Ch5. 靜磁學(Magnetostatics)
5.1 羅倫茲力定律(The Lorentz Force Law)
5.2 必歐沙伐定律(The Biot-Savart Law)
5.3 磁場的散度與旋度(The Divergence and Curl of B)
5.4 磁向量位勢(Magnetic Vector Potential)
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.1 The Lorentz Force Law-磁場(Magnetic Fields)
Bar magnet
Conducting wire
Parallel wires carrying currents in the
same direction attract each other
Parallel wires carrying currents in
opposite direction repel each other
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.2.1 磁力(Magnetic Forces)
  

 
The Lorentz force F  FE  FB  qE  qv  B
Additional speed parallel to B

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.2.2 磁力(Magnetic Forces)
z
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.2.3 磁力(Magnetic Forces)
y( t )   B cos t  A sin t 
E0
tE
B0
z( t )  A cos t  B sin t  C
But the particle started from rest ( y
 (0)  z (0)  0 ), at the origin ( y(0)  z(0)  0 );
E0
y( t )   B cos t  A sin t 
tE
B0
y(0)   B  E  0
z( t )  A cos t  B sin t  C
z(0)  A  C  0
E B0
y ( t )  B sin t  A cos t 
y (0)  A 
E0
0
B0
z ( t )   A sin t  B cos t
z (0)  B  0
A
E0
 C
B 0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
E0
B0
5.1.2.4 磁力(Magnetic Forces)
E0
E
t  E  0 (t  sin t )
B0
B0
E0
z( t )  A cos t  B sin t  C 
(1  cos t )
B0
E0
R
B 0
y( t )   B cos t  A sin t 
y( t )  Rt  R sin t
[ y(t )  Rt ]2  R 2 sin 2 t
z( t )  R  R cos t
[z( t )  R ]2  R 2 cos 2 t
[ y( t )  Rt ]2  [z( t )  R ]2  R 2
This is the formula for a circle, of radius R, whose center (0, Rt, R) travels in
the y-direction at a constant speed
E0
v  R 
B0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.3.1 電流(Currents)
Example 5.3 : A rectangular loop of wire, supporting a mass m, hangs vertically
with one end in a uniform magnetic field B, which points into the page in the
shaded region. For what current I, in the loop, would the magnetic force
upward exactly balance the gravitational force downward?
Fmag  IBa  mg
I
mg
Ba
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.3.2 電流(Currents)
What happens if we now increase the current?
Fmag  IBa  mg
The loop rises a height of h
Wmag  Fmag h  IBah  0
?
quB

Fmag
Fvert  qwB  awB  IaB
qwB
v
w
u
Rise the loop
Fhoriz  quB  auB
Wagent   Fhorizwdt   auBwdt  aB uwdt  IBah
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.3.3 電流(Currents)
Example 5.4
a) A current I is uniformly distributed over a wire of circular cross section,
with radius a. Find the volume current density J.
I
J 2
a
b) Suppose the current density in the wire is proportional to the distance
from the axis, J = ks. Find the total current in the wire.
The current in the shaded path
Jda   J(sd)(ds)  ks 2dsd
3
2

ka
I   Jda     ks 2 dsd  2k  s 2 ds 
3
0
a
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.1.3.4 電流(Currents)
The current crossing a surface S can be written as
 
I   Jda    J  da
S
S
The total charge per unit time leaving a volume V is
 

 J  da   (  J )d
S
V
Because charge is conserved

d

(


J
)
d




d



(
)d
V


dt V
t
V


J  
t
Continuity equation
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.1 穩定電流(Steady Currents)
Continuity equation


J  
0
t
Actually, it is not necessary that the charges be stationary, but only
that the charge density at each point be constant ( be independent
with time).
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.2 穩定電流的磁場(The Magnetic Field of a Steady Current)
Example 5.5
Find the magnetic field a distance s from a long straight wire carrying a
steady current I
s

r


I
ℓ
dℓ
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.3 穩定電流的磁場(The Magnetic Field of a Steady Current)

