Magnetic vector potential
When we derived the scalar electric potential we started with the relation
~ £E
~ = 0 to conclude that E
~ could be written as the gradient of a scalar
r
potential. That wonzt work for the magnetic ¤eld (except where ~j = 0); because
~ is not zero in general. Instead, the divergence of B
~ is zero. That
the curl of B
~ may be written as the curl of a vector that we shall call A:
~
means that B
³
´
~ =r
~ £A
~)r
~ ¢B
~ =r
~ ¢ r
~ £A
~ =0
B
Then the second equation becomes
³
´
³
´
~ £B
~ =r
~ £ r
~ £A
~ =r
~ r
~ ¢A
~ ¡ r2 A
~ = ¹ ~j
r
0
~ = ¡rV
~ is
We had some ¢exibility in choosing the scalar potential V because E
~ (constant) = 0: Similarly here,
not changed if we add a constant to V; since r
~ the gradient of a scalar function, A
~2 = A
~ 1 + rÂ;
~
if we add to A
we have
³
´
~2 = r
~ £A
~2 = r
~ £ A
~ 1 + rÂ
~
~ £A
~1 = B
~1
B
=r
~ ¢A
~ = 0: For suppose this is not true.
With this ¢exibility, we may choose r
Then
³
´
~ ¢ A
~ 1 + rÂ
~
~ ¢A
~ 1 + r2 Â = 0
r
=r
So we have an equation for the function Â
~ ¢A
~1
r2 Â = ¡r
~ 2 whose divergence is zero. Once we
Once we solve this we will have a vector A
~ ¢A
~ = 0 from the start. This is
know that we can do this, we may just set r
~ is
called the Coulomb gauge condition. With this choice, the equation for A
~ = ¡¹0~j
r2 A
(1)
We may look at this equation one component at a time (provided that we use
Cartesian components.) Thus, for the x¡component
r2 Ax = ¡¹0 j x
This equation has the same form as the equation for V
r2 V = ¡
½
"0
and thus the solution will also have the same form:
Z
¹
jx (~r0 ) 0
Ax (~r) = 0
d¿
4¼
R
1
and since we have an identical relation for each component, then
~ = ¹0
A
4¼
Z ~ 0
j (~r ) 0
d¿
R
(2)
Now remember that ~j d¿ corresponds to Id~`; so if the current is con¤ned in
wires, the result is
Z
¹0
Id~`0
~
A=
(3)
4¼
R
At this point we may stop and consider if there is any rule for magnetic ¤eld
analagous to our RULE 1 for electric ¤elds. Since there is no magetic charge,
there is no "point charge" ¤eld. But we can use our expansion
1
X (r0 )l ¡
¢
1
=
P cos µ0
l+1 l
R
r
l= 0
where ~r is on the polar axis. Then
Z
¡
¢
¹ X I
l
~ (~r = r^
A
z) = 0
(r 0) Pl cos µ0 d~`0
l+1
4¼
r
The l = 0 term is
Z
~ 0 = ¹0 I
A
d~`0
4¼ r
R
Since the current ¢ows in closed loops, d~`0 = 0: (This result is actually more
~ ~j = 0; and the lines of ~j also form closed
general, because in a static situation r¢
loops.) This is the result we expected. The next term is
Z
Z
¹0 I
¹0 I
0
0 ~0
~
A1 =
r cos µ d` =
(~r0 ¢ z^) d~`0
4¼r2
4¼r 2
We can use Stokes theorem to evaluate this integral.
