424 



Mr. A. Russell on the Magnetic 



of the filament at R. Let RS be the tangent to the circle 

 at R ? and draw PK and KL perpendicular to OX' and RS 

 respectively. Then, noticing that KL = a— r cos cos <j>, and 



Fisr. 1. 



X and Z are the components of the magnetic force at P due to 

 the current in the circular filament. 



that cosPOR = cos 0cos<j>, so that PR 2 = a 2 -f r 2 — 2arcos0cos <f>, 

 we get by Laplace's formula, 



d%= ^sin PRL . sin KPL 

 PR' 5 



iadfj) PL KL 

 ~PR2 'PR* PL 



ia(a — r cos cos (f))d(f> 

 (a 2 + r 2 — 2ar cos cos <£) 3/2 ' 



— r cos cos <j))d(f> 



and therefore, 



2 Jo (« 2 +^-: 



2a?*cos 6/ cos <£) 3/2 

 Similarly we find that 



y = 2iarfVi- 8inflcos * d * 

 Jo («* + i 



Now (10) can be written 



2ar cos cus $) c 



(10) 



(11) 



H 



dcj) 



(a 2 -r r z — 2a?- cos cos 0) 1 / 2 



^"^oW 



2ar cos cos (j>) 3 / 2 ' 



