)
(b) If this telescopeisused to view a100m tall tower3km
away, whatisthe height of the image of the tower formed by the
objective lens?
(c) Whatisthe height of the final image of the towerifit is
formed at25cm?
Ans.(a)In normal adjustment, the separation between
objective and eyelens
=fo+fe=140+5=145em.
(b)Angle subtended by the 100 m tall tower at 3 km
away is
or
100 1
az:tana=---3 = - rad
3x10 30
Lethbe the height of the image of tower formed by the
objective. Then angle subtended by the image produced
by the objective will also be equal to a and is given by
h h
a=-=-
fo 140
PHYSICS-XII
h 1
-=-
140 30
140 14
h=-= - =4.67cm.
30 3
(c)Magnification produced by the eyepiece is
D 25
m=1+-=1+-=6
e fe 5
:.Height of the final image
14
=hxm= -x6=28 em.
e3
9.36.ACassegrain telescope uses two mirrors as shown in
Fig.9.151.Such a telescopeisbuilt with the mirrors20mm
apart. If the radius of curvature of the large mirror is 220 mm
and the small mirror is 140mm,where will thefinal image of an
object at infinity be?
Ans.The image formed by the larger (concave) mirror
acts as a virtual object for the smaller (convex) mirror.
Parallel rays coming from the object at infinity will focus
at 110 mm from the larger mirror. The distance of the
virtual object for the smaller mirror = 110 - 20 = 90 mm.
For the small convex mirror, we have
u= - 90 mrn,f= - 70 mm,v=?
Using mirror formula,
1 1 1 1 1
=---=-----
vfu-70 - 90
1
315
.. v=-315mm
Thus the image is formed at315mm from the
smaller mirror.
9.37.Light incident normally on plane mirror attached to a
galvanometer coil retraces backward as shown.Acurrent in the
coil produces a deflection of35°of the mirror. Whatisthe
displacement of the reflected spot of light on a screen placed
1.5 m away?
Ans.When the mirror is turned through angle 9, from
positionMtoM',the reected ray turns through angle 29,
-sothat the reected spot moves on the screen from
positionPtoQand
L POQ= 29 = 2x3S = 7°
d
Now tan 29=-
1.5
: r-----:-------~-2.t---r,~1
1.5m I
I
I
MilM
5
Fig. 9.209www3xy°o£n~s•o3myw

773. RAY OPTICS AND OPTICAL INSTRUMENTS
:.Displacement of reected spot on the screen is
d=1.5 tan
29
=1.5xtan 7°
=1.5x0.1228m
=0.1842m=18.4em.
9.38.Figure 9.210 shows an equiconvex lens (of refractive
index 1.50)incontact with a liquid layer on top of a plane
mirror. A small needle with its tip on the principal axisismoved
along the axis until its inverted imageisfound at the position of
the needle. The distance of the needle from the lens ismeasured
to be45.0 cm. Theliquidisremoved and the experiment is
repeated. The new distanceismeasured to be 30.0 cm. Whatis
the refractive index of the liquid ?
QP P'Q'
Fig. 9.210
Ans.Distance of the needle from the lens in the first
case
=Focal lengthFof the combination of the
convex lens and planoconcave lens formed
by the liquid
i.e., F=45 em
Distance measured in second case
=Focal length of the convex lens
i.e.,It=+30cm
9.143
The focal lengthf2of the plano-concave lens is given
by
or
1 1 1
-+-=-
Itf2F
1 1 1
- ---
f2FIt
1 1
45 30
2 - 3 1
--=--
90 90
.. f2=-90em
Now for the equiconvex lens, we have
R,.=R,Rz= -R,f=30 em, 1.1=1.5
Using Lensmaker's formula
7=(1.1-1) [~ - ~]
~ =(1.5-1)[.!+.!]
30 R R
2
=0.5x-
R
or
or R=0.5x2x30em=30 em.
For plano-concave lens, f= -90 emr
For concave surface,R,.= -R=-30 em,
For plane surface,Rz=00
1 [1 1 ]f=(1.1-1) R,.-Rz
1 [1 1]-=(1.1-1) ---
- 90 - 30 00
- 30 1
1.1-1=--=+-
- 90 3
As
or
or
1
1.1=1+-=1.33.
3www5notesdrive5com

774. Text Based Exercises
YPE A :VERY SHORT ANSWER QUESTIONS (1 mark each)
1.To whic
h wavelength of light is our eye most
sensitive ? In which region does this wavelength
lie?
2.A ray of light falls normally on a mirror. What are
the values of angle of incidence and angle of
reection ?
3.A person moves with velocityvtowards a plane
mirror. With what velocity does his image move
towards him ?
4.A mirror is turned through 10°. By what angle will
the reected ray turn ?wwwxnotesdrivexcom

775. 9.144
5.What is the
focal length (or radius of curvature) of a
plane mirror? [Himachal97; CBSED01C]
6.What is the number of images of an object held
between two parallel plane mirrors?
7.An object is held between two plane parallel
mirrors inclined at 45° to each other. How many
images do you expect to see ?
8.What is the minimum size of a plane mirror which
can enable a man to see his full image?
9.How many images of himself can an observer see in
a room whose ceiling and two adjacent walls are
mirrors?
10.What is a spherical mirror? What are its two types?
11.Which spherical mirror is called a divergent
mirror-cancave or convex?
12.Define principal focus of a spherical mirror.
13.Which spherical mirror has(i)a real focus(ii)a
virtual focus ?
14.Which spherical mirror always forms a virtual,
erect and diminished image of an object ?
15.One wants to see an enlarged image of an object in a
mirror. Which type of mirror one should use?
16.Which type of spherical mirror can form a real and
diminished image of an object ?
17.When an object is placed between / and2/of a
concave mirror, would the image formed be(i)real
or virtual and(ii)diminished or magnified ?
[CBSEDISC]
18.Can we obtain image of an object formed by convex
mirror on a screen ? If not, why?
19.Can we photograph a virtual image?
20.A concave mirror has focal length 20 em. Where
should the object-be placed in front of the mirror so
that area of image equal to the size of the object is
formed?
21.What is the angle of incidence, when a ray of light
falls on the spherical mirror from its centre of
curvature?
22.Starting from a large distance, a ame is moved
towards a convex mirror. Comment on how the size
and position of the image change ?
23.A concave mirror, of aperture 4 em, has a point
object placed on its principal axis at a distance of
10 em from the mirror. The image, formed by the
mirror, is not likely to be a sharp image. State the
likely reasons for the same. [CBSESamplePaper 13]
24.What is refraction?. [Haryana01]
25.Define refractive index.
26.Define refractive index in terms of wavelength of
light.
PHYSICS-XII
27.What is meant by relative refractive index of
medium ?
28.State the factors on which the refractive index of a
medium depends.
29.State Snell's law of refraction of light.
[Punjab2000, 04]
30.When does Snell's law of refraction fail ?
31.For which material the value of refractive index is:
(i)minimum and(ii)maximum ?
32.What is lateral shift in refraction?
33.On what factors does the lateral shift depend ?
34.For what angle of incidence, the lateral shift
produced by a parallel sided glass slab is zero ?
35.For what angle of incidence, the lateral shift produced
by a parallel sided glass slabis maximum ?
36.Light of wavelength6000A.in air enters a medium
of refractive index 1.5. What will be its frequency in
the medium? [CBSED 94]
37.When light undergoes refraction, what happens to
its frequency? [CBSEOD 200OC]
38.When light undergoes refraction at thesurface of
separation of two media, what happens to its
wavelength? [CBSEOD 2000C]
39.How does the frequency of a beam of ultraviolet
light change when it goes from air into glass?
[CBSEDOl]
40.Define the term critical angle for a pair of media.
[CBSEDISC]
41.Can total internal reection occur when light
travels from a rarer to a denser medium?
42.Velocity of light in glass 2x108m/s and in air is
3x108m/s. If the ray of light passes from glass to
air, calculate the value of critical angle.[CBSEF 15]
43.Write the value of critical angle for a materialof
refractive index.fi [Himachal93;CBSEF 94]
44.A substance has a critical angle of 45° for yellow
light. What is its refractive index?
[ISCE97;Haryana2000]
45.Which one has a greater critical angle-diamond or
water?
46.A good plane mirror reects about 95% oflight.
What is the percentage of light reectedwhen total
internal reection occurs?
47.Write the relation between the refractive index and
critical angle for a given pair of optical media.
[CBSEOD 09; D 13]
48.When light is incident on a rarer medium from a
denser medium, write the relation between the
critical angle and refractive indices of two media.
[CBSED 07C]www3xy£o~n}s·o3myw

776. RAY OPTICS AND OPTICAL INSTRUMENTS
49.State
the conditions under which total internal
reection occurs. [ISCE99;CBSED 10, 13; F 09]
50.What is an optical fibre? [Haryana04]
51.Which of the two main parts of an optical fibre has a
highervalue of refractive index?
52.Name the physical principle on which theworking
of optical fibres is based.
[ISCE95,2000;Punjab2000, 02]
53.Whatis the main use of optical fibres?
54.Ifallg= ~andallW=~,then what will be the
2 3
value ofwllg?
55.What is a lens?
56.Define optical centre of a lens.
57.What is the deviation produced by a thin lens of a
ray passing through its optical centre?
58.What type of a lens is a tumbler filled with water?
59.Can a lens be used in a medium of which it is made
of?
60.A lens always forms virtual and erect image of the
object irrespective of the position of the object.
What type of lens is this ?
61.What should be the position of an object relative to
a biconvex lens so that it behaves like a magnifying
glass?
62.Where should an object be placed from a convex
lens to form an image of the same size ? Can it
happen in case of concave lens ?
63.Define power of a lens. Give its51units.[CBSEF09]
64.Define one dioptre.
65.If the power of a lens is+5 dioptre, what is its focal
length? [CBSED 94 C]
66.A lens has a power of-2.5 D. What is the focal
length and nature of the lens?
67.Two thin lenses of power+4D and - 2 D are in
contact. What is the focal length of the
combination ? [CBSEOD 09]
68.An object is held at the principal focus of a concave
lens of focal lengthF.Where is the image formed ?
[CBSEOD 03,08]
69.The central portion of a lens is covered with a black
paper. Will the lens form full image of an object ?
70.In Fig. 9.211 given below, path of a parallel beam of
light passing through a convex lens of refractive
index11gkept in a
medium of refractive
index11mis shown. Is(i) Ilm
--~--~~---.---
11g=11mor(ii)11g>11m
or(iii)11g<11 m?
[CBSED02C]
Fig.9.211
9.145
71.A double convex lens, made from a material of
refractive indexIll'is immersed in a liquid of
refractive index112'where112>Ill'What change, if
any, would occur in the nature of the lens?
[CBSESamplePaperDB]
72.A glass lens of reective index 1.5 is placed in a
trough of liquid. What must be the refractive index
of the liquid in order to make the lens disappear ?
[CBSED08,10]
73.A converging lens of refractive index 1.5 is kept in a
liquid medium having samerefractive index. What
would be the focal length of the lens in this
medium ? [CBSED08]
74.How does the power of a convex lens vary, if the
incident red light is replaced by violet light?
[CBSED 08]
75.A diverging lens of focal lengthFis cut into two
identical parts each forming a plano-concave lens.
What is the focal length of each part. [CBSEOD 08]
76.Draw a plot showing the variation of power of a
lens,with the wavelength of the incident light.
[CBSEOD08]
77.Use-lens maker's formula to write an expression for
the refractive index, 11of the material in terms of its
focal lengthf,and the radii of curvatureR,.andRz
of its two surfaces. [CBSEOD 07C]
78.Define linear magnification produced by a
lens/mirror.
79.Three lenses with magnifications 2, 3 and 10 form a
combination. What is its total magnification?
[CBSEF94 C]
80.What is the purpose of adding 'blue' to clothes?
81.What is a prism?
82.Define angle of the prism.
83.Define angle of deviation.
84.What is the efect on a ray of light passing through a
prism?
85.Name the factors on which the angle of deviation
produced by a prism depends.
86.Define angle of minimum deviation.
87.Write the relationship between angle of incidence
'i',angle of prism'A'and angle of minimum
deviation for a triangular prism.[CBSED 13]
88.Write the relation for the refractive index of the
prism in terms of the angle of minimum deviation
and the angle of prism. [CBSEOD 03C;D 10]
89.What is dispersion of light?
[CBSED 93C;Punjab04]
90.State the factors on which dispersive power of a
prism depends.www5notesdrive5com

777. 9.146
91.What
arethefactors onwhich angular dispersion of
a prism depends ?
92.Forwhich colour, therefractive index of prism
material is(i)minimumand(ii)maximum? [D10]
93.Which colour is deviated (i)most (ii)least, on
passing through a prism ?
94.Does the angle of minimum deviation produced by
a prism depend on wavelength ?
95.Out of red and blue lights, forwhich colour is the
refractive index of glass greater? [CBSE OD99C]
96.A glass prism isheld inwater. How is the angle of
minimum deviation affected ?
97.When does a ray passing through a prism deviate
away from its base?
9S.Define dispersive power (for light) of a medium.
99.A monochromatic ray of light is made tofallon a
normal60°prism under minimum deviation
condition. What is the relation between the angle of
incidence and the angle of emergence?[ISCE98]
100.Drawaproperly labelled graph between the angle
of incidence and the angle ofdeviationfor a prism
and show the point of minimum deviation.
[ISCE96;CBSEOD09]
101.How does the angle of minimum deviation ofa
glass prismvary, if the incidentviolet light is
replaced with red light? [CBSEOD08]
102.How does the angle of minimum deviation of a glass
prism of refractive index1.5change, if it is immersed
in a liquid of refractive index1.3?[CBSEOD08]
103.Violet colourisseen at the bottom of thespectrum
whenwhite lightis dispersed by a prism. [D10]
104.State Rayleigh's law ofscattering.
105.What is monochromatic light? Giveone example of
a source of monochromatic light.
106.What is angular size of an object or image?
107.What is simple microscope?
10S.What is the magnification produced by a single
convex lens used as a simple microscope in normal
use ? [ISCE95]
109.If the final image is formedat infinity, what isthe
magnification of asimple microscope?
110.What is the eye-ring of a microscope or a telescope?
111.What is the nature of the final image inacompound
microscope ?
Answers
PHYSICS-XII
112.What can we say about the length of a compound
microscope if the final image is formed atinfinity ?
113.What do you mean by normal adjustment of
telescope?
114.What is thedistance betweenthe objective and the'
eyepiece of a telescope in normal adjustment?
Or
Whatisthe length ofthetelescope in normal
adjustment? [CBSEOD03]
115.Express the angular magnification of an
astronomical telescope in terms of the focal length
of the objective and theeyepiece. [ISCE94]
116.An astronomical telescope set for normal
adjustment has a magnifying power 10. Ifthe focal
length ofthe objective is1.2m,what isthe focal
length ofthe eyepiece? [ISCE01]
117,In which device-microscope or telescope, the
diference inthe focal lengths of thetwo lenses is
larger?
l1S.Aconvex lens of focal length Aiskeptincontact
with a concave lens of focal length12,Findthe focal
length ofthe combination. [CBSEOD 13]
119.Abiconvex lens made of a transparent material of
refractive index 1.25is immersed in water of refractive
index 1.33.Will the lens behave as a converging or a
diverging lens? Give reason. [CBSEOD 14]
120.Redraw the diagram given below and mark the
position ofthecentre of curvature of the spherical
mirrorused in the given set up. [CBSESP15]
B
Principal
Image A axis
A'r-~----------~~-------------=~
Object
B' Fig. 9.212
121.A biconvex lens madeof atransparent material of
refractive index 1.5is immersed inwaterofrefractive
index1.33.Will the lens behave as aconverging or a
diverging lens?Give reason. [CBSEOD 14]
122.A convex lens isplaced in contact withaplane
mirror. A point object at a distance of 20 em on the
axis ofthis combination has its image coinciding
withitselfhat is the focal length ofthelens?
[CBSED 14]
•
1.Our eye is most sensitive towavelength A.=5500A.
This wavelength lies in the yellow-green region of
the visible spectrum.
2.Angle of incidence =0°,
Angleofreflection =0°.
3.The image moves towards theperson withvelocity 2v.www3xy£o~n}s·o3myw

778. RAY OPTICS AND OPTICAL INSTRUMENTS
4.20°, because
the reected ray turns through twice
the angle through which the plane mirror is rotated.
5.Infinity.
6.For parallel plane mirrors,e=0°, therefore,
360
n=--l=ao
e
7.n=360 -1=8 - 1=7images.
45
8.The minimum size{vertical length) of the plane
mirror should be equal to half the height of the man.
9.Six images. The two adjacent walls inclined at 90°
will make three images and the ceiling will repeat
them.
10.A spherical mirror is a reecting surface which
forms part of a hollow sphere. Spherical mirrors are
of two types
(i)cancave mirror and(ii)convex mirror.
11.A convex mirror is called a divergent mirror
because it diverges a parallel beam of light incident
on it.
12.A narrow beam of light parallel to the principal axis
either actually converges to or appears to diverge
from a pointFon the principal axis after reection
from the spherical mirror. This point is called
principal focus of the mirror.
13.(i)A concave mirror has a real focus.
(ii)A convex mirror has a virtual focus.
14.Convex mirror.
15.A concave mirror, because it forms an erect and
enlarged image when the object is placed between
the focus and the mirror.
16.A cancave mirror, when the object is placed beyond
2F,it forms a real and diminished image.
17. (i)Real and
(ii)magnified.
18.No. A convex mirror always forms a virtual image
which cannot be obtained on a screen.
19.Yes,because the rays diverging from the virtual
image are real and can be focused.
20.The object should be placed at 40 cm from the
mirror.
21.A ray of light from the centre of curvature falls
normally on the spherical mirror. So its angle of
incidence is 0°.
22.Size of the image increases and image shifts
towards the pole' of the mirror.
23.For the concave mirror of aperture as large as 4 em,
all the incident rays are not likely to be paraxial.
9.147
24.Refraction is the phenomenon of the change in path
of light as it passes from one transparent medium to
another.
25.The refractiveindex of a medium for a light of
given wavelength may be defined as the ratio of the
speed of light in vacuum to its speed in that
medium.
Rfr..d Speed of light in vacuum
e active m ex=~---:-:-=----,-----
Speed of light in medium
c
Jl=-.
v
26.The ratio of the wavelength of light in vacuum to its
wavelength in a medium is called refractive index
of that medium.
or
'"
Jl=~.
"'med
27.The relative refractive index of medium2with
respect to medium1is defined as the ratio of speed
of light("l1.)in medium1to the speed of light(v2)in
medium2.It is given by
II-5..
'-2 - .
v2
28.Refractive index of a medium depends on
(i)Nature of the medium
(ii)Wavelength of light used
(iii)Temperature
(iv)Nature of surrounding medium.
29.According to Snell's law, the ratio of the sine of the
angle of incidence and the sine of the angle of
refraction is constant for a given pair of media. This
constant is calledrefractive index(u )of second
medium w.r.t. first medium. Mathematically,
sini
-.- =u,a constant.
smr
30.Snell's law of refraction fails when light is incident
normally on the surface of a refracting medium. In
such a situationi=0 and alsor=O. The ratio
sini/sinrbecomes meaningless.
31.(i)Refractive index is minimum for vacuum
(Jl=1).
(ii)Refractive index is maximum for diamond.
32.The sidewise shift in the path of light on emerging
from a refracting medium with parallel faces is
called lateral shift.
33.Lateral shift depends on angle of incidence, the
refractive index and thickness of the refracting
medium.
34.Fori=0°, lateral shift is zero.www5notesdrive5com

