92 HENRY A. EOWLAND 



and 



To find Q', we have 



whence 



and 



fV^AT HAT 



^ _-"). . . (3) 



When L is very large, or s =*/RR' , we have 



Q' = Cf L > and C: 

 in which L / is reckoned from an origin at any point of the rod. 



These equations give the distribution on the part outside the helix; 

 and we have now to consider the part covered by the helix. Let us 



A: c: E 



FIG. 1. 



limit ourselves to the case where the helix is long and thin, so that the 

 field in its interior is nearly uniform. 



As we pass along the helix, the change of magnetic potential due to 

 the helix is equal to the product of the intensity of the field multiplied 

 by the distance passed over ; so that in passing over an elementary dis- 

 tance dy the difference of potential will be &dy. The number of lines 

 of force which this difference of potential causes in the rod will be equal 

 to Qdy divided by the sum of the resistances of the rod in both direc- 

 tions from the given point. These lines of force stream down the rod 

 on either side of the point, creating everywhere a magnetic potential 

 which can be calculated by equation (2), and which is represented by 

 the curves in Fig. 1. In that figure A B is the rod, C D the helix, and 



cPQ' 

 This could have been obtained directly from the equation ,? 9 =Q / r y , and Q/ e from 



Cl-Li' 



dQ' 

 the equation Q f e = -V A L. 



