Page 311 - Engineering Electromagnetics, 8th Edition
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CHAPTER 9  Time-Varying Fields and Maxwell’s Equations    293

                     the curl that is identically zero. Let us therefore tentatively accept (50) as satisfactory
                     for time-varying fields and turn our attention to (49).
                         The inadequacy of (49) is obvious because application of the curl operation to
                     each side and recognition of the curl of the gradient as being identically zero confront
                     us with ∇× E = 0. However, the point form of Faraday’s law states that ∇× E is
                     not generally zero, so let us try to effect an improvement by adding an unknown term
                     to (49),
                                                E =−∇V + N
                     taking the curl,
                                              ∇× E = 0 +∇ × N
                     using the point form of Faraday’s law,
                                                          ∂B
                                                ∇× N =−
                                                          ∂t
                     and using (50), giving us
                                                       ∂
                                             ∇× N =−     (∇× A)
                                                       ∂t
                     or
                                                            ∂A
                                              ∇× N =−∇ ×
                                                            ∂t
                         The simplest solution of this equation is
                                                        ∂A
                                                  N =−
                                                        ∂t
                     and this leads to
                                                           ∂A
                                               E =−∇V −                              (51)
                                                           ∂t
                         We still must check (50) and (51) by substituting them into the remaining two of
                     Maxwell’s equations:
                                                           ∂D
                                               ∇× H = J +
                                                           ∂t
                                                ∇ · D = ρ ν
                     Doing this, we obtain the more complicated expressions
                                                                   2
                                      1                      ∂V   ∂ A
                                        ∇× ∇× A = J + 	 −∇      −
                                      µ                      ∂t    ∂t 2
                     and
                                                       ∂

                                          	 −∇ · ∇V −    ∇ · A = ρ ν
                                                       ∂t
                     or
                                                                     2
                                                               ∂V   ∂ A
                                               2
                                    ∇(∇ · A) −∇ A = µJ − µ	 ∇     +                  (52)
                                                               ∂t   ∂t 2
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