Page:PoyntingTransfer.djvu/8

350 (1) A straight wire conveying a current.

In this case very near the wire, and within it, the lines of magnetic force are circles round the axis of the wire. The lines of electric force are along the wire, if we take it as proved that the flow across equal areas of the cross section is the same at all parts of the section. If AB, fig. 1, represents the wire, and the current is from A to B,



then a tangent plane to the surface at any point contains the directions of both the electromotive and magnetic intensities (we shall write E.M.I. and M.I. for these respectively in what follows), and energy is therefore flowing in perpendicularly through the surface, that is, along the radius towards the axis. Let us take a portion of the wire bounded by two plane sections perpendicular to the axis. Across the ends no energy is flowing, for they contain no component of the E.M.I. The whole of the energy then enters in through the external surface of the wire, and by the general theorem the amount entering in must just account for the heat developed owing to the resistance, since if the current is steady there is no other alteration of energy. It is, perhaps, worth while to show independently in this case that the energy moving in, in accordance with the general law, will just account for the heat developed.

Let $$r$$ be the radius of the wire, $$i$$ the current along it, $$\alpha$$ the magnetic intensity at the surface, P the electromotive intensity at any point within the wire, and V the difference of potential between the two ends. Then the area of a length $$l$$ of the wire is $$2\pi rl$$, and the energy entering from the outside per second is

$\begin{array}{ll} \frac{\mathrm{area\times E.M.I.\times M.I}.}{4\pi} & =\frac{4\pi rl\cdot P\cdot\alpha}{4\pi}\\ \\ & =\frac{2\pi r\alpha\cdot Pl}{4\pi}\\ \\ & =\frac{4\pi iV}{4\pi}\\ \\ & =iV\end{array}$