Field of Circular Currents, 423 



The following theorems can be easily proved by elementary 

 Integral Calculus : — 



o 2 ^#=>-E), (3) 



J 2 ^ #= J (E _ F)+F , . . . (4 ) 



(V W0cos»0 ,._2-f F 2-2F 



J ~S--^--3l^ E -~M^ F '- • (5) 



I 

 1 



rfc£ E 



A 3 1-V 



j; 



'/ 2 sm 2 <f>d<}> _ E _ F 



l„ A 3 ~B(l-k 2 ) P' 



(6) 



772 7.2 "I 1 _p 



s i n ^ A# = ^_i E+ i_l r . . (8) 



The following theorems for transforming the modulus of 

 elliptic functions are of fundamental importance. By their 

 means simple formulas can be found from which, as Legendre 

 has shown, the elliptic functions can be calculated to any 

 required degree of accuracy with extreme ease. The first 

 formula* was given by Landen. 



F = (1 + //)F / .-...-} 

 E=^E'-(1-&')F'J' ' 

 where the modulus of E' and F x is &.', and 



(9) 



k'=(l- Vl-/< 2 )/(l+ s/l-k 2 ) 

 and h = 2 ^Htfcl + V). 



3. Formula for the magnetic force at any point due to the 

 current in a circular filament in rectangular, polar, and 

 hipolar coordinates. 



In fig. 1, BR is the conducting circular filament and 07/ is 

 its axis. It is obvious from symmetry that the lines of force 

 will all be in planes passing through OZ ; . We shall first 

 find the component forces Z and X, parallel to OZ / and OX' 

 respectively, at the point P in the plane Z'OX'. Let the 

 radius of the circular filament be a, let PO be r, and let the 

 angles POX' and X'OR be 6 and $ respectively. We shall 

 consider the force at P due to the current in an element adcj> 



* A proof is given in Todhunter's ' Integral Calculus,' Chapter X. 



