346 PLUCKBR ON THE THEORY OF DIAMAGNETISM. 



solving this force now directed along AP into the direction AT, 

 which touches the circle at the point A, and multiplying by the 

 radius r, we find 



^ cos TAP 



for the moment of rotation sought. Remembering that 

 g2_y2^^2_2rc cos [<f)—a), 



cos TAP=sin PAO=l!ilLMz:^, 



r 



this expression changes immediately into the following : 



c//,r sin (<^ — a) /,^ 



{ r^ -t- c^ — 2rc cos (</> — a) } 



To ascertain the moment of rotation in the same direction due 

 to the repulsion of the same pole P on the other end of the bar 

 of iron B, it is only necessary in the foregoing expression to 

 change the signs of fjt. and a. We thus obtain 



-c^lr^m{(f> + a) ^2) 



{r^ + c^— 2rccos(</> + a)}^ * 



Adding the two expressions (1) and (2) together, we obtain 



r sin(</)-a) sin(</) + c^) \ 



^^ L{^* + c2-2rccos(0-a)}^ [r'^ ■\-c^--2rcco^[(t>-\-a)]^P ^ 



for the total moment of rotation, in so far as it is due to the pole P. 

 The sign of the value of this expression determines the direction 

 of the rotation. The iron bar approaches the pole P when this 

 sign is positive ; and it recedes from the pole when the sign is 

 negative. 



15, To ascertain the moment of rotation, still in the same di- 

 rection, produced by the action of the other pole Q, we will as- 

 sume that this pole lies at the other side of the centre O, and at 

 a distance from it equal to c' ; we need then only to set in the 

 last expression (3) d and (it— (f>) in the places of c and </>, and 

 to change the sign of fju at the same time. In this way we 

 obtain 



r sin(</)-a) sin(</) + a) | 



L{'''^ + ^' + 2rc^cos(</>-a)}^ {^'-f-c'2+2rc'cos((/) + a)}^P ' 



