Page:A Dynamical Theory of the Electromagnetic Field.pdf/31

Let C be the current in the conductor, and let $p',q',r'$ be its components, then

$Xa\delta =Ca\delta xx\left({\frac {dy}{ds}}\mu \gamma -{\frac {dz}{ds}}\mu \beta \right),$

$\left.{\begin{array}{ccc}\mathrm {or} &&X=\mu \gamma q'-\mu \beta r'.\\\\\mathrm {Similarly,} &&Y=\mu \alpha r'-\mu \gamma p',\\\\&&Z=\mu \beta p'-\mu \alpha q'.\end{array}}\right\}$

(J)

These are the equations which determine the mechanical force acting on a conductor carrying a current. The force is perpendicular to the current and to the lines of force, and is measured by the area of the parallelogram formed by lines parallel to the current and lines of force, and proportional to their intensities.

Mechanical Force on a Magnet.

(77) In any part of the field not traversed by electric currents the distribution of magnetic intensity may be represented by the differential coefficients of a function which may be called the magnetic potential. When there are no currents in the field, this quantity has a single value for each point. When there are currents, the potential has a series of values at each point, but its differential coefficients have only one value, namely,

${\frac {d\varphi }{dx}}=\alpha ,\ {\frac {d\varphi }{dy}}=\beta ,\ {\frac {d\varphi }{dz}}=\gamma .$

Substituting these values of $\alpha ,\beta ,\gamma$ in the expression (equation 38) for the intrinsic energy of the field, and integrating by parts, it becomes

$-\sum \left\{\varphi {\frac {1}{8\pi }}\left({\frac {d\mu \alpha }{dx}}+{\frac {d\mu \beta }{dy}}+{\frac {d\mu \gamma }{dz}}\right)\right\}dV$

The expression

$\sum \left({\frac {d\mu \alpha }{dx}}+{\frac {d\mu \beta }{dy}}+{\frac {d\mu \gamma }{dz}}\right)dV=\sum mdV$

(39)

indicates the number of lines of magnetic force which have their origin within the space V. Now a magnetic pole is known to us only as the origin or termination of lines of magnetic force, and a unit pole is one which has $4\pi$ lines belonging to it, since it produces unit of magnetic intensity at unit of distance over a sphere whose surface is $4\pi$.

Hence if $m$ is the amount of free positive magnetism in unit of volume, the above expression may be written $4\pi m$, and the expression for the energy of the field becomes

$E=-\sum \left({\frac {1}{2}}\varphi m\right)dV$

(40)

If there are two magnetic poles $m_{1}$ and $m_{2}$ producing potentials $\varphi _{1}$ and $\varphi _{2}$ in the field, then if ma is moved a distance $dx$, and is urged in that direction by a force $X$, then the work done is $Xdx$, and the decrease of energy in the field is

$d\left({\frac {1}{2}}\left(\varphi _{1}+\varphi _{2}\right)\left(m_{1}+m_{2}\right)\right)$

and these must be equal by the principle of Conservation of Energy.