H = H t ^V^e^. 



458 Prof. J. J. Thomson on the Discharge of Electricity 

 or, since H varies as e tpt , 



dr 2 r dr ' 



where n 2 = 4:7rfitp/(r ; fju being the magnetic permeability of 

 the cylinder, and a its specific resistance. The solution of 

 this equation is 



H=AJ (mr)e^ 



where J denotes Bessel's function of zero order. Since the 

 magnetic force outside the cylinder is not affected by the 

 currents we must have, if a is the radius of the cylinder, 



A J (m<2) = H . 

 Thus 



J (cnr) 

 1 J Q (ma) 



The currents induced in the cylinder are tangential, and if q 

 is the intensity of the current at a distance r from the axis of 

 the cylinder, 



m 



d {uia) 



We shall consider two special cases : first, where the resist- 

 ance is so great that na is a small quantity ; this includes 

 the case of the plumbago crucible and the tube filled with 

 the electrolyte. 



In this case we have approximately 



J (ina) = 1, 



J (mr) = —\mr. 

 Thus 



4irq = — Wre l Pt 



Taking the real part of this expression we find 



tip TT 

 q= r ~ti rsmpt. 



The rate of heat-production per unit length of the cylinder is 



aq^Trr dr, 



: 



