Chapter 5: Magnetostatics, Faraday’s Law,
Quasi-Static Fields
5.1 Introduction and Definitions
We begin with the law of conservation of charge:
Q
3
 d 3x


J
d
x

J

d
a




v

t
t v

   J  t  0 conservation 
h
off charge

B
, J
da
(5 2)
(5.2)
arbitrary volume
Magnetostatics is applicable under the static condition. Hence,

 0 and (5.2) gives
  J  0 [for magnetoststics]
(5.3)
t
Assuming a magnetic force FB is experienced by charge q moving
at velocity v, we define the magnetic induction B by the relation:
FB  qv  B,
which is consistent with the definition in (5.1).
1
5.2 Biot and Savart Law
The Biot
Biot-Savart
Savart law states that the differential magnetic field ddB
B
at point P (see figure) due to a differential current element d  2 in
0 d  2  x12
loop
p1
l
loop
2 is
i given
i
by
b dB 
I2
(5 4)
(5.4)
3
I
4
1
| x12 |
loop 2
P
Thus the total field at P due to I 2 in
Thus,
d 1

d  x
linear superposition, 
loop 2 is: B  0 I 2  2 312 
4
 an experimental fact 
| x12 |
x12
d 2
(1)
I2
Integrating the force on I1 in loop 1 due to I 2 in loop 2, we obtain
F12  I1  d 1  B
(5 7)
(5.7)
0
d 1  ( d  2  x12 )

I1I 2  
d 1  x12
(d 1  d  2 )x12
3

d


4
 2 
 
| x12 |
3
| x12 |
| x12 |3


0
(d 1  d  2 )x12

I1I 2  
4
| x12 |3




d x12
x12 2
   d (
1
)0
x12
2
5.3 Differential Equations of Magnetostatics
and Ampere’s
Ampere s Law
x12  x1  x 2
B
Gauss Law of Magnetism :
x1 x d 2
2
cross
section
0
d  2
xx12
Rewrite (1): B  4 I 2 
of wire
|x12|3
I2
0

Change x1 to x, x 2 to x, and let I 2 d  2  J da d  2  Jd 3 x, we obtain
0
0
x  x 3
1
3



B ( x) 
J
x
d
x
J
x
(
)




(
)
d
x


3
4
4
| x  x |
| x  x |
  a    a    a
x x   1
|x x|
|x x|3
0
J (x)
1
3







[
J
(
x
)]
d
x



4
| x  x  | | x  x | 
0
0
J (x) 3

 
d x
 operates
p
on x
4
| x  x |

  B  0 [Gauss law of magnetism]
B
x
x  x
x
J ( x)
0
(5.16)
(5.17)3
5.3 Differential Equations of Magnetostatics and Ampere’s Law (continued)
Ampere'ss Law : Rewrite (5.16):
Ampere
J ( x) 3


 |xx| d x
0
J (x) 3
0


d x
B ( x) 
 
| x  x |
4
  [J ( x)   |x1x|  |x1x|   J ( x)]d 3 x
   B( x)
   J (x)  |x1x| d 3 x
0
J ( x)

     |xx| d 3 x
1   J ( x) d 3 x    J ( x) d 3 x

 |xx| 
 |xx|
4





0
0
0



0
J ( x)
 [    |xx| d 3 x   J (x)  2 |x1x| d 3 x]
4



4 ( x  x)
  (  a)  (  a)   2a
n
   B ( x)  0 J ( x)
arbitrary (5.22)
da
    B  n da  0  J  n da
 loop




I (through the loop)
 B  d 
d
Ampere'ss law , a much more elaborate 
 Ampere
  B  d   0 I 
(5.25)

 representation of the Biot-Savart law 
4
5.4 Vector Potential
0
J ( x ) 3
Vector Potential : Rewrite (5.16): B(x) 
 
d x
| x  x |
4
 B    A,
where the vector potential A is given by
 0 J ( x ) 3 
A
d x   ,


4 | x  x |
(5 27)
(5.27)
(5.28)
which shows that A may be freely transformed (without changing B)
according to
A  A   (gauge transformation)
(5.29)
W may exploit
We
l it this
thi freedom
f d
by
b choosing
h i a  so that
th t
A  0
(Coulomb gauge)
(5.31)
See proof on previous page.
page
0





  (5.28)    A  40    |Jx(xx)| d 3 x   2   2 ,
 Coulomb gauge requires  2  0 everywhere and hence   const.
5
5.4 Vector Potential (continued)
  B  0 J
Rewrite: 
B    A
     A  0 J
 (  A)   2 A  0 J
Choose the Coulomb gauge (  A  0)
  2 A   0 J
(5 31)
(5.31)
0 J ( x) 3
A
d x