779. 9.148
35.For i= 90°,lateral
shift is maximum and is equal to
the thickness of the slab.
d=tsin(i-r)
cos r
d =t sin(90°-r)=t cos r=t.
max cos r cos r
36.Frequency in air,
e 3x108 14
V= - = 10= 5x10Hz
A6000xlO-
37.
The frequency of light remains same as it travels
form air to the given medium.
:. Frequency in medium=5x1014Hz.
The frequency does not change when light
undergoes refraction.
The wavelength changes when light undergoes
refraction from one medium to another.
Frequency of the ultraviolet light remains unchanged.
The angle ofincidence in the denser medium for
which the angle of refraction in the rarer medium is
900is called critical angle(ie)of the denser medium.
No, it cannot occur.
a~=Speed of light in air= 3x108= ~
gSpeed of light in glass2x108 2
.. 1 2
sml=--=-
ea~ 3
g
.. ie=sin-lG)=41°49'.
..1 1
sinIe=;:; =.fi
:. Critical angle, ie=45°
~ =_1_=_1_=.J2.
sin ie sin45°
Water.
100%.
1
~=-.-..
smIe
. .-1(~2JI=sm -.
e ~1
The necessary conditions for total internal
reection are
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
(i) Light must travel from denser to rarer medium.
(ii) The angle of incidence in the denser medium
must be greater than the critical angle for the
two media.
50.An optical fibre is a thread-like structure of quality
glass or quartz which enables a beam of light to
travel through several kilometres without any
appreciable loss of intensity.
PHYSICS-XII
The value of the refractive index of the core
material is higher than that of the cladding.
The working of an optical fibre is based on the
phenomenon of total internal reection.
The optical fibres are mainly used for transmitting
optical frequencies.
a
w ~g3/2 9
~g=a~w = 4/3=8·
55.A lens is a piece of refracting medium bounded by
two surfaces at least one of which is a curved surface.
51.
52.
53.
54.
56.It is a point situated within the lens through which
a ray of light passes undeviated.
57.0°.
58.It behaves like a biconvex lens.
59.No, it cannot be used as a lens because there would
be no refraction of light.
60.Concave lens.
61.The object should be placed between the optical
centre and the focus of the biconvex lens.
62.The object should be placed at a distance equal to21
from the lens. This cannot happen in a concave lens
which always forms a diminished image.
63.The power of a lens is defined as the reciprocal of
its focal length expressedin metres.
p= 1 100
1(in m) I(incm )
64.One dioptre is the power of a lens whose principal
focal length is1metre.
1 1
65.Focal length, 1=-=-=+0.2m
P +5
1 1
66.Focal length, 1=-= -- = -40em .
P-2.5m
The negative sign shows that the lens is concave.
67.P = ~+Pz=+4 - 2 =+2 D
1 1
1= - =-m =+50em.
P +2
68.Image is formed at infinity.
69.Yes,each part of the lens will form full image. But
the intensity of the image is reduced.
70.~g<~m·
71.The lens would behave as a diverging lens when
immersed in the liquid.
~ = (~: -IJ(~ - ~J
When~2>~1'Itis negative.
72.The refractive index of liquid must be equal to1.5
i.e.,equal to that of glass lens.www3xy£o~n}s·o3myw

780. RAY OPTICS AND OPTICAL INSTRUMENTS
73.7=e~2-1)(~
-~J
=[~: -IJ[~ -~J=(~:~-I)[~ - ~J=O
orf=00.
74.Power of the convex lens increases, because
po: (~ -1)and ~v>~R"
75.Focal length of each partwill be2F.
For original concave lens,
1= (~ _1)(_2 _2)= _2 (~ - 1)
F R R R
For each half lens,
~ = (~ _1)(_2_~) =_ (~-1)
F' R 00 R
On dividing,
F'
-=2 orF'=2F.
F
76.Power of a lens,Po;(~ -1).As ~ of a material
decreases with the increase in wavelengthA.,so the
graph betweenPandAis of type shown in Fig.9.213.
Fig.9.213
Wavelength, A -+
77.By lens maker's formula,
7= (~ -1)[~-~]
1
~=1+ [1 1]'
f---
~ Rz
78.The ratio of the size of the image to the size of the
object is called linear magnification(m).
size of image"2
m= =-
size of object 1;
79.Total magnification,
M="'1xmzx"':3=2x3x10=60.
80.When washed, the clothes get a yellowish tint. Blue
and yellow are complimentary colours and they
give white colour.
81.Any portion of a refracting medium bounded by
two plane faces inclined to each other at a certain
angle is called a prism.
9.149
82.The angle between the refracting faces of a prism is
called angle of the prism.
83.The angle between the incident ray and the
emergent ray is called angle of deviation(0).
84.It bends towards the base of the prism.
85.The angle of deviation produced by prism depends
on(i)Angle of incidence(ii)Material of the prism
(iii)Wavelength of light used(iv)Angle of the
prism.
86.The minimum value of the angle of deviation
sufered by a ray of light on passing through a
prism is called angle of minimum deviation(om)'
87.Angle of minimum deviation,
om=2i -A
.A+Om
sm---
88.~= __-,,;2;---
. A
sm-
2
Dispersion is the phenomenon of splitting of white
light into the constituent colours on passing
through a prism.
(i)Nature of the prism material(iz)Choice of
extreme colours for which dispersive power is to be
measured.
89.
90.
9l.
92.
(i)Angle of prism(ii)Nature of prism material.
Refractive index of prism material is(i)minimum
for red colour(ii)maximum for violet colour.
(i)Violet colour is deviated most(ii)Red colour is
deviated least, on passing through a prism.
Yes,it depends on the wavelength of light.
~B>~R'becauseAB<AR"
When the prism is held in water,
.A+om
sm---
2
93.
94.
95.
96.
.A
sm-
2
Asw~g<a~rso the angle of minimum deviation
decreases in water.
97.This happenswhen the prismisimmersed in a
transparent medium having refractive index
greater than that of the prism material.
98.Dispersive power is defined as the ratio of the
angular dispersion to the mean deviation.
s, -OR
())= .
s
99.In the minimum deviation condition,
Angle of incidence=Angle of emergence.www5notesdrive5com

781. 9.150
100.See Fig.9.117(b)on page 9.67.
101
.As0=(Il -1)Aand IlR<Ilv-so the angle of
minimum deviation decreases when incidentviolet
light is replaced with red light.
.A+om
Sill---
102.Inair, aIlg ~ =1.5
sin-
2
In water
A+ 0'
sin m
wll= 2
g . A
sm-
2
1.5
1.3
103.
Aswllg<allg,soO~<Omi.e.,angle of minimum
deviation decreases when the prism is immersed in
a liquid ofIl=1.3.
Thedeviation produced by a small angled prism,
°=(Il -1)A.AsIlvhas maximum value, so violet
colour is bent must by the glass prism.
According to Rayleigh's law of scattering, the
intensity of light of wavelengthA.present in the
scattered light is inversely proportional to
wavelength "-
Mathematically,
1
ICX:4"'
X
A light of single wavelength is called mono-
chromatic light. The commonly used source of
monochromatic light is a sodium lamp.
It is the angle subtended by the object or image at
the eye when placed at the least distance of distinct
vision.
A simple microscope is a convex lens of short
focal length. It forms a magnified image when the
object is placed between its focus and optical
centre.
104.
105.
106.
107.
D
108. m=1+-.
f
D
109.m=/.
110.The image of the objective in the eyepiece is known
as eye-ring. All the rays from the object refracted by
the objective go through the eye-ring. So it is an
ideal position for our eyes for viewing.
111.Ina compound microscope, the final image is
inverted with respect to the object. It is virtual and
magnified. .
112.The length of the compound microscope is greater
thanfo+fe'
PHYSICS-XII
113.When the final image is formed at infinity, the
telescope is said to be in normal adjustment.
114.Length of telescope in normal adjustment
= fo+fe'
115.When the final image is formed at the least distance
of distinctvision,
116.
m=_fo(l+fe)
fe D
When the final image is formed at infinity,
m=-fo.
fe
Asm=fo(magnitude in normal adjustment)
fe
..f.=fo
em
117.
1.2m
=--=O.12m.
10
In a telescope, the diference in the focal lengths of
the two lenses is larger.
Focal lengthAof convex lens is positive.
Focal lengthf2of concave lens in negative.
Equivalent focal length of the combination is given
by
118.
119.
1 1 1 f2-A
-=-+-=---
f A - f2 Af2
f=Af2
f2-A
Diverging lens, because the light rays diverge on
refraction from rarer to denser medium.
The required ray diagram is shown below:120.
B"
Image
A'
Principal
axis
A
Object
Centre of
curvature
B'
Fig.9.214
121.Converging lens. Light rays get converged on
refraction from denser to a rarer medium.
122.20 em. For explanation, refer to the solution of
Example 83 on page 9.63.www7notesdrive7com

782. RAY OPTICS AND OPTICAL INSTRUMENTS 9.151
"'YPE B :SHORT
ANSWER QUESTIONS (2 or 3 marks each)
1.What is optics? What are its two main branches?
2.Prove that for a concave mirror the radius of
curvature is twice the focal length.[CBSEaD 96)
3.With the help of a suitable ray diagram, derive the
mirror formula for a concave mirror.[CBSEaD09)
4.What is meant by linear magnification of an image?
Using Cartesian sign convention, write the
expression for the linear magnification in terms of
the object and image distance for(a)a concave
mirror and(b)a convex mirror. What is the
meaning of sign of magnification ?
5.What is spherical aberration ? How can it be
removed?
6.Give some practical applications of spherical and
parabolic mirrors.
7.Define refractive index of a material. Give its
physical Significance.
8.Distinguish between absolute refractive index and
relative refractive index of a material. Write a
relation between these refractive indices.
9.Explain the cause of refraction of light.
10.A ray of light bends towards normal as it passes
from air to glass. Give reason.
11.State the principle of reversibility of light. Hence
1 1
prove that112=-2 -.
III
12.Discuss the refraction through a glass-slab and
show that the emergent ray is parallel to the
incident ray but laterally displaced.[Himachal96)
13.A ray of light is incident at angleion a rectangular
slab of thicknesstand refractive indexu ,Show that
the lateral displacement of the emergent ray is
x=tsini[1-2c~s ~ .1/2]
(Il-sm I)
Canxexceedt?
14.For a ray of light sufering refraction through a
combination of three media, show that
1 2 3
112x113xIII=1
Real depth
15.Deduce the relation,Il= ---~--
Apparent depth
16.Explain why does a water tank appear shallower?
17.An object placed at the bottom of a beaker con-
taining water appears to be raised. Why ?
18.Explain, with the help of a diagram, how is the
phenomenon of total internal reection used in
(i)an optical fibre
(ii)a prism that inverts an image without
changing its size.[CBSESamplePaper15)
19.The sun near the horizon appears attened at
sunset and sunrise. Why ?
20.State the conditions for total internal reection of
light to take place at an interface separating two
transparent media. Hence derive the expression for
the critical angle in terms of the speeds of light in
the two media. [CBSED2000)
21.What is critical angle? Give one application of total
internal reection. [Haryana02)
22.What are optical fibres? How are light waves propa-
gated in them? Write their any two uses.
[Himachal02,04; Haryana02,04)
23.Explain briey, with a ray diagram, how a mirage is
formed in deserts. [Haryana01 ;CBSED98C)
24.Why does a diamond sparkle ? Is it a source of
light? [Punjab01,02)
25.Give four advantages of totally reecting prisms
over plane mirrors.
26.Give reasons for the following observations made
from the earth:(i)Sun is visible before the actual
sunrise. (ii)Sun looks reddish at sunset or sunrise.
[CBSED2000,02;on06C)
27.Draw a ray diagram to show the formation of the
image of an object placed between1and21of a
thin convex lens. Deduce the relationship between
the object distance, image distance and focal length
under the conditions stated.
28.Two thin convex lenses ~ and ~ of focal lengthsIt
and12respectively, are placed coaxially in contact.
An object is placed at a point beyond the focus of
lens ~. Draw a ray diagram to show the image
formation by the combination and hence derive the
expression for the focal length of the combined
system. [CBSEon15)
29.Derive the lens formula,1.= ~ - ~for a concave
1v u
lens, using the necessary ray diagram.[CBSEaD 08)
30.Derive the expression for the angle of deviation for
a ray of light passing through an equilateral prism
of refracting angle'A'. [CBSED 93)
31.(a)A ray of light falls on a triangular glass prism
in such a way that the deviation of the emer-
gent ray is minimum for that prism. Draw the
ray diagram for this case and write the relation
between the angle of incidence and angle of
emergence.www5notesdrive5com

783. 9.152
(b)A ray of light falls on a transparent right-
angled isosceles £!ism made from a glass of
refractive index../2.Draw the
ray diagram for
this prism when the incident ray falls normally
on one of the equal sides of this prism.
[CBSEOOSC]
32.Derive the expression for the refractive index of the
material of the prism in terms of the angle of the
prism and angle of minimum deviation.
[CBSEOO6C]
33.Write the relation between the angle ofincidence
(i),the angle of emergence(e),the angle of prism(A)
and the angle of deviation(8)for rays undergoing
refraction through a prism. What is the relation
betweenLiandLefor rays undergoing minimum
deviation? Using this relation, write the expression
for the refractive index(}l)of the material of a prism
in terms ofL Aand the angle of minimum deviation
(8m), [CBSESample Paper OS]
34.Draw a graph to show the variation of the angle of
deviation'8'with that of the angle of incidence'j'
for a monochromatic ray of light passing through a
glass prism of refracting angle'A'.Hence deduce
the relation
. (8m +A)sm---
2
[Haryana 04 ; CBSE0 02C, 04C ; 00 03]
35.Derive an expression for the angle of deviation of a
small prism in terms of the refractive index and the
angle of the prism. [ISCE96]
36.Draw an appropriate ray diagram to show the
passage of a'white ray', incident on one of the two
refracting faces of a prism. State the relation for the
angle of deviation, for a prism of small refracting
angle. [CBSESample Paper 13]
37.What is dispersion of light? Explain it with a ray
diagram. Also explain the cause of dispersion of
light. [Punjab 99C,02]
38.Define the term angular dispersion. Draw the path
of a ray of white light passing through prism and
mark angular dispersion on it. [CBSESP 97]
39.Explain the terms angular dispersion and
dispersive power. How are the two related ?
[Haryana 01]
40.Write the conditions for observing a rainbow. Show,
by drawing suitable diagrams, how one under-
stands the formation of a rainbow.[CBSE00 14C]
41.Draw a neat ray diagram of a simple microscope.
Deduce the formula for its angular magnification
when the image is formed at the least distance of
distinct vision. [ISCE2000]
PHYSICS-XII
42.With the help of a ray diagram, explain the working
of a simple microscope when the image is formed at
infinity. Write an expression for its magnifying power.
43.Draw a ray diagram showing the image formation
bya compound microscope. Obtain expression for
total magnification when the images isformed at
infinity. [CBSE00 14C, lSC]
44.(a)Draw a ray diagram showing the image
formation by a compound microscope.
(b)Derive expression for total magnification when
the image is formed at infinity.
(c)Why is the objective of a compound micro-
scope of short aperture and short focal length ?
Give reason. [CBSE0 13,F13]
45.Draw the course of rays in an astronomical
telescope, when the final image is formed at the
least distance of distinct vision. Also define and
write an expression for the magnifying power in
this position. [CBSE00 09, 13]
46.(a)Draw a labelled diagram of refraction type
telescope in normal adjustment.
(b)Give its two shortcomings over reection type
telescope .
(c)Why is eyepiece of a telescope of short focal
length, while objective is of large focal length ?
Explain. [CBSE0 OS; 00 04;F13]
47.Draw a labelled ray diagram of an astronomical
telescope of the near point adjustment. You are given
three lenses of power0.5D,4D,10D.State, with
reason, which two lenses will you select for
constructing a good astronomical telescope.
[CBSE006C]
48.Two monochromatic rays of light are incident
normally on the faceABof an isosceles right-angled
prismABCThe refractive indices of the glass prism
for the two rays'1'and'2'are respectively1.35and
1.45.Trace the path of these rays after entering
through the prism. [CBSE0014]
A
45
'1'-----i
'2'-----i
B '---------""-'-~ C
Fig. 9.215
49.Draw a labelled ray diagram to show the image
formation in a reecting type telescope. Write its
two advantages over a refracting type telescope. On
what factors does its resolving power depend ?
[CBSE0 06,OS; 00 12]www3xy£o~n}s·o3myw

784. RAY OPTICS AND OPTICAL INSTRUMENTS
SO.(a)Draw a labelled diagram of a reecting type
telescope.
(b)Write two important
advantages justifying
why reecting type telescopes are preferred
over refracting telescopes.
Answers
9.153
(c)The objective of a telescope is of larger focal
length and of larger aperture (compared tothe
eyepiece). Why?Give reasons.[CBSE F 13]
••
1.Refer answer to Q. 1 on page 9.1.
2.Refer answer to Q. 6 on page 9.3.
3.Refer answer toQ.9(a)on page 9.5.
4.Refer answer toQ.11 on page 9.7.
5.Refer answer to Q. 12 on page 9.8.
6.Refer answer to Q. 13 on page 9.8.
7.Refer answers to Q. 16 and Q. 19 on page 9.15.
8.Refer answer toQ.16 on page 9.14. The relative
refractive index of any medium with respect of
vacuum is called its absolute refractive index.
9.Refer answer to Q. 18 on page 9.15.
10.Refer solution to Problem 14 on page 9.103.
11.Refer answer toQ.20 on page 9.15.
12.Refer answer to Q.21 on page 9.16.
13.Refer answer to Q. 22 on page 9.16.
14.Refer answer to Q. 23 on page 9.17.
15.Refer answer to Q. 24 on page 9.17.
16.Refer answer to Q. 24 on page 9.17.
17.Refer answer toQ.24 on page 9.17.
18.(i)See Fig. 9.39(a)and its explanation.
(ii)See Fig. 9.37 and its explanation.
19.Refer answer toQ.26 on page 9.18.
20.Refer to points 21 and 22 of Glimpses and the
solution of Problem 12 on page 9.112.
21.The angle of incidence in the denser medium for
which the angle of refraction in the rarer medium is
90° is called critical angle. Diamonds sparkle due to
the phenomenon of total internal reection.
22.·Refer answer toQ.31 on page 9.24 andQ.32 on
page 9.25.
23.Refer answer toQ.28 on page 9.23.
24.Refer answer toQ.28 on page 9.23. No, diamond is
not a source of light. It accumulates light due to
multiple internal reections.
25.Refer answer toQ.30 on page 9.24.
26.(i)Refer answer toQ.25 on page 9.17.
(ii)Refer solution' to Problem 66 on page 9.108.
27.Refer answer to Q. 42 on page 9.47.
28.Refer answer to Q. 48 on page 9.57.
29.Refer answer to Q. 44 on page 9.48.
30.Refer answer to Q. 51 on page 9.67.
31.(a)See Fig. 9.175(b)See Fig. 9.178.
32.Refer answer to Q. 52 on page 9.67.
33.Refer answer to Q. 52 on page 9.67.
34.Refer answer to Q. 52 on page 9.67 ..
35.Refer answer to Q. 53 on page 9.68.
36.See Fig. 9.118 on page 9.68.
Angle of deviation,
8=(I!-1)A.
37.Refer answer toQ.54 on page 9.68.
38.Refer answer to Q. 56 on page 9.69.
39.Refer answer toQ.56 on page 9.69.
40.Refer answer toQ.65 on page 9.80.
41.Refer answer toQ.79 on page 9.87.
42.Refer answer to Q. 79 on page 9.87.
43.Refer answer toQ.80 on page 9.91.
44.(a)See Fig. 9.146 on page 9.92.
(b)Refer answer toQ.80(b)on page 9.92.
(c)The objective of smaller aperture produces a
highly bright image while the objective of short
focal length produces large angular magni-
fication.
45.Refer answer toQ.82 on page 9.95.
46.(a)See Fig. 9.149 on page 9.97.
(b)Drawbacks of astronomical telescopes:
(i)The large objective lens used is very heavy,
which is dificult to make and support by its
edges.
(it)It is dificult and expensive to make large size
lenses free from chromatic aberration and
distortions.
(c)Whenfa» fe'the telescope will have large
magnifying power.
47.See Fig 9.147 on page 9.95.
For constructing astronomical telescope, the lens of
0.5 D should be used as objective because of its
larger focal length and lens of 10 D should be used
as eyepiece because of its smallest focal length.www5notesdrive5com

