226 Proceedings of the Royal Society 
Here, in the first place, it is obvious that 
- - Sa x V. =^5 
F(Ta) 
for the couple vanishes for a closed circuit of which a x is an ele- 
ment, and the integral of wa x must be a linear and vector function 
of a alone. It is easy to see that in this case 
F(Ta) a (Ta) 3 . 
11. If, again, be an element of a solenoid, and a' an element of 
current, the force is 
pa = — Sc^V. \(/a ' , 
where 
\pa' = Pa' + QaSaa' -I- RVaa' . 
But no portion of a solenoid can produce a force on an element of 
current in the direction of the element, so that 
(pa = Y.a'x<hi 
P '= 0, Q = 0, 
pa = — SajV^Vaa') . 
This must be of - 1 linear dimensions when we integrate for the 
effect of one pole of a solenoid, so that 
T? = M 
K Ta 3 * 
If the current be straight and infinite each way, its equation being 
a = /3 + xy , 
where 
Ty = 1 and S/?y = 0 , 
we have, for the whole force exerted on it by the pole of a solenoid, 
the expression 
so that 
and we have 
ppy 
J C. 
+ GO 
dx 
- 2 pP~'y, 
which agrees with known facts. 
12. Similarly, for the couple produced by an element of a solenoid 
on an element of a current we have 
YaW, , 