  0 I d   r̂
B
4  r 2
  s tan 
s
s
d 
d
2
cos 

d   r̂  d sin   d cos  
B
0I
4 

d   r̂
r2


r

2
1

2
I
ℓ
dℓ
1 cos 2 

2
r
s2
s
d
cos 
2
0I
s
cos   0 I
d


cos d
2


4 1 cos 
s
4s 1
2
0I
I
I

(sin  2  sin 1 )  0 [1  (1)]  0
4s
4s
2s


1   ,  2 
2
2
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.4 穩定電流的磁場(The Magnetic Field of a Steady Current)
The field at 2 due to 1 is
 0 I1
B
2d
into the page

 
F  I d   B
 0 I1
)  d
2d
The force on 2 due to 1 is
F  I2 (
The force per unit length
 0 I1I 2
F
2d
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.5 穩定電流的磁場(The Magnetic Field of a Steady Current)
Example 5.6
Magnetic Field on the Axis of a Circular Current Loop

 0 I d s  r̂  0 I
ds
dB 

2
4 r
4 ( x 2  R 2 )
 0 I ds cos 
Bx   dB cos  

4 ( x 2  R 2 )
cos  
R
( x 2  R 2 )1/ 2
 0 IR
 0 IR 2
Bx 
ds 
2
2 3/ 2 
4( x  R )
2( x 2  R 2 ) 3 / 2
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.6 穩定電流的磁場(The Magnetic Field of a Steady Current)
For surface and volume currents, the Biot-Savart Law becomes

  0 K( r ' )  ˆ
B( r ) 
da '
2

4


   0 J ( r ' )  ˆ
B( r ) 
d'
2

4

Problem 5.8
a) Find the magnetic field at the center of a square loop, which carries a steady
current I. Let R be the distance from center to side.
R
B
0I
I
2 0 I
2
2
(sin  2  sin 1 )  0 (

)
4s
4R 2
2
4R
for
Four sides
s  R, 2  450  1
B4
2 0 I
2 0 I

4R
R
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.7 穩定電流的磁場(The Magnetic Field of a Steady Current)
b) Find the field at the center of a rectangular n-sided polygon, carring a steady
current I. Let R be the distance from center to any side.
0I
0I
0I



B
(sin  2  sin 1 ) 
(sin  sin ) 
sin
4s
4R
n
n
2R
n

s

R
,


  1
for
2
n
n sides
Bn
0I

sin
2R
n
c) Check that your formula reduces to the field at the center of a circular loop,
in the limit n  ∞.
n∞
 
sin  2   sin 1  sin 
n n
n 0 I
 n 0 I   0 I
B
sin 

2R
n 2R n 2R
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.2.8 穩定電流的磁場(The Magnetic Field of a Steady Current)
Problem 5.9
Find the magnetic field at point P for each of the steady current configurations.
The vertical and horizontal lines produce no field at P.
b
I
P
The two quarter-circles
a
0I 1 1
1 0I 0I
B (

)
(  )
4 2a 2b
8 a b
R
I
The two half-lines are the same as one infinite line
P
I
The half-circle:
0I
B
2R
0I
B
4R
Total:
B
0I
2
(1  )
4R

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.1.1 磁場的散度與旋度– 直線電流(Straight-Line Currents)
The magnetic field of an infinite straight wire
 
0I
0I
 B  d    2s d  2s  d  0 I
Notice that the answer is independent of s; that’s because B decreases at the
same rate as the circumference increases.
In fact, it doesn’t have to be a circle; any old loop that encloses the wire would
give the same answer. For if we use cylindrical coordinates (s,,z), with this
current flowing along the z axis,
 0I

ˆ
B

d   dsŝ  sdˆ  dzẑ
2s
2
  0I 1
0I
 B  d   2  s sd  2 0 d   0 I
This assumes the loop encircles the wire exactly once.
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.1.2 磁場的散度與旋度– 直線電流(Straight-Line Currents)
Suppose we have a bundle of straight wires. Each wire that posses through
our loop contributes 0I, and those outside contribute nothing.
 
