Page 177 - Electromagnetics

3.3.4 Uniqueness of the magnetostatic field
Because the uniqueness conditions established for the dynamic field do not apply to
magnetostatics, we begin with the magnetostatic field equations. Consider a region of
space V bounded by a surface S. There may be source currents and magnetic materials
both inside and outside V . Assume (B 1 , H 1 ) and (B 2 , H 2 ) are solutions to the magne-
tostatic field equations with source J. We seek conditions under which B 1 = B 2 and
H 1 = H 2 .
The difference field H 0 = H 2 − H 1 obeys ∇× H 0 = 0. Using (B.44)we examine the
quantity

∇· (A 0 × H 0 ) = H 0 · (∇× A 0 ) − A 0 · (∇× H 0 ) = H 0 · (∇× A 0 )
where A 0 is defined by B 0 = B 2 − B 1 =∇ × A 0 =∇ × (A 2 − A 1 ). Integrating over V
we obtain

(A 0 × H 0 ) · dS = H 0 · (∇× A 0 ) dV = H 0 · B 0 dV.
S V V
Then, since (A 0 × H 0 ) · ˆ n =−A 0 · (ˆ n × H 0 ), we have

− A 0 · (ˆ n × H 0 )dS = H 0 · B 0 dV. (3.160)
S V
If A 0 = 0 or ˆ n × H 0 = 0 everywhere on S,or A 0 = 0 on part of S and ˆ n × H 0 = 0 on the
remainder, then

H 0 · B 0 dS = 0. (3.161)
V
So H 0 = 0 or B 0 = 0 by arbitrariness of V . Assuming H and B are linked by the
constitutive relations, we have H 1 = H 2 and B 1 = B 2 . The fields within V are unique
provided that A, the tangential component of H, or some combination of the two, is
specified over the bounding surface S.
One other condition will cause the left-hand side of (3.160)to vanish. If S recedes to
infinity then, provided that the potential functions vanish sufficiently fast, the condition
(3.161)still holds and uniqueness is guaranteed. Equation (3.135)shows that A ∼ 1/r
2
as r →∞, hence B, H ∼ 1/r . So uniqueness is ensured by the specification of J in
unbounded space.

3.3.5 Integral solution for the vector potential
We have used the scalar Green’s theorem to find a solution for the electrostatic poten-
tial within a region V in terms of the source charge in V and the values of the potential
and its normal derivative on the boundary surface S. Analogously, we may find A within
V in terms of the source current in V and the values of A and its derivatives on S. The
vector relationship between B and A complicates the derivation somewhat, requiring
Green’s second identity for vector fields.
Let P and Q be continuous with continuous first and second derivatives throughout
V and on S. The divergence theorem shows that

∇· [P × (∇× Q)] dV = [P × (∇× Q)] · dS.
V S

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