Page 79 - Electromagnetics Handbook
P. 79
we can use (2.176) to get
∂D
ˆ n × H dS + ˆ n × H dS = J + dV,
∂t
S 1 S 10 V 1
∂D
ˆ n × H dS + ˆ n × H dS = J + dV.
∂t
S 2 S 20 V 2
Adding these we obtain
∂D
ˆ n × H dS − J + dV −
∂t
S 1 +S 2 V 1 +V 2
− ˆ n 10 × H 1 dS − ˆ n 20 × H 2 dS = 0, (2.184)
S 10 S 20
where we have used subscripts to delineate the fields on each side of the discontinuity
surface.
If δ is very small (but nonzero), then ˆ n 10 =−ˆ n 20 = ˆ n 12 and S 10 = S 20 . Letting
S 1 + S 2 = S and V 1 + V 2 = V , we can write (184) as
∂D
(ˆ n × H) dS − J + dV = ˆ n 12 × (H 1 − H 2 ) dS. (2.185)
S V ∂t S 10
Now suppose we use the same volume region V , but let it intersect the surface of
discontinuity (Figure 2.6), and suppose that the large-scale form of Ampere’s law holds
even if V contains points of field discontinuity. We must include the surface current in
the computation. Since J dV becomes J s dS on the surface, we have
V S
∂D
(ˆ n × H) dS − J + dV = J s dS. (2.186)
S V ∂t S 10
We wish to have this give the same value for the integrals over V and S as (2.185), which
included in its derivation no points of discontinuity. This is true provided that
ˆ n 12 × (H 1 − H 2 ) = J s . (2.187)
Thus, under the condition (2.187) we may interpret the large-scale form of Ampere’s law
(as derived from the point form) as being valid for regions containing discontinuities.
Note that this condition is not “derived,” but must be regarded as a postulate that
results in the large-scale form holding for surfaces of discontinuous field.
Similar reasoning can be used to determine the appropriate boundary condition on
tangential E from Faraday’s law. Corresponding to (2.185) we obtain
∂B
(ˆ n × E) dS − −J m − dV = ˆ n 12 × (E 1 − E 2 ) dS. (2.188)
S V ∂t S 10
Employing (2.175) over the region containing the field discontinuity surface we get
∂B
(ˆ n × E) dS − −J m − dV =− J ms dS. (2.189)
S V ∂t S 10
To have (2.188) and (2.189) produce identical results, we postulate
ˆ n 12 × (E 1 − E 2 ) =−J ms (2.190)
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