Page:Thomson1881.djvu/7

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

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

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

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

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

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

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

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

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