Induction in Spheroids. 457 



where D = distance, 



pcos<z = z, 



p sin a = *J x 2 + if = h^/r 2 — 1 sin 0, &c. 

 ... J) 2 = h 2 [> 2 -sin 2 + r' 2 - sin 2 6' —2rr> cos cos 0' 

 -2</r 2 ^1^7 2 ^l . sin sin 6' cos 0^/] . 



Now, putting cos — fju and cos </> — </> / = l, for terms 

 independent of c/>, 



1__ 1 



D 2 ~ /i 2 [(r 2 -l) + (V 2 -l) + f jL 2 + n /2 -2rr'w / 



In order to expand D _1 put fju =1, r' = l. We get, 

 2_ 1 



= ^;(2n + l)P H WQ,0'). 

 /. obviously, 



1 = |s(2 !l +i)P„( /i )P„( /i ')Q»WQ»M- 



3. To find the potential due to a circular current at a 

 point P in any system of harmonics. 



For this, it will be necessary to prove the following lemma. 



4. Lemma. To find the solid angle (©) subtended by any 

 surface at any point. 



Let d$ be the element of surface at any point f , 97, J. 

 Let x, y, z be the coordinates of the point at which the 

 solid angle is to be found. 



Let ^r = angle between the outward-drawn normal at the 

 point and the line joining the two points. 



p = distance between the points, 

 dn = element of outward-drawn normal. 



Then 



cos-v/r. 



-WO- 



