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
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