234] ELECTROMAGNETISM. 463 



For a point in the area of the ring itself, we must divide the ring 

 into two, one within and one without the given point, so that 



TT ( 2 2 - RS) log r = TT (h z - R?) log h 



+ TT {R? log R 2 - A 2 log h - \ (R? - h% 



(17) logr = i{ 



The vector-potential is always, for uniform flow, 



and since this is the flux-function for the induction, by reason of 

 the equation 



- ^- ^ -^, 



we obtain the induction perpendicular to the radius, by differen- 

 tiating according to h, so that 



which agrees with the result (9), in which JRj is equal to zero, 



The electrokinetic energy of the system of currents is, by 

 226 (5), 



and inserting the value of H from (5), 



(19) T=^jjjCwdadbdz-jjjjlnww' logrdadbda'db'dz, 



If we integrate with respect to z from oo to oo , we obtain an 

 infinite result for the energy, but for a finite length I the energy 

 is proportional to I, so that the energy per unit of length of the 

 conductors T/l is given by the above expressions omitting the 

 integration with respect to z. Each point of integration <z, b 

 and a', b' is to traverse the cross-sections of all the conductors. The 

 first integral, containing the constant (7, disappears, since to every 

 current there is a return current, in each of which the same value 

 of C appears, while for the two cross-sections the integral 



wdadb, 



