Page 239 - Engineering Electromagnetics, 8th Edition

CHAPTER 7 The Steady Magnetic Field 221

We thus have succeeded in showing that every result we have essentially pulled
11
from thin air for magnetic fields follows from the basic definitions of H, B, and A.
The derivations are not simple, but they should be understandable on a step-by-step
basis.
Finally, let us return to (64) and make use of this formidable second-order vec-
tor partial differential equation to find the vector magnetic potential in one simple
example. We select the field between conductors of a coaxial cable, with radii of a
and b as usual, and current I in the a z direction in the inner conductor. Between the
conductors, J = 0, and therefore
2
∇ A = 0
We have already been told (and Problem 7.44 gives us the opportunity to check the
results for ourselves) that the vector Laplacian may be expanded as the vector sum of
the scalar Laplacians of the three components in rectangular coordinates,
2 2 2 2
∇ A =∇ A x a x +∇ A y a y +∇ A z a z
but such a relatively simple result is not possible in other coordinate systems. That is,
in cylindrical coordinates, for example,
2 2 2 2
∇ A =∇ A ρ a ρ +∇ A φ a φ +∇ A z a z
However, it is not difficult to show for cylindrical coordinates that the z component
of the vector Laplacian is the scalar Laplacian of the z component of A,or

2
2
∇ A =∇ A z (65)
z
and because the current is entirely in the z direction in this problem, A has only a
z component. Therefore,
2
∇ A z = 0
or
2
2
1 ∂ ∂ A z 1 ∂ A z ∂ A z
ρ + + = 0
2
ρ ∂ρ ∂ρ ρ ∂φ 2 ∂z 2
Thinking symmetrical thoughts about (51) shows us that A z is a function only of ρ,
and thus
1 d dA z
ρ = 0
ρ dρ dρ
We have solved this equation before, and the result is
A z = C 1 ln ρ + C 2
If we choose a zero reference at ρ = b, then
ρ
A z = C 1 ln
b

11 Free space.