# F ebruary 28, 2000

```February 28, 2000
Notes for Lecture #19
Magnetic dipole eld
These notes are very similar to the notes for Lecture #13 on the electric dipole eld.
The magnetic dipole moment is dened by
Z
1
m=
2
with the corresponding potential
d3 r 0 r0
A(r) =
and magnetostatic eld
B(r) =
(
0
4
J(r0);
m ^r
0
4
r2
(1)
(2)
;
)
3^r(m ^r) m + 8 m&AElig;3(r)
r3
3
(3)
:
The last term of the eld expression follows from the following derivation. We note that Eq.
(3) is poorly dened as r ! 0, and consider the value of a small integral of B(r) about zero.
(For this purpose, we are supposing that the dipole m is located at r = 0.) In this case we
will approximate
Z
3
B(r 0) B(r)d r &AElig; 3 (r):
(4)
sphere
First we note that
Z
rR
B(r)d3 r = R2
Z
r=R
^r A(r) d
:
(5)
VdA:
(6)
This result follows from the divergence theorm:
Z
r Vd r =
3
vol
Z
surface
In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate of r A
since r A = x^ (x^ (r A)) + y^ (y^ (r A)) + ^z (^z (r A)): Note that x^ (r A) =
r (x^ A) and that we can use the Divergence theorem with V x^ A(r) for the x
component for example:
Z
Therefore,
Z
rR
r (x^ A)d3r =
vol
(r A)d3r =
Z
r=R
Z
surface
(x^ A) ^rdA =
Z
surface
(A ^r) x^dA:
(A ^r) (x^x^ + y^y^ + ^z^z)dA = R2
Z
r=R
(^r A)d
(7)
(8)
which is identical to Eq. (5). Now, expressing the vector potential in terms of the current
density:
Z
J(r0 )
A(r) = 0 d3 r
(9)
4
jr r0j ;
we can use the identity,
Therefore,
R2
Z
Z
r =R
jr ^r r0j = 43 rr&lt;2 r^0:
&gt;
(10)
d
(^r A)d
= 4R
3
2
Z
d3 r 0
r&lt;
r2
&gt;
r^0 J(r0 ):
(11)
If the sphere R contains the entire current distribution, then r&gt; = R and r&lt; = r0 so that
(11) becomes
Z
Z
4
2
(12)
R
(^r A)d
= 3 d3r0 r0 J(r0) 83 m:
r=R
```