785. 9.154
A
48.
1-
__-+----1 •.•...
2----+---""'T"~
B'-----+.....::::.....I....>oC
For explanation, refer
to the solution of
Problem 3 on Fig.9.216
page9.121.
PHYSICS-XII
49.Refer answer toQ.84on page9.97andQ.86on
page 9.98.
50.(a)See Fig. 9.51on page9.98.
(b)Refer answer toQ.86on page9.98.
(c)The objective of largerfocallength produces
high angular magnification while that of larger
aperture has a high resolving power.
rl\VPEC:LONG ANSWER QUESTIONS (5 marks each)
1.(a)With the help of a suitable ray diagram, derive
themirror formula for a concave mirror.
[CBSEOD09]
(b)Draw a ray diagram to show the image
formation by a concave mirror when the object is
keptbetween its focus and the pole. Using this
diagram, derive the magnification formula for the
image formed. [CBSED 11]
2.By stating the sign conventions and assumptions
made, derive mirror formula for a convex mirror.
[Punjab99,2000; CBSEOD97]
3.(a)For a ray of light travelling from a denser
medium of refractive index"'tto a rarer medium of
refractive index11:2,prove that11:2=sinie'whereicis
"'t
the critical angle ofincidence for the media.
(b)Explain with the help of a diagram, how the
above principle is used for transmission of video
signals using optical fibres.
4.Withthehelp of a ray diagram explain the
phenomenon of total internal reection. Obtainthe
relation between critical angle and the refractive
index of the medium.
Draw ray diagrams to show how a right angled
isosceles prism can be used to
(i)deviate the ray through180°,
(ii)deviate the ray through90°,and
(iii)invert the ray. [CBSEDOlC]
5.A spherical surface of radius of curvatureR
separates a rarer and a denser medium as shown in
Fig.9.217.
Complete the path of the incident ray of light,
showing the formation of a real image. Hence
derive the relation connecting object distance'u',
image distance''d,radius of curvature R and the
refractive indices"'tand11:2of the two media.
Densermedium
C
Fig.9.217
Briey explain, how thefocal length ofaconvex
lens changes,with increase in wavelength of
incident light. [CBSEOD 04]
6.A sphericalsurface'of radius of curvature R'and of
refractive index 1.12is placed inamedium of
refractive index 1.11'where 1.11<1.12' The surface
produces a real image of an object kept infront ofit.
Using appropriate assumptions and sign
conventions, derive a relationship between the
object distance, image distance, R, 1.11and1.12'
Under what conditions this surface diverges a ray
incident on it? [CBSESamplePaper 03]
7.(a)A point object '0'is kept in a medium of
refractive index "'tin front of a convex
spherical surface of radius of curvature R
which separates the second medium of
refractive index 11:2from the first one, as shown
in the figure.
o C
----..-------------.--_.
I----u--+to-- R-.j
Fig.9.218www3xy£o~n}s·o3myw

786. RAY OPTICS AND
OPTICAL INSTRUMENTS
Draw the ray diagram showing the image
formation and deduce the relationship
between the object distance and the image
distance in terms ofr;.,11zand R
(b)When the image formed above acts as a virtual
object for a concave spherical surface
separating the medium11zfromr;.(11z>r;.),
draw this ray diagram and write the similar
(similar to(a))relation. Hence obtain the
expression for the lens maker's formula.
[CBSED 15]
8.Draw a ray diagram showing the formation of the
image by a point object on the principal axis of a
spherical convex surface separating two media of
refractive indicesr;.and11z,when a point source is
kept in rarer medium of refractive indexr;..Derive
the relation between object and image distance in
terms of refractive index of the medium and radius
of curvature of the surface. Hence obtain the
expression for lens-maker's formula in the case of
thin convex lens. [CBSED 09, 14, 14C]
9.Derive expression for the lens maker's formula,i.e.,
7= (~ -1) [~-~]
where the symbols have their usual meanings. State
the assumptions used and the convention of signs
used. [CBSEOD 96C ; Punjab 03 ;
Himachal 99;Haryana 01]
10.Apoint object is placed in front of a double convex
lens (of refractive indexn=11z /r;.with respect to
air) with its spherical faces of radii of curvatureR,..
andRz.Show the path of rays due to refraction at
first and subsequently at the second surface to
obtain the formation of the real image of the object.
Hence obtain the lens-marker's formula for a thin
lens. [CBSEF09, 13; OD 14 C]
11.Draw a ray diagram to show the formation of image
of an object placed between the optical centre and
focus of the convex lens. Write the characteristics of
image formed. Using this diagram, derive the
relation between object distance, image distance
and focal length of the convex lens. Write the
assumptions and convention of signs used. Draw
the graph showing the variation ofvandu.
[CBSED 99C,03]
12.(a)A ray'PQ'of light is incident on the faceABof a
glass prismABC(as shown in Fig. 9.219) and
emerges out of the faceACTrace the path of the
ray. Show that
Li+Le=LA+Lo
where 0 andedenote the angle of deviation and
angle of emergence respectively.
9.155
A
p
L..-----~c
Fig. 9.219
Plot a graph showing the variation of the angle of
deviation as a function of angle of incidence. State
the condition under which Lo is minimum.
(b)Find out the relation between the refractive
index (~) of the glass prism andLAfor the case
when the angle of prism(A)is equal to the angle of
minimum deviation(om).Hence obtain the value of
the refractive index for angle of prismA=60°.
[CBSEOD15]
13.(a)Draw a ray diagram to show refraction of a ray
of monochromatic light passing through a glass
prism.
Deduce the expression for the refractive index of
glass in terms of angle of prism and angle of
minimum deviation.
(b)Explain briey how the phenomenon of total
internal reection is used in fibre optics.
[CBSED 11]
14.Trace the path of a monochromatic ray of light
through a prism of refracting angle'A'.Draw a
graph to show the variation of angle of deviation'8
with the variation of angle of incidence'i',
Deduce the relation
.om+A
sm---
~= __ ~2~
. A
sm-
2
where terms ~,omhave their usual meaning.
[CBSEF08]
15.What is rainbow ? Diferentiate between primary
rainbow and secondary rainbow with a diagram.
Why two observers do not see the same rainbow?
[Punjab 01]
16.You are given two convex lenses of short aperture
having focal lengths 4 em and 8 em respectively.
Which one of these will you use as an objective
and which one as an eyepiece for constructing a
compound microscope ? Draw a ray diagram to
show the formation of the image of a small object
due to a compound microscope. Derive an
expression for its magnifying power.
[CBSEDOlC]www5notesdrive5com

787. 9.156
17.(a)Draw a diagram for the formation of image by
a compound microscope. Define its magni-
fying
power. Deduce the expression for the
magnifying power of the microscope.
(b)Explain: (i)Why must both the objective and
the eyepiece of a compound microscope have
short focal lengths ? (ii) While viewing through
a compound microscope, why should our eyes
be positioned not on the eyepiece but a short
distance away from it for best viewing.
lCBSE F;D 09;OD10]
18.Draw a ray diagram for a compound microscope.
Derive an expression for the magnifying power
when the final image is formed at the least distance
of distinct vision. State the expression for the
magnifying power when the image is formed at
infinity. Why is the focal length of the objective
lens of a compound microscope kept quite small?
ICBSESam ePaper 11]
19.Draw a ray diagram to show the working of a
compound microscope. Deduce an expression for
the total magnification when the final image is
formed at the near point.
In a compound microscope, an object is placed at a
distance of 1.5 em from the objective of focal length
Answers
PHYSICS-XII
1.25 em. If the eye piece has a focal length of 5 em
and the final image is formed at the near point,
estimate the magnifying power of the microscope.
[CBSE D 10]
20.With the help of a ray diagram, explain the formation
of image in an astronomical telescope for a distant
object. Define the term magnifying power of a
telescope. Derive an expression for its magnifying
power when the final image is formed at the least
distance of distinctvision. [CBSEOD2000C]
21.Draw a ray diagram for the formation ofimage of a
distant object by an astronomical telescope in normal
adjustment position. Deduce the expression for its
magnifying power. Write two basic features which
can distinguish between a telescope and a compound
microscope. {CBSED03C;ODMe,14C;F 09]
22.Draw a ray diagram showing the image formation
of a distant object by a refracting telescope. Define
its magnifying power and write the two important
factors considered to increase the magnifying power.
Describe briey the two main limitations and
explain how far these can be minimized in a
reecting telescope. [CBSE F 15]
1.(a)Refer answer to Q. 9(b)on page 9.6.
(b)See Fig. 9.14 on page 9.6.
t:.MPF - A'B' F', therefore
A'B' FB'
MP FP
A'B' FP+ PB'
AP FP
Applying the new Cartesian sign convention, we get
or
A'B'=+~, AB=+~, FP=-I, PB'=v
~=-I+v
~ -I
m= ~=1-v=_~
~ 1 u
(Using mirror formula)
2.Refer answer to Q. 10 on page 9.6.
3.(i) Refer to the solution of Problem 13 on page 9.113.
(ii)Refer answer to Q. 21 on page 9.16.
4.Refer answer to Q. 27 on page 9.22 and Q. 29 on
page 9.23. .
5.Refer answer to Q.36(i)on page 9.32. Also refer to
solution of Problem 24(ii) on page 9.115.
or
'f'
6.Refer answer to Q.36(i)on page 9.32. WhenIII>112'
the given sface diverges the rays incident on it.
7.(a)Refer answer to Q. 37 on page 9.34.
(b)Refer answer to Q. 38 on page 9.40.
8.For refraction at a spherical surface, refer answer to
Q.36 on page 9.31.
For lens maker's formula, refer answer toQ.38 on
page 9.40.
9.Refer answer to Q. 38 on page 9.40.
1.0_Refer answer toQ.38 on page 9.40.
11_Refer answer to Q. 43 on page 9.47. For graph
betweenuandv,see Fig. 9.82.
12(a)Refer answer toQ.51 and see Fig. 9.117(b)on
page 9.67.
(b)Refer answer to Q. 52 on page 9.67.
13.(a)Refer answer to Q. 52 on page 9.67.
(b)Refer answer to Q. 21 on page 9.16.
14.Refer answer to Q. 52 on page 9.67.
15.Refer answer to Q. 65 on page 9.80.
Two observers cannot see the same rainbow.
Primary rainbow is seen when the rays emerging
from the water droplets subtend a mean angle of
41°[=(40°+42°)/2) and secondary rainbow is seenwww3xy£o~n}s·o3myw

788. RAY OPTICS AND OPTICAL INSTRUMENTS
when the emerging
rays subtend a mean angle of
53.sol=(52°+ 55°) /2]. The positions of such
droplets which send these rays, depend on the
position of the observer. Hence two observers at
two diferent positions do not see the rainbow
formed by the same set of droplets.
16.The lens of 4 em focal length should be used as
objective and the lens of 8 ern focal length should be
used as eyepiece of the compound microscope.
Refer answer to Q. 80 on page 9.91.
17.(a)Refer answer to Q. 80 on page 9.91.
(b)Refer answer to Exercise 9.32 on page 9.141.
18.Refer answer to Q. 80 on page 9.91.
m=1110xme=va(1+ J2.J =~(1+ EJ
% fa %-fo t,
Angular magnification(1110) of objective will be
large when%is slightly greater thanfa.Now a
compound microscope is used for viewing very
close objects, so%is small. Consequently,fahas to
be small.
19.Refer answer toQ.80 on page 9.91.
Numerical. Here
fa= 1.25 em,I,=5 em,
%=-1.5 em, ve= -D = -25 em
As
1 1 1
-
faVa%
11 1 1 1
-=-+- =----
vafa%
1.251.5
100 10 300-250
---=
125 15 375
50
375
375
va=-=7.5cm
50
Magnifying power of the compound microscope,
m=VO[l+ DJ= 7.5 (1+ 25)=_5x6=_30.
% I.-1.5 5
9.157
20.Refer answer to Q. 82 on page 9.95.
21.Refer answer to Q. 82 on page 9.95.
The two important diferences between a telescope
and a compound microscope are:
(i)The aperture of the objective of a microscope is
very small while that of the telescope is large.
(ii)Both the lenses of a compound microscope
have short focal length while the objective of a
telescope has large focal length.
22.See Fig. 9.148 on page 9.96.
Magnifying power in the normal adjustment of the
telescope is defined as the ratio of the angle
subtended at the eye by the final image as seen
through the telescope to the angle.subtended at the
eye by the object seen directly, when both the image
and the object lie at infinity.
m=fa
Ie
Factors for increasing the magnifying power:
(i)Increasing focal length of the objective
(ii)Decreasing focal length of the eyepiece.
Limitations of a refracting telescope:
(i)Sufers from chromatic aberration
(ii)Sufers from spherical aberration
(iii)Small magnifying power
(iv)Small resolving power.
Advantages of a reecting telescope:
(i)No chromatic aberration, because mirror
objective is used.
(ii)Spherical aberration can be removed by
paraboloidal mirror.
(iii)Image is bright because there is no loss of
energy due to refraction.
(iv)Large mirror provides an easier mechanical
support over its entire back surface.
'-'YPED :VALUE BASED QUESTIOS(4 marks each)
1.Two students of class XII brought three big plane
mirrors in their classroom for science fair. They
fixed the three mirrors: one at the ceiling and the
other two on the adjacent walls of the room. Every
student was able to see six images of himself/
herself. Students of other classes also came to see
this and felt happy. A student of class X was
determined to know the reason behind it. She went
to the.library, consulted other students and next
day came up with the correct answer.
(a)What values were depicted by the student of
class X?
(b)Give the reason for seeing six images.
2One day Chetan's mother developed a severe
stomach ache all of a sudden. She was rushed to the
doctor who suggested for an immediate endoscopy
test and gave an estimate of expenditure for the
same. Chetan immediately contacted his class
teacher and shared the information with her. Thewww5notesdrive5com

789. 9.158
class teacher arranged for the money and rushed to
the hospital. On realising that Chetan belonged to a
below average income group family, even the
doctor ofered concession for the test fee. The test
was conducted successfully.
Answer the following questions based on the above
information:
(a)Which principle in optics is made use of in
endoscopy?
(b)Briey explain the values reected in the
action taken by the teacher.
(c)In what way do you appreciate the response of
the doctor on
the given situation ?
[CBSEOD 13]
3.Rama was watching a programme on Moon on the
Discovery Channel. He came to know from the
observation recorded on the surface of the moon
that sunrise and sunset are abrupt there and the sky
appears dark from there. He was surprised and
determined to know the reason behind it. He
discussed it with his Physics teacher next day, who
explained him the reason behind it.
(a)What were the values being displayed by
Rama?
(b)Why are sunrise and sunset are abrupt on the
surface of the moon ?
(c)Why does the sky appear dark from the
moon?
Answers
PHYSICS-XII
4.Amit's uncle was finding great dificulty in reading
a book placed at normal place. He was not going to
the doctor because he could not aford the cost.
When Amit came to know of it, he took his uncle to
the doctor. After thoroughly checking his eyes, the
doctor prescribed the proper lenses for him. Arnit
bought the spectacles for his uncle from his pocket
money. Byusing spectacles he could now read with
great ease. For this he expressed his gratitude to his
nephew.
Based on the above paragraph, answer' the
following:
(a)(I)Why does least distance of distinct vision
increase with age?
(il)What type of lens is required to correct this
defect?
(b)What, according to you, are the two values
displayed by Amit towards his uncle?
[CBSE D 13C]
5.Satish was seeing a person wearing a shirt with a
pattern comprising of vertical and horizontal lines.
He was able to see the vertical lines more clearly
than the horizontal ones. He shared his problem
with his friend Ramesh. Ramesh suggested him to
get his eyes checked-up by a doctor immediately.
(a)What value is being displayed by Ramesh
here?
(b)What is this defect due to ?
(c)How is such a defect of vision corrected?
•
1.(a)Determination and critical thinking.
(b)The mirrors on two adjacent walls inclined at
90°will make three images and the ceiling
mirror will repeat them.
2.(a)Endoscopy is based on the phenomenon of
total internal reection of light. AI'ight pipe, a
bundle of optical fibres, is inserted into
stomach. Light transmitted through outer
fibres is scattered by various parts of the
stomach. The reected light coming out of
inner fibres produces a final image with
excellent details.
(b)Empathy, charity, helping and caring.
(c)Doctor displayed sympathy and social
responsibility by ofering concession to
Chetan's poor family.
3.(a)Keen observer and curiosity.
(b)Moon has no atmosphere. There is no
refraction of light. Sunlight reaches moon
straight covering shortest distance. Hence
sunrise and sunset are abrupt.
(c)Moon has no atmosphere. So there is nothing
to scatter sunlight towards the moon. No
skylight reaches the moon surface. Sky appears
dark in the day time as it does at night.
4.(a)(i)Due to stifening of the ciliary muscles, the
eye lens of elderly persons loses exibility and
hence the accommodating power of the eye
lens decreases.
(ii)By using a convex lens of suitable focal
length.
(b)Compassion for others, charity and caring
5.(a)Empathy.
(b)This defect is called astigmatism in which a
person cannot simultaneously see both the
horizontal and vertical views of an object with
the same clarity. It is due to the irregular
curvature of the cornea.
(c)Astigmatism can be corrected by using a
cylindrical lens.www3xy£o~n}s·o3myw

790. Ray Optics and Optical Instruments
GLIMPSES
1.Optics
.It is the branch of physics that deals with
the study of nature, production and
propagation of light. It hastwosub-branches :
ray optics and wave optics.
2.Ray or geometrical optics.It concerns itself with
the particle nature of light and is based on(i)the
rectilinear propagation of light and(ii)the laws
of reflection and refraction of light.
3.Wave or physical optics. It concerns itself with
the wave nature of light and is based on the
phenomena like(i)interference(ii)diffraction
and(iii)polarisation of light.
4.Laws of reflection of light.(i)The incident ray,
the reflected ray and the normal at the point of
incidence al lie in the same plane.
(ii)The angle of incidence'i'is equal to the
angle of reflection' r' i.e.,Li=Lr.
5.Properties of images formed by plane mirrors.
(i)The image formed by a plane mirror is
virtual, erect and lateraly reversed.
(ii)The size of the image is equal to the size of
the object.
(iii)The image is as far behind the mirror as the
object is in front of it.
(iv)The line joining the object and the image is
normal to the plane mirror.
(v)When a plane mirror is rotated through a
certain angle, the reflected ray turns
through twice this angle.
6.Images formed by inclined mirrors. When two
planes mirrors are kept facing each other at an
angleeand an object is placed between them, a
number of images are formed due to multiple
reflections.
Ifeis a submultiple of 180°, then the number of
. f d i 360
lInages orme ISn=--1
e
Ifeis not a submultiple of180°, then the number
of images formed is the integer next higher than
( 3:0 -1). For two paralel plane mirrors,
360
n=-=oo.
o
7.Spherical mirror.It is a mirror whose reflecting
surface forms part of a holow sphere. Spherical
mirrors are oftwotypes:
(i) Concave mirrorin which the reflection of
light takes place from the inner holow surface.
(ii) Convex mirrorin which the reflection of
light takes place from the outer bulged
surface.
8.Definitions in connection with spherical
mirrors.
(i) Pole. Itisthe middle pointPof the spherical
mirror.
(ii) Centre of curvature. Itisthe centreCof the
sphere of which the mirror forms a part.
(iii)Radius of curvature. Itisradius(R)of the
sphere of which the mirror forms a part.
(iv)Principal axis. The linePCpassing through
the pole and the centre of curvature of the mirror
iscalleditsprincipal axis.
(v) Linear aperture. Itisthe diameter of the
circular boundary of the sphericalmirror.
(9.159)wwwlnotesdrivelcom

791. 9.160
(vi)Angular aperture. Itisthe angle subtended by
the boundary of the spherical mirror atitscentre
of curvature C.
(vii)Principal
focus. A narrow beam of light
parallel to the principal axis either actually
converges to or appears to divergefrom a pointF
on the principal axis after reflection from the
spherical mirror. This pointiscalled the
principal focus of the mirror. A concave
mirror has a real focus while a convex
mirror has a virtual focus.
(viii)Focal length. Itisthedistance (f=PF)
between the focus and the pole of the mirror.
(ix)Focal plane. The vertical plane passing
through the principal focus and perpendicular
to the principal axisiscalledfocal plane.When
a paralel beam of light is incident on a
concave mirror at a smal angle to the
principal axis, it is converged to a point in
the focal plane of the mirror.
9.New cartesian sign convention for spherical
mirrors.
(i)Al ray diagrams are drawn with the
incident light traveling from left to right.
(ii)Al distances are measured from the pole of
the mirror.
(iii)Al distances measured in the direction of
incident light are taken positive.
(iv)Al distances measured in the opposite
direction of incident light are taken to be
negative.
(v)Heights measured upwards and perpen-
dicular to the principal axis are taken
positive.
(vi)Heights measured
perpendicular to the
taken as negative.
10.Relation between focal length and radius of
curvature of a spherical mirror.
Focal length=!xRadius of curvature
2
f=:
downwards and
principal axis are
or
In new cartesian sign convention, the focal
length and radius of curvature are taken
negative for a concave mirror and positive for a
convex mirror.
PHYSICS-XII
11.Spherical mirror formula.This gives relation
between object distanceu,image distancevand
the focal lengthfa spherical mirror.
111
-+-=-
uvf
12.Linear or transverse magnification.It is the ratio
of the height of the image to that of the object.
Height of image _h2_V_f_f-v
m= ---------
Height of objecthl uf-uf
(i)IfIml>1,the image ismagnified.
(ii)IfImI<1,the image is diminished.
(iii)IfImI=1,the image is of the same size as the
object.
(iv)Ifmis positive, the image isvirtualanderect.
(v)Ifmis negative, the image is real and inverted.
13.Refraction of light. Itis the phenomenon of
bending of light from itsstraight path when it
passes at an angle from one transparent medium
to another.
14.Laws of refraction of light:
First law.The incident ray, the refracted ray
and the normal at the point of incidence al lie in
the same plane.
Second law.The ratio of the sine of the angle of
incidence and the sine of the angle of refraction
is constant for a given pair of media. This law is
also known asSnell's law of refraction
sini
-.-=11,a constant.
smr
The constant11is caled refractive index of
second medium w.r.t. first medium.
15.Refractive index.Refractive index of a medium
for a light of given wavelength may be defined
as the ratio of the speed of light in vacuum to its
speed in that medium.
Velocity of light invacuum _c
11=Velocity of light in medium --;
It may also be defined as the ratio of the
wavelength of light in vacuum toits wave-
length in that medium.
A
Il=~
Amed
The refractive index of a medium with respect
to vacuumis also caledabsolute refractive
index.wwwAnotesdriveAcom

