498, 499] Mechanical Forces 431 



9V cos 6 cos 6' cos e 



Hence 5-^-7 = - , 



dsds r 



where 0, 0' are the angles between r and ds, ds' respectively, and e as before 

 is the angle between ds, ds, so that 



cos e = cos 6 cos 0' + sin 6 sin & cos (<f> <'), 

 where </>, </>' are the azimuths of ds, ds. 

 From this last equation, we have 



8 2 r sin 6 sin & cos (</> <') 



r 

 and the mutual potential energy w of the two elements now assumes the form 



.., , , , /cose 8 2 r \ 

 w = 11 dsds \- K . ^ . . 



\ -^ \j j rt ' / ' 



- {cos 6 cos 0' + (1 /c) sin sin 0' cos (< <')). 



From this value of w the system of forces can be found in the usual way. 

 The forces acting on the element ds will consist of 



(a) a repulsion along the line joining ds and ds', 



(b) a couple ^ tending to increase 9, 



(c) a couple ^-r tending to increase </>. 



If we take K = 1 we obtain a system of forces originally suggested by 



Ampere. We have 



ii' dsds' n , 



w = -- cos cos 6 , 

 r 



so that the forces are 



77 // *? Ct *? 



(a) a repulsion -- - cos 6 cos 6' along the line joining ds and ds', 



/ 7 j i 



(b) a couple -- sin 6 cos & tending to increase 6, 

 and couple (c) vanishes. 



If we take K = f , we obtain a system of forces derivable from the energy- 

 function 



7? 



w = = -- {sin sin 6' cos (< <') 2 cos cos 0'}, 



which is the same as the energy-function of two magnetic particles of strengths 

 ids and i'ds', multiplied by Jr 8 . Thus force (a) is Jr 2 times the correspond- 

 ing forces for the magnetic particles, while couples (b) and (c) are Jr 2 times 

 the corresponding couples. 



