Field of Helical Currents. 439 



lines o£ force Hinked with the circuit PjPgNgNi pass through 

 the sector N^P. Hence i£ the helix have an integral 

 number of" turns and its ends be joined by a conductor lying- 

 in a plane containing the axis of the helix, we get 



Jo J* {{a + hy-UbAr^e+p^^ 



where fa and fa are the values of i/r at the ends of the helix. 

 Hence, integrating, we get 



»m r^ 2 r , . 



43 ^ cos 20[log (pfa + v/i> 2 ti 2 + A 2 ) 



-log (pt 2 + VpV + A 2 )M 



where A 2 = (a + &) 2 — 4a6 sin 2 0. 



Integrating by parts and noticing that sin 20 vanishes at 

 both limits, we find that 



^M _ C< 2 sin 2 0cos 2 0d0 



***- Jo (p 2 ti 2 + A 2 ) 1 ' 2 (pti+ VpV + A 3 ) "' 

 , _ f^ 2 sin 2 cos 2 / /-^-^ — -rr 2 , \ m 



= 4a6 f'* ™'<W' r ^2 p»i \ dd 



Jo A " I vVf 2 2 + A 2 y/p*fa 2 + A 2 i 



Therefore 

 4a& 



= r P' 2 f 2 t C0S ^ ( # 2 - -£jr\d0, (33) 

 J o l-c 2 sm^ ( R 2 A 3 RiAiJ ' K ' 



where c 2 =4a&/(a + 6) 2 ; R JL 2 = (a + ^ 2 +.P 2 ^i 2 5 



R/ = ( a + 5) 2 + p 2 t 2 2 ; *i 2 = 4a6/R 1 2 ; k 2 2 = 4a&/lV ; 

 A x 2 = 1 - A>! 2 sin 2 (9, and A 2 2 = 1 - & 2 2 sin 2 0. 

 If — ^=^2=^ R 1 =R 2 =R, and P = 4a/>/R 2 , we have 

 M cVf^ sin 2 cos 2 (9 ^0 , u) 



Noticing that 



«" sin 2 cos 2 0=- ^^ + (1-r; 2 sin 2 0) (sin 2 + ^-)i 



