Self -induction of Wires. 15 



Hence the required solution is 



^ 4 ^ ^ sin ma _•** ^ sin nb _J!* 



r= -rloS cos moo e 47T M & . 5 cos ny e 4^* 



ab m n 



or 



4 -r, ^^ sin ma sin n6 , /01 \ 



r=- 7 I U cos w^? cos ny e pt . . . (ZLa) 



ab mn ° 



From this derive the magnetic force by (15a). Thus 



-n- 16-7T -r, ^^ sinma , . e pt 



Mx = j-7 1 o^a sin nt> cos m,£ sm ny- 



a& m *mr+n? 



rr 167T ^ ^^ Sin^6 . . €& 



J± 2 = — r 1 qZa sm ma sm mx cos w?/ — «- — «• 



ab n * m z -\-n z 



The total current, say C, in the prism is given by 

 4,rC=2 P H^.j-sPh*^., 



J—b J —a 



_64tt ^ eP* 

 ~ ab lo ^^V' 



(22a) 



.b)> 



c =^ c »^> • • • • w 



by line integration round the boundary. Or 



m 2 ri 



if C = 4a6r o , the initial current in the prism. 



Since the current is longitudinal, and there is no potential 

 difference, the vector potential is given by E=— A; or, A 

 being the tensor of A, A is got by dividing the general term 

 in the V solution (21a) by — pk ; giving 



. 1 6ttu, ^ v sin ma sin nb ,_. . 



A= — j—ZZ — i~^- y — 9- cos m,r cos ny eP*. . (24a) 

 ab mn\pi l + n z ) J v J 



Since the magnetic energy is to be got by summing up the 

 product ±AT over the section, we find, by integrating the 

 square of T, that the amount per unit length is 



1 — ^r^mVK — • • * (25fl) 



By the square of the force method the same result is 

 reached, of course. We may also verify that Q + T=0, 

 during the subsidence, Q being the dissipativity per unit 

 length of prism. 



The steady-flow resistance per unit length is the L in 