 B  d   0Ienc
I4
I3
I2
I enc : total current enclosed by the integration path
I1
If the flow of charge is represented by a volume current density J, the enclosed
 
current is
Applying Stokes’ theorem
Ienc   J  da
 
 
 
 B  d   (  B)  da  0Ienc  0  J  da


  B  0 J
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.2.1 磁場的散度與旋度(The Divergence and Curl of B)

r
d’
(x’,y’,z’)
(x,y,z)
The Biot-Savart law for the general case of a
volume current reads

   0 J ( r ' )  ˆ
B( r ) 
d'
2

4

 
  0
 




J ( r ' )  ˆ
  B( r ) 
 [
]d'
  (A  B)  B  (  A)  A  (  B)
2

4

 
 
 
ˆ
ˆ
ˆ
  [ J ( r ' )  2 ]  2  [  J ( r ' )]  J ( r ' )  (  2 )



 
  J ( r ' )  0 Because J doesn’t depend on the unprimed variables (x,y,z)
 
 
ˆ
ˆ
  [J( r ' )  2 ]  0
 2  0
  B( r )  0


The divergence of the magnetic field is zero!
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.2.2 磁場的散度與旋度(The Divergence and Curl of B)
 

  0
   0 J ( r ' )  ˆ
J ( r ' )  ˆ
 [
]d'
B( r ) 
d'   B( r ) 
2

2

4

4

 




 


  (A  B)  (B   )A  (A   )B  A(  B)  B(  A)
 
 
 
 
ˆ
ˆ
ˆ  
ˆ
ˆ
  [ J ( r ' )  2 ]  ( 2   ) J ( r ' )  [ J ( r ' )   ] 2  J ( r ' )(  2 )  2 [  J ( r ' )]





=0
=0
 
 
ˆ
ˆ  
ˆ
  [ J ( r ' )  2 ]  [ J ( r ' )   ] 2  J ( r ' )(  2 )




ˆ
3 
3 
  ( 2 )  4 ()  4 ( r  r ' )

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.2.3 磁場的散度與旋度(The Divergence and Curl of B)
 
 
ˆ
ˆ
 [J( r ' )  ] 2  [J( r ' )  ' ] 2




f (x  x' )  
f (x  x' )
x
x '
 
x  x'
The x component [ J ( r ' )   ' ]( 3 )







  (fA)  f (  A)  A  (f )
A  (f )    (fA)  f (  A)
 
 
x  x'
x  x'  
x  x'
[ J ( r ' )   ' ]( 3 )   '[ 3 J ( r ' )]  ( 3 )[ ' J ( r ' )]



 
For steady currents the divergence of J is zero  ' J ( r ' )  0
 
x  x'
x  x'  
[ J ( r ' )   ' ]( 3 )   '[ 3 J ( r ' )]


This contribution to the integral can be written

x  x'  
x  x'  
V  '[ 3 J ( r ' )]d'  S 3 J ( r ' )  da '  0
On the boundary J = 0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.2.4 磁場的散度與旋度(The Divergence and Curl of B)
 
  0
J ( r ' )  ˆ
  B( r ) 


[
]d'
2

4

 
 
  3 
  3  
ˆ
ˆ
  [ J ( r ' )  2 ]  J ( r ' )(  2 )  4J ( r ' ) ()  4J ( r ' ) ( r  r ' )


 
 
  0
  3  
0
J ( r ' )  ˆ
  B( r ) 
 [
]d' 
4J ( r ' ) ( r  r ' )d'  0 J ( r )
2


4

4
 

  B( r )  0 J ( r )
Ampère’s law in differential form
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.1 安培定律的應用(Applications of Ampère’s Law)
Ampère’s law in differential form
Ampère’s law in integral form
 

  B( r )  0 J ( r )
 
 
 
 (  B)  da   B  d  0  J  da  0Ienc
Example 5.8

Find the magnetic field of an infinite uniform surface current K  Kx̂ ,
flowing over the xy plane
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.2 安培定律的應用(Applications of Ampère’s Law)
What is the direction of B?
From the Biot-Savart law
 
BK
Could it have a z-component ?  no (symmetry)
The magnetic field points to the left above the plane and to the right below it
 
 B  d   2B  0Ienc  0K
 0K
ŷ
 
B 2
K
  0 ŷ
 2
B
0K
2
for z  0
for z  0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.3 安培定律的應用(Applications of Ampère’s Law)
Example 5.9
Find the magnetic field of a very long solenoid, consisting of n closely wound
turns per unit length on a cylinder of radius R and carrying a steady current I.
 