Z
Z ³
´
~ £ ~u ¢ ^n dA
~u ¢ d~l =
r
Let ~u = ~c where ~c is a constant vector and  is a scalar function. Then
Z
Z ³
´
~ £ ~c ¢ n
~c ¢ Âd~l =
r
^ dA
Z h³ ´
i
~
=
rÂ
£ ~c ¢ n
^ dA
We may re-arrange the triple scalar product
Z
Z ³ ´
~
~c ¢ Âd~l = ¡~c ¢
rÂ
£n
^ dA
2
This is true for an arbitrary constant vector ~c; so, with  = (~r 0 ¢ z^)
Z
Z h
i
~ 0 (~r0 ¢ z^) £ n
(~r0 ¢ z^) d~l0 = ¡
r
^ 0 dA0
Z h
³ 0
´ ³
´ i
~ £ ~r0 + z^ ¢ r
~ 0 ~r0 £ ^n0dA0
= ¡
^z £ r
Z
= ¡ (0 + z^ £ n
^ 0 ) dA0
Z
Z
0
0
= ¡^z £ n
^ dA =
^n0dA0 £ z^
Note that z^ can come out of the integral because it is a constant. So
Z
¹0 I
¹0
¹0
~
A1 =
n
^ 0 dA0 £ z^ =
m
~ £ z^ =
m
~ £ r^
2
2
4¼r
4¼ r
4¼r2
where
m
~ =I
Z
n
^ 0 dA0
is the magnetic moment of the loop. The corresponding magnetic ¤eld is
h
i
~1 = r
~ £ ¹0 m
B
~
£
r
^
2
µ 4¼r
¶
¹0
3
1 ~
=
¡ 4 ^r £ ( m
~ £ ~r ) + 3 r
£ (m
~ £ ~r)
4¼
r
r
µ
´
³
´´¶
¹0
3
1 ³ ³
~ ~r + m
~ ¢ ~r
=
¡ 3 [m
~ ¡ ^r (m
~ ¢ ^r)] + 3 ¡ m
~ ¢r
~ r
4¼
r
r
¹0
=
(¡3 m
~ + 3~r (~
m ¢ r^) ¡ m
~ + 3m)
~
4¼r 3
¹0
=
[3~r ( m
~ ¢ ^r) ¡ m
~]
4¼r 3
This is a dipole ¤eld. Thus the magnetic equivalent of RULE 1 is :
At a great distance from a current distribution, the magnetic
¤eld is a dipole ¤eld
Here is another useful result:
I
Z ³
Z
´
~
~
~
~
~ ¢ ^n da = ©B
A ¢ d` =
r£A ¢n
^ da =
B
C
S
(4)
S
~ around a curve C equals the magnetic ¢ux through
Thus the circulation of A
any surface S spanning the curve.
~
Boundary conditions for B
We start with the Maxwell equations. Remember, if the equation has a
divergence we integrate over a small volume (pillbox) that crosses the boundary.
3
But if the equation has a curl, we integrate over a rectangular surface that lies
perpendicular to the surface.
~ ¢B
~ =0
So we start with r
Z
I
~
~
~ ¢ dA
~
r ¢ Bd¿ = 0 = B
S
But because we chose h ¿ d; the integral over the sides is negligible, and on
~ 2 = ¡^ndA; so we have
the bottom side dA
³
´
~1 ¡ B
~2 ¢ n
B
^=0
(5)
~ is continuous.
The normal component of B
For the curl equation, we use the rectangle shown:
Then
Z ³
S
³
´
~ £B
~ ¢ dA
~ =
r
I
~ ¢ d~` =
B
C
´ ¡ ¢
~1 ¡ B
~ 2 ¢ ¡^t w
B
³
´ ³
´
~1 ¡ B
~ 2 ¢ ¡^
B
n £ N^
Z
~
¹0~j ¢ dA
Z
^ dh w
¹0 ~j ¢ N
S
µZ
=
¹0 ¹0
=
~ ¢ N^
¹0 K
4
¶
~jdh ¢ N^ w
Rearrange the triple scalar product on the left to get
h³
´
i
~1 ¡ B
~2 £ n
~ ¢ N^
¡ B
^ ¢ N^ = ¹0 K
Since we may orient the rectangle so that N^ is any vector in the surface, we
have
³
´
~1 ¡ B
~ 2 = ¹0 K
~
n
^£ B
(6)
~ has a discontinuity that depends on the
Thus the tangential component of B
~
surface current density K: Crossing both sides with n
^ ; we get an alternate
version:
h
³
´i
~1 ¡ B
~2 £ n
~ £ ^n
n
^£ B
^ = ¹0 K
³
´
h ³
´i
~1 ¡ B
~ 2 ¡ ^n n
~1 ¡ B
~2
~ £ ^n
B
^¢ B
= ¹0 K
But now we may make use of (5) to obtain
³
´
~1 ¡ B
~2 = ¹ K
~ £ ^n
B
0
(7)
What about the vector potential? Remember that for the scalar potential V we were able to show that V is continuous across the surface (in most
~ we ¤rst choose a gauge condition. The Coulomb gauge
cases). When we ¤nd A
condition is
~ ¢A
~=0
r
and then we can use our usual pillbox trick to show that
~ ¢n
A
^ is continuous
(8)
For the tangential component, we make use of equation (4). Then, using the
rectangle,
I
~ ¢ d~` = ©B = B
~ ¢ N^ wh
A
Thus we have
³
C
´ ¡ ¢
~1 ¡ A
~ 2 ¢ ¡^t w
A
~ ¢ N^ wh ! 0 as h ! 0
B
=
~ ¢ ^t is continuous
A
(9)
These two result taken together show that the vector potential as a whole is
also continuous across the boundary.
