Page 230 - Engineering Electromagnetics, 8th Edition

212 ENGINEERING ELECTROMAGNETICS

Thus,
I
V m =− φ
2π
where the constant of integration has been set equal to zero. What value of potential
do we associate with point P, where φ = π/4? If we let V m be zero at φ = 0 and
proceed counterclockwise around the circle, the magnetic potential goes negative
linearly. When we have made one circuit, the potential is −I,but that was the point
at which we said the potential was zero a moment ago. At P, then, φ = π/4, 9π/4,
17π/4,... ,or −7π/4, −15π/4, −23π/4,... ,or

I 1
V mP = 2n − 4 π (n = 0, ±1, ±2,...)
2π
or
1

V mP = I n − 8 (n = 0, ±1, ±2,...)
The reason for this multivaluedness may be shown by a comparison with the
electrostatic case. There, we know that

∇× E = 0

E · dL = 0

and therefore the line integral
a
E · dL
V ab =−
b
is independent of the path. In the magnetostatic case, however,
∇× H = 0 (wherever J = 0)
but

H · dL = I
even if J is zero along the path of integration. Every time we make another complete
lap around the current, the result of the integration increases by I.Ifno current I
is enclosed by the path, then a single-valued potential function may be defined. In
general, however,

a
H · dL (specified path) (45)
V m,ab =−
b
where a specific path or type of path must be selected. We should remember that the
electrostatic potential V is a conservative field; the magnetic scalar potential V m is
8
not a conservative field. In our coaxial problem, let us erect a barrier at φ = π;we

8 This corresponds to the more precise mathematical term “branch cut.”