Let the deflecting magnet be now turned (the position of its 



centre remaining unchanged), so that its axis is horizontal, 



and perpendicular to that of the suspended magnet. In this 



position it exerts no action upon the iron bars ; but tends to 



turn the suspended magnet with a force whose moment we 



shall denote by S. The equation of equilibrium in this case 



is therefore 



U + U'+ S = X sin (u + hri), 



kri being the change of position of the suspended magnet due 

 to the small added force. Hence 



S = X cosukri ; 



and, dividing the equation last found by this, 



V — vv — = — /. 

 S 8 n 



"Now, the deflecting magnet being vertical, and its distance 



considerable as compared with its length, the force which it 



exerts upon the unit of free magnetism at the centre of one of 



the iron bars, in the direction of the joining line, and in the 



perpendicular direction, respectively, are 



2M M . 



— IP cos (p, sin </>, 



in which M denotes the magnetic moment of the deflecting 



magnet, e the length of the line connecting its centre with the 



centre of the iron bar, and <j> the angle which that line makes 



with the vertical.* And the sum of these forces, resolved in 



the vertical direction, is 



M 



— (2 cos2 - sin 2 0). 



But we may consider the quantities e and (and therefore 

 the force exerted by the magnet) to be the same for all points 

 of the bar, the variations of these quantities being of the same 

 order as those neglected in the approximation ; so that 



* Transactions of the Royal Irish Academy, vol. xix. p. 162. 



