54 



Messrs. Boys, Briscoe, and Watson on the 



represent the force due to the current at each point. The 

 rectangular area f g c a, equal to 27r, measures the total force 



Fig. 4. 

 6 



Y 



& & 



7v 



H 



due to a uniform current of the maximum value. The areas 

 a d h and eke are the forces due to the ends of the wires in the 

 wrong direction, and these together 



7T (V 2 . 



sin 6d0=~ 

 It 



1. 



The area h d b e k is the force due to the middle half of the 

 wire in the right direction, and this 



-H" 



sin0d0=£-+l 



The difference of these, which is the area d b e I, is the balance 

 in the right direction, and this = 2. But the force due to a 

 current of the maximum strength over the whole length of 

 the wire is 2ir, so the force in the cranked conductor due to 

 the harmonically distributed current and the opposite effect 



of the ends is — x that due to uniform maximum current over 



7T 



the whole wire. The current is not steady, but varies har- 

 monically with the time, so on this account the value must 

 be halved again, but this halving is exactly balanced by the 

 doubling which results from the fact that the electrostatic 

 exactly equals the electrodynamic effect. 



We have seen that the force was not so great as that due 

 to a steady uniform current of *0017 ampere ; therefore the 

 maximum current in the centre could not, if the oscillations 

 continued uninterruptedly,have reached the value of s/tt x '0017 

 or *003 ampere. Since in the most violent torrent of sparks 

 that can be produced the time during which the oscillations 



