Page:On Faraday's Lines of Force.pdf/74

These expressions would determine completely the motion of electricity in a revolving sphere if we neglect the action of these currents on themselves. They express a system of circular currents about the axis of $y$, the quantity of current at any point being proportional to the distance from that axis. The external magnetic effect will be that of a small magnet whose moment is ${\frac {TR^{3}}{48\pi k}}\omega Isin\theta$, with its direction along the axis of $y$, so that the magnetism of the field would tend to turn it back to the axis of $x$^[1].

The existence of these currents will of course alter the distribution of the electro-tonic functions, and so they will react on themselves. Let the final result of this action be a system of currents about an axis in the plane of $xy$ inclined to the axis of $x$ at an angle $\phi$ and producing an external effect equal to that of a magnet whose moment is $I'R^{3}$.

The magnetic inductive components within the shell are

$I_{1}\sin \theta -2I'\cos \phi$ in $x$ ,

$-2I'\sin \phi$ in $y$

$I_{1}\cos \theta \ \ \ \$in $z$.

Each of these would produce its own system of currents when the sphere is in motion, and these would give rise to new distributions of magnetism which, when the velocity is uniform, must be the same as the original distribution.

$(I_{1}sin\theta -2I'\cos \phi )$ in $x$ produces $2{\frac {T}{48\pi k}}\omega (I_{1}\sin \theta -2I'\cos \phi )$ in $y$,

$(-2I'\sin \phi )$ in $y$ produces $2{\frac {T}{48\pi k}}\omega (2I'\sin \phi )$ in $x$;

$I\cos \theta$ in $z$ produces no currents.

We must therefore have the following equations, since the state of the shell is the same at every instant,

$I_{1}\sin \theta -2I'\cos \phi =I_{1}sin\theta +{\frac {T}{24\pi k}}\omega 2I'\sin \phi$

$-2I'\sin \phi ={\frac {T}{24\pi k}}\omega (I_{1}\sin \theta -2I'\cos \phi$

1. The expression for $p_{2}$ indicates a variable electric tension in the shell so that currents might be collected by wires touching it at the equator and poles.