 
 B  d s   B  d s  B  ds  B
side1
 
 B  d s  B   0 NI
side1
Where N is the number of turns in the length 
N
B   0 I   0 nI

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.4 安培定律的應用(Applications of Ampère’s Law)
Example 5.10
Find the magnetic field of a toroidal coil, consisting of a circular ring around
which a long wire is wrapped.
 
 B  d s  B ds  B(2r )   0 NI
 0 NI
B
2 r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.5 安培定律的應用(Applications of Ampère’s Law)
Problem 5.13
A steady current I flows down a long cylindrical wire of radius a. Find the
magnetic field, both inside and outside the wire, if
(a). The current is uniformly distributed over the outside surface of the wire.
(b). The current is distributed in such a way that J is proportional to s, the
distance from the axis.
a
(a)
I
 
 B  d   2sB  0Ienc
  0 for s  a
B   0I ˆ
 2s  for s  a
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.3.5 安培定律的應用(Applications of Ampère’s Law)
(b)
J  ks
2ka 3
I   Jda   ks(2s)ds 
3
0
3Is
J  ks 
2a 3
a
k
3I
2a 3
 
 B  d   2sB  0Ienc
For s < a
s
 
3 0 I 1 3  0 Is 3
3Is
 B  d   2sB  0 Ienc  0 0 2a 3 2sds  a 3 3 s  a 3
 0 Is 2
B
2a 3
For s > a
a
 
3 0 I 1 3
3Is
 B  d   2sB   0 Ienc  0 0 2a 3 2sds  a 3 3 a   0 I
B
0I
2s
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.3.4. 靜磁學與靜電學之比較(Comparison of Magnetostatics
and Electrostatics)
The divergence and curl of the electrostatic field are
  1
  E  

 0
 E  0

Gauss’s law
The divergence and curl of the magnetostatic field are

 B  0



  B   0 J
Ampère’s law
“The electric force is stronger than the magnetic force. Only when both the
source charge and the test charge are moving at velocities comparable to
the speed of light, the magnetic force approaches the electric force.”
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.1. 磁向量位勢(Magnetic Vector Potential)

E  0

E  V
V: electric scalar potential
You can add to V any function whose gradient is zero
  (V  f )  V  f  V




A: magnetic vector potential
B  0
B  A

You can add to A any function whose curl is zero



  ( A   )    A        A





2
  B    (  A)  (  A)   A  0 J

We will prove that   A  0 :


B  A
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.2. 磁向量位勢(Magnetic Vector Potential)

Suppose that our original vector potential A 0 is not divergenceless

  A0  0

Because we can add to A 0 any function whose curl is zero
 



A  A0  
  A    A0        A0   2
If a function  can be found that satisfies

A  0

     A0
2
Mathematically identical to Poisson’s equation  V  
2

0
In particular, if  goes to zero at infinity, then the solution is
V
1
40

  d '
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.3. 磁向量位勢(Magnetic Vector Potential)

By the same token, if   A 0 goes to zero at infinity, then

1   A0

d'

4

It is always possible to make the vector potential divergenceless
so

A  0





2
2
  B  (  A)   A   A  0 J
This again is a Poisson’s equation
Assuming

J goes to zero at infinity, then

  0 J ( r ' )
A( r ) 
d'

4

For line and surface currents

  0 I ( r ' )
A( r ) 
d'

4


   0 K( r ' )
A( r ) 
da '

4

Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.4. 磁向量位勢(Magnetic Vector Potential)
Example 5.11
A spherical shell, of radius R, carrying a uniform surface charge
, is set spinning at angular
velocity . Find the vector potential

it produces at point r .

r


r'
The integration is easier

if we let r lie on the z
axis, so that  is titled at
an angle .
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.5. 磁向量位勢(Magnetic Vector Potential)
 
 

 0 K( r ' )

2
2
where K  v
A( r ) 
da
'


R

r
 2Rr cos '