792. (Competition Section)
16.Relative refractive index. The relative refrac-
tive
index of medium 2 w.r.t. medium 1 is the
ratio of speed of light(v1)in medium1to the
speed of light(v2)in medium2.
1 v1
112=;-
2
1 sini 112
Il=-- = - =constant
2sinr III
Also
or III sin i=112sinr
17.Principle of reversibility of light. This principle
states that if the final path of ray of light after it
has suffered several reflections and refractions
isreversed, it retraces its path exactly. Itfolows
from this principle that
1 1
1l2=--
2111
i.e.,the refractive index of medium 2w.r.t.
medium1is reciprocal ofthe refractive index of
medium1w.r.t. medium2.
18.Refraction through a rectangular glass slab.A
ray of light on refraction through a glass slab
does not suffer any deviation, i.e.,the incident
and emergent rays are paralel, but the
emergent ray is lateraly displaced w.r.t. the
incident ray. The lateral displacement xon
passing through a glass slab of thicknesstand
refractive index Il is given by
t.(.)"[1 cosi1
x= --smI -r=tsmI - 2 .2.1/2
cosr (Il-sinI)
whereiis angle of incidence
xmax=tsin90°=t
Thus the displacement of the emergentray
cannot exceed the thickness of the glass slab.
19.Refraction through a combination of media.
When a ray of light passes through a combina-
tion of media, the quantity Il sin i remains
constant, where Il is the absolute refractive
index of the medium and ithe angle of
incidence in that medium. Thus
Ilairxsin iair=Ilglassxsin iglass
=Ilwaterxsiniwater
Alsoa" Xco" Xg"=1
rco'rg ra
a
co Ilg
Il=--
ga11co
and
9.161
20.Relation between real depth and apparent
depth.Due to refraction of light, the apparent
depth of an object placed in a denser medium is
Eye
D C
A
t
Apparent 7 ;
depth '>;
i__
I
o
less than the real depth. When an object0,in a
denser medium of thicknesstand refractive
index Il is seen through a rarer medium, its
image is seen at1.It is seen that
Real depth AO
11=-----"'--
Apparent depth AI
t
Also, apparent depth,AI=-
Il
The height through which an object appears to
be raised in a denser medium is caled normal
shift:
:. Normal smft,d=10=AO - AI,=t(1-~)
Total normal shift for compound media
21.Critical angle and total internal reflection. The
angle of incidence in the denser medium for
which the angle of refraction in the rarer medium
is90°is caledcritical angleofthe denser medium
and is denoted byic'When i=ic' r=90°.
sinic1 1
As ---=- or Il=--
sin90° Il sin ic
N
7=90°www3notustryvu3som

793. 9.162
Total internal reflectionis the phenomenon in
which a ray of light traveling at an angle of
incidence greater than the critical angle from a
denser to a rarer medium is totaly reflected
back into the denser medium, obeying the laws
of reflection.
22.Necessary conditions for total internal reflection.
(i)Light must travel from an opticaly denser
to an opticaly rarer medium.
(ii)The angle of incidence in the denser
medium must be greater than the critical
angle for the two media.
23.Relation between critical angle and refractive
. d 1
ilex.Il=-.-. .
SilIe
24.Totaly reflecting
prisms.A right angled
isosceles prism,i.e.,a45° -90° - 45°prism is
caled a totaly reflecting prism. It can be used to
deviate rays through90°or180°.
25.Mirage.It is an optical ilusion observed in
deserts or over hot extended surfaces like a
coaltarred road due to which a traveler sees a
shimmering pond of water some distance ahead
him and in which the surrounding objects like
tree, etc. appear inverted.
26.Optical fibres.Optical fibres consist of
thousands of fine strands of quality glass,
coated with a material of lower refractive index.
Light entering the fibres at one end undergoes
several total internal reflections and finaly
emerges out without any appreciable change in
intensity. A bundle of optical fibres is caled a
light pipe, used in medical and optical
examination and in receiving and transmitting
signals in telecommunication.
27.Lens.A lens is a piece of a refracting medium
bounded by two surfaces, at least one of which
is a curved surface.
Lenses are of two types:
(I)Convex or converging lens.It is thicker at
the centre than at the edges. It converges a
paralel beam of light on refraction through
it. It has a real focus.
(ii)Concave or diverging lens.It is thinner at
the centre than at the edges. It diverges a
paralel beam of light on refraction through
it. It has a virtual focus.
PHYSICS-XII
28.Definitions in connection with spherical lenses :
(i)Centre of curvature.The centre of curvature
of the surface of a lens is centre of the sphere
of which it forms a part. Because a lens has two
surfaces, so it has two centres of curvature.
(ii)Radius of curvature.The radius of the
surface of a lens is the radius of the sphere
of which the surface forms a part.
(iii)Principal axis.It is the line passing through
the two centres of curvature of the lens.
(iv)Principal focus.A narrow beam of light
paralel to the principal axis either converges
to a point or appears to diverge from a
point on the principal axis after refraction
through the lens. This point is caled princi-
pal focus. A lens has two principal focii.
(v) Optical Centre.It is the point situated
within the lens through which a ray of light
passes undeviated.
(vi)Focal length.It is the distance between the
principal focus and the optical centre of the
lens.
(vii)Aperture.It is the diameter of the circular
boundary of the lens.
29.New Cartesian sign convention for spherical
lenses:
(i)Al distances are measured from the optical
centre of the lens.
(ii) The distances measured in the direction of
incident light are taken as positive.
(iii)The distances measured in the opposite
direction of incident light are taken as
negative.
(iv)Heights measured upwards and perpendi-
.cular to the principal axis are taken as
positive.
(v)Heights measured downwards and per-
pendicular to the principal axis are taken as
negative.
In this sign convention, the focal length of a
converging lens is positive and that of a
diverging lens is negative.
30.Refraction through a spherical surface.A
surface which forms part of a sphere of a
transparent refracting material is caled a
spherical refracting surface.
(i) Refraction from rarer to denser medium.When a
ray of light travels from a rarer medium ofwww3notustryvu3som

794. (Competition Section)
refractive inde
x!IIto a denser medium of
refractive index !l2 of a spherical surface of
radius of curvature R, the relation between
object distanceuand image distance vis
!l2_!II =!l2- !II
V u R
Iftherarer mediumis air, then !I 1=1and!l2 = !I,
!I1!I-I
we have - - -=--
v u R
(ii) Refraction from denser to rarer medium. When
the object is placed in a denser medium, the
relation betweenuandvcan be obtained by
interchanging !II and !l2'
&_!l2 =!ll-!l2
V u R
31.Power of a spherical refractingsurface. It isgiven
byp=!l2-!ll=!I-l (for air)
R R
where Rismeasured in metre. The power of a
convex surface is positive and that of a concave
surface is negative.
32.Principal focal lengths of a spherical surface.
(i) First principal local length. It is the distance of a
point from the pole of thesurface atwhich if an
object is placed, the image is formed at infinity.
Firstprincipal focal length, 11=_ !II R
!l2-!II
(ii) Second principal local length. It is the distance
of a point from the pole of the surface at which
theimage of an objectatinfinity is formed.
Second principal focal length, 12= !l2R
!l2 -!II
33.Lens maker's formula. Thisformula relates the
focal lengthIto the refractive index !I and the
radii of curvature R1, ~ofits spherical surfaces.
7=[!l2!1~!ll ] [ ~1- ~]
For the lens placed in air,
1 [1 1]
j=(!I-l) Rl-~.
34.Thin lens formula.Thisformula gives relation-
ship between object distance u,image distance v
and focal lengthIaspherical lens (convex or
concave) of smal aperture.
1 1 1
---=-
v uI
9.163
35.Linear magnification produced by a lens. It is
the ratio of the size of the image formed by a
lens to the size of the object.
M if size of image
agru ication=----~
size of object
m=~=::=_1_= I-v.
hI UI+uI
or
Whenmispositive (or visnegative), the image
is virtual and erect. Whenmis negative (orvis
positive), the image is real and inverted.
36.Power of a lens. The power of a lens is defined as
the reciprocal of its focal length, expressed in metres.
1
P=--
I(m)
51unit of power is m-1, also caled dioptre(D).
One dioptre is the power of a lens whose
principal focal length is1metre.
P=7=(!I-1)[~1- ~].
37.Lens combinations.When lenses are used in
combination, each lens magnifies the image
formed by the preceding lens. The total magni-
fication is equal to the product of the magni-
fications produced by the individual lenses.
m=~x"2x111:3x ....
The combined focal lengthIof two thin lenses
of focal lengths11and12placed in contact is
given by
1 1 1
-=-+-
I11 12
Fornthin lenses in contact,
1 1 1 1
-=-+ - +...+-or P = PI+P2+...+P
111/2 In n
When the two thin lenses are separated by a dis-
tanced,their equivalent focal lengthIis given
111 d
by _=_+_+ __
I1112 11/2
or Power, P = PI +P2+dxPI xP2
38.Prism. Aprism is a portion of a refracting
medium bounded by two planefaces inclined to
each other at a certain angle. The two plane
faces inclined to each other are caled refracting
laces. The line along which the two refracting
faces meet is caledrefracting edge 01 the prism.
The third face of the prism opposite to thewww3notustryvu3som

795. 9.164
refracting edge is caledbase of the prism.The
angle included between
the two refracting faces
is caledangle of prism.
39.Refraction through a prism.When a ray of light
is refracted through a prism, the sum of the
angle of incidenceiand the angle of emergence
i'is equal to the sum of the angle of the prismA
and the angle of deviation0.
A+0=i+i'andA=r+r'
whererandr'are the corresponding angles of
refraction at the two faces.
40.Relation between the refractive index and angle
of minimum deviation.The minimum value of
the angle of deviation suffered by a rayon
passing through a prism is caled the angle of
minimum deviation and is denoted byom .When
a ray of light suffers minimum deviation.
i=i',r=rand0= om
A+Om = i+i =2iori =A+om
2
A
A=r+r=2rorr= -
2
.A+Om
sinism---
Refractive index,Jl=--= 2
sinr . A
sm-
2
41.Deviation produced by a prism of smal angle.It
does not depend on the angle of incidence and
is given byo=(J-l)A
42.Dispersion.The splitting of white light into its
constituent colours when it passes through a
glass prism is caled dispersion. The dispersion
of light occurs because refractive index of prism
material is different for different wavelengths.
43.Angular dispersion.The angular separation
between the two extreme colours (violet and red)
in the spectrum is caled angular dispersion.
Angular dispersion
=Oy -OR =(Jly-1)A-(JlR -1)A=(Jly -JlR)A
44.Dispersion power.It is the ability of the prism
material to cause dispersion and is defined as the
ratio of the angular dispersion to the mean
deviation.
.. Angular dispersion
Dispersion power---"'----'---
Mean deviation
and
(Jl y-1)A-(Jl R-1)A= Jl y - Jl R
(Jl -1)A Jl-1
PHYSICS-XII
Here
45.Pure and impure spectra.The spectrum in
which the component colours of the spectra of
different rays overlap each other and the
various colours are not distinctly seen is caled
an impure spectrum. A spectrum in which there
is no overlapping of colours and different colours
are distinctly seen is caled the pure spectrum.
46.Spectroscope or spectrometer.It is an optical
device used for producing and studying the
spectrum of various light sources. It consists of
three main parts :(i)colimator,(ii)prism table
and(iii)telescope.
47.Spherical aberration.The inability of a lens or
spherical mirror of large aperture to bring the
paraxial and marginal rays of a wide beam of
light to focus at a single point is caled spherical
aberration.
48.Chromatic aberration.The inability of a lens to
bring the light rays of different colours to focus
at a single point is caled chromatic aberration.
Longitudinal chromatic aberration of a lens
=Dispersive power
xfocal length of the lens for mean colour
orI« -fy=coxf
49.Blue colour of the sky. According toRayleigh's
law of scattering,the intensity of light of wave-
lengthApresent in the scattered light is
inversely proportional to the fourth power of
1
wavelength:Io:4"
X
So, blue colour of sunlight is scattered more by
the atmospheric molecules, due to which the
sky appears blue.
50.Rainbow.It is nature's most spectacular display
of the spectrum of light produced by refraction,
dispersion and total internal reflection of sun-
light by several raindrops. It is observed when
the sun shines on rain drops after a shower. An
observer standing with his back towards the
sun observes it in the form of concentric circular
arcs of different colours in the horizon.
Primary rainbowis brighter with its inner edge
violet and outer edge red, sub tending 41° -43°
angle at the observer's eye. Secondary rainbow
is fainter with its inner edge red and outer edge
violet, sub tending 51° -54°angle at the obser-
ver's eye.www3notustryvu3som

796. (Competition Section)
51.Human
eye.It is most important andsensitive
sense organ. The essential parts of a human eye
are sclerotic, cornea, choroid, iris,pupil, crysta-
line lens, ciliary muscles, aqueous humour,
vitreous humour and retina. It isa convex lens
of focal length about 2.5 em.
52.Accommodation.It is theability ofthe eyelens
due towhich it canchange its focal length so
thatimages of objects at various distances can
be formed on the same retina.
53.Range of normal vision. The distance between
infinity and 25 em pointis caled the range of
normal vision.
54.Least distance of distinct vision(0).The mini-
mum distance from the eye, at which the eye
can see the objects clearly and distinctly without
any strain is caled the least distance of distinct
vision. For a normal eye,its value is 25 em.
55.Near point. The nearest point from the eye, at
which an object can be seen clearly bythe eye is
caled its near point. The near point of a normal
eye is at a distance of 25 em.
56.Far point. The farthest point from the eye, at
which an object can be seen clearly bythe eye is
caled thefar point of the eye. For a normal eye,
the far point isat infinity.
57.Power of accommodation. The power of
accommodation of the eye isthe maximum
variation of its power for focussing on near and
far objects. For a normal eye, the power of
accommodation isabout4dioptres.
58.Persistence of vision.The phenomenon of the
continuation of the impression of an image on
the retina for some time even after the light
from the object is cut off is caled persistence of
vision. The impression of the image remains on
the retina for about (1/16)th of a second.
Cinematography works on the principle of
persistence of vision.
59.Rods.These are rod-shaped celsof the retina
that are sensitive to the intensity of light.
60.Cones.These are cone-shaped cels of the retina
that aresensitive to the colours of light.
61.Colour blindness.A personwho cannot
distinguish between various colours but can
see wel otherwise, is said to be colour-blind.
Itisdue to lack of some cones in the retina of
the eyes.
9.165
62.Cataract.It is due to the development of hazy or
opaque membrane over the eyelens which
results in the decrease or loss of vision. It can be
cured by surgery.
63.Common defects of vision.There are mainly four
common defects of vision which can be corrected
bythe use of suitable eye glasses. These are
(I)myopia or near sightedness(il)hypermetropia
or far-sightedness(iil)presbyopia(iv)astigmatism.
64.Myopia or short-sightedness. In this defect a
person can see nearby objects clearly but cannot
see far off objectsclearly. Here, either the eye-
bal becomes too longer or the focal length of the
eyelens becomes too short. It can be corrected by
using a concave lens of suitable focal length.
Focal length of the correcting lens
=Distance of the far point from the eye.
65.Long-sightedness or hypermetropia.In this
defect a person can see the far off objects clearly
but he cannot see nearby objects distinctly.
Here, either the eyebal becomes too short or the
focal length of the eyelens becomes too large. It
can be corrected by using convex lens of
suitable focal length. .
Focal length of correcting lens=~
y-O
wherey=distance of the near point from the
defective eye.
66.Presbyopia. In this defect, a person in old age
cannot read correctly due to the stiffening of the
ciliary muscles and the decrease in flexibility of
the eyelens.
67.Astigmatism.It is defect of vision in which a
person cannot simultaneously see both the
horizontal and vertical views of an object with
the same clarity. It is due to the irregular
curvature of the cornea. It can be corrected by
using a cylindrical lens.
68.Simple microscope.It is a convex lens of short
focal length. When the objectisplaced between
the lens andits focus and the eye is held just
behind the lens, a virtual, erect and enlarged
image isseen. When the finalimage is formed at
the least distance of distinct vision(0),the
magnifying power of the simple microscope is
Angle subtended by the image at
the least distance of distinct vision
m=-------------
Angle subtended by the object at
the least distance of distinct visionwww3notustryvu3som

797. 9.166
m=l=l+D
a I
When the final image is formed at infinity,
m=D,viewing is more
comfortable when the
I
eye is focussed at infinity.
69.Visual angle.The angle subtended by an object
on the eye is caled visual angle. Larger the visual
angle, larger is the apparent size of an object.
70.Compound microscope.It is an optical device
used to see magnified images of tiny objects.
The objective is a convex lens of very short focal
length and of smal aperture. The eyepiece is a
convex lens of relatively larger focal length and
of larger aperture. The difference between the
focal lengths of the eyepiece and the objective is
smal. Its magnifying power is given by
or
m=mo xme
When the final image is formed at the least
distance of distinct vision,
Angle subtended by final virtual
image at distanceDfrom the eye
m=Angle subtended by the object .
at distanceDfrom the eye
or m=l=v0 (1+D)= ~(1+D)
a Uo Ie10 Ie
When the final image is formed at infinity,
L D
m=-x-
10Ie
whereLis the distance between the objective
and the eyepiece.
71.Astronomical telescope.It is used to view
heavenly bodies. The objective is a convex lens
of large focal length and large aperture. The
eyepiece is convex lens of smal focal length and
smal aperture. The difference in the focal
lengths of the two lenses is large. The eyepiece
forms a real, inverted and diminished image.
The eyepiece magnifies this image. The final
image is inverted w.r.t. the object.
When the final image is formed at the least
distance of distinct vision,
Angle subtended by the image
at distanceDfrom the eye
Angle subtended by the
.object at infinity
m=l= _10(1+Ie )
at, D
m=
or
PHYSICS-XII
J,D
Length of telescope, L=10+ue=Iee+D
at infinityWhen the final image is formed
(normal adjustment),
Angle subtended by the
final image formed at00
m- or
- Angle subtended by
the object at00
m=l=_/o
aI,
Length of the telescope in normal adjustment,
L=/o+le
For large magnifying power of a telescope,
clearly
10»Ie'
72.Terrestrial telescope.It is used to see the erect
images of distant earthly objects. It uses an
additional convex lens between the objective
and the eyepiece for erecting the image.
When the final image is formed at infinity, its
ifyi 10magru mg power,m=-
Ie
Length of telescope=10+4I+Ie
whereIis the focal length of the erecting lens.
When the final imageis formed at the least
distance of distinct vision, m=10 (1+Ie ).
IeD
73.Galilee's telescope.It uses a concave lens for the
eyepiece to obtain an erect image of the distant
object. The real,inverted and diminished image
formed by the objective lies at the focus of the
eyepiece. The final image is formed at infinity and
is erect and magnified.
In normal adjustment,m=10
Ie
Length of telescope,L=10-Ie'
74.Reflecting telescope.It uses a concave
paraboloidal mirror of large aperture to view
the distant objects. Both spherical andchromatic
aberrations are minimum.
When the final image is formed at the least
distance of distinct vision,
m=/O(l+le)
IeD
When the final image is formed at infinity,
10 R/2
m=-=--.
IeLwww3notustryvu3som

