Page:On Faraday's Lines of Force.pdf/75

Rh Whence $$\ \ \ \cot\phi=-\frac{TR^3}{24\pi k}\omega$$, $$I'=\frac{1}{2}\frac{\frac{T}{4\pi k}}{\sqrt{1+\left(\frac{T}{4\pi k}\omega\right)^2}}I_1\sin\theta$$.

To understand the meaning of these expressions let us take a particular case.

Let the axis of the revolving shell be vertical, and let the revolution be from north to west. Let $$I$$ be the total intensity of the terrestrial magnetism, and let the dip be $$\theta$$, then $$I cos\theta$$ is the horizontal component in the direction of magnetic north.

The result of the rotation is to produce currents in the shell about an axis inclined at a small angle $$=\tan^{-1}\frac{T}{24\pi k}\omega$$ to the south of magnetic west, and the external effect of these currents is the same as that of a magnet whose moment is

$\frac{1}{2}\frac{T\omega}{\sqrt{\overline{24\pi k}\vert^2+T^2\omega^2}}R^3I\cos\theta$.|undefined

The moment of the couple due to terrestrial magnetism tending to stop the rotation is

$\frac{24\pi k}{2}\frac{T\omega}{\overline{24\pi k}\vert^2+T^2\omega^2}R^2I^2\cos^2\theta$,

and the loss of work due to this in unit of time is

$\frac{24\pi k}{2}\frac{T\omega^2}{\overline{24\pi k}\vert^2+T^2\omega^2}R^2I^2\cos^2\theta$.

This loss of work is made up by an evolution of heat in the substance of the shell, as is proved by a recent experiment of M. Foucault (see Comptes Rendus, XLI. p. 450).