Law of the Induction of an Electric Current upon itself. 21 
element of space parallel to PN. in which the inductive force is 
to be exerted, and let ab be an element of PN. Draw the per-" 
pendicular cp and let cp=r, and let represent’ the angle pea, 
the complement of pac‘or 4 It being ‘understood that ep is 
the origin of the angle and -p the origin of 1, we have in ap- 
plying the formula (AS) citi hi tiie sr 
t=ded! =6 sore: is ‘L=pa=r tan 
a OO 
| D=ca=rsec e : d ger) 
The substitution of these values in (A) gives us; 
- d?J 
Cdq ak 
Aad el ae 
and the integration of this with reference to & gives us for the 
whole inductive force exerted in the elementary space cd’ or ds 
in the direction of the latter by the current in ‘pa 
Lan s= vk ie cos 30'd 6! 
& dt 7 ae 4 
or still dividing by ds we shall retain in what follows only th 
eae aT : f : . 
; coéfficient we calling it the inductive force at the point c parallel 
=, Spa : : i 
with PN, and in the present instance therefore we have 
eR) dtieOdlg fs ei 
ae meee reas | 26 d 6! 
+ ) ds rdt¥ 7 fae 
' inwhich if the limits of integration be constant the value of the _ 
second member is inversely as r. Hence if ‘any two radial lines | 
as cf, cf’ be drawn, the induction in the direction cd’ due to that 
part of the conductor PN cut off by those radial lines is inversely 
Proportional to the distance cp. : oo 
6. Carrying out the integration indicated in the preceding for- 
we have 
mula 
rf! a ‘ “% 
sf cos 76d 0 =4(1/ —7)+4(sin 27/ — sin 27) e 
: : é 
and if, by including a great length of conductor on both sides of 
Paes b ~2 : 
hand h’ are very small ares, we have . 
cos 26d 0= 4a —4(2h/ — sin 2h’) ~4(2h —sin 2h) 
+ 
