Field of an Electrical Current. 203 



the magnetic potential at P due to the current in the fila- 

 ment is 



ai^| (2) 



Let the radius, QV, of the filament be denoted by a ; let 

 PQ = R; ZPQV = <9; then in the typical formula (1) we 

 have p / = R, r = VP, £ = PN = R sin 0, R being, of course, very 

 small compared with a. 



Although our object is to obtain the value of £t correctly to 



the second order of the small quantity — , it will be useful for 



future reference to express the quantities v, &c. as far as the 

 third order. 

 Thus we have 



r =*(l- 5 C05 6>+ ^siu 2 6>+~cos0sin 2 o), . . (3) 

 \ a Ice lec J 7 



J = i{ 1+ 2^ cose+ S (3cos ^- 1) + i£( 5cos ^- 3cos0) }' (4) 



;=i{l+?cos^+g(3cos^-l)+g(5oo S ^-3cos^)}, (5) 



v =^l, . . '. ' (6) 



*'=!> ^ 



where k' is the modulus complementary to h. 



Now, of the two elliptic integrals in (1) the first is one in 

 which the parameter is negative and nearly equal to — 1, 

 while in the second the parameter is positive and very large. 



Observe that - = sinPYQ= V 1 — y 2 , so that (1) can be 

 written r 



+(rt .) A /^n( I |L, i )}. (8 ) 



To deal with the first of these elliptic intervals, let 



' =l-& /2 sin 2 ^, .'. ^/Lz^=k r smf; . (9) 



1 + v 



P2 



