Page:Thomson1881.djvu/7

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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);