6 Prof. J. Larmor on Electromagnetic Induction 



and therefore 



P 2 =A^(-y + Y n outside. 



We have 

 therefore 

 and by (1), 



^_^ = -i!T(I> at the sheet; 

 dr dr a 



47T 



A, 2 !+- 1 R=-a(A + A>; 



47T 



inereiore, writing^ = -—r — , 



■A 



Tims, taking real parts, we find, for a field of force oscil- 

 lating in intensity according to the harmonic law, 



P = A COSKtf.l -J Y„, 



the intensity of the internal field due to the current-sheet will 

 be given by 



p__ a — cos /ct—p sin fct/r\ y 

 ~~~ °~ 1 +p 2 ~ \a) 



= — A cosacos(^— «)(- j Y n , . . (2) 



where tan <x — p. 



For copper at 0° C, R=l(540-f- thickness of the shell in 

 centimetres. 



This solution represents a new and opposite field, with 

 strength reduced by the factor cos a, and phase of variation 

 retarded by the fraction u/'Itt of a complete period. 



For each harmonic term in the potential, the corresponding 

 induced currents How along ilie level lines of that term: for 

 example, in a uniform Held of force the currents circulate round 

 the axis of the field in circles, and produce the same external 

 effect as a simple magnet of moment iFa 3 cos a cos (#£--«) at 

 the centre, with its axis pointing in the stable direction along 

 the lines of force, where P cos kI is the strength of the indu- 

 cing field, and tan a s= 3R / 4nra* % 



