Page:Elementary Text-book of Physics (Anthony, 1897).djvu/360

346 circuit. These statements hold true not only in the case supposed, in which the field is homogeneous, but also when the field contains masses of magnetizable matter which distort the tubes of induction.

293. Energy of a Current in its own Field.—When the circuit is traversed by a current, a magnetic field is present around it, and the circuit possesses energy in consequence of the presence of that field; we may calculate an expression for this energy, if we assume that it is distributed in the tubes of induction around the circuit according to the law developed in § 348. Let the number of unit tubes of induction which pass through the circuit be represented by $$N.$$ The unit of length of each of these tubes contains an amount of energy equal to $$\frac{R}{8\pi},$$ where $$R$$ is the resultant magnetic force. Any one tube therefore contains energy equal to $$\frac{\textstyle \sum \displaystyle R \Delta l}{8 \pi},$$ where $$\Delta l$$ is an element of length of the tube and the summation is extended over the whole tube. But $$\textstyle \sum \displaystyle R \Delta l$$ equals the work done in carrying a unit pole over the whole length of the tube. The tube is a continuous closed tube enclosing the circuit, and the work done in carrying a unit pole over such a closed curve enclosing the circuit is equal to $$4 \pi i$$ (§ 290), so that $$\textstyle \sum \displaystyle R \Delta l = 4 \pi i.$$ The energy of each unit tube is therefore equal to $$\tfrac{1}{2}i,$$ and the energy of all the tubes belonging to the circuit is equal to $$\tfrac{1}{2}N.$$ Therefore, the energy of the circuit, due to its own current, is equal to one half the product of the current and the number of tubes of induction which pass through the circuit. Now we know by experiment that the magnetic force due to a current or the number of tubes of induction which pass through unit area is proportional to the current. Let $$L$$ represent the number of tubes of induction which pass through the circuit when it is traversed by unit current; $$L$$ is called the coefficient of self-induction. Then $$N = Li,$$ and the energy of the circuit equals $$\tfrac{1}{2}Li^2 .$$

294. Energy of Two Circuits.—If two circuits be present in a field, each of them possesses a certain amount of energy due to the magnetic field set up by the other. Let $$N_{1}$$ represent the number