Page:A Dynamical Theory of the Electromagnetic Field.pdf/32

Since the distribution $\varphi _{1}$ is determined by $m_{1}$, and $\varphi _{2}$ by $m_{2}$, the quantities $\varphi _{1}m_{1}$ and $\varphi _{2}m_{2}$ will remain constant.

It can be shown also, as Green has proved (Essay, p. 10), that

$m_{1}\varphi _{2}=m_{2}\varphi _{1}$

so that we get

$Xdx=d\left(m_{2}\varphi _{1}\right)$

$\left.{\begin{array}{l}X=m_{2}{\frac {d\varphi _{1}}{dx}}=m_{2}\alpha _{1},\\\\\mathrm {where} \ \alpha _{1}\ \mathrm {represents\ the\ magnetic} \ \\\mathrm {intensity\ due\ to} \ m_{1}.\ {\mathsf {Similarly}}\\\\Y=m_{2}\beta _{1},\\\\Z=m_{2}\gamma _{1}.\end{array}}\right\}$

(K)

So that a magnetic pole is urged in the direction of the lines of magnetic force with a force equal to the product of the strength of the pole and the magnetic intensity.

(78) If a single magnetic pole, that is one pole of a very long magnet, be placed in the field, the only solution of $\varphi$ is

$\varphi _{1}=-{\frac {m_{1}}{\mu }}{\frac {1}{r}}$

(41)

where $m_{1}$ is the strength of the pole and $r$ the distance from it.

The repulsion between two poles of strength $m_{1}$ and $m_{2}$ is

$m_{2}{\frac {d\varphi _{1}}{dr}}={\frac {m_{1}m_{2}}{\mu r^{2}}}$

(42)

In air or any medium in which $\mu =1$ this is simply ${\tfrac {m_{1}m_{2}}{r^{2}}}$, but in other media the force acting between two given magnetic poles is inversely proportional to the coefficient of magnetic induction for the medium. This may be explained by the magnetization of the medium induced by the action of the poles.

Mechanical Force on an Electrified Body.

(79) If there is no motion or change of strength of currents or magnets in the field, the electromotive force is entirely due to variation of electric potential, and we shall have (§ 65)

$P=-{\frac {d\Psi }{dx}},\ Q=-{\frac {d\Psi }{dy}},\ R=-{\frac {d\Psi }{dz}}$

Integrating by parts the expression (I) for the energy due to electric displacement, and remembering that P, Q, R vanish at an infinite distance, it becomes

${\frac {1}{2}}\sum \left\{\Psi \left({\frac {df}{dx}}+{\frac {dg}{dy}}+{\frac {dh}{dz}}\right)\right\}dV$

or by the equation of Free Electricity (G), p. 485,

$-{\frac {1}{2}}\sum (\Psi e)dV$