Mr. H. A. Rowland on Magnetic Distribution. 261 



tential causes in the rod will be equal to Sjdy divided by the sum 

 of the resistances of the rod in both directions from the given 

 point. These lines of force stream down the rod on either side 

 of the point, creating everywhere a magnetic potential which can 

 be calculated by equation (2), and which is represented by the 

 curves in fig. 1. In that figure AB is the rod, CD the helix, 

 and E the element of length dy. Now, if we take all the ele- 

 ments of the rod in the same way and consider the effect at H F, 

 the total magnetic potential at this point will, by hypothesis 

 No. 1, be equal to the sum of the potentials due to all the ele- 

 ments dy. 



Let AQ' be the number of lines of force produced in the bar 

 at the point E due to the elementary difference of po- 

 tential at that point, S$dy, 

 AQ" be the number of lines of force arriving at the point 



F due to the same element, 

 Q" e be the number of lines passing from bar along 



length AL, 

 p y be the sum of the resistances of the bar in both direc- 

 tions from E, 

 p x be resistance at F in direction of D, 

 y be the distance D E, 

 x be the distance 1) F, 

 b be the distance C D, 

 s" and s' be the resistance of the bar &c. respectively at C 



in the direction of A, and at D in direction of B, 

 § be the magnetizing-force of helix in its interior. 

 Let 



i/RR' + s' \/RW+s» " /R 



'RE/_ S '/' V R' ; 



•RR 7 , 



•EI?—/ ~~ Vrb/ 



2(A'A' f € 2 '-*-l) 

 \- (A"e 2 ^+l)(A'e 2 ^-^ + l) 



AQ'-^- 



Py 



AQ'' = AQ'(AV' + e-^ Af/< g +1 



.ft k^ e rx A- e— rx 



■&AL A e rj — e~ r ' rA i 2r4 , „ _ rt \ , ,*_-r*i ia\ 



