Page 272 - Engineering Electromagnetics, 8th Edition
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254 ENGINEERING ELECTROMAGNETICS
For tangential B,wehave
B t1 B t2
− = K (36)
µ 1 µ 2
The boundary condition on the tangential component of the magnetization for linear
materials is therefore
χ m2
M t2 = M t1 − χ m2 K (37)
χ m1
The last three boundary conditions on the tangential components are much sim-
pler, of course, if the surface current density is zero. This is a free current density,
and it must be zero if neither material is a conductor.
EXAMPLE 8.6
Toillustrate theserelationships with an example, let us assume that µ = µ 1 = 4 µH/m
in region 1 where z > 0, whereas µ 2 = 7 µH/m in region 2 wherever z < 0. Moreover,
let K = 80a x A/m on the surface z = 0. We establish a field, B 1 = 2a x − 3a y +
a z mT, in region 1 and seek the value of B 2 .
Solution. The normal component of B 1 is
B N1 = (B 1 · a N12 )a N12 = [(2a x − 3a y + a z ) · (−a z )](−a z ) = a z mT
Thus,
B N2 = B N1 = a z mT
We next determine the tangential components:
B t1 = B 1 − B N1 = 2a x − 3ay mT
and
(2a x − 3a y )10 −3
B t1
H t1 = = −6 = 500a x − 750a y A/m
µ 1 4 × 10
Thus,
H t2 = H t1 − a N12 × K = 500a x − 750a y − (−a z ) × 80a x
= 500a x − 750a y + 80a y = 500a x − 670a y A/m
and
−6
B t2 = µ 2 H t2 = 7 × 10 (500a x − 670a y ) = 3.5a x − 4.69a y mT
Therefore,
B 2 = B N2 + B t2 = 3.5a x − 4.69a y + a z mT