~ into equation (6):
Finally letzs put A
³
´
~ £ A
~1 ¡ A
~ 2 = ¹0 K
~ £n
r
^
5
~ have a discontinuity. But which ones? Letzs expand
So the derivatives of A
³
´
³
´
~ =n
~ £A
~ = nirA
~ i¡ n
~ A
~
n
^£B
^£ r
^¢r
Then
³
´
³
´³
´
~1 ¡ B
~ 2 = ni r
~ (Ai;1 ¡ Ai;2 ) ¡ n
~
~1 ¡ A
~ 2 = ¹0 K
~
n
^£ B
^¢r
A
(10)
~ is continuous at the surface. So
But we have shown that each component of A
the components of
~ (Ai;1 ¡ Ai;2 )
r
parallel to the surface must be zero. Thus only the normal derivatives remain.
Then the normal component of equation (10) is identically zero, and the only
non-zero components of the boundary condition are the tangential components
³
´³
´
~
~1 ¡ A
~2
~
n
^¢r
A
= ¡¹0 K
(11)
tan
~ satis¤es Laplacezs equation with
Now this is neat. Each component of A
Neumann boundary conditions, and so it must have a unique solution, as we
already proved for V:
Magnetic scalar potential
~ £B
~ = 0 and we may use a
When we have the special case of ~j ´ 0; r
magnetic scalar potenial © ma g . This can be useful if the current is con¤ned to
lines or sheets, because we can create a nice boundary-value problem for © m ag :
~ = ¡ r©
~ m ag
B
~ ¢B
~ = 0 ) r2 © m ag = 0
r
~ m ag is continuous
Bn o rma l continuous ) ^n ¢ r©
´
~1 ¡ B
~2 = ¹ K
~ )n
~ (©m ag 1 ¡ ©m ag 2 ) = ¡¹ K
~
n
^£ B
^£r
0
0
³
(12)
(13)
(14)
Letzs use these boundary conditions to ¤nd the potential due to a spinning
spherical shell of charge. The current is con¤ned to the surface and has the
value
~ = ¾~v = ¾~! £ ~r = ¾!a sin µ Á
^
K
where in the last expression put the z¡axis along the rotation axis. We will
take ¾ to be a constant. The equation for © m ag in the region entirely inside (or
entirely outside) the sphere is (1 with ~j = 0)
r2 © ma g = 0
and because we have azimuthal symmetry, the solution is of the form
©in
=
1
X
C l rl P l (cos µ)
l=1
© ou t
=
1
X
Dl
P l (cos µ)
rl+ 1
l=1
6
We have omitted the l = 0 term because it contributes zero ¤eld inside, and we
know there can be no monopole term outside. What else do we know? At the
boundary, from (13)
¯
¯
@©m ag ,ou t ¯¯
@©m ag ,in ¯¯
¡
= 0
¯
¯
@r
@r
r=a
1
X
r= a
lCl al¡1 P l (cos µ) =
l=1
Cl
=
¡
¡
1
X
(l + 1)
l=1
Dl
P l (cos µ)
al+2
Dl l + 1
a2l+1 l
l>0
(15)
and from (14).
¯
@©m ag, ou t ¯¯
@©m ag,in
¡
¯
@µ
@µ
¯r =a
µ
1
@©m ag, ou t ¯¯
@©m ag,in
¡
¯
a sin µ
@Á
@Á
r =a
1
a
µ
¯
¶
¯
¯
¯
¯r=a ¶
¯
¯
¯
= ¡¹0 ¾a! sin µ
= 0
r=a
The last equation is automatically satis¤ed. Thus the ¤nal condition we need
to satisfy is
1
1
X
X
Dl @
@
P
(cos
µ)
¡
C l al¡1 P l (cos µ) = ¡¹0 ¾a! sin µ
l
al+ 2 @µ
@µ
l=1
l=1
@
Now since P 1 (cos µ) = cos µ and @µ
cos µ = ¡ sin µ; the ¤rst term in the sum is
µ
¶
D1
¡
¡ C1 sin µ
a3
so we may satisfy the boundary conditons by taking
D1
¡ C 1 = ¹0 ¾a!
a3
and all the other C l ; Dl = 0: Then equation (15) gives
D1
D1 2
¹ ¾a4 !
+ 3 = ¹0 ¾ a! ) D1 = 0
3
a
a 1
3
and then
C1 = ¡2
So
© m ag =
giving a ¤eld
~ =
B
(
½
¹0 ¾a!
3
¡ 23 ¹0 ¾a!r cos µ
1
2 a2
3 ¹0 ¾a ! r 2 cos µ
2
3
a
¹0 ¾a! 3r
3
inside
outside
³3 ¹0 ¾a! z^
´
2 cos µ r^ + sin µ ^µ
7
¾
inside
outside
)
Thus the ¤eld inside is uniform and the ¤eld outside is a pure dipole ¤eld. The
dipole moment is
4¼ 4
m=
¾a !
3
The dimensions of m are
4
charge (length)
charge
2
=
£ (length) = current £ area
area
time
time
which is correct. You should verify that you get the same m by summing current
loops.
Compare this solution with Gri¢thsz example 5.11. Which method do you
think is easier?
8