class13.ppt rth ghhhhhhhhhhhh dhhe ethbn tj

rrre t eyh 7r6 rfghn trrrrrhdf

1. Sources of Magnetic FieldsSources of Magnetic Fields
Chapter 30Chapter 30
Biot-Savart LawBiot-Savart Law
Lines of Magnetic FieldLines of Magnetic Field
Ampere’s LawAmpere’s Law
Solenoids and ToroidsSolenoids and Toroids

2. Sources of Magnetic Fields
•Magnetic fields exert forces on moving charges.
•Something reciprocal happens: moving charges give rise
to magnetic fields (which can then exert a force on other
moving charges).
•We will look at the easiest case: the magnetic field
created by currents in wires.
•The magnetism of permanent magnets also comes from
moving charges (the electrons in the atoms).

3. Magnetic Interaction
•A current generates a magnetic field.
•A magnetic field exerts a force on a current.
•Two conductors, carrying currents, will
exert forces on each other.
Rather than discussing moving charges in general,
restrict attention to currents in wires. Then:

4. Biot-Savart Law
•The mathematical description of the magnetic field B due
to a current-carrying wire is called the Biot-Savart law.
It gives B at a selected position.
•A current I is moving all through the wire. We need to
add up the bits of magnetic field dB arising from each
infinitesimal length dl.

€
d
r
B =
μ
0
4π
Id
r
l ׈ r
r
2
Add up all the bits!
r
d l

I
dB

5. Biot-Savart Law
•The mathematical description of the magnetic field B due
to a current-carrying wire is called the Biot-Savart law.
It gives B at a selected position.
•A current I is moving all through the wire. We need to
add up the bits of magnetic field dB arising from each
infinitesimal length dl.

€
d
r
B =
μ
0
4π
Id
r
l ׈ r
r
2
Add up all the bits!
r
d l

I
dB


€
r
r =rˆ r is the vector from dl to
the observation point

6. It turns out that 
 and 
 are related in a simple
way: (

0)
-1/2
= 3x10
8
m/s = c, the speed of light.
Why? Light is a wave of electric and magnetic fields.
The constant 
0 = 4 x 10
-7
T m/A
is called the permeability of free space.
Biot-Savart Law

d
r
B =
μ
0
4π
Id
r
l × ˆr
r
2r
d l

I
dB

7. Example: Magnetic field from a long wire
Consider a long straight wire carrying a current I.
We want to find the magnetic field B at a point P,
a distance R from the wire.
I
R
P

8. Example: Magnetic field from a long wire
Consider a long straight wire carrying a current I.
We want to find the magnetic field B at a point P,
a distance R from the wire.
Break the wire into bits dl.
To do that, choose coordinates:
let the wire be along the x axis,
and consider the little bit dx at a
position x.
The vector r = r r is from this bit
to the point P.
dx
I
R
d l
x
x
0
r
P
^

9. Example: Magnetic field from a long wire
dx
I
R
x
x
0
r

€
d
r
B =
μ
0I
4π
d
r
l xˆ r
r
2
Direction of dB: into page.
+

€
ˆ r

10. Example: Magnetic field from a long wire
dx
I
R
x
x
0
r

€
d
r
B =
μ
0I
4π
d
r
l xˆ r
r
2
Direction of dB: into the page.
This is true for every bit; so we
don’t need to break into
components, and B also points
into the page.
+

€
ˆ r

11. Example: Magnetic field from a long wire
dx
I
R
x
x
0
r

€
d
r
B =
μ
0I
4π
d
r
l xˆ r
r
2
Direction of dB: into the page.
This is true for every bit; so we
don’t need to break into
components, and B also points
into the page.
Moreover, lines of B go around a
long wire.
+

€
ˆ r

12. Example: Magnetic field from a long wire
dx
I
R
x
x
0
r
Moreover, lines of B go around a
long wire. Perspective:
+

€
ˆ r
i
B
P
Another right-hand rule

13. Example: Magnetic field from a long wire
dx
I
R
x
x
0
r

€
d
r
B =
μ
0I
4π
d
r
l xˆ r
r
2
Direction of dB (or B): into page
€
dB=
μ
0
I
4π
dxsinθ
r
2
∴B=dB=∫
μ
0
I
4π
sinθdx
r
2
x=−∞
x=+∞
∫
+

