164 The Rev. H. Lloyd on the mutual Action of permanent Magnets. 



Now the angle q'oo' being very small, we may (in the same order of approxima- 

 tion as before) put oo for oq, noo' for noq', and n'o'o for n'q o ; and accord- 

 ingly, denoting the distances oo' and o'q' by a and r', and the angles noo' and 

 n'o'o by (f) and (f>', we have (5) 



^ 2Mq' ^ ^ Mq' . ^ 



X =■ — 3-^ cos 0, y = — ^ sm ; 



ilf being the moment of free magnetism of the acting magnet, as already defined. 

 Hence the moment of these forces to turn the magnet n's' is 



— ^-12 cos 0sln0' — sin 0cos0'i=: ^3 j sin (0 4" 0') — 3sin (0— 0')> ; 

 and multiplying by dr', and integrating, the total moment is 



^'{sin (0 + 0') - 3 sin (0 - 0')}, (8) 



in which M' denotes the moment of free magnetism of the second magnet, or the 

 value of the integral \q'r'dr., taken throughout its entire length. 



Let us apply this result to the case of the mutual action of two horizontal 

 magnets, the axis of one which, ns, lies in the magnetic meridian, while that of 

 the other, n's', is perpendicular to it (Fig. 3). Such is the position of the magnets 

 in the instruments used in determining the declination, and the horizontal com- 

 ponent of the intensity of the earth's magnetic force. 



The moment of the force exerted by the second magnet on the first is in 



this case (8) 



MM' ,. „ „ ,. 

 -2^j-(l — 3cos2 0); 



since -j- 0' = 90". Hence, that this moment may be nothing, we must have 



cos20 = i. (9) 



Accordingly the mean direction of the first magnet will be undisturbed by the 

 second, when the line connecting their centres is inclined to the magnetic me- 



