and Inductance of a Concentric Main. 541 



By differentiating this equation with regard to t, we 

 find that 



— = — (27rco/m) a bei' ma . i sin cot 



+ i^irco/m) a her 1 ma . i sin cot, (70) 



and hence, solving these equations for ?' cos cot and i sin cot, 

 we get 



m hei' ma T m ber' ma dl 



i Q COS cot = - T77 x-L + n v7 : s^? • (71) 



27ral(ma) IiracoX^nu^ot 



and . . _ 7n her' ma j m bei' ma dl . 



i sin o>* - 2naY(ma) 2iraa>Y(ma) 5*' ' {i) 



Let us now consider the currents in the outer conductor. 

 Let i' be the current density in this cylinder at a distance r 

 from the axis. Let e' be the potential- difference per unit 

 leno-th measured from the distributing station to the alter- 

 nator, i' being considered positive when flowing in the same 

 direction. 



We have, therefore, 



e' = (p/2in%r)(2irrdr.i')--'b4Jfat 

 ^pi'-Wfdt, (73) 



where / =M rfc^^ (74) 



Thus t m: = ._ 2tl (i-v r) 



^ri 'dr. ....... (75) 



Also, since we have supposed that e' does not vary with r, 

 we get from (73) and (75) 



and thus V? , 1 d^ _ m a d'"' ,__. 



> 2 + r-dt ~ co dl [i,) 



By (58), the solution of: (77) may be written as follows: — 

 i' = (A ber ?»? , -f B bei mr+ C ker mr + D kei ??ir) / cos cot 

 + ( — A bei wr-f B ber mr — C kei mr + ~D ker -»ir) / sin &)£, 



.... (78) 



where A^ B, C, and D are constants which have to be 

 determined from the data of the problem. 



