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

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 $Icos\theta$ 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

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

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

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

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

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

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).