8 

 dm dm; 



~7~ 



and the portion of this resolved in the horizontal plane, is 



dm dm' _ _. 



=— x OP. 



P 3 



The moment of this force to turn the suspended magnet is 

 CP x &*L x OP sinOPC = CP x *2!pL * OC sin OCR 



p 3 p 3 



Hence, putting OC = a, CP = r, and OCP = 90° - u, the whole 

 moment of the force of the iron bar is 



[[rdmdm' 

 a cos u ' ' 



Now, PP' 2 = P'0 2 + OP 2 . Or, putting OC'= A, CC'= e, C'P'= r', 



p 2 = (h- r) 2 + a 2 + r 2 - 2ar sin u = e 2 + r 2 + r' 2 - 2 (Ar + ar sin 11). 



Accordingly, expanding — according to the ascending powers 



T T 



of -, — , integrating and making 



[r n dm = M„, JV" dm'= M' n , 

 and observing that, on account of the symmetrical distribution 

 of free magnetism in the magnet and bar, M n and M' n vanish 

 when n is an even number, — we have, for the moment of the 

 force of the iron bar, 



e° I *e 2 LM\ * e*J M\ e 2 J J 



5 . 7 lrM' 5 / ^A 2 \ n M' 3 M s r a 2 . „ n A 2 \ 



+ H— -ir^i 1 -6— ) + 2— -5— j l-9^sm%-3 — ) 



q4l ei\-M\ e 2 J M M\ e 2 e 2 J 



" This formula is unfortunately not convergent, and is, con- 

 sequently, of no use in the present investigation. In fact, 



-=-= is of the same order of magnitude as e 2 , -=— - as e i , and so 



