Page:ThomsonMagnetic1889.djvu/13

Rh so that

$-4\pi wh=k\sqrt{1-\frac{\omega^{2}}{v^{2}}}\frac{d}{dz_{1}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$|undefined

The displacement across any spherical surface must $$=e$$, so that

$\int(xf+yg+zh)d\mathrm{S}=ae;$

and therefore, if $$u<v$$,

$\frac{k}{2\omega}\int_{0}^{\pi}\frac{\sin\theta\ d\theta}{\left(1-\dfrac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}} ;$|undefined

$k=\omega e\left(1-\frac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}.$|undefined

Thus the lines of magnetic forces are circles round the axis of $$z$$ and the magnitude of the force equals

$\frac{\omega e\sin\theta\left(1-\dfrac{\omega^{2}}{v^{2}}\right)}{r^{2}\left\{ 1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\theta\right\} ^{\frac{3}{2}}}$|undefined

which is Mr. Heaviside's result. If $$\omega>v$$, the integral becomes infinite, the displacement will be within a cone of semi-vertical angle $$\sin^{-1}\dfrac{v}{\omega}=\beta$$; we must therefore only integrate within this cone, and the equation to determine $$k$$ is,

$\frac{k}{2\omega}\int_{\pi}^{\sin^{-1}\frac{\omega}{v}}\frac{\sin\theta\ d\theta}{\left(1-\dfrac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}};$|undefined

or

$k=\frac{\omega e\left(1-\dfrac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}}{\cos\beta}\left(1-\frac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}.$|undefined

Thus the magnetic force

$=\frac{\omega e\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{1}{2}}\left(1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}}{\cos\beta\ r^{2}\left(1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}$.|undefined