828 =H. A. Rowland—=Studies on Magnetic Distribution. 
To find p, the ordinary equation for the resistance of a 
derived circuit gives 
R’ 
a (e+RdL) 
p+RdL+—- 
whence oP — 5 (RR! —p?), 
ge? by 
= Asta 1 
and _ f= RRA at] (1) 
To find Q’, we have dQ Pan, 
whence "=a S (dette) py nie aS a 
Q’psL CAL ate 
yl we ees rg rh Per ae P1 
and Sacra = gee er Wie Seat © (3) 
When L is very large, or s= / RR’, we have 
Q’=C 2, and Q’,=C,raLe%, 
in which L, is reckoned from an origin at any point of the rod. 
These equations give the distribution on the part ouside the 
helix, and we have now to consider the part covered by the 
helix. Let us limit ourselves to the case where the helix 18 
long and thin, so that the field in its interior is nearly uniform. 
1 
G 
eas 
s_| Ae 
Ai = ow Dv Bb 
As we pass along the helix, the change of magnetic potential 
due to the helix is equal to the product of the intensity of the 
field multiplied by the distance passed over; so that in passing 
over an elementary distance dy the difference of potential . 
be §dy. The number of lines of force which this difference * 
potential causes in the rod will be equal to $dy 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 oP 
a? 
_ * This could have been obtained directly from the equation T=, and Ve 