€
ˆ r

14. Example: Magnetic field from a long wire
dx
I
R
x
x
0 +


€
∴B=dB=∫
μ
0I
4π
sinθdx
r
2
x=−∞
x=+∞
∫
r=x
2
+R
2
,sinθ=
R
r
B=
μ
0I
4π
Rdx
(x
2
+R
2
)
3
2
x=−∞
x=+∞
∫
=
μ
0
I
4πR
x
(x
2
+R
2
)
1
2
x=−∞
x=+∞
=
μ
0
I
2πR
r
€
ˆ r

15. Force between two current-carrying wires
I
1
I
2
B
2
B
1
Current 1 produces a magnetic
field B
1
=
0
I/ (2d) at the position
of wire 2.
d
This produces a force on current 2:

16. Force between two current-carrying wires
I
1
I
2
B
2
B
1
Current 1 produces a magnetic
field B
1
=
0
I/ (2d) at the position
of wire 2.
d
This produces a force on current 2:
F
2
= I
2
L x B
1

17. Force between two current-carrying wires
I
1
I
2
B
2
B
1
Current 1 produces a magnetic
field B
1
=
0
I/ (2d) at the position
of wire 2.
d
This produces a force on current 2:
F
2
= I
2
L x B
1
F
2

18. Force between two current-carrying wires
I
1
I
2
B
2
B
1
Current 1 produces a magnetic
field B
1
=
0
I/ (2d) at the position
of wire 2.
d
This gives the force on a length L of wire 2 to be:
F
2=I
2LB
1=
μ
0
I
1
I
2
L
2πd
Direction: towards 1, if the currents are in the same direction.
This produces a force on current 2:
F
2
= I
2
L x B
1
F
2

19. Force between two current-carrying wires
I
1
I
2
B
2
B
1
Current I
1 produces a magnetic
field B
1 =
0I/ (2d) at the
position
of the current I
2
.
d
Thus, the force on a length L of the conductor 2 is given by:
F
2
=I
2
LB
1
=
μ
0
I
1
I
2
L
2πd
The magnetic force between two parallel wires
carrying currents in the same direction is attractive .
This produces a force on current I
2
:
F
2 = I
2L x B
1
F
2
[Direction: towards I
1]
What is the force on wire 1? What happens if one current is reversed?

20. Magnetic field from a circular current loop
I
dB
z
dB
perp
B
rz
R
dl

€
d
r
B =
μ
0
4π
Id
r
l ׈ r
r
2
dB=
μ
0
4π
Idl
r
2
dB
z
=dBcosα=
μ
0
4π
Idlcosα
r
2
r=R
2
+z
2
,cosα=
R
R
2
+z
2
Only z component
is nonzero.
along the axis only!

21. I
dB
z
dB
perp
B
rz
R
dl

At the center of the loop
At distance z on axis
from the loop, z>>R
Magnetic field from a circular current loop

€
B=dB
z∫=
μ
0
4π
∫
IR
(R
2
+z
2
)
3
2
dl
B=
μ
0
4π
IR
(R
2
+z
2
)
3
2
dl∫=
B=
μ
0
4π
IR
(R
2
+z
2
)
3
2
2πR=
μ
0
2
IR
2
(R
2
+z
2
)
3
2
€
B=
μ
0I
2R
€
B=
μ
0IR
2
2z
3
along the axis only!

22. The magnetic dipole moment of the
loop is defined as = IA =IR
2
.
The direction is given by the right
hand rule: with fingers closed in
the direction of the current flow,
the thumb points along .
Magnetic field in terms of dipole moment
I
B
z
R
Far away on the axis,

€
B=
μ
0IR
2
2z
3

23. In terms of , the magnetic field on axis (far from
the loop) is therefore
This also works for a loop with N turns. Far from
the loop the same expression is true with the
dipole moment given by =NIA = IR
2
B=
μ
0μ
2πz
3
Magnetic field in terms of dipole moment

24. Dipole Equations
Electric Dipole
 = p x E
U = - p · E
E
ax = (2
0 )
-1
p/z
3
E
bis = (4
0 )
-1
p/x
3

Magnetic Dipole
 =  x B
U = -  · B
B
ax = ( 
0/2 /z
3
B
bis = (
0/4) /x
3

25. Electric fields
Coulomb’s law gives E
directly (as some integral).
Gauss’s law is always true. It
is seldom useful. But when it
is, it is an easy way to get E.
Gauss’s law is a surface
integral over some
Gaussian surface.
Ampere’s Law
Magnetic fields
Biot-Savart law gives B
directly (as some integral).
Ampere’s law is always true.
It is seldom useful. But when
it is, it is an easy way to get B.
Ampere’s law is a line integral
around some
Amperian loop.

