Page:On Faraday's Lines of Force.pdf/37

Now the quantity of the current depends on the electro-motive force and on the resistance of the medium. If the resistance of the medium be uniform in all directions and equal to $k_{2}$,

$\alpha _{2}=k_{2}a_{2},\;\beta _{2}=k_{2}b_{2},\;\gamma _{2}=k_{2}a_{2}$...........(B),

but if the resistance be different in different directions, the law will be more complicated.

These quantities $\alpha _{2},\beta _{2},\gamma _{2},$ may be considered as representing the intensity of the electric action in the directions of $x,y,z$.

The intensity measured along an element $d\sigma$ of a curve is given by

$\epsilon =l\alpha +m\beta +n\gamma$,

where $l,m,n$ are the direction-cosines of the tangent.

The integral $\int _{}{}\epsilon d\sigma$ taken with respect to a given portion of a curve line, represents the total intensity along that line. If the curve is a closed one, it represents the total intensity of the electro-motive force in the closed curve.

Substituting the values of $\alpha ,\beta ,\gamma ,$ from equations (A)

$\int _{}{}\epsilon d\sigma =\int _{}{}(Xdx+Ydy+Zdz)-p+C.$

If therefore $(Xdx+Ydy+Zdz)$ is a complete differential, the value of $\int _{}{}\epsilon d\sigma$ for a closed curve will vanish, and in all closed curves

$\int _{}{}\epsilon d\sigma =\int _{}{}(Xdx+Ydy+Zdz),$

the integration being effected along the curve, so that in a closed curve the total intensity of the effective electro-motive force is equal to the total intensity of the impressed electro-motive force.

The total quantity of conduction through any surface is expressed by

$\int _{}{}edS$,

where

$e=la+mb+nc$,

$l,m,n$ being the direction-cosines of the normal,

$\therefore \;\int _{}{}edS=\iint adydz+\iint bdzdx+\iint cdxdy$,

the integrations being effected over the given surface. When the surface is a closed one, then we may find by integration by parts

$\int _{}{}edS=\iiint \left({\frac {da}{dx}}+{\frac {db}{dy}}+{\frac {dc}{dz}}\right)dxdydz$.