4 | x  x |
((5.32))
Note:
(5.32) is valid in unbounded (infinite) space, i.e. the volume of
integration must include all currents. If there is a boundary surface,
the currents on the boundary must be accounted for by application
6
of boundary conditions (See example in Sec. 5.12.)
5.4 Vector Potential (continued)
A Comparison of Electrostatics and Magnetostatics :
Magnetostatics
Electrostatics
Definition of E:
FE  qE
x
Definition
D
fi i i off B:
FB  qv  B

x
Coulomb's law:
1  ( x )( x  x ) 3
d x
E( x ) 

3
4 0
| x  x |



  E   0
E  0


 E  da  q  0  E  d   0
G
Gauss's
' llaw
of electrostatics
J
x
x
Biot-Savart law:
0
x  x 3
B( x ) 
J ( x ) 
d x

3
4
| x  x |



B  0
  B  0 J


 B  da  0  B  d   0 I
Gauss's Law Ampere's law
of magnetism
7
5.6 Magnetic Field of Localized Current
Distribution, Magnetic Moment
Magnetic (Dipole) Moment :
A  (B  C)  B( A  C)  C( A  B)
0 J (x) 3
1
1 x  x

A
d
x

 3   [Eq.
[
(5),
( ) Ch.
Ch 4]

4 | x  x |
| x  x | | x | |x|
 1
1
J
 0 [  J (x)d 3 x  3 x   xJ (x)d 3 x  ]
x

4 | x |  | x | 
x
0
Proved on next page.
page
 1  x[ xJ ( x)]d 3 x
2
0
Proved on p.185 under the
0  x[xJ (x)]d 3x


1. J is localized
3
|x|
8
 within volume
0 m  x  If x is far  conditions:  of integration


4 | x |3 from source.
2 J  0
2.
where m  12  x  J (x)d 3 x [magnetic (dipole) moment]
(5.55)
(5.54)
Note
N
t : In
I (5
(5.54),
54) m iis ddefined
fi d with
ith respectt to
t a point
i t off reference.
f
Here, it coincides with the origin of the coordinates (x  0).
8
5.6 Magnetic Field of Localized Current Distribution, Magnetic Moment (continued)
Problem : Prove the relation  J ( x) d 3 x  0 under the conditions:
  J  0 and J is localized within volume of integration.
Proof : Since J  0 outside the volume of integration, we may extend
the volume of integration to  without changing the integral value.



3
(
)
J
x
d
x   dx  dy  dz(J x e x  J y e y  J z e z )

Consider the x-component first:
3
e x   J ( x) d x 



J x dx
 dyy  dz 





J x



  dy  dz  x x

J



  dy  dz  x( xx
dx

J y J z
 y  z )dx



xJ x 



J x
x
dx
x
   x  Jd 3 x  0
The insertion of these 2 terms will not
change the value of the integral because
Similarly, the y - and
z -components
components also vanish.
vanish   ( J y )dy  J y   0 &   ( J z )dz  J z   0
 y
 z


3
Thus,  J (x) d x  0
9
5.6 Magnetic Field of Localized Current Distribution, Magnetic Moment (continued)
Anti - symmetric unit tensor ( ijk ): (used on pp.185
185 and pp.188)
188)
0 , if two or more indices are equal

 ijk  1 , if i, j , k is
i an even permutation
t ti off 1,
1 2,3
23
Levi-Civita symbol 
1 , if i, j , k is an odd permutation of 1, 2,3
E
Examples
l : 112  0,
0 123  1,
1 132  1,
1  312  1
( A  B)i    ijk
j A j Bk , (  A )i    ijk
j
jkk
  ( A  B)    ijk
ijk

xi
   ijk

ijk
j 
 A j Bk 
A j
xi
   kij Bk

ijk 
jkk
Bk   ijk A j
A j
xi
  jik A j

x j
(2)
Ak
Bk 
xi 

Bk 
xi 

 B  (  A )  A  (  B)
10
5.6 Magnetic Field of Localized Current Distribution, Magnetic Moment (continued)
Example 1 of magnetic moment: plane loop
I
3
1
I
m  2  x  J (x)d x  2  x  d 
d



2( area )
x
da
(5.57)
 m  I  (area )

loop
m is normal (by right hand rule) to the plane of the loop.
Example 2 of magnetic moment: a number of charged particles
i motion
in
ti
angular momentum
J   qi vi (x  xi )
L  M x v
i
i
qi
3
1
1