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898. Wave Optics
GLIMPSES
1.Nature
of light. The phenomena like inter-
ference, diffraction and polarisation establish
the wave nature of light. However, the pheno-
mena like black body radiation and photo-
electric effect establish the particle nature of
light. de Broglie suggested thatlight has a dual
nature i.e.,it can behave as particles as well as
waves.
2.Wavefront. A wavefront is defined as the
continuous locus of all such particles of the
medium which are vibrating in the same phase at
any instant. In case of waves travelling in all
directions fromla point source, the wavefronts
are spherical in shape. When the source of light
is linear in shape, the wavefronts are
cylindrical. At very large distances from the
source, a portion of spherical or cylindrical
wavefront is plane wavefront.
3.Ray.An arow drawn perpendicular to a
wavefront in the direction of propagation of a
wave is called a ray.
Two general principles are valid for rays and
wavefronts :
(I)Rays are normal to wavefronts.
(ii)The time taken to travel from one wave-
front to another is the same along any ray.
4.Huygens' principle of secondary wavelets.
Huygens' principle is the basis of the wave theory
of light. It tells how a wavefront propagates
through a medium. It is based on the following
assumptions:
(i)Each point on a wavefront acts as a source
of new disturbance called secondary waves
or wavelets.
(ii)The secondary wavelets spread out in all
directions with thespeed of light in the
given medium.
(iii)The wavefront at any later time is given by
the forward envelope of the. secondary
wavelets at that time.
S.Effect on frequency, wavelength and speed
during refraction. When a light wave travels
from one medium to another, its frequency
remains unchanged but both its wavelength
and speed get changed, depending on the
refractive index of the refracting medium.
6.Interference of light waves.When light waves
from two coherent sources travelling in the
same direction superpose each other, the
intensity in the region of superposition gets
redistributed, becoming maximum at some
points and minimum at others. This pheno-
menon is called interference of light.
7.Constructive and destructive interference.If
path differencep=nAor phase difference_
<p=2nt;the two waves are in same phase and so
add up to give maximum of intensity. Thisis
calledconstructive interference.
Ifp=(2n -1)A/2 or<p=(2n -1)7r,the two super-
posing waves are out of phase, the resultant
amplitude is equal to difference between their
individual amplitudes and hence intensity is
minimum. This is calleddestructive interference.
s.Young's double slit experiment. In Young's
double slit experiment, two identical narow
slits51and52are placed symmetrically with
respect to narow slit 5 illuminated with
(10.101)wwwsnotesdrivescom

899. 10.102
monochromatic light
.Theinterference pattern
is obtained on an observation screen placed at
large distance D fromSIandSz·
The position of nth bright fringe from the centre
of screen is
DA
x=n-
n d
Theposition of nth dark fringe from the centre
ofthescreen is
DA
x' =(2n-1)-
n 2d
Fringe widthis the separation between two
successive bright or dark fringes and is given by
p=DA
d
9.Resultantamplitude and intensity of interfering
waves. Ifalanda2are the amplitudes andIIand
12are the intensities of two coherent waves
having phase difference cp,then their resultant
amplitude and intensity at the point of
superposition are given by
a =~ai+ ~+2al~coscp
andI=II+12+2 .~II12coscp
If amplitude of each wave is aoandintensity 10 '
then
I =2ka~(1+coscp)=210 (1+coscp)
=410 cos2 ~
2
Thetenn2 ~1112coscpis calledinterference term.
(t)When coscpremains constant with time,
the two sources are coherent. The intensity
will be maximum at points for which.
coscp=+1and minimum at points for
which coscp= -1.
(it)When coscpvaries continuously with time
so that its average value is zero over the
time interval of measurement, the resultant
intensity at all points will beII+12,No
interference fringes are observed. The
sources are incoherent.
10.Ratio of intensity at maxima and minima of an
interference pattern. Ifaland ~ are the ampli-
tudes of two interfering waves, then the ratio
between the intensities at maxima and minima
will be
PHYSICS-XII
a%,where r=-1.=-1.,is the amplitude ratio of
~ 12
two waves. IfWIandw2are the widths of the
two slits, then
11.Coherent sources. Two sources of light which
continuously.emit light waves of same
frequency (or wavelength) with a zero or
constant phase difference between them, are
called coherent sources. Two independent
sources of light cannot act as coherent sources,
they have to be derived from the same parent
source.
12.Conditions for substained interference:
(I)The two sources should continuously emit
waves of same frequency or wavelength.
(il)The two sources of light should be coherent.
(iil)The amplitudes of the interfering waves
should be equal.
(iv)The two sources should be narow.
(v)The interfering waves must travel nearly
along the same direction.
(vi)The sources should be monochromatic.
(viI)The interfering waves should be in the
same state of polarisation.
(viii) The distance between the two coherent
sources should be small and the distance
between the two sources and the screen
should be large.
13.Fresnel'sbiprism method. Here two coherent
sources are obtained from an incoherent source,
by refraction. A biprism is essentially a single
prism with an obtuse angle of 179°, but behaves
as a combination of two acute angled prisms
placed base to base, each with a refracting angle
1°
of about-.
2
14.Lloyd's single mirror method. In this method,
an illuminated slit and its reflected image serve
as two coherent sources. In contrast to Young's
double slit and Fresnel's biprism methods, here
the central fringeis dark.www;notesdrive;com

900. WAVE OPTICS(Competition Section)
15.Displacement of interference fringes.When a
thin transparent sheet of thicknesstand
refractive indexIlis inserted in the path of one
of the interfering beams, the extra path
difference introduced is
fip= Lengthtin transparent sheet
- Lengthtin air
or fip=Ilt-t=(Il-1)t
:. Net path difference for any point on the screen
xd
=v-(Il-I) t
For the
central point of the screen,
xd D
V-(Il-I)t=O orx="d(Il-I)t
Thus the shift in the central bright fringe and
hence shift of any other fringe is
fix=D(Il-1)t=l!(u-1)t
d A
16.Interference in thin films.A soap film or thin
film of oil spreadover-water shows beautiful
colours, when seen in the reflected light. This is
due to interference between light waves
reflected by the upper and lower surfaces of.
thin films, as shown in the figure below. The ray
reflected from the upper denser surface of thin
film suffers a phase change of 1tor path
difference of 1./2.
Reflected rays
Transmitted rays
Reflected system.The path difference between
the two consecutive rays reflected from the
upper and the lower surfaces of a thin film of
refractive indexIland thicknesstis given by
A
p=2.lltcosr-2"
A
For maximum intensity. 2Iltcosr= (2n+1)-
2
For minimum intensity.2Iltcosr=n'A..
10.103
Transmitted system.
For maximum intensity. 21ltcos r = nA
For minimum intensity.
A .
2Iltcosr=(2n+I)-, wheren=O,1,2,3 ....
2
17.Diffraction of light. The phenomenon of
bending of light around the comers of small
obstacles or apertures and their consequent
spreading into the regions of geometrical
shadow is called diffraction of light.
18.Diffraction at a single slit.A plane wave of
wavelength Aon passing through a narow slit
of widthasuffers diffraction producing a
central bright fringe (a =Oo~ flanked on both
sides by minima and maxima. The intensity of
secondary maxima decreases with the increase
in distance from the centre.
For nth minimum :
asinan=nA, n=1,2,3, ....
For nth secondary maximum:
asinan=(2n+I)!:,n=I,2,3, ....
2
Angular position of nth minimum,
a=nA
na
Distance of nth minimum from the centre of the
screen,
DA
x =n-
n a
Angular position of nth secondary maximum,
a'=(2n+I)!:.
n 2a
Distance of nth secondary maximum from the
centre of the screen,
DA
x~=(2n+I)-
2a
Width of a secondary maximum,
13= DA
a
Width of central maximum,
130=213= 2 DA
a
Angular spread of central maximum on either
side of the centre of the screen is
A
a=±-
awww5not~szriv~5yom

901. 10.104
Total angular spread of the central maximum is
2e=2,,-
a
For diffraction to
be more pronounced, the size
of the slit should be comparable to the wave-
length of light used.
19.Diffraction at a circular aperture.For diffraction
of light at a circular aperture of diametera,the
angular spread of central maximum is
e=1.22"-
a
If D is the distance at, which the effect is
observed, then
Linear spread,x=De
Areal spread,x2=(Det
20.Fresnel's distance. It is the distance at which the
diffraction spread of a beam becomes equal to
the size of the aperture. Ifais the width of the
aperture, then
a2
DF=-
"-
The ray optics is valid for a distance D<DF.
21.Diffraction grating. It is an arangement of a
very large number of very narow, equidistant
and parallel slits. The diffraction pattern has
the central principal maximum of maximum
intensity and a number of higher order intensity
maxima whose intensity decreases with the
increase ofn;the order of the spectrum. The
direction ofnthprincipal maximum is given by
(a+b)sinen=n,,-
wheren=0,1,2,3 ....
This equation is known asgrating law.Here
(a+b)is calledgrating element,whereais width
of each slit andbis the width of opaque space
between two consecutive slits.
22.Limit of resolution.The smallest linear or
angular separation between two point objects at
which they can be just resolved by an optical
instrument is called the limit of resolution of the
instrument.
23.Resolving power: It is the ability of an optical
instrument to resolve or separate the images of
two nearby point objects so that they can be
PHYSICS-XII
\
distinctly seen. It is equal to the reciprocal of the
limit of resolution of the optical instrument.
24.Diffraction as a limit on resolving power.All
optical instruments like lens, telescope, micro-
scope, etc. act as apertures. Light on passing·
through them undergoes diffraction. This puts
the limit on their resolving power.
25.Rayleigh's criterion for resolution. The images
of two point objects are just resolved when the
central maximum of the diffraction pattern of
one falls over the first minimum of the
diffraction pattern of the other.
26.Resolving power of a microscope.The resolving
power ofa microscope is defined as the
reciprocal of the smallest distancedbetween
two point objects at which they can be just
resolved when seen in the microscope.
. 12/lsine
R.P. of a mIcroscope=-= ~--
d "-
whereeis half the angle of cone of light from
each point object and/lis the refractive index of
the medium between the object and the objective.
The factor/lsineis callednumerical aperture
(N.A.).
27.Resolving power of a telescope.The resolving
power of a telescope is defined as the reciprocal
of the smallest angular separation'de'between
two distant objects whose images can be just
resolved by it.
1 D
R.P. of a telescope= - = --
de 122 "-
where D is the diameter of the telescope
objective and "-is the wavelength of light used.
28.Resolving power of the human eye. The human
eye can see two point objects distinctly if they
subtend at the eye, an angle equal to one minute
of arc. This angle is called the limit of resolution
of the eye. The reciprocal of this angle equals
the resolving power of the eye.
29.Polarisation of waves.A transverse wave in
which vibrations are present in all possible
directions, in a plane perpendicular to the direc-
tion of propagation, is said to be unpolarised. If
the vibrations of a wave are present in just one
direction in a plane perpendicular to thewww;notesdrive;com

902. WAVE OPTICS(Competition Section)
direction of propagation, the wave is said to be
polarised or plane polarised. The phenomenon
of restricting the oscillations of a wave to just
one direction in the transverse plane is called
polarisation.
30.Unpolarised light.Akind of light in which the
electric field vector takes all possible directions
in the transverse plane, rapidly and randomly,
during the time of measurement is called
unpolarised light. For example, the
light of the
sun, candle light, etc.
31.Plane polarised light. If the electric field vector
vibrates just in one direction perpendicular to
the direction of wave propagation, the light is
said to be linearly polarised. In a linearly
polarised wave, the vibrations at all points, at
all times, lie in the same plane, so it is also called
a plane polarised wave.
32.Polariser. Adevice that plane polarises the
unpolarised light passed through it is called a
polariser. For example, a tourmaline crystal,
nicol prism, polaroid, etc.
33.Law of Malus.This law states that when a beam
of completely plane polarised light is passed
through an analyser, the intensity'I'of the
transmitted light varies directly as the square of
the angle'9'between the transmission direc-
tions of polariser and analyser.
2
I=Iocose
where10is the maximum intensity of
transmitted light.
34.Plane of polarisation.The plane passing
through the direction of wave propagation and
perpendicular to the plane of vibration is called
the plane of polarisation.
35.Plane of vibration.The plane containing the
direction of vibration and the direction of wave
propagation is called the plane of vibration.
36.Brewster angle. The angle of incidence at which
a beam of unpolarised light falling on a trans-
parent surface is reflected as a beam of comp-
letely plane polarised light is called polarising
or Brewster angle. It is denoted byip'
37.Brewster law.This law states that the tangent of
the pol arising angle of incidence of a trans-
parent medium is equal to its refractive index.
Il=tanip
10.105
38.Nicol prism.It is an optical device based on the
phenomenon of double refraction which is used
for producing and analysing plane polarised
light. It consists of two pieces of calcite cut with
a 68° angle and stuck together with Canada
balsam.
39.Polaroids.These are thin commercial sheets
which make use of the property of selective
absorption (dichroism) to produce an intense
beam of plane polarised light. Polaroids are
used in sunglasses, camera filters, wind screens
and car head lights of motor cars to reduce glare
of light reflected from shiny surfaces, etc.
40.Optical activity. Substances which can rotate
the plane of polarisation of light are called
optically active substances while the pheno-
menon is called optical activity.
41.Specific rotation.It is the angle through which
the plane of polarisation rotates when plane
polarised light is passed through one decimetre
length of solution containing one gram of the
substance per cnr'. The measurement is done at
a given temperatureT,using sodium light (the
D-line).
Specific rotation
Observed angle of rotation in degrees
Length of the tube in decimetrex Grams of
substance in1em3of solution
Te
[s]D=-
Ixc
42.Doppler effect. It is the phenomenon of the I
apparent change in the frequency of light due to
the relative motion between the source and
observer. The apparent frequency v' is given by,
When source moves towards the observer,
velocityvis taken+veand when it moves away
from the observer,vis taken -ve.
43.Doppler shift. The apparent change in the
frequency of light due to Doppler effect is called
Doppler shift.
. v
(i)~v=±-v
c
(ii) ~A.=+~ .A..
cwww5not~szriv~5yom

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964. Dual Nature of Radiation and Matter
G
LIMPSES
1.Electron. Itis an elemeary particle having a
negative charge of 1.6x10-19C and mass
9.1x10-31kg.
Work function.The minimum amou of
energy required by an electron to just escape
from the metal surface is known as work
function of the metal. It is denoted by WOo
3.Electron emission. The phenomenon of emission
of electrons from a metal surface is called
electron emission. It is of the following types :
(I)Thermionic emission. Here electrons are
emitted from the metal surface with the
help of thermal energy.
(iz)Fieldorcold cathode emission. Electrons
are emitted from a metal surface by
subjecting it to a very high electric field.
(iil)Photoelectric emtsston. Electrons are
emitted from a metal surface with the help
of suitable electromagnetic radiations.
(iv)Secondary emission. Electrons are ejected
from a metal surface by striking fast
moving electrons over it.
4.Kinetic energy gained by an electron. When an
electron is accelerated from rest through a
poteial difference of Vvolts, the gain in its
kinetic energy is.
eV.=l mv2
-i >2
..
5.Electron volt (ey).It is the kinetic energy gained
byanelectron when it is accelerated through a
potential difference.of 1 volt.
1eV=1.6x10-19J,1MeV=1.6x10-13J
The work function of a metal is generally
measured in electron volt (eV).
6.Photons.According to Planck's quaum theory
of radiation, an electromagnetic wave travels in
the form of discrete packets of energy called
quaa. One quaum of light radiation is called
a photon. The main features of photons are as
follows:
(I)A photon travels with the speed of light.
(iz)The frequency of a photon does not change
as it travels from one medium to another.
(iil)The speed of a photon changes as it travels
through differe media due to the change
in its wavelength.
(iv)The rest mass of a photon is zeroi.e.,a
photon canot exist at rest.
(v)Energy of a photon, E=hv=hc .
A
(VI)Momeum of a photon, p=me=hv=!!..
eA
(viI)From Einstein's mass-energy relationship,
the equivale massmof a photon is given by
E =m?=hvorm= ~ .
7.Photoelectric effect. The phenomenon of
emission of electrons from a metal surface,
when electromagnetic radiations of sufficiely
high frequency are incide on it, is called
photoelectric effect. The photo (light)-generated
electrons are called photoelectrons.
Alkali metals like Li, Na, K, Ce show photo-
electric effect with visible light. Metals like Zn,
Cd,Mg respond to ultraviolet light.
Photoelectric effect involves the conversion of
light energy io electrical energy. It follows the
law of conservation of energy. It is an instan-
taneous process.
(11.61)wwwSnotesdriveScom

965. 11.62
8.Photoelectric curre. The curre
ntconstituted
byphotoelectrons is called photoelectric
curre. Itsvaluedepends on :
(I)the iensity oflight~
(ii)the poteial difference applied between
the two electrodes, and
(iil)the nature of the cathode material.
9.Cut off or stopping poteial. Itis the mini-
mumvalue of the negative poteial that must
be appl~ed to theanode of photo-cell tomake
thephotoelectric current zero. It is denoted by
Va'Itsvalue depends on :(i)the frequency of
incide light, and(il)the nature of the cathode
material. Fora given frequency of incident light,
itis independe of its iensity. The stopping
poteial isdirectly related to the kinetic energy
of the emittedelectrons.
1 ?
~ax=2'mV~ax=eVa
10.Threshold frequency.Theminimum value of
the frequency of incide radiation below which
thephotoelectric emission stops altogetheris
called threshold frequency. Itis denoted by VO'
and is a characteristic of the metal.
11.Laws of photoelectric effect. (I)For a given
metal and a radiation of fixed frequency, the
rate of emission of photoelectrons is
proportional to the intensity of incide
radiation. (i/)Forevery metal, there is a certain
minimum frequency below which no
photoelectrons are emitted, howsoever high is
theiensity of incident radiation. This
frequency iscalledthreshold frequency. (iii)For
the radiationof frequency higher than the
threshold frequency, the maximum kinetic
energy ofthe photoelectrons is directly
proportional tothe frequency of incide
radiation and is independe of the iensity of
incide radiation. (iv) The photoelectric
emissionis aninstantaneous process.
12.Failure of wave theory to explain photoelectric
effect. Theclassical wave theory of radiation
could not explain the main features of photo-
electric effect.Itspicture of coinuous absorption
ofenergy from radiation could not explain (I)the
independence of Kmax on intensity, (il)the existence
of threshold frequency v0and(iil)theinstant-
aneous nature of the phenomenon.
PHYSICS-XII
13.Einstein's theoryof photoelectric effect.
Einstein explained photoelectric effect withthe
help of Planck's quaum theory. When a
radiation offrequency v is incide on a metal .
surface, it is absorbed in the form of discrete
packets of energy called quaa or photons. A
part of energy hv of a photon is used in
removing the electron from the metal surface
andremainingenergy is used in giving kinetic
energy to the photoelectron. Einstein's
photoelectric equation is
1 2
~ax=2'mvmax =eVo =hv-Wo=h(v-Yo)
whereWois thework function ofthe metaland
Vois the threshold frequency.
All the experimeal observations can be
explained on the basis of Einstein's photo-
electric equation.
14.Compton scattering. It is the phenomenon of
increase in the wavelength of X-ray photons
which occurswhen these radiations are scattered
on striking an electron. Then difference in the
wavelength of scattered and incide photons is
calledCompton shift, which is given by
h
/).'k=- (1 -cos ~)
711l
where ~ is the angle of scattering'of the X-ray
photon and1110is the rest mass of the electron.
15.Photocell. It is an arrangeme which converts
lightenergy io electric energy. It works on the
principleofphotoelectriceffect. It is used in
cinematography for the reproduction ofsound.
Photo-cells are used to operate various corol
systems and in light measuring devices.
16.Dual nature of radiation. Light has dualnature.
It manifests itself as a wave in diffraction, inter-
ference, polarisation, etc.,while it shows
particle nature in photoelectric effect, Compton
scattering, etc.
17.Dual nature of matter.Asthere is complete
equivalence between matter (mass) and radia-
tion(energy) and the principle of symmetry is
always obeyed, de Broglie suggested that
moving particles like protons, neutrons,
electrons, etc.should beassociated with waves
known as deBroglie wavesand their wavelength
is called de Broglie wavelength. ThedeBrogliewww5notwsvr~vw5uom

