# Page:Thomson1881.djvu/7

 $F=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx^{2}}\frac{1}{R}+\frac{\mu ep}{3}\frac{2}{R}$, $G=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dy}\frac{1}{R}$, $H=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dr}\frac{1}{R}$.

Now if α, β, γ be the components of the magnetic induction at the point (x, y, z),

 $\alpha=\frac{dH}{dy}-\frac{dG}{dz}=0$, $\beta=\frac{dF}{dz}-\frac{dH}{dx}=-\frac{\mu epz}{R^{3}}=\mu ep\frac{d}{dz}\frac{1}{R}$, $\gamma=\frac{dG}{dx}-\frac{dF}{dy}=\frac{\mu epy}{R^{3}}=-\mu ep\frac{d}{dy}\frac{1}{R}$.

Hence we see, by symmetry, that if the sphere move with velocity q parallel to the axis of y, the corresponding values would be

 $\alpha=-\mu eq\frac{d}{dz}\frac{1}{R}$, $\beta=0$, $\gamma=\mu eq\frac{d}{dx}\frac{1}{R}$;

and if it moved with velocity r parallel to the axis of z, the corresponding values would be

 $\alpha=\mu er\frac{d}{dy}\frac{1}{R}$, $\beta=-\mu er\frac{d}{dx}\frac{1}{R}$, $\gamma=0$.

Hence, if p, q, r be the components of the velocity of the centre of the sphere parallel to the axes of x, y, z respectively, the components of magnetic induction are

 $\alpha=\mu e\left(r\frac{d}{dy}\frac{1}{R}-q\frac{d}{dz}\frac{1}{R}\right)$, $\beta=\mu e\left(p\frac{d}{dz}\frac{1}{R}-r\frac{d}{dx}\frac{1}{R}\right)$, $\gamma=\mu e\left(q\frac{d}{dx}\frac{1}{R}-p\frac{d}{dy}\frac{1}{R}\right)$;