Li
 m  2  x  J (x )d x  2  qi xi  vi  
i
i 2M
i i
i
(5.58)
i
e

L
2M
2M
if qi / M i  e / M for all particles.
(5.59)
L: total angular momentum
11
5.6 Magnetic Field of Localized Current Distribution, Magnetic Moment (continued)
Dipole Field : (valid far from the source)
0 mx
Rewrite (5.55) : A 
4 |x|3
0
x 

 B   A 
m 3 
|x| 
4

  x3
|x|


1   x  x 
 13
3
|x|
|x|
3  x  3x  0
|x|3
|x|5
(5.55)
m is a constant.)
0 (



0


x
x
x
x

 0 [m   3  3   m   3 
  m   m 
 3]
|x| |x|
|x|
4
 |x|

 
x
x
x 
 0   mx x 3  m y y 3  mz z 3    ( A  B)  (B ) A  ( A )B
 A(  B)  B(  A )
|x|
|x|
|x| 
4 
0 


3
x
 ex


 mx  3  x 5    y    z  

|x| 
4 
 |x|

x
0 3n(n  m)  m
magnetic
g
dipole
p 

n


(5 56)
(5.56)
3


fie
ld
|x|
4


|x|
0

12
5.6 Magnetic Field of Localized Current Distribution, Magnetic Moment (continued)
As in the case of the electric dipole moment, here we characterize
a localized current distribution by a constant quantity, the magnetic
moment m, which turns an otherwise complicated field calculation
(
(see,
ffor example,
l Sec.
S 55.5)
5) iinto
t a simple
i l one (with
( ith limited
li it d validity.)
lidit )
Consider, for example, a circular loop carrying current I . Using
(5 57) we hhave m  I  a 2e z ((see fi
(5.57),
figure).
) H
Hence, th
the dipo
di le
l field
fi ld is
i
z
0 3n(n  m)  m
n

e
r
B
(5.56)
e
3
r
4
|x|
m  I  a 2e z

r
0
2 3e r (e r  e z )  e z

I a
4
r3
a
(e z  e r cos   e sin  )
I
0
2
cos


sin
i

e
e
r


I a2
(5.41)
3
4
r
Wh r  a, the
When
th dipole
di l field
fi ld is
i a goodd approximation
i ti off the
th total
t t l
13
field [see Jackson (5.40).]
5.7 Forces and Torque on and Energy of a Localized
Current Distribution in an External Magnetic Induction
Magnetic Force in External Field :
F   J (x)  B(x)d 3 x
(5 12)
(5.12)
Expanding B : [see lecture notes, Ch. 4, Appendix A, Eq. (A.4)]
B(x)  B(0)  (x 
)B(0)  

This implies
p
"After differention of B(x),
) set x in results to 0," i.e.
 y  y B(x) 
 z  z B(x) 
(x )B(0)  x  x B(x) 
x 0
x 0

 x 0
 F  [  J (x)d 3 x]  B(0)   J (x)   (x )B(0)  d 3 x  

(proved in Sec. 5.6))
 0 (p
  J (x)   (x )B(0)  d 3 x    (m  B)
 
U ,
See derivation on pp.188
pp.188-189
189
where U  m  B  potential energy.
(5.69)
(5.72)
14
5.7 Forces and Torque… (continued)
Magnetic
g
Torque
q in External Field:
N   x  f (x)d 3 x
)]d 3 x
  x  [J (x)  B
(
x

B ((0))  ( x  )B (0)
( )    B ((0))
  x  [J (x)  B(0)]d 3 x
2
  [ x J (x)]
2 x

2
2
0



 J ( x )    x   x     J ( x )
(5 13)
(5.13)
 2x  J (x)



  [B(0)  x]J (x)d 3 x  B(0)  x  J (x) d 3 x
2
=  [B(0)  x]J (x)d 3 x  12 B(0)    [ x J (x)]d 3 x