4

x̂
ŷ
ẑ
  
v   r ' 
 sin 
0
 cos 
R sin ' cos ' R sin ' sin ' R cos '
 R[ (cos  sin ' sin ' ) x̂  (cos  sin ' cos ' sin  cos ' ) ŷ  (sin  sin ' sin ' )ẑ]
2
Because
2
 sin ' d'  
0
cos ' d'  0 We just consider

v   R sin  cos ' ŷ
0
 
 0 ( R sin  cos ' ) 2
A( r ) 
R sin ' d' d'ŷ

2
2
4
R  r  2Rr cos '

 0 R 3 sin 
sin ' cos '

(
d') ŷ
2
2
2
R  r  2Rr cos '
0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.6. 磁向量位勢(Magnetic Vector Potential)
u  cos ' , the integral becomes
Letting

sin ' cos '
I
R  r  2Rr cos '
2
0
2
1
d' 

1
u
R  r  2Rru
2
2
du
1
u
1
2
2
3/ 2
 [
R 2  r 2  2Rru 
(
R

r

2
Rru
)
]
2 2
Rr
3R r
1
 [

1
(R  r  Rru )
2
2
R

r
 2Rru ]
2 2
3R r
1
2
2
1
2
2
2
2
[(
R

r

Rr
)
R

r

(
R

r
 Rr )( R  r )]
2 2
3R r

r
2r
3R 2

2R
If the point r lies outside the sphere, Then R < r.  I 
3r 2
If the point
lies inside the sphere, Then R > r. 
I
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.7. 磁向量位勢(Magnetic Vector Potential)

 
 0 R 3 sin 
 0 R 3 sin 
sin ' cos '
A( r )  
(
d') ŷ  
Iŷ
2
2
2
2
R  r  2Rr cos '
0

r
2r
If the point
lies inside the sphere, Then R > r.  I 
3R 2

2R
If the point r lies outside the sphere, Then R < r.  I 
2
3
r
 
(

 r )   r sin ŷ
Noting that
For the point inside the sphere
 
 0 R 3sin 
 0 R 3sin  2r
 0 R  
A( r )  
Iŷ  
ŷ

(  r )
2
2
2
3R
3
For the point outside the sphere
 
 0 R 3sin 
 0 R 3sin  2R
0 R 4  
A( r )  
Iŷ  
ŷ 
(  r )
2
3
2
2
3r
3r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.8. 磁向量位勢(Magnetic Vector Potential)

We revert to the original coordinates, in which  coincides
with the z axis and the point r is at (r,,)
For the point inside the sphere
   0 R    0 R
A( r ) 
(  r ) 
r sin ˆ
3
3
 

2 R
2 R
B    A( r )  0
(cos r̂  sin ˆ )  0
ẑ
3
3
The magnetic field inside this spherical shell is uniform!
For the point outside the sphere
   0 R 4    0 R 4 
A( r ) 
(  r ) 
sin ˆ
3
2
3r
3r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.1.9. 磁向量位勢(Magnetic Vector Potential)
Example 5.12
Find the vector potential of an infinite solenoid with n turns per
unit length, radius R, and current I
 
  0 I ( r ' )
We cannot use A( r ) 
d' because the current itself extends

4

to infinity.
 

 

Notice that
 A  d    (  A)  da   B  da   
Since the magnetic field is uniform inside the solenoid : B  0 nI ẑ
 
 
  0 nIs
2
A
ˆ
For s < R
 A  d   A(2s)   B  da  0nI (s )
2
For an amperian outside the solenoid
 
 
2
A

d


A
(
2

s
)

B

d
a


nI
(

R
)
0


  0 nIR 2
A
ˆ For s > R
2s
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.2.1. 總結;靜磁邊界條件(Summary; Magnetostatic
Boundary Conditions)

J
 
   0 J ( r ' )  ˆ
B( r ) 
d'
2

4

 
  0 J ( r ' )
A( r ) 
d'

4



 A   0 J


  B  0 J
2

A
 
A  B

B
?
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.2.2. 總結;靜磁邊界條件(Summary; Magnetostatic
Boundary Conditions)

B  0

 
 (  B)d   B  da  0

Babove
 B below
V
For an amperian loop running perpendicular to the current
 
//
//
B

d


(
B

B
above
below )  0 I enc  0 K

//
//
Babove
 Bbelow
 0K
Perpendicular to the current
For an amperian loop running parallel to the current
 
