Page:ThomsonMagnetic1889.djvu/12

 So that if the sphere is placed in a magnetic field the force acting upon it is the same as that on a current $$\tfrac{1}{3}e\omega\cos pt$$.

When the sphere is moving with a uniform velocity $$\omega$$, equations (3) become

$-\frac{1}{\mu\mathrm{K}}\nabla^{2}\mathrm{F}=-\omega^{2}\frac{d^{2}\mathrm{F}}{dz^{2}}+\omega\frac{d^{2}\psi}{dx\ dz};$|undefined

whence

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz^{2}}\left(1-\frac{\omega^{2}}{v^{2}}\right)=0,$|undefined

where $$v$$ is the velocity of propagation of electrodynamic action through the dielectric. If we put

$z=\left\{ 1-\frac{\omega^{2}}{v^{2}}\right\} ^{\frac{1}{2}}z',$|undefined

this equation becomes

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz_{1}^{2}}=0.$|undefined

With similar equations for $$b$$ and $$c$$, we see that a solution of these equations is

$\begin{array}{l} a=k\dfrac{d}{dy}\dfrac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ b=-k\dfrac{d}{dx}\dfrac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ c=0; \end{array}$|undefined

where $$k$$ is a constant. Since, if $$\mu=1$$,

$4\pi\frac{df}{dt}=\frac{dc}{dy}-\frac{db}{dz},$

and

$\frac{df}{dt}=-\omega\frac{df}{dz},$

we have

$-4\pi\omega f=-b=k\frac{d}{dx}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$|undefined

Similarly

$-4\pi\omega g=+a=k\frac{d}{dy}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}},$|undefined

and

$-4\pi\omega\frac{dh}{dz}=\frac{db}{dx}-\frac{da}{dy}=k\frac{d}{dz_{1}^{2}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};$|undefined