Feb 28, 2000
Notes for Lecture
Notes for Lecture #19
Magnetic dipole field
These notes are very similar to the notes for Lecture #13 on the
electric dipole field.
The magnetic dipole moment is defined by
m = |
1 2
|
|
ó õ
|
d3r¢ r¢ ×J(r¢), |
| (1) |
with the corresponding potential
and magnetostatic field
B(r) = |
m0 4p
|
|
ì ï í
ï î
|
r3
|
+ |
8p 3
|
m d3(r) |
ü ï ý
ï þ
|
. |
| (3) |
The last term of the field expression follows from the following
derivation. We note that Eq. (3) is poorly defined 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
B(r » 0) » |
æ è
|
|
ó õ
|
sphere
|
B(r) d3r |
ö ø
|
d3(r). |
| (4) |
First we note that
|
ó õ
|
r £ R
|
B(r) d3r = R2 |
ó õ
|
r = R
|
|
^ r
|
×A(r) dW. |
| (5) |
This result follows from the divergence theorm:
|
ó õ
|
vol
|
Ñ·V d3r = |
ó õ
|
surface
|
V · dA. |
| (6) |
In our case, this theorem can be used to prove Eq. (5) for
each cartesian coordinate of Ñ×A
since Ñ×A = [^(x)] ( [^(x)] ·(Ñ× A) ) + [^(y)] ( [^(y)] ·(Ñ×A) ) + [^(z)] ( [^(z)] ·(Ñ×A) ).
Note that [^(x)] ·(Ñ×A) = -Ñ·([^(x)] ×A) and that we can use the
Divergence theorem with
V º [^(x)] ×A(r)
for the x- component for example:
|
ó õ
|
vol
|
Ñ·( |
^ x
|
×A) d3r = |
ó õ
|
surface
|
( |
^ x
|
×A) · |
^ r
|
dA = |
ó õ
|
surface
|
(A × |
^ r
|
) · |
^ x
|
dA. |
| (7) |
Therefore,
|
ó õ
|
r £ R
|
(Ñ×A) d3r = - |
ó õ
|
r = R
|
(A × |
^ r
|
) ·( |
^ x
|
|
^ x
|
+ |
^ y
|
|
^ y
|
+ |
^ z
|
|
^ z
|
) dA = R2 |
ó õ
|
r = R
|
( |
^ r
|
×A) dW |
| (8) |
which is identical to Eq. (5). Now, expressing the vector
potential in terms of the current density:
A(r) = |
m0 4 p
|
|
ó õ
|
d3r |
J(r¢) |r - r¢|
|
, |
| (9) |
we can use the identity,
|
ó õ
|
dW |
|r - r¢|
|
= |
4 p 3
|
|
r < r > 2
|
|
^ r¢
|
. |
| (10) |
Therefore,
R2 |
ó õ
|
r = R
|
( |
^ r
|
×A) dW = |
4 pR2 3
|
|
ó õ
|
d3r¢ |
r < r > 2
|
|
^ r¢
|
×J(r¢). |
| (11) |
If the sphere R contains the entire current distribution, then
r > = R and r < = r¢ so that (11) becomes
R2 |
ó õ
|
r = R
|
( |
^ r
|
×A) dW = |
4 p 3
|
|
ó õ
|
d3r¢ r¢ ×J(r¢) º |
8 p 3
|
m. |
| (12) |
File translated from TEX by TTH, version 2.20.
On 28 Feb 2000, 16:59.