CASE OF A SINGLE POLE. 569 



which gives 



Z = --^- F= mU 



m 



~ 



The force is perpendicular therefore to the direction of the 

 magnetic trail; it is the same as if there were a mass equal to 



m U 2 



- ._ TT . in the plane m its direction. 

 4 (R + U)^ 



We may suppose again that the force F is produced by an in- 

 finitely small magnet situate in the plane. Applying Gauss' formula 

 (154), for instance, we find that this magnet is situate behind the 

 projection of the pole, at a distance x, defined by the equation 



the magnetisation being parallel to the direction of the motion. 

 The moment of this magnet would moreover be easy to calculate. 



The component X is a maximum, for a given value of R, when 

 u=i.2 f jR; it is zero for R = 0and R = co. 



The component Z tends to move the pole from the plate; it 



increases with the velocity, and tends towards the value , when 

 the velocity tends towards infinity. 



586. In the case of a uniform rectilinear motion parallel to 

 the plane, we may consider the phenomena in still another 

 manner. 



The magnetic ribbon which starts from the point B (Fig. 125), 



consists of two magnetic lines, the density of which is , and 

 the horizontal distance uSf. 



Let us denote by s the distance M'B from any point M', and 

 by x' the distance M'K. For an element ds of the ribbon the 

 magnetic moment or' is 



u 

 ?3' = m ds = mds sin a . 



As x' s sin a, we have &' = mdx. 



