82 Dr. J. W. Nicholson on the Effective 



but not that defined by x- The vectors are given by 

 X=Y=0, *= 7 =0, 



*-$£ *—»*§ ■ ■ ■ w 



The electric force is along 6 increasing, or parallel at any 

 point to the corresponding tangent of the central curve. The 

 magnetic force is along <f> increasing. At the surface of any 

 wire of the helical system, both forces are tangential, as they 

 should be. 



Solution as a Fourier Series. 



If ds x is an element of arc along p increasing, 



so that p 1 = 1 . Moreover, 



\oj) = V5*) + Uw + feiM 



= ( — p sin <£ cos 6 -f f> sin a sin # cos <£) 2 

 + (/3 sin <£ sin # + /> sin a cos # cos <f>) 2 + p 2 cos 2 a cos 2 = p* 

 on reduction, andp 2 =/3 _1 . Similarly 

 p 3 = \a 2 sec 2 a. + 2ap cos <j> + p 2 (cos 2 <j) + sm 2 asin 2 <f>)\~ s . . (12) 



If the wire be thin, p 3 = a~ 1 cos a, and we are led to the 

 usual solution for the straight wire of great length. 

 The differential equation may be reduced to 



3 3* , 13> , z , . x 3+ 1 3/>. , 13* l 3j* n m v 



0p r Op pO<p 2 r OppzOp pO(j>PzO<l> 



Now 



1 3p?> __ a p sin $ + /o 2 sin $ cos <£> cos 2 a 



Pi 3<£ ~~ « 2 sec 2 a + 'lap cos + p 2 (cos 2 <f> + sin 2 * sin 2 <£) 



2 



= ° sin cos 2 a — ^-g- sin 20 cos 4 a + (14) 



a 



on expansion in powers of p/a, replacing the trigonometric 

 functions by those of multiple angles. Again, 



P . |e> _. _ P cos 2 a cos $ _ £_ C0S 2 a ( 2 - 3 cos-* -* cor* cos 26). . (15) 

 p 2 3/? « 2a- v r/ v / 



We shall not retain p 3 /« 3 or higher powers. 



