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292                ENGINEERING ELECTROMAGNETICS


                                        D9.5. The unit vector 0.64a x + 0.6a y − 0.48a z is directed from region 2
                                        (	 r = 2,µ r = 3,σ 2 = 0) toward region 1 (	 r1 = 4,µ r1 = 2,σ 1 = 0).
                                        If B 1 = (a x − 2a y + 3a z ) sin 300t Tat point P in region 1 adjacent to the
                                        boundary, find the amplitude at P of: (a) B N1 ;(b) B t1 ;(c) B N2 ;(d) B 2 .

                                        Ans. 2.00 T; 3.16 T; 2.00 T; 5.15 T

                                        D9.6. The surface y = 0isa perfectly conducting plane, whereas the region
                                                                                        8
                                        y > 0 has 	 r = 5, µ r = 3, and σ = 0. Let E = 20 cos(2×10 t −2.58z)a y V/m
                                        for y > 0, and find at t = 6 ns; (a) ρ S at P(2, 0, 0.3); (b) H at P;(c) K at P.
                                                     2
                                        Ans. 0.81 nC/m ; −62.3a x mA/m; −62.3a z mA/m

                                     9.5 THE RETARDED POTENTIALS
                                     The time-varying potentials, usually called retarded potentials for a reason that we
                                     will see shortly, find their greatest application in radiation problems (to be addressed
                                     in Chapter 14) in which the distribution of the source is known approximately. We
                                     should remember that the scalar electric potential V may be expressed in terms of a
                                     static charge distribution,

                                                                    ρ ν dν
                                                            V =            (static)                  (45)
                                                                  vol 4π	R
                                     and the vector magnetic potential may be found from a current distribution which is
                                     constant with time,
                                                                     µJ dv

                                                             A =            (dc)                     (46)
                                                                   vol 4πR
                                        The differential equations satisfied by V ,
                                                               2      ρ ν
                                                             ∇ V =−       (static)                   (47)

                                     and A,
                                                                2
                                                               ∇ A =−µJ    (dc)                      (48)
                                     mayberegardedasthepointformsoftheintegralequations(45)and(46),respectively.
                                        Having found V and A, the fundamental fields are then simply obtained by using
                                     the gradient,

                                                              E =−∇V     (static)                    (49)
                                     or the curl,
                                                               B =∇ × A   (dc)                       (50)
                                        We now wish to define suitable time-varying potentials which are consistent with
                                     the preceding expressions when only static charges and direct currents are involved.
                                        Equation (50) apparently is still consistent with Maxwell’s equations. These
                                     equations state that ∇ · B = 0, and the divergence of (50) leads to the divergence of
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