416 Prof. E. Edlund on the Theory 



situated in the same horizontal plane as the pole. The sine 

 of the angle which the direction of motion makes with the 

 line of junction between the element of the magnet and the 

 element of the circuit (the angle designated by /3 in the for- 

 mula (6)) will be in this case (fig. 4) equal to cos t= (u — t f ) 



cos u(R + r cos u) rsiri 2 u 



~~ ~\/R 2 + r 2 + 2Rr coTS + V^ 2 "+ "r 8 + 2RrcosM" 



Consequently sin/3 



R cos u + r 



VR 2 + ^ 2 + 2Rrcos?* 



The angle denoted by ^ in the formula (6) is equal to zero, 

 and its cosine is equal to 1. The inductive action of the 

 element of the magnet Mrdu situated at the point c upon the 

 element of circuit As at rest will be then 



M?* 2 ?;(E cos u + r)Asdu 



(R 2 + r 2 + 2Rrcosu)? 



If we multiply by 2 the integral of this expression between 

 the limits zero and ir, we shall have the value of the induction 

 which the sum of the elementary poles situated in the peri- 

 phery of the magnet is able to produce by the rotation of the 

 magnet round its axis in the element at rest A,s. 



2Mr\A 5 r* = *_( Reos» + r>fe 



J„=»(R 2 + »* 2 + 2Ercos«)* 



= M A f" = ' ( R2 + y2 + 2Rrcos M )<fa + (r 2 -~R 2 )du 

 l=o ~ (R 2 + r 2 + 2Rroos«.)i 



Putting in this expression 1 — 2 sin 2 -^- in place of cos u, and (/> 

 instead of ^, and lastly sin ® for ^ , \ we have 

 2MrvAs f<P=i <ty 



rvAs f <P=f 



R + r Jv=o Vl— sin 2 © sin 2 



2MrvAs.{r~R) f<P=| dcf> 



+ (R + O 2 X=o (l-sm^sin 2 ^' 

 from which we obtain the elliptical integral 



2rMtA s [ lr LF'(0)+^E'(0)]. . . (8) 

 2. If we suppose that the element of the circuit only is in 



