252 ON THE DETERMINATION OF THE HORIZONTAL 



Now, if D denote the distance between the centres of the two 



magnets, 



a? = D* + r z 2x/r cos i. 



Wherefore, developing the inverse powers of a in series ascending 

 according to the powers of ^, stopping at the inverse fifth power 

 of the distance, and substituting in the expression for the moment 

 above given, it becomes 



This being the moment of the force exerted by the deflecting 

 magnet upon a single particle, m, of the suspended magnet, the 

 moment of the force exerted upon the entire magnet is obtained 

 by multiplying by dr, and integrating between the limits r = l, 

 I being half the length of the suspended bar.* The magnetism 

 being supposed to be distributed symmetrically on either side of 

 the centre of the suspended magnet, and the axis of suspension to 

 pass through that centre, we have- 



f *i 

 -l 



mr z dr = 0. 



( + l r+l 



mrdr, m^dr, by M.' and 



Jf s ', the expression for the moment of the whole force is 



When there is equilibrium, this moment must be equal to that 

 of the force exerted by the earth upon the suspended magnet. 

 Let X be now taken to denote the horizontal component of the 

 earth's magnetic force. The moment of the force exerted by that 

 component upon the particle m of the suspended magnet is 



mXr sin u ; 



* "We have here assumed that the effect of a magnet is the same, with respect to 

 any point at a moderate distance, as if the magnetic elements in each section perpen- 

 dicular to its axis were all concentred in the axis ; or, in other words, that the 

 integration with respect to the other dimensions of the bar introduces no new term 

 into the integral. There is no difficulty in proving that such is the case, the magnetic 

 elements being supposed to be distributed symmetrically with respect to the axis. 