  12  B(0)  [x  J (x)]d 3 x
[Using the formula at the
bottom of p.185, replacing
x with B(0).]
 m  B(0)
2
  s x J ( x)da  0
J is localized.
 J  0 on surface
(5.71)15
5.7 Forces and Torque… (continued)
A Comparison between Electric and Magnetic Potential Energy,
Force, and Torque in External Field :
Potential energy
Force
U   p  E (4.24) F  U
U   m  B (5.72)
(5 72) F  
U
Torque
N  pE
N  mB
 p +
Both p and m tend to orient along the
positive field direction under the action of
the torque (see figures on the right). This
I 
results in a state of minimum potential
m
energy. In this state, p reduces E, whereas
I
m enhances B.
Questions:
(1) How does a permanent magnet attract a piece of iron?
(2) How does it attract another permanent magnet?
E
B
16
5.7 Forces and Torque… (continued)
Force in Self -Consistent Field;; Magnetic
g
Pressure and Tension :
A self-consistent field is the combined field
generated by the source J under consideration
and
d the
h externall source (if present).
) Thus,
Th using
i
  B  0 J , we may express the magnetic force
density f (force / unit volume) in terms of B.
B-field
a solenoid
f  J  B  1 (  B)  B
lines
0
( a  b )  ( a   ) b + ( b   )a + a  (   b ) + b  (   a )
2
B
  2   1 (B )B [[see pp. 320)]
)]
0 


0 
a uniform
if
electron beam
((3))
magnetic tension force density,
magnetic pressure
as if a curved B-field line tended
B-field
force density
B
to become a straight line.

lines
I regions
In
i
where
h J  0,
0 we have
h
f 0
uniform current 17
[pressure and tension force densities cancel out].
5.8 Macroscopic Equations, Boundary
Conditions on B and H
Macroscopic Equations : To be more general, we move the
point of reference for m from x  0 to x  x0 and write
p
( x  x 0 )  ( x  x 0 )
1
1


  [See Sec. 4.1]
3
x  x
x  x0
x  x0
Sub. this relation into How about Jfree?
J
x
0 J ( x) 3
A
d x [(5.32)]

4 | x  x |
we obtain
vanishes
i h only
l if J is
i localized
l li d
within the volume of integration.

0  J (x)d 3 x 0 m (x0 )  (x  x0 )
A

 ,
3
4 | x  x0 |
4
| x  x0 |
where m(x0 )  12  ( x  x0 )  J ( x) d 3 x.
J
x
0
x  x 0
x  x0
x
x 0
0
x
(4)
18
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
To proceed, we consider the orbital motion of atomic/molecular
electrons, which can collectively give rise to a permanent or induced
magnetization M (total magnetic moment / unit volume) given by
M ( x)   N i  m i 
(5.76)
i
volume density of
type i molecules
magnetic moment per type i molecule
averaged over a small volume
As will be shown in (5.79), a current density (J M ) is associated
with M. In addition, there is also a current density due to the flow of
free charges, which we denote by J ffree (Jackson denotes it by J in
Sec. 5.8). By the principle of linear superposition, we may write
A (x)  A free (x)  A M (x)),
where A free and A M are due to J free and J M , respectively.
0 J free
f ( x) 3
Obviously, A free (x) 
d x

4 | x  x |
19
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
For A M , we have the expression
p
for M, but not yyet for J M . So
we approximate A M by the dipole term in (4).
3

J
(
x
)
d
x  0 m ( x 0 )  ( x  x 0 )
0  M
A M ( x) 

 ,
3
4
| x  x0 |
4
| x  x0 |
where
h we hhave sett  J M (x)d 3 x  0 because
b
J M is
i formed
f
d off currentt
loops of atomic dimensions (  volume of integration). Under this
condition m is independent of the point of reference because
condition,
m(x0 )  12  (x  x0 )  J M (x)d 3 x
0


 (5.54)
 12  x  J M (x)d 3 x  12x0   J M (x)d 3 x  m(0).
To represent A M by the dipole term
term, we must have x  the
dimension of the dipole. So, we divide the source into infinistesimal
p moment is MV ,
volumes. In each small volume V , the dipole
which generates a small A M at x given by
20
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
A M ( x ) 
0 M (x)( xx) V
|xx|3
4
V
where we have replaced
p
the notation x0 with x. This
gives
x
x
0 M (x)( xx) 3
d x

3
|xx|
4
0
Volume of integration
  M (x)  |x1x| d 3 x includes
all sources.
4
0
0 M ( x) 3
( x) 3

  |xx| d x     M
d
x
x
x

|
|
 
4
4 

A M ( x) 
   a    a     a
3
v   Ad x   s n  Ada
0 M ( x) 3
  |xx| d x
4
( x)
  s n  M
da  0
|xx|
(M  0 on S )
What if M ≠ 0 on S?
Question: Does this relation still hold as x  x ?
21
Griffith
1/4
6.2 The Field of a Magnetized Object
621B
6.2.1
Bound
dC
Currents
t
Suppose we have a piece of magnetized
material (i.e. M is given). What field does
this object produce?
The vector potential of a single dipole m is
0 m  rr̂
A(r ) 
4 r 2
IIn the
th magnetized
ti d object,
bj t each
h volume
l
element
l
t carries
i a
dipole moment Md’, so the total vector potential is
0 M (r)  rˆ
d 
A(r ) 
2

