Mr Searle, On the Coefficient of Mutual Induction, etc. 403 



§ 8. Turning now to the circle, we have to find the magnetic 

 potential at points on its axis, due to unit current, in the form 



> + #I + # 2 



3 ' ' 



We notice at once that g = 0, since at a, great distance the current 

 acts as a small magnet ; the leading term consequently varies 

 inversely as the square of the distance. Hence the term Y will 

 not appear in the expression for M. 



I now suppose that the centre of the circle coincides with 0. 

 If c be the radius of the circle, and if the current in it appear 

 from a point T, on the " axis " of the circle, to circulate in the same 

 direction as the hands of a watch, then the magnetic potential 

 at a point on the axis, for unit current, is 



K = -2tt 1- 



vr 2 + 



1 c 2 1.3c 4 1.3.5c 6 



+ 



= _ 2tt - — + 



\2r 2 2 . 4 r 4 2 . 4 . 6 r 6 



Comparing (11) with (1) we have 

 5 r o = 5 , 2 = 5 f 4 = ...=0 



(r>c)... (11). 



(12). 



| 9. Remembering that M = — W, where W is here the 

 mutual potential energy when a unit current flows in each coil, we 

 have by (6), (10) and (12) 



M = 4nrc 



- c sin (Ui cos <j> — v x sin <£) 



13 1 



— ^—. • K c 3 sin 3 6 (w 3 cos 3$ — v 3 sin 3<£) + 



= 4ttc 2 (- l)™" 1 zp 



(2m)! 



sin-* 



2 2w m!m!(2m-l) 



/pim— l c 2m— *\ 



cos (2m - 1) <f> sin (2m - 1) y ^^ + ^J 

 — sin (2m — !)</> cos (2m -1)7 



where m ranges from 1 to 00 , and c< a and c < b. 

 VOL. XI. PT. V. 



am— 1 Jj-2m—i 



29 



