Page 76 - Electromagnetics Handbook
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Figure 2.5: Derivation of the electromagnetic boundary conditions across a thin contin-
                        uous source layer.


                        The parameter   describes the “width” of the source layer normal to the reference
                        surface.
                          We use (2.156)–(2.157) to study field behavior across the source layer. Consider a
                        volume region V  that intersects the source layer as shown in Figure 2.5. Let the top and
                        bottom surfaces be parallel to the reference surface, and label the fields on the top and
                        bottom surfaces with subscripts 1 and 2, respectively. Since points on and within V are
                        all regular, (2.157) yields

                                                                                    ∂D


                                    ˆ n 1 × H 1 dS +  ˆ n 2 × H 2 dS +  ˆ n 3 × H dS =  J +  dV.
                                                                             V      ∂t
                                  S 1           S 2            S 3
                        We now choose δ = k  (k > 1) so that most of the source lies within V .As   → 0
                        the thin source layer recedes to a surface layer, and the volume integral of displacement
                        current and the integral of tangential H over S 3 both approach zero by continuity of
                        the fields. By symmetry S 1 = S 2 and ˆ n 1 =−ˆ n 2 = ˆ n 12 , where ˆ n 12 is the surface normal
                        directed into region 1 from region 2. Thus


                                                   ˆ n 12 × (H 1 − H 2 ) dS =  J dV.          (2.182)
                                                 S 1                   V
                        Note that
                                                  δ/2           δ/2

                                       J dV =        J dS dx =     f (x, ) dx  J s (r, t) dS.
                                     V         S 1  −δ/2       −δ/2          S 1
                        Since we assume that the majority of the source current lies within V , the integral can
                        be evaluated using (2.181) to give


                                                  [ˆ n 12 × (H 1 − H 2 ) − J s ] dS = 0,
                                                 S 1
                        hence

                                                     ˆ n 12 × (H 1 − H 2 ) = J s .




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