Page:On Faraday's Lines of Force.pdf/73

The expressions for $\alpha _{0},\beta _{0},\gamma _{0}$ due to the magnetism of the field are

$\alpha _{0}=A_{0}+{\frac {I}{2}}ycos\theta$,

$\beta _{0}=B_{0}+{\frac {I}{2}}(\ sin\theta -xcos\theta )$,

$\gamma _{0}=C_{0}-{\frac {I}{2}}ysin\theta$,

$A_{0},B_{0},C_{0}$ on being constants: and the velocities of the particles of the revolving sphere are

${\frac {dx}{dt}}=-\omega y,{\frac {dy}{dt}}=\omega x,{\frac {dz}{dt}}=0$

We have therefore for the electro-motive forces

$\alpha _{2}=-{\frac {1}{4\pi }}{\frac {d\alpha _{0}}{dt}}=-{\frac {1}{4\pi }}{\frac {I}{2}}cos\theta \omega x,$

$\beta _{2}=-{\frac {1}{4\pi }}{\frac {d\beta _{0}}{dt}}=-{\frac {1}{4\pi }}{\frac {I}{2}}cos\theta \omega y,$

$\gamma _{2}=-{\frac {1}{4\pi }}{\frac {d\gamma _{0}}{dt}}=-{\frac {1}{4\pi }}{\frac {I}{2}}sin\theta \omega x.$

Returning to equations (1), we get

$k\left({\frac {db_{2}}{dz}}-{\frac {dc_{2}}{dy}}\right)={\frac {d\beta _{2}}{dz}}-{\frac {d\gamma _{2}}{dy}}=0$,

$k\left({\frac {dc_{2}}{dx}}-{\frac {da_{2}}{dz}}\right)={\frac {d\gamma _{2}}{dx}}-{\frac {d\alpha _{2}}{dz}}={\frac {1}{4\pi }}{\frac {I}{2}}sin\theta \omega$,

$k\left({\frac {da_{2}}{dy}}-{\frac {db_{2}}{dx}}\right)={\frac {d\alpha _{2}}{dy}}-{\frac {d\beta _{2}}{dx}}=0.$,

From which with equation (2) we find

$a_{2}=-{\frac {1}{k}}{\frac {1}{4\pi }}{\frac {I}{4}}sin\theta \omega z$,

$b_{2}=0$,

$c_{2}={\frac {1}{k}}{\frac {1}{4\pi }}{\frac {I}{4}}sin\theta \omega x$,

$p_{2}=-{\frac {1}{17\pi }}I\omega \{(x^{2}+y^{2})cos\theta -xzsin\theta \}$.