382 Permanent Magnetism 



The equations to be solved are 



[CH. xi 



_ 



dy dz 



dF dH 



dz dx ^dydz\rr 



dx dy dz 2 \r) ' 



and the simplest solution, similar to that given by equations (372), is 



F= -(-}, 



G 



8/1 



= LL 



dx W ' 



The components of vector-potential for a magnet parallel to the axes of 

 x or y can be written down from symmetry. In terms of the coordinates 

 x, y f , z' of the magnetic particle, this solution may be expressed as 



d /I 



#=0. 



445. Let us superpose the fields of a magnetic particle of strength I/JL 

 parallel to the axis of x, one of strength m^ parallel to the axis of y, and 

 one of strength n/j> parallel to the axis of z. Then we obtain the vector 

 potential at x, y, z due to a magnetic particle of strength //, and axis (I, m,n) 

 at x, y', z in the forms 



'dz dy) r 



F = - 



m =-, n^- f 



dz oy 



( d 7 d \ l 

 G = - fjbi n= I ^- - 



\ ox dz] r 



->(*4'i" 



8 



^-, 

 dx 



) 



i *V 



1 d^'jr 



(377). 



8 8\1 



^ , m ^ , - 

 8?/ 80; / r) 



The number of lines of induction which cross the circuit from a magnetic 

 particle is accordingly 



ds 



~T~ ** ~T~ 



ds ds 

 which may be written in the form 



dx dy 



ds' ds' 



I, 



m, 



dz 

 ds 



n 



ds, 



dx\rj' dy\rj' dz\r. 



the integral being taken round the circuit in the direction determined by the 

 rule given on p. 376. 