26. Draw an “Amperian loop”
around the sources of
current.
The line integral of the
tangential component of B
around this loop is equal to

o
I
enc
:
Ampere’s Law

€
r
B ∫•d
r
l =μ
0
I
enc
I
2
I
3
Ampere’s law is to the Biot-Savart law exactly
as Gauss’s law is to Coulomb’s law.

27. Draw an “Amperian loop”
around the sources of
current.
The line integral of the
tangential component of B
around this loop is equal to

o
I
enc
:
Ampere’s Law

€
r
B ∫•d
r
l =μ
0
I
enc
I
2
I
3
Ampere’s law is to the Biot-Savart law exactly
as Gauss’s law is to Coulomb’s law.
The sign of I
enc comes
from another RH rule.

28. Ampere’s Law - a line integral
blue - into figure
red - out of figure
I
1
I
3
I
2
a
c
b
d

€
r
B ⋅d
r
l
a
∫ =
r
B ⋅d
r
l
b
∫ =
r
B ⋅d
r
l
c
∫ =
r
B ⋅d
r
l
d
∫ =

29. Ampere’s Law - a line integral

€
r
B ⋅d
r
l
a
∫ =μ
0
I
1
r
B ⋅d
r
l
b
∫ =
r
B ⋅d
r
l
c
∫ =
r
B ⋅d
r
l
d
∫ =
blue - into figure
red - out of figure
I
1
I
3
I
2
a
c
b
d

30. Ampere’s Law - a line integral

€
r
B ⋅d
r
l
a
∫ =μ
0
I
1
r
B ⋅d
r
l
b
∫ =μ
0I
2
r
B ⋅d
r
l
c
∫ =
r
B ⋅d
r
l
d
∫ =
blue - into figure
red - out of figure
I
1
I
3
I
2
a
c
b
d

31. Ampere’s Law - a line integral

€
r
B ⋅d
r
l
a
∫ =μ
0
I
1
r
B ⋅d
r
l
b
∫ =μ
0
I
2
r
B ⋅d
r
l
c
∫ =μ
0
−I
3()
r
B ⋅d
r
l
d
∫ =
blue - into figure
red - out of figure
I
1
I
3
I
2
a
c
b
d

32. Ampere’s Law - a line integral

€
r
B ⋅d
r
l
a
∫ =μ
0
I
1
r
B ⋅d
r
l
b
∫ =μ
0
I
2
r
B ⋅d
r
l
c
∫ =μ
0
−I
3()
r
B ⋅d
r
l
d
∫ =μ
0
I
1
−I
3( )
blue - into figure
red - out of figure
I
1
I
3
I
2
a
c
b
d

33. Ampere’s Law on a Wire
i
What is magnetic field
at point P ?
P

34. Ampere’s Law on a Wire
What is magnetic field
at point P? Draw Amperian
loop through P around current
source and integrate B · dl
around loop
i
B
P
dl
TAKE ADVANTAGE OF SYMMETRY!!!!

35. Ampere’s Law on a Wire
i
B
What is magnetic field
at point P? Draw Amperian
loop through P around current
source and integrate B · dl
around loop
P
Then
dl

€
r
B ∫•d
r
l =Bdl∫=B(2πr)=μ
0I
TAKE ADVANTAGE OF SYMMETRY!!!!

36. B=
μ
0I
2πr
Ampere’s Law for a Wire
What is the magnetic field at point P?
Draw an Amperian loop through P,
around the current source, and
integrate B · dl around the loop.
Then:
i
B
P
dl

€
r
B ∫•d
r
l =Bdl∫=B(2πr)=μ
0I

37. A Solenoid
.. is a closely wound coil having n turns per unit length.
current flows
out of plane
current flows
into plane

38. A Solenoid
.. is a closely wound coil having n turns per unit length.
current flows
out of plane
current flows
into plane
What direction is the magnetic field?

