Experiments with Alternating Currents, 237 



The p's cancel — this is an important part of the analysis — 

 and we get 



du E cos a , . . , /n . 



df=— Y~ C0S O +a ) ( 6 ) 



If we assume that the value of pk is large in comparison with 

 that of n 2 —p 2 , a. will he small, and cos a will approximately 

 equal unity. 



Let us suppose that n >p, so that a, is positive. 



Let e h = the back E.M.F., due to the forced oscillation of 

 the particle. 



Let K = a multiplier which will increase with the mag- 

 netization of the particle, i. e. with the strength of the field. 



We have 



-T-r E cos a , . ,_. 



e b =K — j- cos{pt + a) (7) 



Let i = current along the bismuth, 

 r = true resistance of the bismuth ; 



. f KE cos a , x , . 1 / 



z = < excospt r cos(pt + a) >lr. 



To obtain a rough result we may put a = 0, 



»= {( e i-^) cos ^}/ n 



Let r a = apparent resistance, 

 KE\ 





Let F = strength of magnetic field : for the sake of mathe- 

 matical simplicity let us suppose that the magnetization of 

 the bismuth particle varies as the square root of F. 



We We R ^ ^ 



Let q = a multiplier ; we obtain 



r a =r(l + q¥). . (8) 



This deduction diverges from the observed facts ; its chief 

 interest lies in the result that, so long as k is great compared 



with — , the frequency of the alternating current has 



Jr 



