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



The natural course is to assume the azimuth of the magnet c, and thence 

 determine the place of its centre. Let us suppose, then, that the plane of the 

 magnet c is parallel to the magnetic meridian, or that 



The equations (J, 8) then give, 



cos 2j3 = — ^, sin 2a = ; 



and these two equations, together with (6), solve the problem. As we cannot 

 have 7 = 0, a = 0, simultaneously, there are two solutions, namely : 



7=0, a = 90°, 1 



S = 54° 44'. 

 7 = 90°, a = 0, J ^ 



The corresponding arrangements of the magnets are represented in Figs. 1 

 and 2. 



Again, if the plane in which the magnet c is constrained to move be perpen- 

 dicular to the magnetic meridian, or 



f = 90°, 



the equations (^, 8) are then reduced to 



sin 2^ = 0, cos 2a = ^ ; 



which, in conjunction with (6), furnish the two solutions : 



7 = 0, /3 = 90°," 



, a = 35° 16'. 

 7 = 90°, 13 = 0, 



These arrangements are represented in Figs. 3 and 4. 



In estimating the comparative merits of these four arrangements, we should 

 observe that the magnet c is usually much less massive, and therefore less 

 powerful than either of the other two ; and, accordingly, that the arrange- 

 ments represented in Figs. 1 and 4, in which the distance, ab, of the stronger 

 magnets is the shortest side of the triangle abc, are, on that account, in- 



