﻿204 Mr. G. H. Bryan on Electromagnetic Induction in 

 the negative side is 



<«i' <mj_ <m ... 



dzdi +Si dz* ~~dzdt' ■ ■ ■ ■ ( l0 > 



where R 1 . ,„ 



R= ftK3? \ (16) 



Integrating (15), the value obtained for ft' is found to be 

 the potential due to magnetic poles or images on the positive 

 side of the sheet, and these are outside the region over which 

 the solution is required to hold, as they should be. We have 



V {0-*o) 2 + (y-yoY + (*-*o)Y 



-i 

 -i 





V^ (*-«b) 1, + (y-yo) 2 + (*-*o) 2 } 

 /W 



• J (tf-4'o) 2 + (y-«/o) 2 + (*-*o--R< + Rt) s 



£ 



Rf{j) Tz VH*-^o) 2 +(y-yo) 2 +(~-^-R< + RT) 2 } <? ' r ' 



The first term represents the effect of a pole equal and 

 opposite to the given pole at the point (# , y , z ), while the 

 second term shows that the effect of the action of the pole 

 between the times r and t + St is represented at time t by a 

 magnet of strength Hf{T)Sr situated at the point (% , y , 

 z -\-~R(t — t)). The potential at the positive side is found 

 from the relation (5), or 



When the pole is in motion the results thus arrived at 

 enable us to plot out the moving images contributed during 

 every time element St of the motion in the well-known way, 

 or they may be obtained as follows : — 



9. If, instead of a single magnetic pole, we are dealing 

 with a magnetic distribution on the positive side of the sheet 

 such that the volume-density of magnetism of the point 

 Oo */o<Zo) # is ¥(x ?/ z t\, we must write F(z y z () dd' dy dz 

 for f(t) in the expressions for fl and II'. Hence we have 