//
//
B

d


(
B

B
)  0 Ienc  0
above
below

//
//
Babove
 B below
Parallel to the current



Babove  Bbelow  0 (K  n̂)
Where
n̂ is a unit vector perpendicular to the surface, pointing “upward”.
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.2.3. 總結;靜磁邊界條件(Summary; Magnetostatic
Boundary Conditions)
The vector potential is continuous across any boundary
 


 A  d   (  A)  n̂da   B  n̂da  
For an amperian loop of vanishing thickness
0


Aabove  Abelow
//
//
Babove
 Bbelow
x̂
ŷ
ẑ

A
(  A) y   / x  / y  / z 
 B
 0K
z
A
0
0 y



A above A below

  0 K
n
n
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.3.1. 向量位勢的多極展開(Multipole Expansion of the
Vector Potential)
A multipole expansion which is an approximate formula and valid at
distant points) for the vector potential of a localized current distribution.

r

'

r'
I


dr '  d '
1
1
1  r' n

  ( ) Pn (cos ' )
2
2

r  (r ' )  2rr ' cos ' r n  0 r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.3.2. 向量位勢的多極展開(Multipole Expansion of the
Vector Potential)
  0I 1  0I  1

n
A( r ) 
d ' 
(r ' ) Pn (cos ' )d '

n 1 

4 
4 n  0 r
  0I 1  1
 1
1 
2 3
2
A( r ) 
[  d '  2  r ' cos ' d '  3  (r ' ) ( cos ' )d '  ....]
4 r
r
r
2
2
monopole

 d '  0
dipole
quadrupole
The magnetic monopole tern is always zero

B  0
 

 
0I
0I
A dip ( r ) 
r ' cos ' d ' 
(r̂  r ' )d '
2 
2 
4r
4r

 
Let v  cT , where c is a constant vector.



 

c
T

d

'

(

'

c
T)  n̂ ' da '


 v  d '   ('v)  n̂' da '
area
area
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.3.3. 向量位勢的多極展開(Multipole Expansion of the
Vector Potential)






 '(cT)  ( 'c)T   ' T  c  ( 'c)T  c   ' T   c   ' T




c
T

d

'


n̂
'

(
c


'
T
)
da
'


c


  (' T  n̂' )da '
area


Let
T  r̂  r '
 Td '    (' T  n̂' )da '
area
 

 (r̂  r ' )d '    [' (r̂  r ' )  n̂' ]da '
area
 
 (r̂  r ' )d '    (r̂  n̂' )da '   r̂ 
area
area

n̂
'
da
'


r̂

a

area
 1  
 
1  
a   r 'd '   n̂ 'da '
(r̂  r ' )d '   r̂  (  r 'd ')

2
2
 
 

0I
0I
0I 
A dip ( r ) 
(r̂  r ' )d ' 
( r̂  a ) 
(a  r̂ )
2 
2
2
4r
4r
4r

 
 m  r̂

 
where m  I da  Ia
A dip ( r )  0 2

4 r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.3.4. 向量位勢的多極展開(Multipole Expansion of the
Vector Potential)
Example 5.13
Find the magnetic dipole moment of the “Bookend-shaped” loop shown
in Figure below. All sides have length w, and it carries a current I.
w
w
w
The wire could be considered the superposition of two plane square
loops shown in Fig. 5.53 of text book.
The combined (net) magnetic dipole moment is

m  Iw 2 ŷ  Iw 2 ẑ
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
5.4.3.5. 向量位勢的多極展開(Multipole Expansion of the
Vector Potential)
The magnetic dipole moment is independent of the choice of origin.
The magnetic field of a (pure) dipole is easier to calculate if we put the
dipole moment at the origin and let it point in the z-direction.

 
 m  r̂
A dip ( r )  0 2
4 r
 


Bdipole( r )    A( r ) 
1
r 2 sin 
 
 m sin  ˆ
A dip ( r )  0

2
4 r
r̂
rˆ
 / r  / 
0
0
r sin ˆ
 / 
 0 m sin 
r sin (
)
2
4r
0m

(2 cos r̂  sin ˆ )
3
4r
This is identical in structure to the field of an electric dipole!
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung