Magnetostatics
Biot-Savart Law
 
 

B
(r
)
J
The magnetic field
at point r due to a current density (r ) is
 
 
1   
r  r'
B(r )   dr ' J (r ' )    3
c
| r  r '|
Here c is the speed of light. We note that the magnetic field has the same dimensions as the
electric field.
Exercise: Find the magnetic field due to an infinite straight wire carrying a constant current I.
Solution: In cylindrical coordinates, in which the wire is put along z, the current density is

J  zˆI (  '2 )
Since the current is along the z-direction, then
 
zˆ  (r  r ' )  zˆ  ˆ (    ' )  ˆ . We hence
have

 
I
1
B(r )  ˆ  dz '
c
( z  z' )2   2


and the magnetic field depends only on


3/ 2
I
1
 ˆ  dz '
c
z '2   2



3/ 2

2I
ˆ
c
.
Exercise: Find the magnetic field due to a circular ring (of radius a) carrying a constant current,
on the symmetry axis.
Solution: We put the ring in the x-y place, and then the symmetry axis is the z-direction. In
cylindrical coordinates, the current is along the
 
r  r '  zzˆ  a̂
̂

direction. Then,

ˆ  (r  r ' )  zˆ  azˆ
1
Biot-Savart law gives
2
 
I
azˆ  zˆ
I
a2
B(r )   ad
 zˆ
2
2 3/ 2
c 0
c a2  z 2
a z




3/ 2
2
and the magnetic field depends only on z.
Exercise: Find the magnetic field at the center of a wire square of side a carrying a current I.
Solution: Let us calculate the contribution from the side along the x-direction. Putting the
coordinate origin at the center of the square, we have
 /4
 I a/2
xˆ   xxˆ  (a / 2) yˆ 
I 2
I 2
ˆ
ˆ
B
dx

z
d

cos


z
2
3/ 2
c a/ 2
c a / 4
ca
x 2  (a / 2) 2


Calculating the contributions of the other three sides in the same way, the final result is

I 2
B  zˆ
4 2
ca
Let us now calculate the magnetic field due to this square current loop on the z axis. Using



I d   ( zˆz  r ' )
B( z )  

c
| zˆz  r ' |3
We have the contributions of the four segments of the wire as follow:
a/2
a/2

I
xˆ  ( zˆz  xˆx  yˆ (a / 2)) I
yˆ  ( zˆz  xˆ (a / 2)  yˆy)
B( z )   dx 2

dy
c a / 2
( z  (a / 2) 2  x 2 ) 3 / 2 c a/ 2
( z 2  (a / 2) 2  y 2 ) 3 / 2
I

c
a / 2
a / 2
xˆ  ( zˆz  xˆx  yˆ (a / 2)) I
yˆ  ( zˆz  xˆ (a / 2)  yˆy)
dx

dy
2
2
2
3
/
2
 ( z  (a / 2)  x )
c a/ 2
( z 2  (a / 2) 2  y 2 ) 3 / 2
a/2
(note the integration limits!). Collecting terms, this expression becomes
2

2I
B( z ) 
c
xˆ  yˆ (a / 2)
2I
dx

2
2
2
3
/
2
 ( z  (a / 2)  x )
c
a / 2
a/2
a/2

a / 2
dy
yˆ  ( xˆ (a / 2))
( z  (a / 2) 2  y 2 ) 3 / 2
2
Now we see that the resulting magnetic field is along the z-direction. Moreover, the two integrals
are identical, and therefore
a/2

4I a
1
2 Ia
1
a/2
ˆ
B( z ) 
zˆ  dx 2

z
2
c 2 a / 2 ( z  (a / 2) 2  x 2 ) 3 / 2
c
z 2  (a / 2) 2
z 2  2(a / 2) 2

I 2a 2
ˆ
B( z )  z
c z3
In particular, for z>>a, this becomes
Forces between current carrying conductors


The force exerted by a magnetic field B on a charge q moving with velocity v is
 

vB
F q
c


d
I

I
d
Hence the force experienced by a current element
1
1 1` is
 I 

dF  1 d 1  B
c
When the magnetic field is due to another (closed) current loop we have

II
F12  1 22
c


 
d 1  d  2  r12
  | r12 |3

3
Exercise: Find the force acting between two circular current loops, of radii a and b respectively,
placed one on the top of the other such that the z-axis passes through both centers. The
distance between the two centers is d.
Solution: Let us use polar coordinates, such that for the lower ring

r1  xˆa cos 1  yˆa sin 1  arˆ1
and

d 1  ( xˆa sin 1  yˆ a cos 1 )d1  aˆ1d1
Then

r2  xˆb cos 2  yˆb sin 2  zˆd  bcos(2  1 )rˆ1  sin( 2  1 )ˆ1   dzˆ
and

d 2  ( xˆb sin  2  yˆb cos  2 )d 2  bd 2  sin(  2  1 )rˆ1  cos( 2  1 )ˆ1 
Thus,


(d  2  r12 )  (bd 2 )(b  a cos( 2  1 )) rˆ1  d cos( 2  1 ) zˆ 
and



d 1  (d  2  r12 )  (ab)( d1d 2 )(  zˆd cos(1   2 )  rˆ1 (b  a cos(1   2 )))
But since
r122  d 2  a 2  b 2  2ab cos(1   2 )
we can do the integrations separately on
1
and
1  2 .
We are left with the contribution of
the z-direction alone. Hence,
2

I1 I 2
cos 
F12   zˆ 2 dab2  d
c
d 2  a 2  b 2  2ab cos  3 / 2
0
4
Magnetostatics and Ampere’s law
The Biot-Savart law is

  1  J (r' )  (r  r' )
B(r )   dr '
 
c
| r  r ' |3
  
 
  (r ' )( r  r ' )
E (r )   dr '
 
.)
| r  r ' |3
(Compare with
We can write the above equation in the form
 
  1
 J (r ' )
B(r )     dr '  
c
| r  r '|
and then
  
  B(r )  0
Next we consider the rotor of the magnetic field:
 
   1 
 J (r ' )
  B(r )       dr '  
c
| r  r '|
Using

 

  
  
    V     V   2V
which holds for any vector, we find
  
 1 
1    
1
1   
  B(r )    dr ' J (r ' )       dr ' J (r ' ) 2    
c
| r  r '|
c
 | r  r '| 
1    
1
4  
    dr ' J (r ' )  '   
J (r )
c
| r  r '|
c
By integrating by parts the first term we have
5
  
  


 'J (r ' )
4  1
  B( r ) 
J (r )    dr '  
c
c
| r  r '|
Now we use the continuity equation, by which

 (r )   
   J (r )  0
t
to finally obtain

  1 E 4 
 B 

J
c t
c
(For a stationary situation, the time derivative does not appear. Also, compare this equation with
 
  E  4 ).
Applying Stokes theorem, the surface integral
 
4
ˆ


B

n
da

S
c

J
  nˆda
S
becomes
  4
C B  d   c

4
ˆ
J

n
da

I
S
c
which is, again, Ampere's law.

Exercise: Find the magnetic field due to a surface current K density on a plane perpendicular to
the z-axis.
Solution: Let us denote the direction of the current density on the plane as the x-direction.
Applying Ampere's Law
6
  4
B
C  d   c

4
ˆ
J

n
da

I
S
c
for a rectangular loop of a length L along y and a tiny (goes to zero) length c along z, we obtain
B y ( z  0) L  B y ( z  0) L 
4
KL
c
Therefore,
2

K, z  0
  yˆ
c
B
2
 yˆ
K, z  0
c

Conclusion: The tangential component of the field has a discontinuity at a current-carrying
surface.
Exercise: Find the magnetic field due to a conducting infinite slab perpendicular to the zdirection, located in the region  a  z  a , in which there is a current density
Solution: Firstly, the field cannot depend on x or on y. Then by

J  xˆJ .
 
  B  0 we find that
dBz / dz  0 and hence Bz  0 . (It cannot be simply a constant, being the same close and far
 

away from the slab.) Secondly, by   B  ( 4 / c) J we find that
dB y
dz
 (4 / c) J
Then,
By ( z)  
4
Jz
c
a z a
Now we use continuity (there is no discontinuity in the tangential component of the field, because
there is no finite surface current density) to find
7
By ( z)  
4
c
 Ja

( Ja)
za
z  a
Note that a constant magnetic field does penetrate into a conductor. The same result can be
obtained from Ampere's law.
Exercise: Find the magnetic field produced by a very long solenoid, of a circular cross-section,
carrying a current I.
Solution: We use cylindrical coordinates, and take the solenoid to be along the z-direction. The
wire winds around it densely, such that there are n loops per unit length. The surface current
density is then

K  nI̂
For an infinitely long solenoid, the field cannot depend on z and by symmetry it cannot depend on
 
 ; However, it may depend on  . By   B  0 we find that there is no field component along
 

̂ By   B  (4 / c) J , the field is directed along z. So we write

B(  )  zˆB(  )


If we choose to integrate B  d  along a loop which lies outside the solenoid, we will obtain zero.
Hence, the magnetic field outside the solenoid is zero. Now let us integrate along a loop with one
arm (of length L) outside the solenoid, the other inside. Then we find

4
B  zˆ
In
c
(inside).
8
Exercise: Find the magnetic field produced by an infinite cylindrical wire, of radius a, carrying a
current density which is a function of

 , J (  )  zˆJ (  ) .
Solution: By the similar considerations as in the previous example, we know that

B (  )  ˆB (  )
Integrating along a circular loop we find
4
B2 
c

 J (  ' )2' d '
0
so that

2
B(  ) 
J (  ' )2' d '
c 0
When
 a
the integral gives the total current in the wire, I. Hence outside the wire
B(  ) 
2I
c
as has been also found from the Biot-Savart law. When

J is independent of  , the field
increases within the wire, and decreases outside.
The vector potential
Let first summarize our results so far, and compare them to electrostatics.
  4 
 B 
J
c
 
 B  0
 
  E  4
 
 E  0
9


B curls around I E diverges from q

We now introduce the vector potential A such that
  
B   A
with the (arbitrarily) chosen gauge
 
 A  0
(Coulomb gauge)
For example, to have a constant magnetic field along the z-direction, we may take
 1   1
A  B  r  B xˆy  yˆ x 
2
2
Ampere’s law now takes the form of a vectorial Poisson equation
   

 
4  
J    (  A)   2 A  (  A)   2 A
c
and then

  1  J (r ' )
A(r )   dr '  
c
| r  r '|
 
 
(note that   A  0 because by the continuity equation for a stationary situation 'J  0 .)
From the definition of the vector potential, and using Stokes theorem we find that the magnetic
flux through an area S is given by the contour integral around the loop closing that area


 

 
ˆ
ˆ
   B  nda     A  nda   A  d 
This yields that the tangential components of the vector potential are continuous.
10
Exercise: Find the vector potential of a wire (of zero thickness) of length L carrying a current I.
Solution: We take the current along the z-direction, and use cylindrical coordinates. From the

  1  J (r ' )
general formula, A(r )   dr '   , we know that the vector potential is along the zc
| r  r '|
direction as well. In this special case, the integral becomes
L/2
I
Az (  ) 
dz
c  L/ 2
1
2I

c
z2   2
L/2

dz
0
1
2I

ln( z  z 2   2
c
z2   2
L/2
0
2
2
2 I L / 2  ( L / 2)  

ln
c

For
L
2
,
Az (  )  
2I
ln 
c
(up to an unimportant constant).
Exercise: Find the vector potential of a wire of length L and radius a carrying a current I.
Solution: We take the current along the z-direction, and use cylindrical coordinates. Here again
the vector potential is along the z-direction. Therefore, Poisson equation gives
4
 Az  
J,
c
2
Since
 I
 2
where J  a

 0
a
 a
Az  Az (  ) (it cannot have an angular dependence) we have
11
 I

1 d  dAz 
4
 
  
J where J  a 2
 d  d 
c

 0
a
 a
.
This equation can be integrated in the two regimes, to give
I 
Az      C1 ln   C
ca
A  D1 ln   D
2
where
a
 a
C1 , C , D1 , and D are constants. The magnetic field is given by
 2 I C1



  
dAz
 ca 2

B    A  ˆ
 ˆ 
D1
d




We therefore find that
a
 a
C1  0 . The normal component of the magnetic field is continuous and
therefore
D1  
2I
c
The tangential component of the vector potential is continuous and therefore

I
 C  D1 ln a  D
c
giving
D
I 2I

ln a  C
c c
12
The final result, up to the constant C (in the vector potential, but not in the magnetic field!) is
then
I 
Az      C
ca
 2I 
B
ˆ
c a2
a
I

Az   1  2 ln   C
c
a
 2I 1
B
ˆ
c 
 a
2
The magnetic dipole
We look for the asymptotic behavior of the vector potential created by a current density
(confined in space). We start from the expression

  1  J (r ' )
A(r )   dr '  
c
| r  r '|
and introduce there

1
1 rˆ  r '
    2  .....
| r  r '| r
r
,
r  r '
The first term of the expansion gives
 
  
1
A(r )   dr ' J (r ' )
rc
But this is zero. To see this property, we write




 
  

 

J i (r ' )    r 'i J (r ' )  r 'i   J (r ' )    r 'i J (r ' )

13
where we used the continuity equation, by which





  
  J (r ' )  0 . Now we apply Gauss's law

 
 
 
ˆ
d
r
'


r
'
J
(
r
'
)

da
'
n

r
'
J
(r ' )  0
i
i


(since the current density does not go out of the region where it is confined).
In other words, the volume integral of the steady current density vanishes:
 
d
r
 J (r )  0
Thus the lowest order in the expansion the vector potential is
 
  

1
A(r )  2  dr ' J (r ' )rˆ  r '
cr
Let us now define the magnetic dipole,

   
1
m
d
r
' r ' J (r ' )
2c 
We will now prove that the lowest order of the expansion of the vector potential can be written
in the form

 
m  rˆ
A(r )  2
r
First, we write the result obtained above for the vector potential explicitly,




1
1
A j (r )  2  dr ' J j (r ' )rˆ  r '  2
cr
cr


ˆ
r
d
r
'
J
(
r
 i  j ' ) r 'i
i
14
Secondly, we show that

 dr ' r '
i



J j (r ' )    dr ' r ' j J i (r ' )
This is accomplished as follows. By Gauss's theorem,




 
 
 
ˆ
d
r
'


r
'
r
'
J
(
r
'
)

da
'
n

r
'
r
'
J
(r ' )  0
i
j
i
j


Writing explicitly the left-hand-side (remembering the continuity equation!)


 
 




d
r
'


r
'
r
'
J
(
r
'
)

d
r
'
r
'
J
(
r
'
)

r
'
J
(
r
' )  0
i
j
i
j
j
i


So, we find that




1
1
A j (r )  2  dr ' J j (r ' )rˆ  r '  2
cr
cr
1

2cr 2
where
 ijk

i

 rˆ  dr ' J
i
j

( r ' ) r 'i
i



ˆ
d
r
'
(
J
(
r
'
)
r
'
r

J
(
r
' )r ' j rˆi ) 
j
i i
i



   
1
ˆ
ri ijk  dr ' r ' J (r ' )
k
2cr 2
is the Levi-Civita tensor (it is 1 for i=1, j=2, and k=3, and any other cyclic permutation,
-1 for all other combinations, and 0 if two or more of the indices are equal). We can now write
this result in the form

1
A j (r )  2
r

rˆ mk
ijk i
i
and in vector notations, this becomes

 
m  rˆ
A(r )  2
r
Remember that the magnetic dipole is

   
1
m
d
r
' r ' J (r ' ) .
2c 
15
The magnetic field in this approximation is


 
  
3rˆ(m  rˆ)  m
B(r )    A(r ) 
r3
Exercise: Find the magnetic dipole moment of a planar current loop, carrying the current I.
  

J
(
r
'
)
d
r
'

Id
 , and the magnetic dipole is
Solution: In this case

   
1
1   1
m
d
r
'
r
'

J
(
r
'
)

I
r 'd   ISnˆ
2c 
2c 
c
where S is the area enclosed by the current loop and n̂ is a unit vector normal to it.
This is general. For example, for a circular current loop of radius a, we have


r '  xˆa cos   yˆa sin  , and d   ( xˆa sin   yˆ cos  )d ,
giving
 
r 'd   zˆa 2 (cos 2   sin 2  )d
Exercise: A current loop, carrying a current I, is made of a wire of the shape of square with side
a. Find the magnetic field on the axis perpendicular to the current loop, and passing through its
center.
Solution: We have already solved this problem, and found that for z>>a (the loop lies in the x-y

I 2a 2
plane), the result is B( z )  zˆ
. We now obtain the same result from the magnetic dipole.
c z3
We now know that the magnetic moment of this current loop is directed along z, and has the
m  Ia 2 / c . Indeed, since m is along z and the magnetic field is required on the z axis,


 

3rˆ(m  rˆ)  m

B
( z )  2m / z 3 which is our result above.
the result B ( r ) 
gives
just
3
r
magnitude
16
Exercise: Find the magnetic dipole of a thin disk of radius a carrying surface charge  and
rotating with angular velocity  around the z axis which passes through the center of the disk.
Solution: We use the formula

  
1
m
d
r
r  J (r ) . In our case the current is confined to the
2c 
disk, and is given by
 
J (r )  ˆr ( z )
Since
rˆ  ˆ  zˆ , we find
2

1

a4
2
m  zˆ
d drr r   zˆ 
2c 0 0
c
4
a


Magnetic dipole in a (constant) magnetic field


A magnetic dipole m is placed in a magnetic field B (which is not created by the dipole). The
magnetic dipole is created by a current loop carrying a current I. The force on each small element
of the wire carrying the current is
 I  
dF  dr  B
c
This force creates a torque,

N , given by


 I 

 
N   r  dF   r  dr  B
c

Let us write the integrand here in a different form, using (remember that the magnetic field is
constant)
17
  

 
  
d [r  (r  B)]  r  (dr  B)  dr  (r  B)
Then,




 
  
  
I
I
N
r

(
d
r

B
)

d
r

(
r

B
)

d
[
r
 ( r  B )]
2c 
2c 
The last integral vanishes (it is a complete derivative). The other terms become



   

 
  

I
I  


N
r

(
d
r

B
)

d
r

(
B

r
)


B

r

d
r


B
m  m B
2c 
2c 
(For any three vectors,
     
  
a  (b  c )  b  (c  a )  c  (a  b )  0 )
Forces and potential energy
 
 
J
(r
)
B
When a current density,
, is placed in an external magnetic field, (r ) , the total force on
the current density is given by
 1    
F   dr J (r )  B(r )
c
In case the magnetic field varies slowly enough over the region where the current is flowing, we
may expand it to obtain the dominant terms:

 
Bi (r )  Bi (0)  (r  ) Bi ...
and thus
18
Fi 





  

1

B
(
0
)
d
r
'
J
(
r
'
)

d
r
'
J
(
r
'
)(
r
'


)
B
(
r
)  ...

ijk
k
j
j
k


c jk
As we have found above, the volume integral of the current density vanishes. By the same vector
algebra as above, the second term gives
 
Fi    ijk (m  ) j Bk
jk
so that

   
  
 
  
F  (m  )  B  (m  B)  m(  B)  (m  B)
It follows that the potential energy of a dipole moment in an external magnetic field is
 
U  m  B
Magnetic fields in media

 
m  rˆ
We have found above that the vector potential due to a magnetic dipole is A( r ) 
. Let us
r2
now consider a material ("medium") placed in a magnetic field. The moving bound charges in the
material are affected by that magnetic field, and as a result we may imagine that each atom or
molecule acquires a tiny magnetic dipole moment. Let us denote that magnetic dipole distribution
 
M (r ) . This is the magnetization of the material. If in
 
addition there is "true" current density J (r ) flowing in the material, the total vector potential
(magnetic dipole per unit volume) by
created will be
19

 
 
  1   J (r ' )
M (r ' )  (r  r ' ) 
,
A(r )   dr '     c
  3

c
|
r

r
'
|
|
r

r
'
|


The second term here can be written in the form
 
 
  1 
  M (r ' )  (r  r ' ) 
  

1   


d
r
'

d
r
'
M
(
r
'
)


'



d
r
'
  'M (r ' )
 | r  r ' |  
  | r  r' |3  
|
r
 r '|


using integration-by-parts, and assuming that the surface contribution may be neglected. Then

 
  1   J (r ' )  c'M (r ' ) 

A(r )   dr ' 
 

c
|
r

r
'
|


This means that we may define now an effective medium current density,
 
  
J m (r )  c  M (r ) .
As a result,
  4 
 
 B 
J  4  M ,
c
which is put in the form
  4 
 H 
J with
c
(Compare with
 

H  B  4M
 

D  E  4P ).
As was the case in electrostatics, we also have here


B  H
where



M  H
is the magnetic permeability, and

is the magnetic susceptibility, and both are
properties of the medium.
20
The boundary conditions on the magnetic fields at the interface between two media are: The


B are continuous, the tangential components of H have a discontinuity if

K
the interface carries a finite surface current density
,
normal components of


4 
nˆ  ( H 2  H 1 ) 
K
c
21