Page 238 - Engineering Electromagnetics, 8th Edition

220 ENGINEERING ELECTROMAGNETICS

Because we are concerned only with steady magnetic fields, the continuity equa-
tion shows that the first term of (63) is zero. Application of the divergence theorem
to the second term gives

µ 0 J 1
∇ 2 · A 2 =− · dS 1
4π S 1 R 12
where the surface S 1 encloses the volume throughout which we are integrating. This
volume must include all the current, for the original integral expression for A was an
integration such as to include the effect of all the current. Because there is no current
outside this volume (otherwise we should have had to increase the volume to include
it), we may integrate over a slightly larger volume or a slightly larger enclosing surface
without changing A. On this larger surface the current density J 1 must be zero, and
therefore the closed surface integral is zero, since the integrand is zero. Hence the
divergence of A is zero.
In order to find the Laplacian of the vector A, let us compare the x component
of (51) with the similar expression for electrostatic potential,

µ 0 J x dν ρ ν dν
A x = V =
vol 4πR vol 4π 0 R
We note that one expression can be obtained from the other by a straightforward
change of variable, J x for ρ ν , µ 0 for 1/ 0 , and A x for V .However, we have derived
some additional information about the electrostatic potential which we shall not have
to repeat now for the x component of the vector magnetic potential. This takes the
form of Poisson’s equation,

2 ρ ν
∇ V =−
0
which becomes, after the change of variables,
2
∇ A x =−µ 0 J x
Similarly, we have
2
∇ A y =−µ 0 J y
and
2
∇ A z =−µ 0 J z
or

2
∇ A =−µ 0 J (64)
Returning to (60), we can now substitute for the divergence and Laplacian of A
and obtain the desired answer,
∇× H = J (28)
We have already shown the use of Stokes’ theorem in obtaining the integral form of
Amp`ere’s circuital law from (28) and need not repeat that labor here.