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



This gives the positive part of Q" e . To find the negative part, 

 change x into b—x, A' into A", and A" into A', and then change 

 the sign of the whole. 



When the helix is symmetrically placed on the bar, we have 

 s' = s", A^A"; whence, adding the positive and negative parts 

 together, we have 



Qlt _ V^L _J ~ A f e r{b -x) _ r*\ ( 5 ) 



We ~2^M7AV-*-l le 6 h ' ' l J 



which gives the number of lines of induction passing out from 

 the rod along the length AL when the helix is symmetrically 

 placed on the rod. 



To get the number of lines of induction passing along the rod 

 at a given point, we have 



Q»=|VA=|^i^ + l-e»- S «-) + C'» J (6) 

 where 



C"'=- 



r (AV 6 -l)(v/RR'-s') 



When the bar extends a distance 1/ out of both ends of the 

 helix, so that 



we have 



Qm= $ (**-!) (e» L '-l)_ 

 r (e rb € 2rL ' + l)2\/ftW 



It may be well, before proceeding, to define what is meant by 

 magnetic resistance, and the units in which it is measured. If 

 fjb is the magnetic permeability of the rod, we can get an idea of 

 the meaning of magnetic resistance in the following manner. 

 Suppose we have a rod infinitely long placed in a magnetic field 

 of intensity ^ parallel to the lines of force. Let Q' be the num- 

 ber of lines of inductive force passing through the rod, or the 

 surface-integral of the magnetic induction across its section ; also 



let a be the area of the rod. Then by definition u= —=. If L 



is the length of the rod, the difference of potential at the ends 

 will be L§ ; hence 



and R in the formulas becomes 



L CtfJb 