4
r
22
Griffith 2/4
Vector potential and Bound Currents
Can the equation be expressed in a more illuminating form,
as in the electrical case? Yes!
By exploiting the identity,
1 rˆ
  2
r r



1
 yˆ 
 zˆ  )
x
y
z  ( x  x) 2  ( y  y) 2  ( z  z) 2
xˆ ( x  x)  yˆ ( y  y)  zˆ ( z  z )
rˆ


(( x  x) 2  ( y  y) 2  ( z  z ) 2 )3/ 2 r 2
(xˆ 
0
1
The vector potential is A(r ) 
M (r)  ( )d 

4
r
U i th
Using
the product
d t rule
l   ( fA)  f  A  f (  A)
and integrating by part, we have

A(r )  0
4

 0
4
M (r)
 1 





[


M
(
r
)]
d




[
]
d




r
r


How? See next page.
 1 
 0 1




M
r
d

[M (r)  nˆ ]da
[
(
)]



23
 r
 4 r
Griffith 3/4
Gauss's law
 (  E)d   E  da
v
S
 (  ( v  c))d  c  (  v )d
v
 v
Let E  v  c, 
  ( v  c)  da  c   v  da
S
S
Since c is a constant vector, so  (  v )d    v  da
v
S
24
Griffith
4/4
Vector potential and Bound Currents
0 1
0 1
A(r ) 
[  M (r)]d  
[M (r)  nˆ ]da


4 r
4 r
J b    M (r)
K b  M (r)  nˆ 
volume current
surface current
bound currents
With these definitions
definitions,
0
A(r ) 
4

v
Jb
0

d 
r
4

S
Kb
da
r
The electrical analogy
volume charge density  b    P
surface charge density σ b  P  nˆ
25
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
Th
Thus,
A (x)  A free (x)  A M (x)
0 J free (x)    M (x) 3

d x

4
| x  x |
p
in Sec. 5.3, we have  2 A   0 J ,
For comparison,
0 J (x) 3
which has the solution: A( x) 
d x

4 | x  x |
(5 78)
(5.78)
((5.31))
(5.32)
In (5.31) and (5.32), J represents the current due to both free
and bound (atomic) electrons, whereas in (5.78) contributions from
free and bound electrons are separated into two terms.
p
g (5.78)
(
) and (5.32),
(
), we find that the bound electrons
Comparing
contribute to A (x) through a magnetization current density (J M )
given by
(5.79)
J M    M,
which is the macroscopic exhibition of the atomic currents.
26
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
Hence, by
y separating
p
g free and bound electrons,   B( x)  0 J (x)
[(5.22)] can be written
  B  0 ( J free    M )
(5.80)
Defining a new quantity called the magnetic field H :
 Effects of the atomic 
H  1 B  M ,  currents are implicit
p
in H.
0

we obtain from (5.80) the macroscopic version of (5.22):
  H  J free
f
Question: Does H have a physical meaning?
(5.81)
(5 82)
(5.82)
Diamagnetic Paramagnetic
Diamagnetic,
Paramagnetic, and Ferromagnetic Substances :
The counterpart of (5.81) in electrostatics is D   0E  P [(4.34)].
In Sec.
Sec 4.3,
4 3 it is shown that,
that for small displacement of the bound
electrons, we have the linear relations:
(4 36)
(4.36)
P   0  e E

(4.37), (4.38) 27
D   E, with    0 (1   e )
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
However, the magnetic properties of materials are such that M is
not always proportional to B, depending on the type of the material.
We summarize, without derivation, possible relations between B and H.
1. For diamagnetic and paramagnetic substances, M is proportional to B and we express the linear relation as
  0
   0  M  B, paramagnetic 
M
B 
(5)

0
   0  M  B, diamagnetic 
Substituting M into H  1 B  M, we get the linear relation:
0
B   H,
(5.84)
where
h  is
i called
ll d the
th magnetic
ti permeability.
bilit
Question : The plasma is diamagnetic. Why?
B
2 For
2.
F the
h fferromagnetic
i substance
b
we
H
have a nonlinear relation (see figure):
B  F (H ),
)
(5 85)
(5.85)
which exhibits the hysteresis phenomenon shown in the figure.
28
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
Boundary Conditions :
(i)   B  0  v   Bd 3 x   s B  da  0
 (B 2  B1 )  n  B1  B2
(ii)   H  J free
n : unit normal pointing
from region 1 into region 2
n
B2
region 2
    H  da   J free  da
region 1 B
1
( LHS )   H  d  (see lower figure)
n
d
d
 (H 2  H1 )  (n  n)L
K free
 n  [n  (H 2  H1 )]L
n
d
a  (b  c )  b  (c  a )
( RHS )   J ffree  nda  K free
 nL
f
 n  ( H 2  H1 )  K free
Special case: K free  0  H t 2
t: tangential to surface
((5.86))
pillbox of
infinitesimal
thickness
hi k
 L  

rectangular
loop of
infinitesimal
height
K free : surface
f
currentt off
free charges (unit: A/m) (5.87)
 H t1
(6)
29
5.8 Macroscopic Equations, Boundary Conditions on B and H (continued)
Application to a Solenoid :
i
l
i
n turns
unit length

H in  Bin / 
permeable
material
 Bout
Approximate the manetic field by that of
an infinite solenoid.
solenoid So,
So H in = constant.
constant
 H  dl  I free  H inl  nil
 H in  ni  Bin   H in   ni
 Bout  Bin   ni
B-field lines
Question : " Bout   ni " implies that
filling the solenoid core with   0
material (while keeping i at a constant
value) can greatly enhance Bout . Why?30
5.9 Methods of Solving Boundary-Value
Problems
ob e s in Magnetostatics
g e os cs
We put the basic equations :   B  0 and   H  J free (5.90)
in forms suitable for 2 types of boundary - value problems.
Type 1 : Linear medium with   const (in each region).
(a) Equation for vector potential (with or without J free )
B   H    A  H  1   A
   H  1     A  1 [(  A )   2 A ]  J free
  2 A    J free  use Coulomb gauge,   A  0
(b) Equation for scalar potential (only for J free  0)
  B  0    H  0 and   H  0  H   M
  2 M  0
(7)
(5.93)
(8)
Typically, we use (7) or (8) to solve for A or  M in each uniform
region and find the coefficients by applying conditions (5
(5.86)
86) and
(5.87) on the boundary. An example will be provided in Sec. 5.12. 31
5.9 Methods of Solving Boundary-Value Problems in Magnetostatis (continued)
Discussion:
In a vacuum medium, we have
 2 A    0J free
In a uniform- medium, we have
 2 A    J free .
I 
I
m
(5.31)
B
(7)
Hence, the effect of    0 medium is to increase the ability of
J free to produce B by a factor of  /  0 (see figure above).
In electrostatics, we have
E
 free
 p +
2
  
(vacuum medium)
(1.13)
0
 free
2
and
    f
(uniform dielectric medium)
(4.39)
Hence, an    0 medium reduces the ability of  ffree to produce
E by a factor of  /  0 (see figure above).
32
5.9 Methods of Solving Boundary-Value Problems in Magnetostatis (continued)
Type 2 : Hard ferromagnets (permanent magnet, M given, J free = 0)
(a) Vector potential
  H    ( B  M )  0
real current
0
   B  0  M  0 J M , where J M    M [see (5.79)]
B    A    B      A  (  A)   2 A  0 J M
 J (x )
  2 A   0 J M  A(x)  40  |xMx| d 3 x,
((5.102))
(b) Scalar potential
 M is a mathematical
  H  0  H  
 M tool, not real charge.
  B  0  (H  M )  0   2 M    H    M (5.95)

where  M  
  M (effective magnetic charge density)
 ( x)

(5
(5.96)
96)

(x ) 3

  M  41  |xMx| d 3 x   41  |xM
x| d x   a    a + a 

(x ) 3

 41  M (x)  |x1x| d 3 x   41    |M
xx| d x
(5.98)
33
5.9 Methods of Solving Boundary-Value Problems in Magnetostatis (continued)
Effective
ff
magnetic
g
surface
f
charge
g densityy  M :
Rewrite (5.96):   M    M
(5.96)
 v   Md 3 x   s M  da   v  M d 3 x ((see ppillbox below))
0
M


n surface area  A
 (M 2  M1 )  nA   M A
M2  0
M
 M  nM
(5.99)
M1  M
a mathematical tool thickness  0
Surface current density K M due to magnetization M:
Here  by the same algebra  ,
In Sec. 5.8,
  H  J free
  M  JM
real current
 K free  n  (H 2  H1 )
n
K free
H2
H1
real current
 KM
0
M


 n  (M 2  M1 )  M  n
KM
n M2  0
M1  M
34
5.10 Uniformly Magnetized Sphere
Consider a ppermanent magnet
g with magnetization
g
:
B
M 0e z , r  a
r P
M
0
, ra

