Page 189 - Electromagnetics

stored within the solid wire. The result is
µ 0 I 2 1
b
W m
= µ r ln + .
l 4π a 4
3.3.7 Magnetic field of a permanently magnetized body
We now have the tools necessary to compute the magnetic field produced by a perma-
nent magnet (a body with permanent magnetization M). As an example, we shall find
the field due to a uniformly magnetized sphere in three different ways: by computing the
vector potential integral and taking the curl, by computing the scalar potential integral
and taking the gradient, and by finding the scalar potential using separation of variables
and applying the boundary condition across the surface of the sphere.
Consider a magnetized sphere of radius a, residing in free space and having permanent
magnetization

M(r) = M 0 ˆ z.

The equivalent magnetization current and charge densities are given by

J M =∇ × M = 0, (3.194)
ˆ
J Ms =−ˆ n × M =−ˆ r × M 0 ˆ z = M 0 φ sin θ, (3.195)
and

ρ M =−∇ · M = 0, (3.196)
ρ Ms = ˆ n · M = ˆ r · M 0 ˆ z = M 0 cos θ. (3.197)
The vector potential is produced by the equivalent magnetization surface current.
Using (3.137)we find that

π π ˆ
µ 0 J Ms µ 0 M 0 φ sin θ

A(r) = dS = sin θ dθ dφ .
4π S |r − r | 4π −π 0 |r − r |

Since φ =−ˆ x sin φ + ˆ y cos φ , the rectangular components of A are

ˆ

sin φ
" # π π M 0 cos φ sin θ
−A x µ 0 2
= a sin θ dθ dφ . (3.198)

A y 4π −π 0 |r − r |
The integrals are most easily computed via the spherical harmonic expansion (E.200)for
−1
the inverse distance |r − r | :
∞ Y nm (θ, φ) r
" # n n π π
−A x 2

< sin φ 2 ∗
= µ 0 M 0 a sin θ Y (θ ,φ ) dθ dφ .
nm
n+1
A y 2n + 1 r > −π 0 cos φ
n=0 m=−n

Because the φ variation is sin φ or cos φ , all terms in the sum vanish except n = 1,

m =±1. Since

3 − jφ 3 jφ
Y 1,−1 (θ, φ) = sin θe , Y 1,1 (θ, φ) =− sin θe ,
8π 8π
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