540 Dr. A. Russell on the Effective Resistance 



is #, I is the total current flowing in the inner conductor, 

 and \' x is the sum of the currents flowing in the tube of the 

 outer conductor whose inner radius is b and outer radius 

 is x. 



By differentiating </> with respect to r, we get 



and hence, by (60), 



P *r = — ).'&**> (62) 



and finally, by differentiating, 



-ft r 2 r ^r p ^t 

 Writing 



m , = i^ = 8*V (64) 



/> P 



where /is the frequency of the alternating currents, we see 

 that the value of i is given by (58). 



Let us suppose that the current-density along the axis of 

 the cylinder is given by i = i coscot. At the axis r is zero ; 

 and since ker is infinite both and D must be zero, as 

 otherwise the current-density would be infinite. Since 

 also ber0=l and bei0 = 0, we have A = z and B = 0, 

 and thus 



i=i Q ber mr cos cot — i bei mr sin cot. . . . (65) 



It is easy to verify by the help of the equations 



(Vber mr~dr = (r/m) bei' mr . . . . (6Q) 



and JV bei mr~$r = — (r Jm) her' mr . . . (67) 



that this value of i, which is the same as that given by 

 Kelvin, satisfies (62). It therefore gives the solution of 

 the problem of finding the current-density at any point in the 

 substance of the inner conductor. 

 From (65) we find that 



l r = 2ir ir~dr 



= (2irjm) r bei' mr . i cos cot + (27r/m) r ber' mr . i sin cot. (68) 



Putting r equal to a in this equation, we get for the total 

 current, 



I = (2ir/m) a bei' ma . io cos cot -f (27r/m) a ber' ma . i sin cot. (69) 



