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for the line integral of the magnetic intensity $2\pi r\alpha$ round the wire is $4\pi \times$ current through it, and $Pl=V$.

But by Ohm's law $V=iR$ and $iV=i^{2}R$, or the heat developed according to Joule's law.

It seems then that none of the energy of a current travels along the wire, but that it comes in from the nonconducting medium surrounding the wire, that as soon as it enters it begins to be transformed into heat, the amount crossing successive layers of the wire decreasing till by the time the centre is reached, where there is no magnetic force, and therefore no energy passing, it has all been transformed into heat. A conduction current then may be said to consist of this inward flow of energy with its accompanying magnetic and electromotive forces, and the transformation of the energy into heat within the conductor.

We have now to inquire how the energy travels through the medium on its way to the wire.

(2.) Discharge of a condenser through a wire.

We shall first consider the case of the slow discharge of a simple condenser consisting of two charged parallel plates when connected by a wire of very great resistance, as in this case we can form an approximate idea of the actual path of the energy.

Fig. 2.

Let A and B, fig. 2, be the two plates of the condenser, A being positively and B negatively electrified. Then before discharge the sections of the equipotential surfaces will be somewhat as sketched. The chief part of the energy resides in the part of the dielectric between the two plates, but there will be some energy wherever there is electromotive intensity. Between A and B the E.M.I. will be from A to B, and everywhere