492 Dr. W. F. Gr. Swann or the Electrical 



2ne Ye 



Thus, writing; ir+~ — — — = v 2 , we have to integrate (19) 



° m m ° 



between the limits given by 



v ( 1 Xx<?# „ , X t g« , 2 , 1 , 



c= : < 1 ^-cos-"ur — - — - sin" vr > anuc = », 



cosy (_ mv zmv 1 J 



then replace x by RcostJt and integrate from R = to 

 R = ^ , and finally integrate from i/r = to yjr = 7r/2. If SH 

 is the result of the first integration, we can write 6fl = 



c^r dR . dyjr I \ 1 + — ^- -f —-^-s tan 2 ifr } sin i/r cos ilr e" 



E \ 



e- hmc ~c ?> dc 



vfco&y? 



+ ^ .dR . ^ ( e- R ' A sin ^ . cos ^ . e~^ c2 c 3 <7c, 



v$ cos y 



where <E> is written for 1 — ^-^r cos 2 -\/r — -^— ^sin 2 <r, and 



mir tmv 



where in the second integral we have omitted the terms 

 multiplied by X^aj/mc 2 , since such terms would only con- 

 tribute quantities of the second order, owing to the fact that 

 the limits of integration only differ by first-order quantities. 

 Again, for the same reason we can write the second 

 integral as 



^€-**sin yjr cos f . e -W/cos^_J^ f^± e Z\ | C os 2 ^+ ^±\ 

 -/V • ■ cos 4 y L mv J [ - J 



so that writing x — R cos yjr and cos yjr = z~ l we have 



+ *£%£)*—]■ 



It is hardly necessary here to give the full analysis ; the 

 result of performing these integrations is 



The current density i 2 due to the flow from right to left is 

 obtained by writing — X x for X 1 in the above, and yu for v 



2t]C Ye 



where u? = ir+ ^—4- — . The resultant current density is 



m m 



then equal to i 1 — i 2 . If we write hmn 2 + 2rjeh = a 2 we have 

 hmv = a 2 — heYj and Junjuu 2 = a 2 -\- JieY. 



