Field of an Electrical Current. 207 



Now it is a known result that when U is small, 



4 V 2 / 4 \ 

 K=logJ + T (logJ-l), . . . (25) 



E = l+f(log|-i), (26) 



and these enable us to verify the above value of XI. Thus 

 for any point, P, in the plane of the circle and within its 

 circumference, X2 = 27r; and if in (21-) we put = 0, we get 



n = 2.-(l-E).-g(2K-E)| 



= 2tt, 



as we see from (25) and (26), since, to the second order, 



k n =— 2 . Similarly for any point, P, in the plane of the 



circle and outside its circumference, 11 = ; and this we find 

 to be the case by putting = 7r. 



To find the value of O in its final form in terms of R and #, 

 we must substitute the value of k! in K and E. Now 



(27) 



k'= 



R 



P ~ 



Rf R 



lOL C let 



e+ 



R 2 



^- 2 (3cos 2 l9- 



-»},. 



and if we denote log -^ by L 



, we 



have 









K= 



= L— ^-cos# + 



R 2 



lb'a 2 



(L + l- 



-4cos 2 0), . 





E = 



=l+g(L-i). 











(28) 



(29) 



Now in (24) K occurs only in terms of the first and second 

 order, and therefore its value need be taken to the first order 

 only, i. e. 



K = L-^cos0; (30) 



and hence we have finally 



a = 27r-2[6>+^sin6'4- T |^(6L-5)sin6>cos6'"l, (31) 



which is the expression for the conical angle correct to the 

 second order of small quantities. 



As a test of the correctness of this value of Q we should 

 find that \7 2 H = 0, as far as quantities of the second order. 

 To apply this test, express X2 in terms of the columnar co- 

 ordinates of P. Let the distance of P from the central axis 



