Page 176 - Electromagnetics

With (3.118)we can also write (3.150)as

ˆ n 12 × (B 1 − B 2 ) = µ 0 (J s + J Ms1 + J Ms2 ) (3.153)
where J Ms =−ˆ n × M is the equivalent magnetization surface current density.
We may also write the boundary conditions in terms of the scalar or vector potential.
Using H =−∇ m , we can write (3.150)as
m1 (r) = m2 (r) (3.154)

provided that the surface current J s = 0. As was the case with (3.36), the possibility of
an additive constant here is generally ignored. To write (3.151)in terms of m we first
note that B/µ 0 − M =−∇ m ; substitution into (3.151)gives
∂ m1 ∂ m2
− =−ρ Ms1 − ρ Ms2 (3.155)
∂n ∂n
where the normal derivative is taken in the direction of ˆ n 12 . For a linear isotropic material
where B = µH we have
∂ m1 ∂ m2
µ 1 = µ 2 . (3.156)
∂n ∂n
Note that (3.154)and (3.156)are independent.
Boundary conditions on A may be derived using the approach of § 2.8.2. Consider
Figure 2.6. Here the surface may carry either an electric surface current J s or an equiv-
alent magnetization current J Ms , and thus may be a surface of discontinuity between
differing magnetic media. If we integrate ∇× A over the volume regions V 1 and V 2 and
add the results we find that

∇× A dV + ∇× A dV = B dV.
V 1 V 2 V 1 +V 2
By the curl theorem

ˆ n × A dS + −ˆ n 10 × A 1 dS + −ˆ n 20 × A 2 dS = B dV
S 1 +S 2 S 10 S 20 V 1 +V 2
where A 1 is the field on the surface S 10 and A 2 is the field on S 20 .As δ → 0 the surfaces S 1
and S 2 combine to give S. Also S 10 and S 20 coincide, as do the normals ˆ n 10 =−ˆ n 20 = ˆ n 12 .
Thus

(ˆ n × A) dS − B dV = ˆ n 12 × (A 1 − A 2 ) dS. (3.157)
S V S 10
Now let us integrate over the entire volume region V including the surface of discontinuity.
This gives

(ˆ n × A) dS − B dV = 0,
S V
and for agreement with (3.157)we must have

ˆ n 12 × (A 1 − A 2 ) = 0. (3.158)
A similar development shows that

ˆ n 12 · (A 1 − A 2 ) = 0. (3.159)
Therefore A is continuous across a surface carrying electric or magnetization current.

© 2001 by CRC Press LLC