HI, A. Rowland—Studies on Magnetic Distribution. 329 
either side of the point, creating everywhere a nana poten- 
tial which can be calculated by equation (2), and which is rep- 
resented 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 elements of the rod in the same way and consider 
the effect at HF, the total magnetic potential at this point will, 
by hypothesis No. 1, be equal to the sum of the potentials due 
to all the elements dy. 
Let 4Q’ be the number of lines of force produced in the bar 
at the point E due to the elementary difference of 
potential at that point, Hdy 
AQ” b mber of ae _ force arriving at the 
point F due to the same element, 
Q’. be the number of lines passing from bar along 
length aL, 
f, be the sum of the resistances of the bar in both direc- 
tions from K, 
p, be resistance at F in direction of D, 
° 
s’ and s’ be the resistance of the bar, &c., respectively 
_ C in the direction of A, and at t D in direction 
f B. 
H ie the magnetizing-force of helix in its interior, 
Let 
Re RY satis ene: 
~A/RR 8” VW RRs” RB 
pe ee fe 
* (As? 41) (Nae) 
aga, 
reer Osten a 
: Anette, 
- “2 Ray Arb ends, 
Cron un _ HAL i, Airs JE tp 2rb ry 
ne : a AQ"). oh A’A" "1S, (Ae ee”) dy, 
ae — SAL AL A t ‘ re : 
WRT Aa Weert epee]. 
This gives the positive part of Q’,. To find the negative 
change x into b—a, A’ into A” and A” into A’, and then 
change the sign of the whole. : 