966. DUAL NATURE OF RADIATION AND MATTER (Competition Section) 11.63
wavelength of a particle of massmmoving with
velocityvis given by
A=!!.=~
Pmv
wherehis the
Planck's consta. The de Broglie
wavelength is independe of the charge and
nature of the material particles. It has signi-
ficaly measurable values for sub-atomic
particles like electrons, protons,etc.,due to
their small masses. For macroscopic objects of
everyday life, the de-Broglie wavelength is
extremely small, quite beyond measureme.
18.Davisson and Germer experime. This electron
diffraction experime has verified and con-
firmed the wave-nature of electrons.
19.de Broglie wavelength of an electron. The
wavelength associated with an electron beam
accelerated through a poteial difference ofV
volts is given by
A=h= 1.227 nm
.J2meV..JV
20.Electron microscope.It is a device that exploits
the wave-nature of electrons to provide high
resolving power. It is used to investigate the
structural details of bacteria, viruses, etc. It has
proved to be a powerful tool of investigation for
research in science, technology, metallurgy,
industry, medicine, etc.
JEEAdvance ;
Multiple Choice Questions (a) I (b)
with one correct answer
p
1.In a photoelectric experime anode poteial is
q
plotted against plate curre.
T
I
V V
(e)
V
(a)Aand Bwill have same intensities while B and
Cwill have differe frequencies
(b)BandCwill have differe iensities whileA
and Bwill have differe frequencies
(e)Aand B will have differe iensities while B
andCwill have equal frequencies
(d)Band C will have equal iensities whileAand
Bwill have same frequencies. [lIT 2004]
2.Photoelectric effect experimes are performed
using three differe metal platesp. qand r having
work functions4>p=2D eV, 4>q=2.5eV and 4>,=3DeV,
respectively. Alight-beam coaining wavelengths of
550 nm, 450 nm and 350 nm with equal iensities
illuminates each of the plates. The correctI-Vgraph for
the experime is [lIT 2009]
q
p
V V
3.The maximum kinetic energy of photoelectrons
emitted from a surface when photons of energy6eV
fall on it is 4 eV. The stopping poteial, in volt, is
(a)2 (b) 4
(c)6 (d)10 [lIT1997]
4.Ametal surface is illuminated by light of two
differe wavelengths 248 nm and 310 nm. The
maximum speeds of the photoelectrons corresponding
to these wavelengths are ~ andUz,respectively. If the
ratio ~: Uz= 2 : 1 andhe= 1240eV nm, the work
function of the metal is nearly
(a)3.7 eV (b)3.2 eV
(e)2.8 eV (d)2.5 eV [JEEAdv. 14]wwwDnotesdriveDcom

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994. GUIDELINES T
ONCERT EXERCISES
12.1.Choose the corect alternative from thecluesgiven at
the end of eachstatement:
(a) The size of the atom in Thomson's model is....the
atomic size inRutherford's model.
-(much greater than,no different from,
much less than).
(b) In theground stateof electrons areinstable
equilibrium, while in electrons always
experience a net force.
(Thomson's model, Rutherford's model).
(c) A classical atom based on is doomed to collapse.
(Thomson's model, Rutherford's model)
(d) An atom has a nearlycontinuous massdistribution
in but has a highly non-uniform mass
distribution in .
(Thomson's model, Rutherford's model).
(e) The positively charged part oftheatompossesses
mostof the mass of the atom in .
(Rutherford's model, both the models).
Ans. (a)no different from
(b)Thomson's model, Rutherford's model
(c)Rutherford's model
(d)Thomson's model, Rutherford's model
(e)both the models
12.2.Suppose you are givenachance to repeat the
alpha-particle scattering experiment using a thin sheet of solid
hydrogen in place of the gold foil. (Hydrogen is a solid at
temperatures below14K).What results do you expect?
Ans.Hydrogen nuclei (or protons) are much lighter
than a-particles. So a-particles are not scattered by solid
hydrogen. They pass throughsolid hydrogen almost
undeflected from their paths.
12.3.What isthe shortest wavelength present inthe
Paschen series of spectral lines ?
Ans.For shortest wavelength of Paschen series,'1.=3,
11z=oo
or
: =R[3;-~ ]= ~
s
A._~_ 9
5-R - 1.097 x107
=8.2041x10-7 m=8204.1 Awww.notesdrive.com

995. 12.28
12.4.A diffe
renceof2.3 eV separates two energy levels in an
atom. Whatisthe frequency of radiationemitted when the atom
makesa transition from the upper level to thelowerlevel?
Ans.Here E = 2.3 eV = 2.3 x1.6x1O-19J
As E=hv
:.Frequency,
E2.3x1.6x10-19
v=-=-------..,---
h 6.6x10-34
=5.6x1014Hz.
12.5.Theground state energy of hydrogen atom is-13.6eV.
What are the kinetic and potential energies of the electron in his
state?
Ans.Total energy, E =- 13.6 eV
K.E.= - E= -(-13.6) =13.6 eV
P.E. =-2 K.E. = - 2x13.6 = -27.2 eV.
12.6.Ahydrogen atom initially in the ground level absorbs
a photon, which excites ittothen=4level. Determine the
wavelength and frequency ofphoton. [CBSE 00 14C]
Ans.Energy of an electron in the nth orbit of H-atom,
En= -1~6 eV
Energy in the ground(n= 1)level,
11= - 13~6 = -13.6eV
1
Energy in the fourth(n=4)level,
E4 =-13/ = - 0.85eV
4
tJ.E= E4 -11
=-0.85 -(- 13.6) = 12.75eV
= 12.75 x1.6x10-19J
hc
AstJ.E=hv =-
/..
..Wavelength,
/..=~= 6.63x10-34x3xl08m
tJ.E 12.75x1.6x10-19
= 0.975xl0-7 m = 975A
c 3x108
Frequency, v=-= 7
x0.975x10-
= 3.077" x1015Hz.
12.7(a)Using the Bohr's model calculate the speed of the
electronina hydrogen atominthen=l,.2and3levels.
(b) Calculate the orbital periodineach of these levels.
PHYSICS-XII
Ans.(a)Speed of the electron in Bohr's nth orbit is
21tke2 q
v=--=-
" nh n
Speedof the electron in Bohr's first(n= 1)orbit is
2rtke2
q=-
h
2x3.14x9x109x(1.6 x10-19)2
6.63x10-34
= 2.186x106ms-I
v2=q=1.093x106 ms"!
2
v3=3.= 0.729x106ms-I.
3
(b)Orbital period of electron in Bohr's firstorbit is
21t1.
11=-
q
2x3.14x0.53x10-10
---2-.-18-6-x-l~06~--s
= 1.52x10-I6S
As T"=n311
12=(2)3x1.52x10-16
= 12.16 x10-16 =1.22x10-I5 s
'13= (3)3x1.52x10-16
=41.04x10-16= 4.10x10-15s.
12.8.The radius of the innermost electron orbit of a
hydrogen atomis5.3x10-11m:What are the radii of the n=2
andn=3orbits? [CBSE 0 14C]
Ans.Here
1.=5.3x10-11m
As rn=ifl.
r2=22x5.3x10-11= 2.12x10-10 m
r3= 32x5.3x10-11= 4.77x10-10 m.
12.9.A12.75eVelectron beam isused tobombard gaseous
hydrogen at room temperature. What series of wavelengths will
be emitted?
Ans.Here tJ.E= 12.75eV
Energy of an electron in nth orbit of hydrogen atom is
En=-1~6 eV
In ground state, n=l,.
11= - 13~6= -13.6eV
1
Energy of an electron in the excited state after
absorbing a photon of 12.75 eV energy becomes
En= - 13.6+12.75 = - 0.85 eVwww4nottssrxvt4ro°

996. ATOMS
r?-= -13.6=-~
=16orn=4
En -0.85
Thus the electron gets excited ton=4 state.
Total number of wavelengths in emission spectrum
=n(n-1)=4x3=6
2 2
The possible emission lines are shown in Fig. 12.23.
4
3
- 0.85 eV
-1.51 eV+
1
2 -3.4 eV
-13.6eVn=l
Fig.12.23
Emitted wavelength,
he 6.6x10-34 x3x10B
Aif= --- = -------
Ei-Ef Ei-Ef
A_ 19.8x10-26
43- (_ 0.85+ 1.51)x1.6x10 19
=28.409x10-7m=28409A
A_ 19.8x10-26
42 - (_ 0.85 + 3.4)x1.6x10-19
=4.8529x10-7m=4852.9A
A= 19.8x10-26 =19.8x10-7
41 (_ 0.85 +13.6)x1.6x10-19 12.75x1.6
=0.9706x10-7m=970.6A
A _ 19.8x10-26 _ 19.8 x10-7
32- (_ 1.51 + 3.4) x1.6x10-19-1.89x1.6
=6.5476x10-7m=6547.6A
19.8x10-26
----m
Ei-Ef
19.8x10-7
0.66x1.6
19.8x10-7
2.55x1.6
19.8x10-26 19.8x10-7
A- ----
31- (_ 1.51 + 13.6) x1.6x10-19-12.09x1.6
=1.0236x10-7m=1023.6A
19.8x10-26 19.8x10-7
A- -----
21-(-3.4+ 13.6)x1.6xlO 19-10.2x1.6
=1.2132x10-7m=1213.2A
12.29
12.10.In accordance with the Bohr's model, findthe
quantum number that characterises the earth's revolution
aroundthe sun in an orbit of radius15x1011m with orbital
speed3 x104m/s.(Mass ofearth=6.0x1024kg)
Ans. According toBohr's quantisation condition of
angular momentum,
Angular momentum of the earth around thesun,
nh
mvr=-
21t
21tmvr
n=--
h
2x3.14x6.0x1024x1.5x1011x3x104
6.6x10-34
=2.57x1074•
12.11. Answer thefollowing questions, which helpyou
understand thedifference between Thomson's model and
Rutherford's model better :
(a)Isthe average angle of deflection ofa-particles by a thin
gold foil predicted byThomson's model much less, about
the same, or much greater thanthatpredicted by
Rutherford's model?
(b)Isthe probability of backward scattering (i.e., scattering
of a-particles at angles greater than90°)predicted by
Thomson's model much less, about the same, or much
greater than that predicted by Rutherford's model?
(c)Keeping other factors fixed, itisfound experimentally
that forsmallthickness i,the number of a-particles
scattered at moderate angles is proportional tot. What
cluedoesthislineardependence on tprovide ?
(d)Inwhich model isitcompletely wrong to ignore
multiple scattering for the calculation of average angle
ofscattering of a-particles bya thin foil?
Ans.(a)About the same. Thisisbecause we are
considering theaverage angle of deflection.
(b)Muchless,because there isno such massive
core (nucleus) in Thomson's model as in Rutherford's
model.
(c)Thissuggests thatscattering is predominantly due
to a single collision, because the chance of a single
collision increases linearly with the number of the target
atoms, and hence linearly with the thickness ofthe foil.
(d)In Thomson model, positive charge is distributed
uniformly in the atom. Sosingle collision causes very little
deflection. The observed average scattering angle can be
explained onlybyconsidering multiple scattering. Hence
it iswrong toignore multiple scattering in Thomson's
model.
12.12.The gravitational attraction between electron and
proton in a hydrogen atom isweaker thanthecoulumb
attraction by a factor of about10-40. An alternative way ofwww4nottssrxvt4ro°

997. 12.30
looking att
his factistoestimatethe radius ofthe first Bohr0lit
ofa hydrogen atom iftheelectron and proton were bouru by
graritational attraction. You will find the answerinteresting.
Ans.The radius ofthe first orbit in Bohr's model is
given by
If instead of electrostatic attraction between electron
and proton, we consider the atom bound by gravitational
Gmpme 2
force-?-- ,then the termkeshould be replaced by
Gmpme' The radius of the first Bohr orbit in a
gravitationally bound hydrogen atomwill be
h2
r,c--.,-----"
o -41t2Gm m 2
pe
4x9.87x6.67x10-11 x1.6725 x10-27 x(9.1 x10-31)2
=1.194x1029 m",,1.2 x1029m
This radius is much greater than the estimated size of
the wholeuniver e.
12.13.Obtain an expression for the frequency of radiation
emitted when a hydrogen atomde-excites from level n tolevel
(n-1).For large n,show that this frequency equals the classical
frequency of recclutic., ufthe electron in the orbit.
[CBSE Sample Paper 111
Ans.From Bohr's theory, the frequency v of the
radiation emitted when anelectron de-excites from level
"2to level '\ is given by
v=21t2meZ2e4[_1_ - _1_]
h3 ,\2 "22
Given '\= ": 1, "2=n
v=21t2meZ2e4[ __1__~]
.. h3 (n-l)2 r?
=21t2me z2e4 [~ -(n-1)2]
h3 (n_1)2 ~
21t2mk2Z2e4(2n -1)
h3 (n- 1)2n2
For largen,2n-1=:2nandn -1=:n,andfor hydrogen
Z=1
21t2mk2 e4 2n 41t2 mk2 e4
v= h3 x~ . ~=n3h3
Now in Bohr's model,
Velocity of electron in nthorbit =~
21tmr
and
di fth b' ~h2
ralUS0norIt= 2 2
41tmke
PHYSICS-XII
Thus orbital frequency of electron in nthorbit is
v 1 nh
v=-=--x--
c21tr21tr21tmr
which is same as obtained inequation(1).
Hence for large value of n,the classical frequency of
revolution of electron in nth orbit issame as that obtained
from Bohr's theory.
12.14.Classically, anelectroncan be in anyorbit around
thenucleus of an atom. Then whatdetermines thetypical
atomic size? Why isan atom not, say,thousand times bigger
thanitstypical size ?Thequestion had greatly puzzled Bohr
before hearived at his famous model of the atom that you have
learnt in the text. Tosimulate what he might well have done
before hisdiscovery, let us playas follows with the basic
constants of nature andseeifwe can get a quantity with the
dimensions of length thatisroughlyequal to the known size of
an atom ("" 10-10 m)
(a) Construct a quantity with the dimensions of length from
the fundamental constants e,me'andc.Determine itsnumerical
value.
(b) You will find that the lengthobtained in (a)ismany
orders of magnitude smaller than the atomic dimensions.
Further, itinvolves c.Butenergies ofatomsaremostly in
non-relativistic domain where c isnotexpected to play any role.
Thisiswhatmay have suggested Bohr todiscardcand loak: fur
'something else' to get the right atomic size. Now,thePlanck's
constant h had already made its appearance elsewhere. Bohr's
great insight lay in recognising thath,me'andewill yield the
right atomic size. Construct a quantity with the dimension of
length fromh,me'andeand confirm thatitsnumerical value
hasindeedthe correct order of magnitude.
An()Th .. ke2
.s.ae quantity ISm?or
...(1) Also,
e2 9x109x(1.6 x10-19)2
---,,= m
41tEomc2 9.1x10-31 x(3x108)2
=2.8xlO-1S m
This length is much smaller than the typical atomic
size (""10-10 m).www6not~szriv~6yom

998. ATOMS
(6.6x10-
34)2
= m
4x9.87x9x109x 9.1x1031x(1.6x10-19)2
=5.26x10-11m
::::.0.53x10-10mor0.53A
The length is ofthe order of atomic size (- 10-10 m).
12.15.Thetotal energy of an electron in the firstexcited
state ofthehydrogen atomisabout-3.4eV.
(a) What isthe kinetic energy of the electron in this
state?
(b) What isthe potential energy of theelectron in this
state?
(c)Which of the answers above would change ifthe
choice of the zero of potential energyischanged?
[CBSE D14C]
Ans.K.E. ofan electron innthorbit,
T=.!kZe2
2r2
P.E. ofan electron innthorbit,
V=_kZe2 =_2T
r
Total energy,
E=T+V=T-2T=-T
(a)Kinetic energy,
T=-E=-(-3.4)=3.4eV.
(b)Potential energy,
V=-2T=-2x3.4=-6.8 eV.
(c)Ifthezero of the potential energy is chosen
differently, kinetic energy does not change. Potential
energy and hence total energy will be affected.
12.16.IfBohr's quantisation postulate (angular momentum
=nhlZn)isabasic law of nature, itshould be equally valid for
or
the caseof planetary motion also. Why then do we never speak of
quantisation of orbits of planets around the sun?
12.31
Ans.Angular momenta associated with planetary
motion are incomparably large relative tohiZn,For
example, angular momentum of the earth in its orbital
motion is ofthe order to 1070 hi 2n. In terms of the
Bohr's quantisation postulate, this corresponds to a
very large value of n(ofthe order of 1070).For such large
values ofn,the differences in the successive energies
and angular momenta of the quantised levels of the
Bohr model are so small compared to the energies and
angular momenta respectively of the levelsthat one
can, for all practical purposes, consider thelevels
continuous.
12.17.Obtain the first Bohr's radius and the ground state
energy of a'muonichydrogen atom' (i.e.,an atomin which a,
negatively charged muon (fl-)of mass about207meorbits
around a proton).
Ans.In Bohr's model, the radius of nth orbitis
1
roc-
m
Now in a muonic hydrogen atom, a negatively
charged muon (fl-)ofmass207merevolves around a
proton,
Therefore, wecanwrite
If.=me=~
re mil207me
r"=_1_xr=_1_x0.53xlO-10m
r-207e207
=2.5x10-13m
Energy of electron in nth orbit,
2n2mez2e4
E=- ~h2.
"
When allother factors are fixed, Eocm
11..~ 207m,
-=-'-!::.....=---
Eeme me
11..=207Ee=-207 x13.6eV
::::.-2.8 keY.www4nottssrxvt4ro°

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1061. NUCLEI
GI DELINESTONCERT
EXERCISES
You may find the following data useful in solving the exerces:
e =
1.6xlO-19C
1/ (4m:o)=
9x109 Nm2/C
1 MeV =
1.6x10-13J
1 year = 3.154x107s
mH = 1.007825 amu
13.1.(a) Two stable otopes of lithium~Liand~Lihave
respective abundances of 7.5% and92.5%. Theseotopes have
masses 6.01512 amu and 7.01600 amu respectively. Find the
atomic weight oflithium.
(b)Boron has two stable otopes,1~Band1~B.Their
respective masses are 10.01294 amu and 11.00931amu,and the
atomic weight of boron is10.811 amu. Find the abundances of
1~Band1~B.
Ans.The atomic weight of lithium is
m(Li)=7.5x6.01512 +92.5x7.01600
100
45.1134+648.98 694.0934
100 100
:::::6.941 amu.
(b)Suppose the natural boron contains x%of1~B
isotope and(100-x)%of I~B isotope. Then,
Atomic mass ofnatural boron
=Weighted average of themasses
oftwo isotopes
xx10.01294+(100-x)x11.00931
10.811 =---,---.-----'------'----
100
1081.1 =-0.99637x+1100.931
x=19.831 =19.9
0.99637
:.Relative abundance of 1~Bisotope =19.9%.
Relative abundance of 1~Bisotope =80.1%.
13.2.Thethree stable otopes of neon: NiO, Ne21, Ne22have
respective abundances of90.51%, 0.27% and9.22%. The atomic
masses ofthethree otopes are 19.99amu, 20.99 amu and
21.99amu respective/yo Obtainthe average atomic mass of neon.
Ans.The average atomic mass of neon is
or
or
90.51x19.99+0.27x20.99+9.22x21.99
m(Ne)=--------1-00--------
1736.89+5.67+202.75
100
-1945.31 _ 20 18
- -. amu.
100
13.57
m(~He) = 4.002603 amu
N = 6.023x1023per mole
k =
1.381x10-23JK-1
1amu =
931.5 MeV
mn = 1.008665 amu
13.3.Obtain the binding energy of a nitrogen nucleus(IiN)
from the following data:
mH=1.00783amu ;mil=1.00867amu ;mN=14.00307 amu.
Give your answer in MeV.
Ans.TheliNnucleus contains 7 protons and
7 neutrons.
Mass of 7 protons =7x1.00783=7.05481 amu
Mass of 7neutrons =7x1.00867=7.06069 amu
Total mass =14.11550amu
Mass ofliNnucleus =14.00307 amu
Mass defect, Sm=0.11243 amu
B.E.of nitrogen nucleus
=0.11243x931.5 =104.7MeV.
13.4.Obtain the binding energy of the nuclei ~ Fe and
~~ Biinunits of MeV from the following data:
mH=1.007825 amu
mil=1.008665amu
m( ~Fe)=55.934939 amu
m(2~Bi)=208.980388 amu
1amu=931.5 MeV
Which nucleus has greater binding energy per nucleon ?
Ans. The ~Fe nucleus contains 26 protons and
30 neutrons.
Mass of 26 protons
=26x1.007825=26.203450 amu
Mass of 30 neutrons .
=30x1.008665=30.259950 amu
=56.463400 amu
=55.934939 amu
Sm=0.528461amu
Total mass
Mass of~Fe nucleus
Mass defect,
B.E.of~Fe nucleus
=Smx931.5 MeV =0.528461 x931.5
=492.26 MeVwww7notesdrive7com

