moving charges and magnetism class 12 pdf download

NCERT Solutions for Moving Charges and Magnetism Class 12
Class 12 Physics typically covers the topic of moving charges and magnetism, which is an essential part of electromagnetism.
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1. MOTION OF CHARGES AND MAGNETISM
Reading Materials
1. Write a brief description about the invention of magnetic eects by a current carrying
conductor.
In 1820, the Danish physicist Hans Christian Oersted noticed that a current in a straight
wire caused a noticeable deection in a nearby magnetic compass needle. It is noticeable
when the current is large and the needle suciently close to the wire so that the earth's
magnetic eld may be ignored. Reversing the direction of the current reverses the orien-
tation of the needle. The deection increases on increasing the current or bringing the
needle closer to the wire. Iron lings sprinkled around the wire arrange themselves in
concentric circles with the wire as the centre. Oersted concluded that moving charges or
currents produced a magnetic eld in the surrounding space.
Figure 1:
2. State and explain Lorentz force.
Lorentz force is the force acting on a charge in a combined electric and magnetic eld.The force on an electric charge q moving with a velocity v can be written as
F=q
h
~
E+~v
~
B
i
=Felectric+Fmagnetic
The Lorentz force depends on the following factors
(a) It depends on q, v and B (charge of the particle, the velocity and the magnetic eld).
Force on a negative charge is opposite to that on a positive charge.
(b) The magnetic force q [ vÖB ] includes a vector product of velocity
(c) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving
charge feels the magnetic force.
3. Show diagrammatically, the direction of F,v and B in the case of Lorentz force.
Figure 2:
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2. 4. Derive an equation for the magnetic force experienced by a current carrying conductor
placed in a magnetic eld B?
Consider a rod of a uniform cross-sectional area A and length l. Let the number density
of these charge carriers in it be n. Then the total number of charge carriers in it isnAl.
For a steady current I in this conducting rod, letvddrift velocity . In the presence of an
external magnetic eld B, the force on these carriers is:
F= (charge)v dB
F= (nAl)qvdB
where q is the value of the charge on a carrier. Now current
I=nAqvd
Thus,
F= [(nqvd)Al]B=I1B
where l is a vector of magnitude l, the length of the rod, and with a direction identical
to the current I
5. What are permittivity and permeability?
Electric permittivity0is a physical quantity that describes how an electric eld aects
and is aected by a medium. It is determined by the ability of a material to polarise in
response to an applied eld, and thereby to cancel, partially, the eld inside the material.
Similarly, magnetic permeability0is the ability of a substance to acquire magnetisation
in magnetic elds. It is a measure of the extent to which magnetic eld can penetrate
matter.
6. Describe the motion of a charged particle in uniform and non-uniform magnetic eld?
In the case of motion of a charge in a magnetic eld, the magnetic force is perpendicular
to the velocity of the particle. So no work is done and no change in the magnitude of the
velocity is produced(though the direction of momentum may be changed).
Consider motion of a charged particle in a uniform magnetic eld. First consider the case
of v perpendicular to B. The perpendicular force, q(vÖB) , acts as a centripetal force
and produces a circular motion perpendicular to the magnetic eld. The particle will
describe a circle if v and B perpendicular to each other.
Figure 3:
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3. If velocity has a component along B(ie, velocity is not purpendiculat to B), this component
remains unchanged as the motion along the magnetic eld will not be aected by the
magnetic eld, since F in that direction is zero There fore the particle will execute a
helical motion
7. Derive an expression for the radius of the circular path followed by a charged particle
crossing a uniform magnetic eld normally.
Consider a charged particle q is projected into a uniform magnetic eld B with an initial
velocity v perpendicular to the magnetic eld. It gets trapped in a circular path. The
force exerted by the eld provides the necessary centripetal force.
The magnetic force is,
Fm=qvB
The direction of this force will be perpendicular to both B and v. This force provide the
centripetal force for the charge to move in circular motion. Ler r be radius of circular
path.
