} ***- w 



Energy in the Electromagnetic Field. 21 



The current of! energy at the point is 



S= ~K 2 + «2 2 + %cl\<H cos 20}* 



07T 



= gj | l + * 2 -2*cos ^2r 2 2 j . . . (19) 



The lines of flow of energy are determined by the condition 

 that at any point (r 2 , 0) the inclination to the direct rays 

 is <£, where <p is given by equation (18). From this we can 

 get the differential equation to the lines of flow in terms of 

 r 2 and 0, or, dropping the suffix, r and 0. It is seen from 

 fig. 3 that 



, dr'-addd 

 tan </> = dri -add' 



Fig. 3. 



r+dr, 

 6-fdd 



But r' = r sin 20 = 2r0 and ?- 1 =rcos 20 = r approximately.] 



,, , , 2(rd0 + 0dr) -a0d0 , 9ft v 



Hence tan 6= — 7 ™ , . . . {IV a) 



T dr — a dv 



or 



tan*(l-«^)=2* + (2r-aflg. . . (20) 



From (18) and (20), we get 



d0 

 dr 



(2r-a*)+20=(l-ag) 



2K0( y K-QOS- c 2rO^) 



1 + K 2- 2 k cos ~2r0 2 

 c 