z
 M vanishes everywhere M
a

discontinuous!
except on the surface.
M ( x) 3
nM ( x)
1
1

 M  4  |xx| d x  4  s |xx | da
d 

(3.70)
by (5.95)
4


= 3 Y 10( ,  )
1
1  4 r 


2
r
3
|
x

x
|

r

M 0a 2

   1 M a 2 r cos 
 4  d  |cos
xx| 3 0 r2
[Y1,1 ( ,  )Y1,1 ( ,  )
H
 13 M 0 r cos   13 M 0 z , r  a
Y10 ( ,  ) Y10 ( ,  )



(5.104)
 1
3 cos
ra
 43 cos
 3 M 0 a r 2 ,
Y11 ( ,  )Y11 ( ,  )]  
Inside: H in   1 M
3


Bin  0 H in  0M  32 0M ( H in  Bin )
Outside: dipole field with dipole moment m  4 a3 M.
35

3
5.12 Magnetic Shielding, Spherical Shell of Permeable
Material in a Uniform Field
B0 
Consider a spherical  -shell in an external B0 .
r
im 
m
l
a





P (cos  ) e
r
b
 2 M  0   M   l 1   l m



im



r
Ql (cos  )  e




Eq. (8)
 
 H r cos   
l
(5.117)
0
l

1 Pl (cos  ) , r  b

H
r
cos


r
0
l 0


l
gives the
1
(

r
((5.118))

 M  
l   l r l 1 ) Pl ((cos  ), a  r  b
external B0 .
l 0

   l r l Pl (cos  ),
r  a 
(5.119)

M
M

l
l 0
  




b
b
boundary conditions
 M
  M
H   M (5.93) 

 a 

 H t 2  H t1 (6)    a 
B  0 H (outside)  + 
   

B   H (inside)   B1  B2 (5.86)   0 rM    rM 
b
b

The shell is assumed to
 M
   M


0 r
be a linear medium.
 r a 
a  36
5.12 Magnetic Shielding, Spherical Shell of Permeable Material in a Uniform Field (continued)
Boundary conditions result in solutions for the coefficients:
 l  l   l   l  0 if l  1

(2  1)(  1)(b3 a3 )

3
 
0
0
H  b H0

 1 (2  1)(   2)2 a3 (  1)2 0
(5.121) 





 

0
0
b3 0
  0
& 



9 
9 0
 (5.122) 
0
1 
H

H
0
0
3 
3


2
a
a

(2  1)(   2)2 3 (  1)
2  (1  3 )
0
0
0

b
b

Bin  as  , implying that   0
B  B + dipole field
0
materials tend to "absorb"
absorb B-field
field lines and
thereby provide the shielding effect. High-
materials can have  / 0 as high
g as 103  106.
When   0 , B reduces to B0 everywhere,
i.e. a static megnetic field penetrates into the
shell as if there were no shell present (even if the
shell is made of good conductor, such as copper).
out
0
Bin: uniform field
(  1/  if   0 )
37
5.15 Faraday’s Law of Induction
The Biot
Biot-Savart
Savart (or Ampere's)
Ampere s) law relates the magnetic field to
electrical current. Faraday then discovered experimantally that
time-varying magnetic flux through an electrical circuit could
induce an electric field around the circuit. This not only provided
the first link between electric and magnetic fields, but also led to a
new mechanism
h i to
t generate
t the
th E-field,
fi ld i.e.
i a time-varying
ti
i B-field.
fi ld
Referring to the figure, let loop C
loop C
be an electrical circuit (as in Faraday
Faraday'ss
B( x , t )
original experiment) or any closed path
n
d
da
p
((a ggeneralization of the original
g
in space
observation with immense consequences).
Faraday's law states
S : an arbitrary surface 


B
 c E  d    s t  nda,  bounded by loop C  (5.141)


where
h E iis the
h electric
l i field
fi ld at d  in
i the
h frame
f
iin which
hi h d 
 is
i at
rest, and B is the magnetic induction in the lab frame.
38
5.15 Faraday’s Law of Induction (continued)
R i (5.141):
Rewrite
(5 141)
 c E  d    s Bt  nda
Assume loop C is at rest in the lab
frame, then E  E (electric field in the
lab frame) and (5.141) becomes
loop C
B(x, t )
d
n
(5.141)
da
 c E  d    s Bt  nda integral form of Faraday's law 
((9))
where both E and B are lab-frame quantities.
(9) can be
b written
itt (by
(b Stokes's
St k ' th
theorem:  c E  d   s   E  nda
d )
s   E  nda   s Bt  nda
Thus,
  E   B  differential form of Faraday's
y law)) 
((5.143))
t
39
5.16 Energy in the Magnetic Field
To find the energy associated with a magnetic field,
field we evaluate
the work needed to establish the current J (x), which produces the
magnetic field. We break up J (x) into a network of thin loops. In
the build-up process, an E field will be induced by B / t. The
rate of work done by E within each loop is
 is the cross section of the loop (same as integration over the area
Jackson's  ). J &  may vary along d . encircled by the loop

