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

Rh 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 $$k_2$$,


 * $$\alpha_2=k_2a_2,\;\beta_2=k_2b_2,\;\gamma_2=k_2a_2$$...........(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

$\int_{}{}e dS$,

where

$e = la + mb + nc$,

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

$\therefore\;\int_{}{}e dS=\iint adydz+\iint bdzdx+\iint cdxdy$,

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

$\int_{}{}e dS=\iiint \left (\frac{da}{dx}+\frac{db}{dy}+\frac{dc}{dz}\right ) dxdydz $.