456 THE ELECTROMAGNETIC FIELD. [PT. III. CH. XI. 



For the pair of sides AB, C'D', substituting the square of their 

 distance apart, a 2 + / 2 2 > for a 2 and changing the sign, we have 



The portions for the pairs of sides BG, B'C', and BC, D'A', are ob- 

 tained from these by changing ^ into 1 2 . We have then considered 

 j ust half of the two circuits, so that, adding these four parts and 

 multiplying by two, we obtain the value of the inductance 



M = 8 a - 



The attraction of the two circuits for each other when traversed 

 by unit current is obtained by differentiating this expression by a. 



233. Pair of Parallel Circles. If the circuits are circles 

 of radii R lt R 2) their planes being perpendicular to the line 



joining their centers, of length a, we 



may put 

 IR L_ I so 1 = R l cosfa, # 2 = ^2 cos < 2 , 



a VL\ I 



T/J = R! sin fa , 2/2 = ^2 sin fa> 

 z\ =0, 2 2 = a, 



cos fa - R 2 cos </> 2 ) 2 + (A sin X - R. 2 sin 

 cos (fa - fa), 



M 



=/T 



Jo Jo 



cos (fa - fa) R^dfadfa 



V a 2 + ^ 2 + ^ 2 2 - ^^i^ 2 cos (0! - fa) ' 

 The integration with respect to fa amounts merely to multipli 

 cation by 2-7T. If we put 



fa-fa = 2^ 7T, d (fa fa) = 2ctyr, COS (fa -fa) = - COS 2l|r, 



the integral becomes 