dWloop
n
a thin
hi
   J  E  d    s J  

E

n
da

dt
loop
B / t
 s J  n  B da
t
St k ' th
Stokes's
thm.
S
Work done within each loop to generate  B :
d  (d   J )

 Wloop  s J  n  
B da  s J    A  nda  : cross section of loop
  A
  J  A  d 
 loop  A  Jd 3 x
J  d   J 
d   Jd 3 x
d 3x
(10) 40
5.16 Energy in the Magnetic Field (continued)
As shown in  Wloopp  loopp  A  Jd 3 x [(10)], the work done within
each loop is an integral over the volume of the loop. Thus, an integration over all space gives the total work done to generate  B :
Assume the rate of build-up  0  H obeys the static law   H  J.
Otherwise, the static law breaks down and there will be radiation loss.
 W    A  Jd 3 x    A  (  H )d 3 x
  H  (   A ) d 3 x     (H   A ) d 3 x




B
  s ( H A )da0
  H   Bd 3 x  12   (H  B)d 3 x

Assume linear medium: B   H or B  μ  H
(5 144)
(5.144)
For this integral
to vanish,
vanish the
volume of integration must be .
Total work done to bring the field up from 0 to the final value B :
B conservation
ti off energy, this
thi is
i 
W  12  (H  B)d 3 x  By
(5.148)
 the total magnetic field energy. 
gy density]
y]
((11))
 w  12 H  B [[field energy
Note: w  12 H  B  12 (  H j )  (  B j )  12  ( H j  B j )
j
j
j
41
5.17 Energy and Self- and Mutual Inductances
Assume linear relation between J and A
for nonpermeable
3
3
 W    A  Jd x  12   ( A  J )d x
(μ0) medium (5.144)
0 J (x) 3 
3
1
A
(
x
)
d x (5.32)

 W  2  A  Jd x
(5.149)
4  |xx|

J (x)J (x)
f N currentfor
t
 0  d 3 x  d 3 x
(5.153)
8
|xx|
carrying circuits
N N
0 N 3 N 3  J (xi )J (xj ) 1 N
2

 d xi   d x j |x x |  2  Li I i    M ij I i I j , (5.152)
8 i
i
j
1
j 1
i 1
i 1 j i
self-inductance
for a thin wire
d  i d i 
0
3
3 J ( xi )J ( xi )  0

Li 
d xi C d xi
i
|xi xi |  4  C i  C i |xi xi | 
4 Ii2 C i
where
(5.154)
d  i d j 
J (xi )J (xj )  0
0
3
3
M ijj 
d xi C d xj

(5.155)

C
4 Ii I j i
j
|xi xj |  4 C i C j |xi xj | 
mutual inductance (Mij = Mji)
for thin wires
42
5.17 Energy and Self- and Mutual Inductances (continued)
0 J (x) 3 
A ( x) 
d x
4  |xx|
(5 32)
(5.32)
 Vector potential at circuit i due to current in circuit j :
J (xj ) 3
0
(12)
A ij (xi )  4  C j |x x | d xj
i
j
From (12) and (5.155), we obtain M ij  I 1I C Aij (xi )  J (xi )d 3 xi
i j i
Assume J flows along wire d  of negligible cross section da
 J (xi )d 3 x i  J  dad   I i d 
B
magnetic flux from circuit j
passing through circuit i
ij


 M ij  I1  C Aij  d   I1  s (  Aij )  nda  I1 Fij
(5.156)
i
i
j
j
j
ij: induced voltage in circuit i due
d
d
  ij   Fij   M ij I j
to current variation in circuit j.
dt
dt
Th “–”
The
“ ” sign
i implies
i li that
th t the
th induced
i d d ij tends
t d to
t drive
d i a currentt in
i
circuit i to inhibit the flux change caused by circuit j (Lenz’s law). 43
Homework of Chap. 5
Problems: 1,, 3,, 6,, 11,, 13,,
18, 19, 20, 22, 30
44