A^r — .y mkral A'$,^ = 0. 



In order to caloiilate tlie inipiiLse in tlie first circuit, we siiall 

 divide the first conductor info conducfiug tubes, whicli eacli of tiieni 

 again consists of circuits. Let tlie conducting tubes, wliic!! are con- 

 centric and cylindrical in the iron, have a radius r there and a 

 thickness dr. When we then give them dimensions proportional to 

 this in the other parts of the conductor, the resistance of sucli a 

 tube will be: 



R' 



w = . ir, 



2r dr 

 The increase of the induction flux through the surface surrounded 

 by every circuil belonging to the conducting tube, amounts to: 

 Ji 



1 



A J/,. 



/ ( dr . Ais. = mlU(( (B'~r'). 



The quantity of electricity set in motion in the conductoi', now 

 becomes, when we make use of the mode of calcnlalion explained 

 in ^ 3, which finds expression in (14) : 



M 

 _ AM,. mlkla r mlklaR^ 



\Mr mlkla C 



= (A" - 7'^) 



e.w 2R'Wc'J ' 



8T^, 



With introduction of the angle of twisting '/ this becomes: 



e,= ' (22) 



Hence from (21) and (22) we find really 



^ W^ ^r: fi W 



in agreement witii (J 8'). 



If k is positive, then the sense in which the impulse takes place, 

 is in leflhand cyclical order with the current /. 



As A^,, = {), the impulse in the second conductor is zero. 



We may assume that the circuits run parallel to the axis over 

 the greater part of the length. The direction of the current can, 

 however, be diffeient for different circuits. In this case we shall be 

 allowed to use the formula (13) for the real current. It follows 

 from this that the circuits wlierc the motion of electricity is zero, 

 will lie on a cylinder surface, (he radius r of which is given by 

 the equation : 



c 



23* 