The centripetal force is,
Fc=
mv
2
r
Since centripetal force = magnetic force,
Fc=Fm
mv
2
r
=qvB
r=
mv
qB
Figure 4:
8. What is pitch? Derive an equation for the same?
For the circular path of a particle,r=
mv
qB
The larger the momentum, the larger is the radius and bigger the circle described. If!
is the angular frequency, thenv=!r.
So!= 2f=
qB
m
, which is independent of the velocity or energy. Here f is the frequency
of rotation. The independence of f from energy has important application in the designof a cyclotron .The time taken for one revolution isT=
2
!
.
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4. If there is a component of the velocity parallel to the magnetic eld, it will make the
particle move along the eld and the path of the particle would be a helical one. The
distance moved along the magnetic eld in one rotation is called pitch p. We have
p=vjjT=
2mvjj
qB
The radius of the circular component of motion is called the radius of the helix.
9. Explain the working of velocity selector.
consider the simple case in which electric and magnetic elds are perpendicular to each
other and also perpendicular to the velocity of the particle.
Figure 5:
The force acting on the charge is,
F=Fe+Fm
The electric force is
Fe=qE
The magnetic force is
Fm=qV B
If the electric and magnetic forces are in opposite directions the total force on the charge
is zero and the charge will move in the elds undeected. Then
Fe=Fm
qE=qvB
v=
E
B
This condition can be used to select charged particles of a particular velocity out of a beamcontaining charges moving with dierent speeds. The crossed E and B elds, therefore,serve as a velocity selector.
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5. 10. State and explain Biot-Savart Law ?
Consider a XY carrying current I. Consider an innitesimal element dl of the conductor.
The magnetic eld dB due to this element is to be determined at a point P which is
at a distance r from it. Letbe the angle between dl and the displacement vector r.
According to Biot-Savart's law, the magnitude of the magnetic eld dB is proportional
to the current I, the element length |dl|,and inversely proportional to the square of the
distance r
Its direction is perpendicular to the plane containing dl and r
~
dB/
I
~
dl~r
r
3
~
dB=
0
4
I
~
dl~r
r
3
The magnitude of the eld is
~
dB=
0
4
Idlsin
r
2
where
0
4
= 410
7
TmA
1
and0= permeability of free space(vacuum).
11. Compare the dierences and similarities between Biot-Savart law and Coulomb's law?
(a) Both are long range, since both depend inversely on the square of distance from the
source to the point of interest.
(b) The electrostatic eld is produced by a scalar source, namely, the electric charge.
The magnetic eld is produced by a vector source I dl
(c) The electrostatic eld is along the displacement vector joining the source and the eld
point. The magnetic eld is perpendicular to the plane containing the displacement
vector r and the current element I dl.
(d) There is an angle dependence in the Biot-Savart law which is not present in the
electrostatic case.
12. Derive an expression for the magnetic eld due to a current-carrying coil at a point on
its axis.
Consider a circular coil having radius a and centre O from which current I ows in
anticlockwise direction. The coil is placed at YZ plane so that the centre of the coil
coincide along X-axis. P be the any point at a distance x from the centre of the coil
where we have to calculate the magnetic eld. Let dl be the small current carrying
element at any point A at a distance r from the point P where
x=
p
a
2
+r
2
and the angle between r and dl is 90°.
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6. Figure 6: Magntic eld due to a circular coil carrying current
Then from Biot-Savart law, the magnetic eld due to current carrying element dl is
~
dB=
0
4
Idlsin
x
2
Here= 90
o
~
dB=
0
4
Idl
x
2
The net magnetic eld at P is,
B=
I
dBsin
=
I
0
4
Idl
x
2
sin
=
0
4
I
x
2
sin
I
dl
But,sin=
r
p
a
2
+r
2and
H
dl= 2rthe circumference of the coil.
B=
0
4
I
x
2
r
p
a
2
+r
2
B=
0Ir
2
2(a
2
+r
2
)
3=2
13. State Ampere's circuital law? Derive expression for magnetic eld due to a straight
current carrying conductor.
