SOME EQUIVALENCE THEOREMS OF ELECTROMAGNETICS 111 

 only the first two terms 



gi&p' cos .? ^ 1 ^ j^p> COS t? = 1 + i0p' sin 9 cos {,p - <p'). (31) 



We need the second term because the integral of the first vanishes. 

 Integrating the second term, we obtain 



ilSPe-'^^ sine r" , f 2. 



= -^iiS(^)2 - a^)P t sin0. 



8 r 



The magnetic current is uniform around the axis and there is no 

 accumulation of magnetic charge anywhere ; hence the second term in 

 the expression for H as given by (11) vanishes. Therefore 



iJ^ ^ _ ^coe(3{b^ - a^) ^-^ sin 6. (33) 



At a great distance from the source the wave tends to become plane so 

 that in the radiation field the electric intensity is perpendicular to OA 

 and to H and is given by 



'^H^= UOttH^. (34) 



According to Poynting the radiated power is the real part of the 

 following integral 



W = ^ 11 EeH^*r^ sin 6 dd d<p watts, (35) 



2 Jo I 



where H^* is the complex number conjugate to H^. Substituting from 

 (31) and (32), we obtain 



3 / A2 _ ^2 \ 2 /»?r /'2ir 



^^^(^V^'^ ^' ' "'"'^^^ ' ^^ 



r s\n^d dd C 

 Jo Jo 



/,2 _ ^2 \ 2 



360 ' ~^ ' ^' ''■^"'' ^^^^ 



Introducing from (21) the expression for P and designating by 5 the 

 area of the opening, we have 



^=^/-^W" watts. (37) 



-(^0''^" 