1062. 13.58
56 492.26
B.E
./nucleon of 26Fe =---s6= 8.79 MeV.
Now, the 2~ Bi nucleus contains 83 protons and
126 neutrons.
Mass of 83 protons
= 83x1.007825 = 83.649475 arnu
Mass of 126 neutrons
= 126x1.008665 = 127.091790 arnu
= 210.741265 amu
= 208.980388 amu
Total mass
Mass of 2: Bi nucleus
Mass defect,
B.E. of 2:Bi nucleus
Jim= 1.760877 arnu
= 1.760877 x931.5
=1640.3 MeV
B.E./nucleon of 2%Bi = 1640.3 = 7.85 MeV
209
Clearly, ~Fe has a greater B.E. per nucleon. In fact, it
is the maximum value. or
13.5.Agiven coin has a mass of3.0 g. Calculate the nuclear
energy that would be required to separate all the neutrons and
protons from each other. For simplicity assume that the coinis
entirely made of ~Cu atoms (of mass62.92960amu~ The
masses of proton and neutron are1.00783amu and
1.00867 amu, respectively.
Ans.The ~Cu nucleus contains 29 protons and
34 neutrons.
Mass of 29 protons = 29x1.00783 = 29.22707 arnu
Mass of 34 protons = 34x1.00867 = 34.29478 amu
Total mass = 63.52185 amu
Mass of ~Cu nucleus = 62.92960 amu
Mass defect, Jim= 0.59225 arnu
B.E. of ~Cu nucleus
= 0.59225x931.5 MeV = 551.5032 MeV
Number of atoms in 63 g of Cu
= Avogadro's number = 6.023x1023
:.Number of atoms in 3 g of Cu
6.023x1023x3 = 2.868x1022
63
Energy required to separate all the neutrons and
protons from each other of 3 g copper coin
= 551.5032 x2.868x1022= 1.582x1025MeV.
13.6.Write nuclear reactn equatns for
(i)a-decay of ~ Ra (i) a-decay of 2~Pu
(iii)rr-decay of:~P (iv)p--decay of2~Bi
(v)p+-decay of1~C (vi)p+-decay of ~ Tc
(vi) Electron capture of1~Xe
PHYSICS-XII
Ans.
(i)
(i)
(ii)
(iv)
(v)
(vt)
(vii)
226Ra ~ 222Rn +4He
88 86 2
242pu ~ 238U +4He
94 92 2
32p~ 32S+e- +V
15 16
210B~ 21Opo +«:+V
83 84
llC~ llB+e+ +v
6 5
97Tc~ 97Mo +e++v
43 42
120Xe+e+ ~ 1201+v
54 53
13.7.A radactive otope has a half-life of T years. How
long will it take the activity to reduce to (a)3.125%, (b)1%of
its original value?
Ans.(a)~ =.!i = 3.125 =J..
RoNo100 32
~=or=ororn=5
t=nT=5Tyears.
R N 1
(b)Ro=No= 100
Required time,
t= 2.303 logNo= 2.303Tlog 100
A. N 0.693
2.303x2xT
--0-.6-9-3 - '" 6.65 Tyears.
13.8.The normal activity of living carbon-containing
matterisfound to be about15decays per minute for every gram
of carbon. Th activity ares from the small proportnoj
radactive C14 present with the ordinary carbon otopeC1.
When the organm dead, its interactn with the atmosphere
(which maintains the above equilibrium activity) ceases andits
activity begins to drop. From the known half life(=5730years)
of C14, and the measured activity, the age of the specimen can be
approximately estimated. Thisthe principle ofct4dating
used in archaeology. Suppose a specimen from Mohenjodaro
gives an activity of9decays per minute per gram of carbon.
Estimate the approximate age of the Indus-Valley civilatn.
Ans.Given normal activity,
Ro= 15 decays min-1
Present activity,
R= 9 decays min -1,
1J.{2= 5730 years
Since activity is proportional to the number of
radioactive atoms, therefore,
N -At
R N 0e -At
-=-=---=e
RoNo No
9 -At At15
or -=eore=-
15 9www3yz•p°o~tvp3nzx

1063. NUCLEI
Taking natural
logarithms,
logeAt=log 15
e e9
5
or Atlogee=2.303log 10-=2.303x0.2218
3
t=_0._51_09_
A
'T' _0.693
11/2-A
t_0.5109 _0.5109 'T'
- - x11/2
0.693/ 11/2 0.693
0.5109x5730
--0-.-69-3-- years=4224years.
13.9.Obtain the amount of~Co necessary toprovide a
radactive source of 8.0mCi strength. The half-life of ~Cois
5.3years.
Ans.HereR=8.0 mCi
=8.0x10-3x3.7xio"dis s-l
=29.6x107dis s-l
11/2=5.3 years =5.3x3.16x107s
R = AN =0.693 .N
11/2
R11/2 29.6x107x5.3x3.16x107
N=-- =----------
0.693 0.693
=7.15xUy6atoms
As 60 g of cobalt contains 6.023 x1023atoms, so the
amount necessary to obtain a source of the required
strength
or [.: loge e=1]
As
But
60x7.15x1016 -6
= 23=7.123x10g.
6.023x10
13.10.The half-life of : Sris28years. Whatisthe
dintegratn rate of 15mg of th otope?
Ans.Here 11/2=28 years=28x3.154x107s
m=15 mg=0.015 g,M =90
Number of atoms in 0.015 gsample of ~Sr,
N=.!!!..xAvogadro's number
M
0.015x6.023x1023atoms
90
Activity ofthe sample,
0.693 0.693 x0.015x6.023 x1023
R= AN=--. N= 7
11/2 28 x3.154x10x90
=7.877x101°disintegrations/second
=7.877x1010'Bq
7.877x101°. .
= d0o=2.13ci,
3.7x1
13.59
13.11.Obtain approximately the rat of the nuclear radi of
thegold otope 1~Au and the silver otope1~Au. Whatisthe
approximate rat of their nuclear mass densities ?
Ans.AsR = Ro Al/3, whereRo=1.1x1O-15m
R(197Au) (197)1/3_
.. Re07Ag) =107 -1.23
Since the nuclear mass density is independent of the
size of the nucleus, so
Pnu(Au)=1
Pnu(Ag)
13.12.Find the Q-value and the kinetic energy of the
emitted a.-particle in the a.-decay of
(a)z:Ra (b) ~ Rn.
Given m(Z:Ra)=226.02540 amu,
m(~Rn)=222.01750 amu,
m ( ~ Rn)=220.01137amu,
me~ Po)=216.00189amu.
Ans. (a)22~Ra ~ ~Rn +~He+Q
Q=[m (22~Ra ) -m (~Rn)-m (~He )] c2
=[226.02540 - 222.01750 - 4.00260] x931.5 MeV
=0.0053x931.5 =4.937MeV
A-4
I\.=j\Q
226-4
=-- x4.937=4.85MeV.
226
(b)2~Rn~ 2;!PO+~He+Q
Q=[m (22~Rn ) -m (2~po ) -m (~He )] c2
=[220.01137 - 216.00189 - 4.00260]x931.5 MeV
=0.00688x931.5
=6.41MeV
A-4
Ka=j\Q
220-4
= -- x6.41=6.29MeV.
220
13.13.Theradionuclide 11Cdecays accordingto
llC ~ llB+e++v:1112=203min.
The maximum energy of the emitted positronis0.960MeV.
,
Given the massvalues:
me~C) =11.011434amu
me;B) =11.009305 amu
me=0.000548 amu
Calculate Qand compare itwith the maximum energy of the
positronemitted.www7notesdrive7com

1064. 13.60
Ans.I~C ~
I~B+e"+v+Q
where Q is the energy released in the decay process. It is
given by
Q = [mN(I~C) -mN(~B)-me]c2
To express Q-value in terms of atomic masses, wehave
tosubtract6mefrom the atomic mass ofcarbon and5me
from the atomic mass of boron to get the corresponding
nuclear masses. So we get
Q = [m (11C) -6me-m (11B)+5me- me]~
= [ me~C) -me~B) -2me]c2
= [11.011434 -11.009305 - 2x000548]amu x~
MeV
= 0.001033amu x931.5--
amu
= 0.9622 MeVoc0.96MeV.
Q=Ed+Ee+t;
The daughter nucleus is much heavier than e+andv,
so its energyEd=O.When the energy of neutrino is
minimum, the energy of position is maximum and Ee ::::.Q.
13.14.Thenucleus Ne23decays by l3--emsion. Write down
theI3--decay equation anddetermine themaximum kinetic
energy of the electrons emitted from the following data:
m(liNe)=22.994466amu,m(1213Na) =22.989770amu.
[CBSE D 08]
Ans.The I3--decayof ~Ne may be represented as
23N 23N'0 -Q
10e ~ 11 a+_Ie+v+
Ignoring the rest mass of neutrino, the expression for
the kinetic energy released may be written as
Q =[mN(~Ne)-mN(~fNa)-me]c2
=({mN(~gNe) +lOme} -{mN(~Na)+11mellc2
= [m@Ne)-m(~Na)]~
r. ~=931.5 MeV / amu]
= [22.994466 - 22.989770] x931.5 MeV
,
= 0.004696x931.5 MeV =4.374 MeV.
As 23Na is massive, thekinetic energy released is
mainly shared by electron-positron pair. When the
neutrino carries no energy, the electron has a maximum
kinetic energy equal to 4.374 MeV.
13.15.TheQvalue ofa nuclear reactn A +b ~C+d
defined by Q=[mA+mb-me- md] ~
where the masses refer to nuclear rest masses. Determine
from the given data whether the following reactns are
exothermic or endothermic. [CBSE OD 14C]
(i)1H+3H~2H+2H
11 1 1
(i) 12C+12C ~ 20Ne=t-4He
6 6 10 2
Atomic massesare given to be :
m(~H)=1.007825amu
m(~H) = 2.014102 amu
PHYSICS-XII
m(f H) = 3.016049 amu
me~H)=12.000000 amu
m(~gH)=19.992439 amu
m(iH)=4.002603 amu
Ans.(i)~H+fH~ ~H +~H
Q=[m(~H) + m(fH) - {m(~H) + m( ~H)}] ~
= [(1.097825+ 3.016049)- 2x2.014102] x931.5 MeV
=(4.023874-4.028204)x931.5 MeV
= -0.00433x931.5 MeV = -4.033 MeV
Negative value of Q indicates thatthe reaction is
endothermic.
(ii)12C+ 12C ~ 20Ne+ 4He
6 6 W 2
Q=[2me~C) - {m(~gNe) +m(~He)}] ~
=[2x12.000000-(19.992439+4.002603)]
x931.5 MeV
=(24-23.995042)x931.5 MeV
= 0.004958 x931.5MeV = 4.618MeV
Positive value of Q indicates that the reaction is
exothermic.
13.16.Suppose, we think of fission of a ~: Fenucleus into
two equal fragments, ~: AI.Isthe fsn energetically possible?
Argue byworking out Qoftheprocess.Given
m(~Fe)= 55.93494 amu and m(~: AI) = 27.98191 amu:
Ans.~Fe ~ ~:AI+~:AI +Q
Q=[m(~:Fe)-2m(~:AI)]c2
= [55.93494 -2x27.98191]x931.5 MeV
=-0.02888 x931.5 =-26.90 MeV
Asthe Q-value is negative, the fission isnot possible
energetically.
13.17.The fsion properties of ~: Pu are very similar to
thoseof~U.Theaverage energy released per fissn is180MeV.
Howmuchenergy, in Me V, isreleased ifall the atoms in1kg of
pure 2~~Pu undergo fission ?
Ans.Number of atoms present in 239g of2~~pu
= 6.023x1023
:.Number of atoms present in 1 kg or 1000 g of~~Pu
6.023x1023 x1000 24
------- =2.52x10
239
Energy released per fission = 180 MeV
Total energy released
=2.52x1024x180 MeV=4.54x1026Me V.
13.18.A1000 MWfissn reactor consumes halfofits fuel
in5.00y.Howmuch~~Udid it contain initially? Assume the
reactor operates 80%of the time,thatall the energy generated
arises from the fsn of 2~~Uand that th nuclide isconsumed
onlyby the fissn process.www4{}tr»q°vvr4p}z

1065. NUCLEI
Ans.Power of the
reactor, P=1000 MW=109W
Time of power generation,
t=5y=5x3.154x107s
Total energy generatedin 5y with 80%on-time
=80%ofPt=0.8x109x5x3.154x107J
Energy generated in each fission of ~ U
=200 MeV=200x1.6x1O-13J
Numberofatoms in 235gof~U=6x1023
6x1023
235
Number of atoms in 1g of ~~U
Energy generated per gram of ~~U
200x1.6x10-13x6x1023 1
235 Js
The amount ofZ:{Uconsumed in 5y with 80% on-time
Total energy generated
Energy generated per gram
0.8x109x5x3.154x107 x235
200x1.6x10-13x6x1023g
=1544x106g=1544 kg.
Thisamountishalfthefuel taken initially.
:.Mass of~Utaken initially = 3088kg.
13.19.Howlonganelectric lamp of100 Wcan bekept
glowing by fusn of 2.0 kg of deuterium ?The fusn reactn
can be taken as
iH+iH~ ~He+n+32MeV.
Ans.Number of atoms present in2 gofdeuterium
=6x1023
Number of atoms presentin 2.0 kg or 2000 g of
deuterium
Ox1023x2000=6x1026
2
Energy released in the fusion of 2 deuterium atoms
=3.2MeV
Total energy released in the fusion of 2.0 kg of
deuterium atoms
=3.2x6 x 1026 =9.6x1026MeV
2
=9.6x1026x1.6x1O-13J
=15.34x1013J
Energy consumed by the bulb persecond
=100J
Time for which the bulbwill ;;low
15.34xHy3 15.34xHy1
100 s =3.15x107years
=4.9x104years.
13.61
13.20.Calculate theheight of the potential barrier for a head
on collisn of two deuterons. (Hint. Theheight of the potential
barrier isgiven by theCoulomb repulsion between the two
deuterons when they just touch each other. Assume that they
can betaken ashardspheres ofradius 2.0fm.)
Ans.Charge on each deuteron,
e= 1.6x1O-19C
Radius of deuteron,
R=2.0 =2.0x10- 15m
The Coulomb barrier is given by
1e2
U=--.-
47t Eo 2R
9x109x(1.6x10-19)2
2x2x10-15 J
=9x109x{1.6x10-19)2 keV =360 keY.
4x10-15x 1.6x10-16
13.21. From the relatn R=RoA1I3, whereRoisa
constant and Aisthemass number of a nucleus, show that the
nuclear matter density isnearlyconstant(i.e.,independent
ofA).
Ans.Refer answer toQ.6 on page 13.4.
13.22.For the p+(positron) emissn from anucleus, there
isanother competing process known as electron capture i.e.,
electronfromaninner orbit(say the K -Shell) iscaptured by the
nucleus and a neutrino isemitted:
-AX Ay
e+ z~ Z-1 +v
ShowthatifP+emissn isenergetically allowed, electron
capture isnecessarily allowed but not vice versa.
Ans.Consider the two competing processes:
Positronemissn:
1X~ z~y+e++v+Q1
Electron capture:
e-+1X~ z_~y+v+Q2
The energy changes in the two processes are :
Q1=[mN(1X)-mN(z-1Y) - me] ~
=[m{~X) -2me -m{z_~Y)+ (2-l)m. - me] ~
=[m(1X) -m{z_~Y) - 2me] ~
Q2=[mN(1X)+me- mN (Z:1Y)] ~
=[m{1X)-m(z_~Y)]~
This meansQ1>0 impliesQz>0 butQ2>0 does not
necessarily imply Q1>O.Thus if positron emission is
energetically allowed, electron captureis necessarily
allowed but not vice versa.
13.23.In aperiodic table,theaverageatomic massofmagnesium
isgivenas24.312u. Theaverage value isbasedon their relativewww7notesdrive7com

1066. 13.62
natur
al abundance on Earth. The three isotopes and their
masses areiiMg(23.98504u), ~ Mg(24.98564 u),i~Mg
(25.98259 u).The natural abundance ofiiMgis78.99"/0 by
mass. Calculate the abundances of the other two otopes.
Ans.The abundance of iiMg (23.98504 u) is 78.99%.
Then abundance of ~Mg (24.98564 u) is
[100 -(x+ 78.99 u)flo, wherex%is the abundance of
i~Mg (25.98259).
Average atomic mass of Mg
= Weighted average ofmasses of isotopes
78.99x23.98504+[100-(x+ 78.99)]
24.312 = x_2_4_.9_85_64_+_x--=['--25_.9_8_2_59~]
100
2431.2 = 1894.5783 +2498.564 + xx0.99615- 1973.6157
4404.8157 - 4393.1423
or x=--------
0.99695
= 11.6734 = 11.71
0.99695
.. x= 11.71%
Abundance of i~Mg = 11.71%
Abundance of ~Mg = 100-(11.71+ 78.99) =9.3%.
13.24. The neutronseparatn energyisdefined to be theor
energy required to remove a neutron from a nucleus. Obtain the
neutronseparatnenergies of the nuclei ~~Ca andi~Al from
the following data:
mn= 1.008665amu;
m(~Ca)=39.962591 amu;
m(~~Ca)=40.962278 amu;
m(i~AI)=25.986895amu;
m(i~AI)=26.981541amu.
Ans.Neutron separation energySnof anucleus ~Xis
given by or
Sn=[mN(A-iX) +»; -mN(~X)] c?
=[{mN(A-iX)+ Zme} +=; -(mN(~X)+ Zme}] c?
= [m(A-iX)+mn - m(~X)]c?
Here mNrefers tonuclear mass while mrefers to or
atomic mass.
:.Neutron separation energy of ~~Ca is givenby
Sn(~~Ca) =[m(~Ca) +mn- m(~~Ca)]c?
=[39.962591+ 1.008665 - 40.962278] x931 MeV
.[By using c?= 931 MeV / amu]
= 0.008978 x931 MeV
=8.36MeV
PHYSICS-XII
Neutron separation energy ofi~AI is given by
Sn(i~AI)=[m(i~AI) +mn- m(i~AI)] c2
= [25.986895 + 1.008665 - 26.981541]
x931 MeV
= 0.014019x931 MeV
=13.05 MeV.
13.25.Asource contains two phosphorous rad nuclides
i~P(1J./2 = 14.3 d) and~P(11/2 =25.3d} Initially, 10%of
the decays come fromilP.How long one must wait until 90%
doso?
Ans. Clearly, the source has initially 90%ofi~p
radionuclides and 10% of~pradionuclides. Finally, say
aftertdays, the source has 10% ofi;P radionuclides and
90%of~P radionuclides.
:.Initial number of i~Pnuclides = 9x
Initial number ofilpnuclides = x
Final number ofi~pnuclides =y
Final number of~Pnuclides =9y
1 1
As ~ =G)"=G)l/2=(2)-li/2
N=No (2)li/2
Forfirst isotope,
y=9x(2)
1
14.3
For second isotope,
or
9y=x(2)25.3
On dividing, we get
11(_1__1)
9=-(2) 14.3 25.3
9
111
81=214.3 x25.3
11t
log 81= log 2
14.3x25.3
1.9085 =11tx0.3010
14.3x25.3
1.9085x14.3x25.3 d
t= =208.5 ays.
11x0.3010
or
13.26. Undercertain circumstances, a nucleus can decay by
emitting a particle more massive than an aparticle. Consider
thefollowing decay process:
223Ra~ 209Pb+14C
88 82 6
223Ra ~ 219Rn +4He
88 86 2
Calculate theQ-values for these decays and determine that
both areenergetically allowed.www3yz•p°o~tvp3nzx

