SELF-INDUCTION OF TWO PARALLEL WIRES. 123 



On the other hand, the intrinsic energy of a current (524) is 



equal to - LI 2 . Replacing W by this expression, we find that the 

 2 



coefficient of induction of two parallel wires traversed by the same 

 current in opposite directions, is, for unit length of each, 



(9) L = 2[/K /.R;-^/.R a -V/.R;]. 



Unless the field contains iron, we might put /u = //, = // = i, which 

 gives 



(10) L= 2 [/.R?-/.R 2 -/.R 2 ] = 2 /.--. 



760. If the sections of the two conductors are circles of radii a 

 and a', the centres of which are at the distance <, we have 



(I!) 



aa 



This expression only differs materially from that found above 

 (756) when the wires are very close to each other. When the wires 

 have the same diameter and are in contact, we have b = 

 and therefore 



(12) L = 2(/.4 + -) =37726. 



This is the least value which the coefficient of self-induction 

 could have for unit length in the case of a wire folded on itself, like 

 those used in resistance boxes. 



As the wires should be separated by an insulating substance, if, as 

 above (723), y is the radius of the bare wire, and z the thickness of 

 the insulating layer, we should make b = 2 (y + z), which gives for the 

 minimum value 



This coefficient could be considerably reduced if we replaced the 

 wires by very thin plates of metal. 