The line integral of magnetic eld around a closed loop is equal to0times the total
current passing through the loop.
I
B:dl=0I
Consider a straight conductor carrying a current I. P is a point at a distance r from the
conductor. To nd the magnetic eld at P imagine a circular loop passing through P
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7. with conductor at the center. Let B be the magnetic le at P.
Then as per Ampere's circuital law,
I
B:dl=0I
I
Bdlsin=0I
I
Bdlsin90
o
=0I
I
Bdl=0I
B
I
dl=0I
I
dl= 2r, the circumfrence of loop
B2r=0I
B=
0I
2r
14. Suggest a method to nd out the direction of magnetic eld produced by a current carrying
straight conductor?
Right-hand thumb rule : Hold the wire in your right hand with your extended thumb
pointing in the direction of the current. The direction of the curled ngers give the
direction of magnetic eld.
15. What is a solenoid? Derive an equation for the magnetic eld produced by a solenoid?
Solenoid consists of a long wire wound in the form of a helix where the neighboring turns
are closely spaced. So each turn can be regarded as a circular loop. The net magnetic
eld is the vector sum of the elds due to all the turns.
Figure 7:
Consider a rectangular Amperian loop abcd. Along cd the eld is zero . Along transversesections bc and ad, the eld component is zero. Thus, these two sections make no con-tribution.Let the eld along ab be B. Thus, the relevant length of the Amperian loop is, L = h
Let n be the number of turns per unit length, then the total number of turns is nh. The
enclosed current is, I (n h), where I is the current through each turn in the solenoid. From
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8. Ampere's circuital law
BL=0I
Bh=0I(nh)
B=0I
The direction of the eld is given by the right-hand rule. The solenoid is commonly used
to obtain a uniform magnetic eld.
16. What is a toroid, Derive an equation for the magnetic eld produced by a toroid?
The toroid is a hollow circular ring on which a large number of turns of a wire are closely
wound. It can be viewed as a solenoid which has been bent into a circular shape to close
on itself.
Figure 8:
It can be viewed as a solenoid which has been bent into a circular shape to close on itself.It is shown in the gure carrying a current I. We shall see that the magnetic eld in the
open space inside (point P) and exterior to the toroid (point Q) is zero. The eld B inside
the toroid is constant in magnitude for the ideal toroid of closely wound turns.
We shall now consider the magnetic eld at S. From Ampere's law,L = 2r.
The current enclosed I is (for N turns of toroidal coil) N I.
B(2r) =0NI
B=
0NI
2r
n=
N
2r
;no of turns per unit length
B=0nI
17. Derive an expression for the force acting per unit length between two long straight parallel
current-carrying conductors.
Force per unit length between two long straight parallel conductors: Suppose two long
thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum
(or air) carrying currentsI1andI2respectively. It has been observed experimentally that
when the currents in the wire are in the same direction, they experience an attractive force
(g. a) and when they carry currents in opposite directions, they experience a repulsive
force (g. b). Let the conductors PQ and RS carry currentsI1andI2in same direction
and placed at separation r.
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9. Figure 9:
Consider a currentelement `ab' of length L of wire RS. The magnetic eld produced by
current-carrying conductor PQ at the location of other wire RS
B1=
0I1
2r
According to Maxwell's right hand rule, the direction ofB1will be perpendicular to the
plane of paper and directed inwards. Due to this magnetic eld, each element of otherwire experiences a force. The direction of current element is perpendicular to the mag-netic eld; therefore the magnetic force on length L of the second wire is
F=B1I2Lsin90
o
F=
0I1
2r
I2L
The force acting per unit length on the conductor
f=
F
L
=
0I1I2
2r
18. Dene one Ampere?
The ampere is the value of that steady current which, when maintained in each of thetwo very long, straight, parallel conductors, placed one metre apart in vacuum, would
produce on a force equal to210
7
newtons per metre on each other.
19. Show that a rectangular current carrying coil experiences a torque when it is placed in a
magnetic eld.