1067. NUCLEI
Ans.The Q-value
for the first decay process is
given by
Q=m(~Ra) -me~Pb) -me:q
=[223.01850 -208.981107 -14.0003241amu xr?
= 0.034109 x931.5MeV =31.85 MeV
For thesecond process,
Q=m(~Ra) - m(~Rn) - m(~He)
= [223.01850 - 219.00948 - 4.002601amu xr?
=0.00642x931.5 = 5.98 MeV
AstheQ-value is positive in both cases, soboth decay
processes are energetically possible.
13.27.Consider the fsion of 1~Uby fast neutrons. Inone
fsion event, no neutrons are emitted and thefinalstable end
products, after the beta-decay of the primary fragments, are
1:~Ceand:Ru Calculate Qfor this fsn process. The
relevant atomic and particle masses are
m(1~U) = 238.05079 amu
me~Ce)= 139.90543amu
m(:Ru)= 98.90594 amu
mn=1.00867 amu
Ans.The fission may be represented as
238U+In ~ 140Ce +99Ru+10°e+Q
92 1 58 44 -1
The Q-value forthe process is
Q=[mN(~U) +m" - mNe~Ce)
- mN (:Ru) - 10 melr?
In terms of atomic masses, we can write
Q=[(m(~~U) - 92me}+mn- {m(I~Ce) - 58me}
-{m(:Ru) - 44me}-10melc2
= [mei~U) +mn- me:~Ce) - m(:Ru)l r?
= [238.05079+1.00867-139.90543-98.905941
amuxr?
MeV
= (239.05946 - 238.81137) amu x931.5--
amu
= 0.24809x931.5 = 231.09 MeV
=231.1 MeV.
13.28.Consider theD-T reactn (deuterium-tritium
fusn) given by the equatn:
2H+3H ~4He+n
112
(a)Calculate the energy released inMeVin this reactn
from the data:
m(iH)= 2.014102amu
m(lH)=3.016049amu
m(i He)= 4.002603 amu
m"=1.00867 amu
13.63
(b) Consider the radius of both deuterium and tritium to be
approximately 2.0fm.What the kinetic energy neededto
overcome theCoulomb repulsn?To what temperature must
the gases be heated toinitiate the reactn?
Ans. (a)The net reaction is
2H+3H~ 4He+1n+Q
1 1 2 0
:.Energy released in the reaction is
Q=[mN(iH)+mN(lH)-mN(~He)-mnlr?
Adding and subtracting 2me'we get
Q =[{mN(iH)+me}+{mN(lH+me)}
-{mN(~He)+2me} - mnlr?
Q = [m(iH) +maH) - m(~He) -mnlr?
= [2.014102 +3.016049
- 4.002603 - 1.00867] x931.5 MeV
= 0.018883 x 931 MeV =17.58 MeV.
(b)Distance betweenthe nuclei when they almost
touch other is
d=2r= 2x1.5x10-15m = 3 x1O-15m
:.Repulsive potential energy of the two nuclei when
they almost touch other
_1_ M2
4nso' d
9x109x(1.6x10-19)2
3x1015 J
= 7.68x10-14J [.:'It=q2= 1.6x1O-19q
K.E.required for one fusion event
=Average thermal K.E. available with
the interacting particles
3
=2x-kT=2kT
2
2kT= 7.68x1O-14J
T= 7.68xlO-14J =1.85x109K.
2x1.38x10 23JK 1
13.29.Obtain the maximum kinetic energy of (3-particles
and the radiatn frequencies correspondingtoy-decays in the
decay scheme shown in Fig. 13.21.You are given that,
m(AJ98) = 197.968233amu
m(HgI98) = 197.966760 amu
or
:"'-r-"""t:"'-1.088 MeV
~~...,..-"-- 0.412MeV
[CBSE003C, 08C]Fig. 13.21
'---"---0www7notesdrive7com

1068. 13.64
Ans.The frequencies
ofy-radiation will be equal to the
corresponding energy diferences divided by Planck's
constant h.
v=Ez-E;.
h
. (1.088 - 0) x1.6x10-13 20
••V (y1) = 34 = 2.627 x10 Hz
6.63x10-
(0.412 - 0)x1.6x10-13 19
v(Y2)= 34 =9.949xl0 Hz
6.63x10
v(Y3)= (1.088-0.412) xl:x10-I3 =1.632x1019Hz
6.63x10
TheP1--decay can be represented as
1~~Au ~ l':Hg +_~e+Q(P1- ) +Q(Y1)
whereQ(Y1)= 1.088 MeV.
:. Maximum kinetic energy ofPIparticle is
~ax(~)=[me~AU) -{m(l':Hg) + 1.088}] ~
931.5
[Neglecting the rest mass ofp-particle]
= [197.968233 - (197.966760 +1.088)] x931.5 MeV
931.5
.,' - =Q31.5 MeV / amu]
= (0.001473 x931.5 -1.088) MeV
= (1.372 -1.088) MeV =0.284 MeV
TheP;:decay canbe represented as
l~~Au ~ 19:0Hg + _~e+ Q(P;) +Q(Y2)
whereQ(Y2)= 0.412 MeV.
.,Maximum kinetic energy of P;:particle is
J((P-)=[m(198Au)_{m(198Hg) + 0.412 }]~
"max2 79 80 931.5
=[197.068233-(197.966760+ ~':~~)] x 931.5 MeV
=[0.001473 x 931.5 -0.412]MeV
= [1.372 - 0.412] MeV = 0.960 MeV.
13.30.Calculate and compare the energy released by (a)
fusn of1.0kg of hydrogen deep within the Sun and(b)the
fsn of1.0kg of235Uina fsn reactor.
Ans.(a)In the sun, 4 hydrogen nuclei combine to form
a helium nucleus with the release of 26 MeV of energy.
Thenet fusion reaction is
4~H~ ~He +2e++ 26 MeV
Number of atoms in 1 g ofiH = 6x1023
Number of atoms in 1 kg or 1000 g ofiH
= 6x1023x1000 = 6x1026
The energy released by 1 kg of hydrogen
26x6x1026
---- =39x1026MeV.
4
PHYSICS-XII
(b)Number of atoms in 235 g of235U= 6x1023
Number of atoms in 1 kg or 1000 g of235U
6x1023 x1000
235
Energy released in per fission of235U= 200 MeV
Energy released infission of 1 kg of 235U
6x1026x200 26
-----=5.1xl0 MeV.
235
Thus the energy released in fusion of 1 kg of hydrogen
fusion is about 8 times that of energy released in fission of
1 kg 235u.
13. 1Suppose India has a target of producing by 2020
A.D.,200,000 MWofelectric power, 10percent ofwhich isto
beobtained from nuclear power plants. Supposewe are given
that,on average theefficiency of utilatn(.e.,conversn to
electric energy) of thermal energy produced in a reactoris25%.
How much amount of fsnable uranium would ourcountry
need per year at the turn ofth century? Take the heat energy
per fsn ofU235to be about 200 MeV. Avogadro's Number
N= 6.023x1023mol-I.
Target of power by 2020 A.D.
= 2x105MW=2x1011W
Power required from nuclear power plants
= 10% of 2x1011W
=~x2xl(P=2xlOlOW
100
Energy required from nuclearpower plants per
year
= Powerxtime
= 2xUyOx365.25x24x60x60J
= 6.312xio"J
Energy released per fission = 200 MeV
Available electrical energy per fission of235Unucleus
= 25% of 200 MeV
= ~x200 MeV = 25x2x1.6x10-13J
100
= 8x1O-12J [.,'1 MeV=1.6 x 10 13Jl
Number of235Ufissions required per year
6.312x1017 28
= 12 = 7.89x10
8x10-
Required number of moles of 235U
7.89x1028 7.89x1028
= Avogadro's number =6.023x1023
= 13.1x104
Mass of 235U required
= Number of molesxmass number
= 13.1 x104x235 g
= 3078.5x104g'" 3.078 x104kg.www3yz•p°o~tvp3nzx

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1170. GIDELINESTONCERT EXERCISES
14.1In an n-type silicon
,which of the following statement
istrue:
(a) Electrons are majority carriers and trivalent atoms
are the dopants.
(b) Electrons are minority carriers and pentavalent
atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms
are the dopants.
(d) Holes are majoritycarriers and trivalent atoms are
the dopants.
Ans.(c)In ann-typesilicon, holes are minority carriers
and pentavalent atoms are the dopants.
14.2.Which of the statements given in Exercise14.1istrue
for p-type semiconductors?
Ans.(d)In ap-typesemiconductor, holes are majority
carriers and trivalent atoms are the dopants.
14.3.Carbon, silicon and germanium have four valence
electrons each. These are characterised by valence and
conduction bands separated by energy band gap respectively
equal to (Eg)c' (Eg)Si and (Eg)Ge' Which of the following
statements istrue? [AIIMS 13]
(a)(Eg)Si «Egb «Eg)c
(b) (Eg)c «Egb >(Eg)Si'
(c)(Eg)c >(Eg)Si >(Egb·
(d) (Eg)c =(Eg)Si =(Egb.
Ans.(c)The energy band gap is largest for carbon, less
for silicon and least for germanium.
14.4.In an unbiased p-n junction, holes diffuse from the
p-region to n-region because
(a) freeelectrons in the n-region attract them.
(b) they move across the junction by the potential
difference.
(c) hole concentration in p-type regionismore as
compared to n-region.
(d) All the above.
Ans.(c)In an unbiasedp-njunction, holes diffuse from
the p-region to n-region because hole concentration in
p-region is more as compared to n-region.
14.5.When a forward biasisapplied to ap-njunction,it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lawers the potential barrier.
(d) none of the above.
Ans. (c)When a forward bias is applied to a p-n
junction, it opposes the potential barrier and hence
lowers it.
14.6.For transistor action, which of the follawing statements
are correct:
(a) Base, emitter and collector regions should have
similar size and doping concentrations.
(b) The base region must be very thin and lightly
doped.
(c) The emitter junctionisforward biased and collector
junctionisreverse biased.
(d) Both the emitter junction as well as the collector
junction are forward biased.
Ans. (b)and(c)are correct.
14.7.For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b)ishigh at high and low frequencies and constant in
the middle frequency range.
(c)islaw at high and low frequencies and constant at
mid frequencies.
(d) none of the above.
Ans.(c)The voltage gain is low at high and low
frequencies and constant in the middle frequency range.www.notesdrive.com

1171. 14.88
14.8.In half-wave rectification, whatisthe
output frequency
if the input frequencyis50Hz.Whatisthe output frequency of
a full-wave rectifier for the same input frequency?
Ans.Input frequency=50 Hz
In half-wave rectification, only one ripple is obtained
per cycle of the output.
:.Output frequency of a half-wave rectifier
=Input frequency=50Hz
In full wave rectification, two ripples are obtained per
cycle of the output.
:.Output frequency of full wave rectifier
=2xInput frequency=2x50=100Hz.
14.9.For a CE-transistor amplifier, the audio signal voltage
across the collector resistance of2ko.is 2V. Suppose the current
amplification factor of the transistoris100,find the input
signal voltage and base current, if the base resistanceis1ko..
Ans. HereRc=2kn=20000., RB=1kn=10000.,
~ =100,Vo=2 V
(i)Voltage gain,
Vo=~Ra
V; Ri
or
~=100x 2000
V; 1000
..Input signal voltage,Vi=0.01V.
(ii) ~=Ie=Vo /Rc .
IB IB
. . Base current,
Vo 2 5
I = - = = 10-A= 10j.lA.
B~Rc100 x 2000
14.10.Two amplifiers are connected one after the otherin
series (cascaded). The fir~ amplifier has a voltage gain of10and
the second has a voltage gain of20.If the input signalisomV,
calculate the output a.c. signal.
Amplifier 1 Amplifier 2
rlG111----1 G2 ~
Fig.14.200
Ans.Here ~= 10, G2 =20,Vi=0.01V,Vo=?
Total gain,
or
Vo=~G2Vi
=10x20 x 0.01 = 2 V.
PHYSICS-XII
14.11.A p-n photodiodeisfabricated from a semiconductor
with a bandgap of2.8eV. Can it detect a wavelength of
6000 nm? [CBSE 005]
Ans.Energy required to cross the band gap,
Eg=2.8eV
Wavelength of incident photon,
A.=6000nm=6 x 1O-6m
Energy of incident photon,
E=hc
A.
6.63x10-34 x3x108=0.207 eV
6x10-6
AsE<Eg,thep-njunction cannot detect the radiation
of wavelength6000nm.
14.12. The number of silicon atoms per m3 is5x1028. This
isdoped simultaneously with5x1022atoms perm3ofArsenic
and5x1020per m3 atoms of Indium. Calculate thenumber of
electrons and holes. Given that ni =15x1016m-3. Isthe
material n-type or p-type?
Ans.Here ND=5x1022m-3,
NA=5x1020m -3,ni=1.5x1016m -3.
For the semiconductor to remain electrically neutral,
Also
ND -NA=ne-'1z
nnh=rf
e ,
...(1)
(ne+'1z)2=(ne- '1z)2+2ne'1z
2 2
=(ND-NA) +4ni
ne+'1z=~(ND - NA)2 +4n/
...(2)
Adding equations(1)and(2),we get
1 I 2 2
ne=2.(ND-NA)+V( ND -NA)+4ni
=-.![(5x1022 -0.05x1022)
2
+~(4.95 x1022)2 +4x(1.5 x1016)2]
'"-.![4.95x1022+~(4.95 x1022)2]
2
=4.95x1022m-3
2
n
'1z=-'-
ne
(1.5x1016 )2
4.95x1022
2.25x1032
4.95x1022
=4.5x109m-3,
Asne>'1z,the material is ofn-type.www6notzsyrivz6xom

1172. SEMICONDUCTOR DEVICES AND DIGITAL CIRCUITS
14.13.In an intrinsicsemiconductor
,the energy gap Eg is
12eV.Its hole mobility is very much smaller than electron
mobility and is independent of temperature. What is the ratio
between conductivity at 600Kand that at300K?Assume that
thetemperature dependence of intrinsic carrierconcentration ni
is expressed as
ni=noexp [ ~:~]
where nois a constant and kB= 8.62 x10-5eVr1 is the
Boltzmann constant.
Ans.AsIle»Ilh'and,for an intrinsic semiconductor
ne=1\=ni
Conductivity is given by
0'=e (neIle+1\Ilh)
=eni(Ile+Ilh)::: enille
Here all the pre-exponential terms are assumed
independent of temperature. So we can put a constant
0'0=elleno
and express the conductivity as
0' = 0'0 exp [ ~:~ ]
NowEg= 1.2 = 0.6eV, kB=8.62x10-5eVK-1
2 2
..0' (600 K) = 0'.0 exp [ - O.~ ]
8.62x10-x600
0' (300 K) = 0'0 exp [ -O.~ ]
8.62x10-x300
Hence
[ - 0.6 ]
exp 8.62x10-5 x600
[ - 0.6 ]
exp 8.62x10-5 x300
[ 0.6(1 1)]
= exp 8.62 x105300 - 600
[0.6x105]
= exp 8.62x600
= exp(1l.6) = 1x105.
This shows that the conductivity of a semiconductor
increases rapidly with the rise in temperature.
0'(600 K)
0'(300 K)
14.89
14.14.In a p-n junction diode, the currentIcan be
expressed as1= 10exp (~ -IJ'where10is called the
2kBT
reverse saturationcurrent, V is the voltage across the diode
and is positive for forward bias and negative for reverse bias,
andIisthe current through the diode, kBis the Boltzmann
constant(8.6 x 10-5 eVIK) andTis the absolute temperature. If
for a given diode10 = 5x10-12A andT= 300K,then:
(a) What will be the forward current at a forward
voltage of0.6V?
(b) What will be the increase in the currentifthe
voltage across the diode is increased to0.7V?
(c)What is the dynamic resistance?
(d) What will be the currentifreverse bias voltage
changes from1V to2V?
Ans.The currentIthrough a junction diode is given as
where 10 = 5x10-12A,T= 300 K,
kB= 8.6 x10-5eVK-1
= 8.6 x10-5 x1.6x10-19JK-1.
(a)WhenV= 0.6 V
eV 1.6x10-19x0.6
kBT=8.6x1.6xl0 24 x 300
=~=23.26
8.6x3
= 5x10-12[exp (23.26) - 1] A
= 5x10-12[1.2586x1010- 1] A
= 5x10-12x1.2586x1010A
= 0.06293 A.
(b)WhenV= 0.7 V,
eV _1.6x10-19x0.7 _ 2713
- 24-.
kBT8.6x1.6x10
= 5x10-12[exp (27.13) - 1] A
= 5 x 10-12 [6.07 x 1011-1] A
:::5x10-12x6.07x1011 A = 3.035 A
Increase in current,
M= 3.035 - 0.06293 = 2.972 A.www=notesdrive=com

1173. 14.90
(c)For!:o
.V=0.7 - 0.6 =0.1 V, M=2.972 A
Dynamic resistance,
sv 0.1
rd=M=2.972=0.0336n.
(d)For both the voltages, the current Iwillbe almost
equalto10,showing almost infinite dynamic resistance in
the reverse bias.
1= - 10=-5x10-12 A.
PHYSICS-XII
14.17.Write thetruth table for a NAND gateconnected as
given in Fig. 14.203.Hence identify the exact logic operation
carried out by this circuit.
Ans. Refer to the solution of Example 51on
page14.60.
14.15.You are given the two circuits as shown in
Fig.14.201. Show that the circuit(a)actsasORgatewhile theFig. 14.203
circuit (b) acts as AND gate. [CBSE OD 11]
:-:I>-
(i)
y
(iz)
Fig. 14.201
Ans.Refer to the solution of Example 50on
page14.60.
14.16.Youare given twocircuits as shown in Fig.14.202,
which consist of NAND gates. Identify the logic operation
carried out by the two circuits. [CBSE D 09C, 11; OD 11]
(a)
_~A
Ae
r--- .....
y
(b)
Fig. 14.202
Ans.Refer to the solution of Example 52on
page14.60.
14.18. Writethetruth table for circuit given in Fig. 1'
below consisting of NOR gatesand identify thelogic opt:.
(OR,AND, NOT) which this circuit isperforming.
Ans. Refer to the solution of Example 48 on
page14.59.
Fig. 14.204
14.19.Write thetruth table forcircuit given in Fig.14.205
consisting of NOR gates only. Identify the logic operations
(OR,AND, NOT) performed by the twocircuits.
[CBSE D 09C]
(a)
A
B
(b)
Fig. 14.205
Ans. Refer to the solution of Example 49on
page14.59.www5notzsyr»vz5xom

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1240. 15.44
GUIDEL
INES TONCERT EXERCISES
15.1.Which of the following frequencies will be suitable for
beyond the horizon communication using sky waves?
(a)10 kHz (b)10 MHz
(c)1 GHz (d)1000 GHz
Ans.(b)10 MHz. Frequency of 10 kHz will require
very large radiating antenna while frequencies1GHz and
1000 GHz will penetrate the ionosphere and cannot be
reflected by it.
15.2.Frequencies intheUHF range normally propagate by
means of:
(a)Ground waves (b)Skywaves
(c)Surface waves (d)Space waves
Ans.(d)Space waves. Frequencies in the UHF(0.3to
3GHz) range normally propagate by means of space
waves. Their sky wave reflection from ionosphere is not
possible.
15.3.Digital signals (i)do not provide a continuous set of
values (ii)representvalues asdiscrete steps(iii)can utilize
binary system and(io)can utilize decimal as well as binary
systems. Which of the followingistrue?
(a)(i)and(ii)only
(b)(ii)and(iii)only
(c)(i), (ii)and(iii)but not(iv)
(d)All of(i),(ii),(iii)and(iv)
Ans.(c).Digital signals cannot utilize decimal system
which represents a continuous set of values.
15.4.Is itnecessary for a transmitting antenna to be at the
same height as that of the receiving antennafor line ofsight
communication?A TVtransmitting antenna is81m tall. How
much service area it can cover ifthereceiving antennaisat the
ground level? [CBSE F 15]
Ans.No, it is not necessary for a transmitting antenna
to be at the same height as that of the receiving antenna
for LOS communication. The heights should be such that
the receiving antenna is capable of directly intercepting
the signals radiated by the transmitting antenna.
Service area covered by the transmitting antenna,
2
A=7UiT =1t(2Rf7)
=22x2x6.4x106x81m2
7
=3258km2.
15.5.Acarrier wave of peak voltage12V is usedto transmit
a message signal. Whatshould be thepeakvoltage of the
modulating signal in order to have a modulation index of75% ?
[CBSEOD 10]
PHYSICS-XII
A
~=~
Ac
75
A=~A=- x12 V=9 V.
"m CIaO
15.6.Amodulating signal is a square wave as shown /;, u:
Ans.As
'"
~1
;>
.5
...--
1 2
tinecnds0s
Fig. 15.49
The carrier wave isgiven by c(t)=2sin(81tt)volts.
(i)Sketch the amplitude modulated waveform
(ii)Whatisthemodulation index?
Ans.(i)The amplitude modulated wave is shown
below:
3~~-r----------~--r-------~
-3
2
r
.l!l
'0
>
0
.5
\
-
'J-1
-2
o0.20.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time inseconds ~
Fig. 15.50
~ =~n=1 V=0.5.
Ac2 V
15.7.For an amplitude modulated wave, themaximum
amplitude is foundtobe10V while the minimum amplitude is
found to be2V. Determine the modulation index, ~. What
would be thevalueof ~ifthe minimum amplitude is zero volt?
(ii)
Ans. 11=~ax - ~in =10 -2=~ =~.
~ax +~in 10+2 12 3
If ~=0,then ~=~ax =1
~ax
This is true for all values of~ax'www7notesdrive7com

1241. COMMUNICA
TION SYSTEMS
15.8.Dueto economic reasons, only the upper sideband of
anAMwaveistransmitted, but atthe receiving station, there is
a facility for generating the carrier. Show thatifa device is
available which can multiply two signals, thenit ispossible to
recover the modulating signal at the receiver station.
Ans.Let, for simplicity, thereceived signalbe
~ cas(ooc+OOm)t
The carrier signal available at thereceiving stations is
Accas00/
15.45
Multiplying the two signals, we get
~ Accas(ooc+oom)teas00/
=~Ac[cos(200c+00 )t+cas00tj
2 n m
Ifthis signal ispassed through a low pass filter, we
can record the modulating signal ~AccasooJ.
2wwwCnotesdriveCcom

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