5 
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=Xsin (u+ kn), 
kn’ being the change of position of the suspended magnet due 
to the small added force. Hence 
- S=X cosukn'; 
and, dividing the equation last found by this, 
Ro ph 8, 
S e. 
‘< 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 
V 
perpendicular direction, respectively, are 
M. 
. cos, — we 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 ¢ the angle which that line makes 
with the vertical.* And the sum of these forces, resolved in 
the vertical direction, is 
= (2 cos? ¢ — sin? @). 
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 negletted in the approximation; so that 
* Transactions of the Royal Irish Academy, vol. xix. p. 162. 
