Page 234 - Engineering Electromagnetics, 8th Edition

216 ENGINEERING ELECTROMAGNETICS

In each of these two expressions the vector character is given to the current. For the
filamentary element it is customary, although not necessary, to use IdL instead of
I dL. Since the magnitude of the filamentary element is constant, we have chosen
the form which allows us to remove one quantity from the integral. The alternative
expressions for A are then

µ 0 K dS

A = 4πR (50)
S
and

µ 0 J dν
A = (51)
vol 4πR
Equations (47), (50), and (51) express the vector magnetic potential as an inte-
gration over all of its sources. From a comparison of the form of these integrals with
those which yield the electrostatic potential, it is evident that once again the zero ref-
erence for A is at infinity, for no finite current element can produce any contribution
as R →∞.We should remember that we very seldom used the similar expressions
for V ; too often our theoretical problems included charge distributions that extended
to infinity, and the result would be an infinite potential everywhere. Actually, we cal-
culated very few potential fields until the differential form of the potential equation
was obtained, ∇ V =−ρ ν / ,or better yet, ∇ V = 0. We were then at liberty to
2
2
select our own zero reference.
The analogous expressions for A will be derived in the next section, and an
example of the calculation of a vector magnetic potential field will be completed.

D7.8. A current sheet, K = 2.4a z A/m, is present at the surface ρ = 1.2in
free space. (a) Find H for ρ> 1.2. Find V m at P(ρ = 1.5,φ = 0.6π, z = 1) if:
(b) V m = 0at φ = 0 and there is a barrier at φ = π;(c) V m = 0at φ = 0 and
there is a barrier at φ = π/2; (d) V m = 0at φ = π and there is a barrier at φ = 0;
(e) V m = 5V at φ = π and there is a barrier at φ = 0.8π.

2.88
Ans. a φ ; −5.43 V; 12.7 V; 3.62 V; −9.48 V
ρ

D7.9. The value of A within a solid nonmagnetic conductor of radius a car-
rying a total current I in the a z direction may be found easily. Using the
known value of H or B for ρ< a, then (46) may be solved for A. Select
A = (µ 0 I ln 5)/2π at ρ = a (to correspond with an example in the next sec-
tion) and find A at ρ =:(a)0;(b) 0.25a;(c) 0.75a;(d) a.

Ans. 0.422Ia z µWb/m; 0.416Ia z µWb/m; 0.366Ia z µWb/m; 0.322Ia z µWb/m