Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/186

154 The sign of this expression is reversed if we reverse the direction in which we measure $$\beta^\prime$$. It must therefore represent either a force in the direction of $$\beta^\prime$$, or a couple in the plane of $$\alpha$$ and $$\beta^\prime$$. As we are not investigating couples, we shall take it as a force acting on $$\alpha$$ in the direction of $$\beta^\prime$$.

There is of course an equal force acting on $$\beta^\prime$$ in the opposite direction.

We have for the same reason a force Rh acting on $$\alpha$$ in the direction of $$\gamma^\prime$$, and a force Rhacting on $$\beta$$ in the opposite direction.

514.] Collecting our results, we find that the action on $$ds$$ is compounded of the following forces,

Let us suppose that this action on $$ds$$ is the resultant of three forces, $$Rii^\prime\, ds\, ds^\prime$$ acting in the direction of $$r$$, $$Sii^\prime\, ds\, ds^\prime$$ acting in the direction of $$ds$$, and $$S^\prime ii^\prime\, ds\, ds^\prime$$ acting in the direction of $$ds^\prime$$, then in terms of $$\theta$$, $$\theta^\prime$$, and $$\eta$$,

In terms of the differential coefficients of $$r$$

In terms of $$l$$, $$m$$, $$n$$, and $$l^\prime$$, $$m^\prime$$, $$n^\prime$$,

where $$\xi$$, $$\eta$$, $$\zeta$$ are written for $$x^\prime - x$$, $$y^\prime - y$$, and $$z^\prime - z$$ respectively.

515.] We have next to calculate the force with which the finite current $$s^\prime$$ acts on the finite current $$s$$. The current $$s$$ extends from $$A$$ where $$s = 0$$, to $$P$$, where it has the value $$s$$. The current $$s^\prime$$ extends from $$A^\prime$$, where $$s^\prime = 0$$, to $$P^\prime$$, where it has the value $$s^\prime$$.