Magnetic Energy of two moving Point Charges. 193 



We are now in a position to solve the general case. 



The component velocities of e x at A (fig. o) are Wj, v u w l7 



Fiff. 3. 



and those of e 2 at B u 2 , v 2 , io 2 - The origin is taken at C the 

 middle point of AB, the axis of y being along GB. 



r x 2 = r 2 + yc + ~ > and 



c 2 

 r 2 — r 2 — yc + -r-, where AB = c, as before. 



The components of magnetic force due to e Y are 



ViZ 



«i=^i 



A=«i 



— »i(y+|) 





M l 



and those due to £ 2 are 



(y + !)-i>i* 



v-«fr(y-|) 



« 3 = ^2 



ft = «j 



w; 2 <£ — w 2 c; 

 r 3 



72=^2 



M2 ( y ""l) ~~ V 



PAi7. 1%. S. 6. Vol. 32. No. 188. i%. 1916. 



