The Method of Moments in Electromagnetics

A Brief Review of Electromagnetics 7we write this asÖ´Ö¡Eµ Ö ¾ E ¾ E J (2.20)where is the wavenumber, Û Ô ¯ ¾. Substituting (2.3) into the abovewe getÖ ¾ E · ¾ E J ÖÕ · (2.21)¯Employing the equation of continuityallows us to obtainÖ¡J Õ (2.22)Ö ¾ E · ¾ E J½(2.23)¯Ö´Ö¡JµSince Maxwell’s Equations are linear, we can consider J to be a superposition ofpoint sources distributed over some volume. Therefore, if we know the response ofa point source, we can solve the original problem by integrating this response overthe volume. We now make use of this idea to convert (2.23) into an integral equation.Since (2.23) comprises three separate scalar equations, let us consider just the xcomponent, which isÖ ¾ Ü · ¾ Ü ´Â Ü · ½ ¾ Ö¡Jµ (2.24)ÜWe now introduce the Green’s function ´r r ¼ µ, which satisfies the scalar Helmholtzequation [1]Ö ¾ ´r r ¼ µ· ¾ ´r r ¼ µ Æ´r r ¼ µ (2.25)and assuming that ´r r ¼ µ is known, we can obtain Ü via Ü´rµ δr r ¼ µÂ Ü´r ¼ µ· ½Generalizing to the full vector form, we writeE´rµ δr r ¼ µ ¾Ü Ö¼ ¡ J´r ¼ µJ´r ¼ µ· ½ ¾ Ö¼ Ö ¼ ¡ J´r ¼ µr ¼ (2.26)r ¼ (2.27)where the integral is performed over the support of J. By a similar derivation, theradiated magnetic field due to magnetic current M and charge Õ Ñ isH´rµ ¯Î´r r ¼ µM´r ¼ µ· ½ ¾ Ö¼ Ö ¼ ¡ M´r ¼ µr ¼ (2.28)To use these equations, we must now find the solution to (2.25) and obtain´r r ¼ µ.