Mutual Induction -of a Circle and a Coaxial Helix, tjr. 203 



the limits of x being x\ and x' z , so that the axial length of the helix 

 is x'% x'i. 



Let the equations to a coaxial circular cylindrical current sheet bo 



y = a cos 0, z = a gin 0, 



its ends being at distances x } , x-> from the origin. 



Let 7A be the current in the helix, and let 7 be the current per unit 

 length of the uniform current sheet. Then if M' is the potential 

 energy of the helical current and the current sheet we have 



A f 



= 7A7 cos 0d0[/(# r 2 x\) /(#^ ^)+/(#\ #2) /(Xi #0] 

 P Jo 



whore f(z) EE z Ic 



2 



The integral cos<j)f(z)d(j> is easily reducible to elliptic integrals 



J/o 

 of standard form. 



It is to be noticed that f(z) =f(z). 



If the axial length of the helix = 2?, and the axial length of the 

 cylindrical sheet = 2?n, and if x equals the distance between their 

 mean planes, we have 



M ' = 



A 



cos cZ0 [ / (, + 1 -f- m) / (x + Z m) 



(9). 



Similarly for two coaxial cylindrical current sheets v/e have 



cos 



Z w)-/(a; Z + m)]... (10), 



w lie re 7, 7' are the currents per unit lent^th in the sheets. 



14. To find the force betvv-eeu a helical current and a coaxial 

 circular cylindrical uniform current sheet we have to differentiate 

 the potential energy as expressed in equation (9) with regard to x. 



