Page:A history of the theories of aether and electricity. Whittacker E.T. (1910).pdf/269

 electrodynamic energy is being given up, and yet the operator is doing positive work. The explanation of this apparent paradox is that the energy derived from both these sources is being used to save the energy which would otherwise be furnished by the battery, and which is expended in Joulian heat.

Thomson next proceeded to show that the energy which is stored in connexion with a circuit in which a current is flowing may be expressed as a volume-integral extended over the whole of space, similar to the integral by which he had already represented the energy of a system of permanent and temporary magnets. The theorem, as originally stated by its author, applied only to the case of a single circuit; but it may be established for a system formed by any number of circuits in the following way:—

If Ns denote the number of unit tubes of magnetic induction which are linked with the sth circuit, in which a current is is flowing, the electrokinetic energy of the system is $$\textstyle \frac{1}{2}\sum_sN_si_s$$; which may be written $$\textstyle \frac{1}{2}\sum_rI_r$$, where Ir, denotes the total current flowing through the gap formed by the rth path unit tube of magnetic induction. But if H denote the (vector) magnetic force, and H its numerical magnitude, it is known that (1/4π)&int; Hds, integrated along a closed line of magnetic induction, measures the total current flowing through the gap formed by the line. The energy is therefore $$\textstyle \frac{1}{8\pi}\sum\int Hds$$, the summation being extended over all the unit tubes of magnetic induction, and the integration being taken along them. But if dS denote the cross-section of one of these tubes, we have BdS = 1, where B denotes the numerical magnitude of the magnetic induction B: so the energy is $$\textstyle \frac{1}{8\pi}\sum BdS\int Hds$$; and as the tubes fill all space, we may replace $$\textstyle \sum dS\int ds$$ by $$\textstyle \iiint\ dx\ dy\ dz$$. Thus the energy takes the form $$\textstyle \frac{1}{8\pi}\iiint BH\ dx\ dy\ dz$$, where the integration is extended over the whole of space; and since in the present case B = μH, the energy may also be represented by $$\textstyle \frac{1}{8\pi}\iiint \mu H^2\ dx\ dy\ dz$$.