220 MAGNETISM 



M 

 If = the point P is in the axis of the magnet and V = . 



The intensity along the axis is 



dV 2M 



If = 90, i.e. if P is in a line through C perpendicular to the 

 axis, the intensity at P parallel to the axis is 



1 <*V Msinfl = M 



r dO " 7- 3 ~ r 3 



d\ 

 and is in the direction in which increases. The component ^ 



along r vanishes, since cos = 0. 



These are the values which we obtained in the last chapter for 

 the end-on and broadside-on positions. the values on which (iaiiss 

 founded his proof of the inverse square law. 



Vector resolution of a small magnet. It M be re- 

 garded as a vector, drawn in the direction of the axis the potential 



-^ at a point P is the potential of the resolute of M in the 



direction of r, and if M be resolved like a force into any number of 

 components, M r M 2 , &c., making r 2 &c -. with ;. the potential 

 of the components will be 



M! cos Q 1 + M 2 cos fl, 4- . . . 



+ M 2 cos^ 2 -f . . . _ 





since the resolved part of M equals the smn of the resolved parts 

 of its components. 



Hence the potential of M equals the sum of the potentials of'iK 

 vector components. 



This result follows also from the consideration that a magnet 

 AB (Fig. !()()) may be replaced by two magnets AC, CH. \\itli 

 equal pcles. For the two magnets HC. CA have equal and opposite 

 poles at C, neutralising each other, and are equivalent therefore to 

 m at A and ;// at B. 



The direction and magnitude of the intensity at any 

 point in the field of a small magnet. Let AB(Fig. Hi7) be tin- 

 magnet with moment M ; resolve it into 1)K along CP with moment 

 M cos 0, and FG perpendicular to CP with moment M sj n 0. 



The intensity due to I)K i^ along CP and i " 

 ACP = and CP = r. 