39. A Solenoid
.. is a closely wound coil having n turns per unit length.
current flows
out of plane
current flows
into plane
What direction is the magnetic field?

40. A Solenoid
Consider longer and longer solenoids.
Fields get weaker and weaker outside.

41. Apply Ampere’s Law to the loop shown.
Is there a net enclosed current?
In what direction does the field point?
What is the magnetic field inside the solenoid?
current flows
out of plane
current flows
into plane

42. Apply Ampere’s Law to the loop shown.
Is there a net enclosed current?
In what direction does the field point?
What is the magnetic field inside the solenoid?
current flows
out of plane
current flows
into plane
€
BL=μ
0NI⇒B=nμ
0I
wheren=N/L
L

43. Gauss’s Law for Magnetism
For electric charges
Gauss’s Law is:
because there are single electric charges. On the other hand, we
have never detected a single magnetic charge, only dipoles. Since
there are no magnetic monopoles there is no place for magnetic
field lines to begin or end.
€
E∫•dA=
q
ε
0
€
B∫•dA=0
Thus, Gauss’s Law for
magnetic charges must be:

44. Laws of Electromagnetism
We have now 2.5 of Maxwell’s 4 fundamental
laws of electromagnetism. They are:
Gauss’s law for electric charges
Gauss’s law for magnetic charges
Ampere’s law (it is still incomplete as it only
applies to steady currents in its present form.
Therefore, the 0.5 of a law.)

45. Magnetic Materials
The phenomenon of magnetism is due mainly to the
orbital motion of electrons inside materials, as well
as to the intrinsic magnetic moment of electrons (spin).
There are three types of magnetic behavior in bulk
matter:
Ferromagnetism
Paramagnetism
Diamagnetism

46. Magnetic Materials
Because of the configuration of electron orbits in atoms,
and due to the intrinsic magnetic properties of electrons
and protons (called “spin”), materials can enhance or
diminish applied magnetic fields:
r
Bapplied

47. Magnetic Materials
Because of the configuration of electron orbits in
atoms, and due to the intrinsic magnetic properties
of electrons and protons (called “spin”), materials
can enhance or diminish applied magnetic fields:

€
r
B
applied

€
r
B
internal

48. Magnetic Materials
Because of the configuration of electron orbits in
atoms, and due to the intrinsic magnetic properties
of electrons and protons (called “spin”), materials
can enhance or diminish applied magnetic fields:

€
r
B
int
=κ
M
r
B
app

€
r
B
applied

€
r
B
internal

49. Magnetic Materials

M is the relative permeability
(the magnetic equivalent of 
E
)
Usually 
M is very close to 1.
- if 
M
> 1, material is “paramagnetic” - e.g. O
2
- if 
M
< 1, material is “diamagnetic” - e.g. Cu
Because 
M is close to 1, we define the
magnetic susceptibility 
M
= 
M
- 1
r r
B B
Mappint=κ

50. Magnetic Materials
Hence:
For paramagnetic materials 
M
is positive
- so B
int
> B
app

For diamagnetic materials 
M
is negative
- so B
int
< B
app
Typically, 
M
~ +10
-5
for paramagnetics,

M ~ -10
-6
for diamagnetics.
(For both 
M is very close to 1)

€
r
B
int=κ
M
r
B
app=(1+χ
M
)
r
B
app

51. Magnetic Materials
Ferromagnetic Materials:
These are the stuff permanent magnets are
made of.
These materials can have huge susceptibilities:

M
as big as +10
4

52. Magnetic Materials
Ferromagnetic Materials:
These are the stuff permanent magnets are made
of.
These materials can have huge susceptibilities:

M as big as +10
4

But ferromagnets have “memory” - when you
turn off the B
app, the internal field, B
int ,
remains!

53. Magnetic Materials
Ferromagnetic Materials:
These are the stuff permanent magnets are
made of.
These materials can have huge susceptibilities:

M
as big as +10
4

But ferromagnets have “memory” - when you
turn off the B
app, the internal field, B
int ,
remains!
permanent magnets!