Consider a rectangular coil of length l and breadth b carrying a current I placed in a uni-
form magnetic eld B.be the angle between plane of rectangular coil and magnetic eld.
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10. Figure 10:
The the force acting on side PQ,
F1=BIlsin=BIlsin90
o
=BIl
The the force acting on side RS,
F2=BIlsin=BIlsin90
o
=BIl
Applying the Flemming Left Hand Rule we nd the directions of these forces are opposite
to each other.
As the force acting on the sides PQ and RS are equal and opposite along dierent lines
of action they constitute a couple. Hence the rectangular coil experiences a torque.
=Forcedistance
=BIlbsin
=BIAsin
where A =lb, area of coil.
=BIAsin
Let N be the number of turns of the coil, then
=NIABsin
20. Derive an equation for the magnetic dipole moment of a revolving electron?
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11. The electron performs uniform circular motion around a stationary heavy nucleus of
charge +Ze. This constitutes a current I, where,
I=
e
T
T is the time period of revolution. Let r be the orbital radius of the electron, and v the
orbital speed. Then,
T=
2r
v
Substituting for T in rst equation,
I=
ev
2r
There will be a magnetic moment, usually denoted byl,associated with this circulating
current
l=Ir
2
=
evr
2
=
e
2me
(mevr)
=
e
2me
l
where l = angular momentum of electron. From Bohr's second postulatel=
nh
2
where n
ia an integer.
l=
e
2me
nh
2
l=
nhe
4me
For n =1, we get
l=
eh
4me
= 9:2710
24
Am
2
This value is called theBohr magneton.
21. Explain the construction and working of a moving coil galvanometer with the help of a
diagram.
Construction
The moving coil galvanometer is made up of a rectangular coil that has many turns and it
is usually made of thinly insulated or ne copper wire that is wounded on a metallic frame.
The coil is free to rotate about a xed axis. A phosphor-bronze strip that is connected
to a movable torsion head is used to suspend the coil in a uniform radial magnetic eld.
Essential properties of the material used for suspension of the coil are conductivity and a
low value of the torsional constant. A cylindrical soft iron core is symmetrically positioned
inside the coil to improve the strength of the magnetic eld and to make the eld radial.
The lower part of the coil is attached to a phosphor-bronze spring having a small number
of turns. The other end of the spring is connected to binding screws.
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12. The spring is used to produce a counter torque which balances the magnetic torque and
hence help in producing a steady angular deection. A plane mirror which is attached
to the suspension wire, along with a lamp and scale arrangement is used to measure the
deection of the coil. Zero-point of the scale is at the centre.
Figure 11: Moving coil galvanometer
WorkingThe torque acting on a coil in a magnetic eld is given by,
=NIABsin
Since the eld is radial by design, we have takensin= 1as= 90
o
always.The mag-
netic torque NIAB tends to rotate the coil. A spring S provides a counter torquek
that balances the magnetic torque NIAB; resulting in a steady angular deection. In
equilibrium
k=NIAB
where k is the torsional constant of the spring; i.e. the restoring torque per unit twist.
I=
k
NAB

I=K
where
K=
k
NAB
is called Galvanometer constant.
I/
22. How will you convert a galvanometer to ammeter?
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13. Figure 12: Conversion of Galvanometer to Ammeter
A galvanometer is converted to ammeter by connecting a low resistance S called shunt
resistance in parallel to it. Let G be the resistance of galvanometer, S the resistance of
shunt resistor,Igthe maximum safe current that can pass through the galvanometer. Let
I be the current to be measured. Then the current owing through the shunt resistor is
IIg.
The potential dierence across galvanometer and the shunt are equal.
(IIg)S=IgG
S=
IgG
IIg
23. How will you convert a galvanometer to voltmeter?
A galvanometer is converted to voltmeter by connecting a high resistance R in series toit.
Figure 13: Conversion of Galvanometer to Voltmeter
Let V be the potential dierence to be measured. then
V=IgG+IgR
IgR=VIgG
R=
V
Ig
G
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