Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/302

Let the coordinates of the pole at the time $t$ be

$\xi ={\mathfrak {u}}t$,⁠$\eta =0$, ⁠ $\zeta =c+{\mathfrak {w}}t$.

The coordinates of the image of the pole formed at the time $t-\tau$ are

$\xi ={\mathfrak {u}}(t-\tau )$, ⁠ $\eta =0$, ⁠ $\zeta =-(c+{\mathfrak {w}}(t-\tau )+R\tau )$,

and if $r$ is the distance of this image from the point ($x$, $y$, $z$),

$r^{2}=(x-{\mathfrak {u}}(t-\tau ))^{2}+(z+c+{\mathfrak {w}}(t-\tau )+R\tau )^{2}$.

To obtain the potential due to the trail of images we have to calculate

${\frac {d}{dt}}\int _{0}^{\infty }{\frac {d\tau }{r}}$.

If we write

$Q^{2}={\mathfrak {u}}^{2}+(R-{\mathfrak {w}})^{2}$,

$\int _{0}^{\infty }{\frac {d\tau }{r}}={\frac {1}{Q}}\log\{Qr+{\mathfrak {u}}(x-{\mathfrak {u}}t)+(R-{\mathfrak {w}})(z+c+{\mathfrak {w}}t)\}$,

the value of $r$ in this expression being found by making $\tau =0$.

Differentiating this expression with respect to $t$, and putting $t=0$, we obtain the magnetic potential due to the trail of images,

$\Omega ={\frac {1}{Q}}{\frac {Q{\frac {{\mathfrak {w}}(Z+C)-{\mathfrak {u}}x}{r}}-{\mathfrak {u}}^{2}-{\mathfrak {w}}^{2}+R{\mathfrak {w}}}{Qr+{\mathfrak {u}}x+(R-{\mathfrak {w}})(z+c)}}$.

By differentiating this expression with respect to $x$ or $z$, we obtain the components parallel to $x$ or $z$ respectively of the magnetic force at any point, and by putting $x=0$, $z=c$, and $r=2c$ in these expressions, we obtain the following values of the components of the force acting on the moving pole itself,

$X=-{\frac {1}{4c^{2}}}{\frac {\mathfrak {u}}{Q+R-{\mathfrak {w}}}}\left\{1+{\frac {\mathfrak {w}}{Q}}-{\frac {{\mathfrak {u}}^{2}}{Q(Q+R-{\mathfrak {w}})}}\right\}$,

$Z=-{\frac {1}{4c^{2}}}\left\{{\frac {\mathfrak {w}}{Q}}-{\frac {{\mathfrak {u}}^{2}}{Q(Q+R-{\mathfrak {w}})}}\right\}$.

665.] In these expressions we must remember that the motion is supposed to have been going on for an infinite time before the time considered. Hence we must not take ${\mathfrak {w}}$ a positive quantity, for in that case the pole must have passed through the sheet within a finite time.

If we make ${\mathfrak {u}}=0$, and ${\mathfrak {w}}$ negative, $X=0$, and

$Z={\frac {1}{4c^{2}}}{\frac {\mathfrak {w}}{R+{\mathfrak {w}}}}$,

or the pole as it approaches the sheet is repelled from it. If we make ${\mathfrak {w}}=0$, we find $Q^{2}={\mathfrak {u}}^{2}+R^{2}$,

$X=-{\frac {1}{4c^{2}}}{\frac {{\mathfrak {u}}R}{Q(Q+R)}}$⁠ and ⁠$Z={\frac {1}{4c^{2}}}{\frac {{\mathfrak {u}}^{2}}{Q(Q+R)}}$.