Page 179 - Electromagnetics
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This can be put in slightly different form by the use of (B.8). Note that

                                             (A × B) · (C × D) = A · [B × (C × D)]
                                                             = (C × D) · (A × B)
                                                             = C · [D × (A × B)],
                        hence

                                                                     1


                                         ˆ n · P × (∇ × Q) =−c · ∇      × (ˆ n × A) .
                                                                     R
                        The other surface term is given by
                                                     c                   c          c
                                                                !

                               ˆ n · [Q × (∇ × P)] = ˆ n ·  × (∇ × A) = ˆ n ·  × B =−  · (ˆ n × B).
                                                     R                  R          R
                        We can now substitute each of the terms into (3.162)and obtain
                                    J(r )                                               1


                             µc ·       dV − 4πc ·   A(r )δ(r − r ) dV − c ·  [ˆ n · A(r )]∇  dS
                                  V  R              V                     S            R
                                                       1                        1


                                          =−c ·    ∇      × [ˆ n × A(r )] dS + c ·  ˆ n × B(r ) dS .
                                                 S     R                       S R
                        Since c is arbitrary we can remove the dot products to obtain a vector equation. Then
                                                                                  1
                                            µ     J(r )     1    "


                                     A(r) =           dV −        [ˆ n × A(r )] ×∇   +
                                           4π  V   R       4π   S                 R
                                            1                         #
                                                                     1

                                         +    ˆ n × B(r ) + [ˆ n · A(r )]∇  dS .              (3.163)
                                            R                        R
                          We have expressed A in a closed region in terms of the sources within the region and
                        the values of A and B on the surface. While uniqueness requires specification of either
                        A or ˆ n × B on S, the expression (3.163)includes both quantities. This is similar to (3.56)
                        for electrostatic fields, which required both the scalar potential and its normal derivative.
                          The reader may be troubled by the fact that we require P and Q to be somewhat well
                        behaved, then proceed to involve the singular function c/R and integrate over the singu-
                        larity. We choose this approach to simplify the presentation; a more rigorous approach
                        which excludes the singular point with a small sphere also gives (3.163). This approach
                        was used in § 3.2.4 to establish (3.58). The interested reader should see Stratton [187]
                        for details on the application of this technique to obtain (3.163).
                          It is interesting to note that as S →∞ the surface integral vanishes since A ∼ 1/r
                                   2
                        and B ∼ 1/r , and we recover (3.135). Moreover, (3.163) returns the null result when
                        evaluated at points outside S (see Stratton [187]). We shall see this again when studying
                        the integral solutions for electrodynamic fields in § 6.1.3.
                          Finally, with

                                                               1


                                                       Q =∇       × c
                                                               R
                        we can find an integral expression for B within an enclosed region, representing a gen-
                        eralization of the Biot–Savart law (Problem 3.20). However, this case will be covered in
                        the more general development of § 6.1.1.




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