Page:Elementary Text-book of Physics (Anthony, 1897).djvu/289

§ 348] $$\frac{n^2} = \frac{(n - 1)^n}{2n^2} = \frac{1}{2} \left( 1 - \frac{1}{n} \right) = \frac{1}{2},$$ if $$n$$ be supposed to be very large. The work done in forming the magnetic field is therefore $$\textstyle \sum \displaystyle \tfrac{1}{2}mV.$$

Now, to show how this energy may be distributed in the field, we may consider any one of the magnets which give rise to the field as being the origin of $$4\pi I a$$ unit tubes of induction, the magnets being thought of as bar magnets. The energy of this magnet is, by the previous proposition, equal to $$\tfrac{1}{2}m(V_{a} - V_{S}),$$ where $$V_{n}$$ and $$V_{S}$$ are the potentials at the places occupied by its poles; the pole $$m$$ is equal to $$Ia$$ (§ 341). The difference of potential $$V_{n} - V_{S}$$ equals $$\textstyle \sum \displaystyle R \Delta l,$$ where $$R$$ is the force along a line of force in the field passing outside the magnet from its north to its south pole, and $$\Delta l$$ is an element of that line, the summation being extended over the whole line. If, therefore, we suppose each unit tube of induction which proceeds from the magnet to contain an amount of energy equal to $$\textstyle \sum \displaystyle \frac{R \Delta l}{8\pi},$$ the energy contained in the bundle of tubes belonging to the magnet will equal the energy of the magnet. We may therefore consider the energy of the magnet as distributed throughout the field, in such a way that each unit length of a unit tube of induction contains $$\frac{R}{8\pi}$$ units of energy. The tubes of induction here considered are those which exist outside the magnets. It has already been shown that the number of tubes of induction which pass through unit area is equal to the induction, or that $$N = F = (1 + 4\pi k)R = \mu R.$$ Hence the energy in unit length of a tube of induction may be expressed by $$\frac{N}{\mu 8\pi}\cdot$$

The energy in unit volume of the field may be determined by considering a small cylinder of length $$l$$ and cross-section $$s$$ placed in the field with its end surfaces normal to the lines of induction. The number of tubes of induction which pass through the end surfaces is $$Ns = \mu Rs,$$ and the energy contained in the length $$l$$ of each of these tubes is $$\frac{Nl}{\mu 8 \pi} = \frac{Rl}{8\pi}\cdot$$ The energy contained