542 PROFESSOR W. F, AYRTON. MR. T. MATHER AND MR. F. E. SMITH: 



one turn of the line of flow through that point. Hence, if the distance of the point 

 from the axis is A + h, the current density is equal to Ki[(A. + hy +p :t ~]~ 1 ' 2 , where K is 

 a constant, Zirp is the pitch of the helix, and t is the total current through the wire. 

 To find u, the current per unit area of the section by a plane containing the axis of 

 the helix, we must multiply the current density by cos a, where a is the slope of the 

 line of flow. Hence 



........ (5). 



The constant K is to be determined from the condition that the total current in 

 the wire is i. When, as in the case of the coils of the ampere balance, p is small 

 compared with A + h, it will be sufficient to take the first two terms of the expansion 

 of (5) in powers of p\ and to write 



u = Ki {T^-(iwl'- -" < 6 >- 



If h, y be the co-ordinates of the point relative to axes through the centre of the 

 section parallel and perpendicular to the axis, we can write 



h = p cos 4>, y = p sin <j> (7). 



Thus, if R is the radius of the wire, 



fli rijt rV. riw | i -,2 i 



up dp dd> = KM { -r r *- ,r, \ pdp dd). 

 oJo Jo Jo [A + pcosd> (A + p cos <j>) 3 ) * 



Now, since p is less than A, 



f 2 * d<f) _ 2:r 



and two applications of the reduction formula 



P* d<f> 1 A , _ ^ \ f 2 ' ^ 



Jo (A+p cos <) m+1 A\ mdp) Jo (A+pcos <) m 



give 



)o(A+pcos<^) 3= 2?r 12 (A a -y) B/3 ~ 2 (A 3 -/) 2 )" 8 ]' 

 On integrating with respect to p, we find 



irR. 



~A~ 



