﻿460 Mr. W. M . llicks on the Self-induction 



Hence 4 (T v' t'f^ 1 



An ~(4n2-l)7rS l " n " " ;? ( (3) 



(4ti 2 — l)7rS^ - 



This is the general solution for any case in terms of F. 



4. We proceed to apply the foregoing to special cases. 

 Consider, first, the case of a current flowing in a uniform 

 circular ring of any size. The parts of the current all flow in 

 circles, and the equipotential surfaces will be planes through 

 the straight axis. The lengths of conductor which a current 

 uses between two such surfaces will be proportional to the 

 arc between — i. e, to p. Hence, in order that each element 

 of current may produce equal fall of potential between the 

 two planes, the current-density must vary inversely as p, i. e. 

 a=b/p, where b is some constant. Let I denote the total 

 current ; then, integrating over a cross section, it will be 

 found that I = 2 7t5{R- */(B*-r 9 )}. 



Now r = Esin«, a = Rcosa; (4) 



whence I = 27r6R(l — cos a), 



or _ I K 



°"~27r /3 R(l-cosa) W 



Equation (1) now becomes 



dty ,d?jr_ldjr_ 

 dz* + dp* p dp~ ^ °> 

 a particular solution of which is 

 /=_4tt 2 ^ 2 



-~ 4 " 1WI (tS [T.F.i.3]. 



This has to be expanded in a series of the form 



It is clear that there are no terms in sin nv. Hence 



tfr-^ ai a ( n sin 2 v cos nv . 

 Z F n = — \nr%o? —77; 3— dv 



2 n Jo (C-c) 



47r 2 6a 2 d C n cos nv — ^ (cos n + 2 v + cos n — 2 v ) ^, 

 S - chij " V'(C— c) 



47 ^^f {Q,-i(Qn +2 +Q«- 2 )} [T.F.ii.i] 



