364 The Magnetic Field of a Circular Current. 



Putting ^r=7r— x, this becomes 



sin 2 £ + 1 a .. V^&tZ^ 



iW 1 - 



2 / „v ( 2 2 + ^sin 2 y : 



Z P Vl-^sin^J " * 



p' 2 

 where, as before, p = PB, p' = PA, and k 2 = 1 — ^ . If we put 



^=2o>, this becomes 



7T 





dco 



p& Jz 2 + x 2 sm 2 2co> 



where A = V 1— k 2 sin 2 co. 



To reduce this to elliptic integrals, we must resolve the 

 fraction 1/z 2 + x 2 sin 2 2o> into two fractions. It is easily found 

 that 



z 2 + x 2 sin 2 2w = £ 2 + 4# 2 (sin 2 « — sin 4 w) 



= ( s/z 2 + x 2 + x—2x sin 2 eo)(s/ z z + x'—x + 2x sin 2 &>) . 



Let v denote the sine of the angle between PO an^ 

 the axis of the current (or PN) ; then the expression, after 

 resolution into partial fractions, becomes 



|^J o V A+ "7A~Al-vH-2vsin 2 ft) + l+^^Tsm 2 ©)^- 



The portion of this expression which has A in the deno- 

 minator is at once the sum of two complete elliptic integrals 

 of the third kind ; and the portion which has A in the nume- 

 rator is easily reduced to the same form. The result is 



z C^/r + a r — a\ 



where N= 1— v + 2vsin 2 ft>, and W is the value of N when v 

 is changed to —v. Hence we have two elliptic integrals of 



the third kind, one with (t^— ? &) for parameter and mo- 

 dulus, and the other with fyr^, &)• In the usual notation^ 



then, we have for the complete expression of the conical 

 angle subtended by the circuit at P the value 



H^Mi^smS-:.*)}:, 



