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

Rh The expressions for $$\alpha_0, \beta_0, \gamma_0$$ due to the magnetism of the ﬁeld are


 * $$\alpha_0=A_0+\frac{I}{2}y cos\theta$$,


 * $$\beta_0=B_0+\frac{I}{2}(\ sin\theta-x cos\theta)$$,


 * $$\gamma_0=C_0-\frac{I}{2}y sin\theta$$,

$$A_0, B_0, C_0$$ 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


 * $$a_2=-\frac{1}{k}\frac{1}{4\pi}\frac{I}{4}sin\theta \omega z$$,


 * $$b_2=0$$,


 * $$c_2=\frac{1}{k}\frac{1}{4\pi}\frac{I}{4}sin\theta\omega x$$,


 * $$p_2=-\frac{1}{17\pi}I\omega\{(x^2+y^2)cos\theta-xzsin\theta\}$$.