Page 188 - Electromagnetics

Note the similarity between (3.189)and (3.86). We now manipulate (3.189)into a
form involving only the electromagnetic fields. By Ampere’s law
1
W m = A · (∇× H) dV.
2 V
Using (B.44)and the divergence theorem we can write
1 1
W m = (H × A) · dS + H · (∇× A) dV.
2 S 2 V
We now let S expand to infinity. This does not change the value of W m since we do not
2
enclose any more current; however, since A ∼ 1/r and H ∼ 1/r , the surface integral
vanishes. Thus, remembering that ∇× A = B,wehave
1
W m = H · B dV (3.190)
2
V ∞
where V ∞ denotes all of space.
Although we do not provide a derivation, (3.190)is also valid within linear materials.
For nonlinear materials, the total energy required to build up a magnetic field from B 1
to B 2 is
1 B 2
W m = H · dB dV. (3.191)
2
V ∞ B 1
This accounts for the work required to drive a ferromagnetic material through its hystere-
sis loop. Readers interested in a complete derivation of (3.191)should consult Stratton
[187].
As an example, consider two thin-walled, coaxial, current-carrying cylinders having
radii a, b (b > a). The intervening region is a linear magnetic material having perme-
ability µ. Assume that the inner and outer conductors carry total currents I in the ±z
directions, respectively. From the large-scale form of Ampere’s law we find that

0, ρ ≤ a,

ˆ
H = φ I/2πρ, a ≤ ρ ≤ b, (3.192)

0, ρ > b,

hence by (3.190)
1 2π b µI 2
W m = dz ρ dρ dφ,
2 0 a (2πρ) 2
and the stored energy is
b
W m I 2
= µ ln (3.193)
l 4π a
per unit length.
Suppose instead that the inner cylinder is solid and that current is spread uniformly
throughout. Then the field between the cylinders is still given by (3.192)but within the
inner conductor we have
Iρ
ˆ
H = φ
2πa 2
by (3.169). Thus, to (3.193) we must add the energy
2 2
W 1 2π a µ 0 I ρ µ 0 I 2
m,inside
= ρ dρ dφ =
2 2
l 2 0 0 (2πa ) 16π

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