Inductance and magnetic energy
Continuing with the analogy of the preceding section, we would expect that corresponding to the mechanical momentum $p=mv$, whose rate of change is the applied force, there should be an analogous quantity equal to $\selfInd I$, whose rate of change is $\voltage$. We have no right, of course, to say that $\selfInd I$ is the real momentum of the circuit; in fact, it isn’t. The whole circuit may be standing still and have no momentum. It is only that $\selfInd I$ is analogous to the momentum $mv$ in the sense of satisfying corresponding equations. In the same way, to the kinetic energy $\tfrac{1}{2}mv^2$, there corresponds an analogous quantity $\tfrac{1}{2}\selfInd I^2$. But there we have a surprise. This $\tfrac{1}{2}\selfInd I^2$ is really the energy in the electrical case also. This is because the rate of doing work on the inductance is $\voltage I$, and in the mechanical system it is $Fv$, the corresponding quantity. Therefore, in the case of the energy, the quantities not only correspond mathematically, but also have the same physical meaning as well. We may see this in more detail as follows. As we found in Eq. (17.16), the rate of electrical work by induced forces is the product of the electromotive force and the current: \begin{equation*} \ddt{W}{t}=\emf I. \end{equation*}  Replacing $\emf$ by its expression in terms of the current from Eq. (17.34), we have \begin{equation} \label{Eq:II:17:36} \ddt{W}{t}=-\selfInd I\,\ddt{I}{t}. \end{equation}  Integrating this equation, we find that the energy required from an external source to overcome the emf in the self-inductance while building up the current4 (which must equal the energy stored, $U$) is \begin{equation} \label{Eq:II:17:37} -W=U=\tfrac{1}{2}\selfInd I^2. \end{equation}  Therefore the energy stored in an inductance is $\tfrac{1}{2}\selfInd I^2$. Applying the same arguments to a pair of coils such as those in Figs. 17–8 or 17–9, we can show that the total electrical energy of the system is given by \begin{equation} \label{Eq:II:17:38} U=\tfrac{1}{2}\selfInd_1I_1^2+\tfrac{1}{2}\selfInd_2I_2^2+\mutualInd I_1I_2. \end{equation}  For, starting with $I=0$ in both coils, we could first turn on the current $I_1$ in coil $1$, with $I_2=0$. The work done is just $\tfrac{1}{2}\selfInd_1I_1^2$. But now, on turning up $I_2$, we not only do the work $\tfrac{1}{2}\selfInd_2I_2^2$ against the emf in circuit $2$, but also an additional amount $\mutualInd I_1I_2$, which is the integral of the emf [$\mutualInd(dI_2/dt)$] in circuit $1$ times the now constant current $I_1$ in that circuit. Suppose we now wish to find the force between any two coils carrying the currents $I_1$ and $I_2$. We might at first expect that we could use the principle of virtual work, by taking the change in the energy of Eq. (17.38). We must remember, of course, that as we change the relative positions of the coils the only quantity which varies is the mutual inductance $\mutualInd$. We might then write the equation of virtual work as \begin{equation*} -F\,\Delta x=\Delta U=I_1I_2\,\Delta\mutualInd\quad(\text{wrong}). \end{equation*}  But this equation is wrong because, as we have seen earlier, it includes only the change in the energy of the two coils and not the change in the energy of the sources which are maintaining the currents $I_1$ and $I_2$ at their constant values. We can now understand that these sources must supply energy against the induced emf’s in the coils as they are moved. If we wish to apply the principle of virtual work correctly, we must also include these energies. As we have seen, however, we may take a short cut and use the principle of virtual work by remembering that the total energy is the negative of what we have called $U_{\text{mech}}$, the “mechanical energy.” We can therefore write for the force \begin{equation} \label{Eq:II:17:39} -F\,\Delta x=\Delta U_{\text{mech}}=-\Delta U. \end{equation}  The force between two coils is then given by \begin{equation*} F\,\Delta x=I_1I_2\,\Delta\mutualInd. \end{equation*}  Equation (17.38) for the energy of a system of two coils can be used to show that an interesting inequality exists between mutual inductance $\mutualInd$ and the self-inductances $\selfInd_1$ and $\selfInd_2$ of the two coils. It is clear that the energy of two coils must be positive. If we begin with zero currents in the coils and increase these currents to some values, we have been adding energy to the system. If not, the currents would spontaneously increase with release of energy to the rest of the world—an unlikely thing to happen! Now our energy equation, Eq. (17.38), can equally well be written in the following form:  \begin{equation} \label{Eq:II:17:40} U=\frac{1}{2}\,\selfInd_1\biggl(I_1+\frac{\mutualInd}{\selfInd_1}\,I_2\biggr)^2+ \frac{1}{2}\biggl(\selfInd_2-\frac{\mutualInd^2}{\selfInd_1}\biggr)I_2^2. \end{equation} 
  \begin{align} U=\;\;\frac{1}{2}\,\selfInd_1&\biggl(I_1+\frac{\mutualInd}{\selfInd_1}\,I_2\biggr)^2\notag\\[.75ex] \label{Eq:II:17:40} +\;\frac{1}{2}&\biggl(\selfInd_2-\frac{\mutualInd^2}{\selfInd_1}\biggr)I_2^2. \end{align}     That is just an algebraic transformation. This quantity must always be positive for any values of $I_1$ and $I_2$. In particular, it must be positive if $I_2$ should happen to have the special value \begin{equation} \label{Eq:II:17:41} I_2=-\frac{\selfInd_1}{\mutualInd}\,I_1. \end{equation}  But with this current for $I_2$, the first term in Eq. (17.40) is zero. If the energy is to be positive, the last term in (17.40) must be greater than zero. We have the requirement that \begin{equation*} \selfInd_1\selfInd_2>\mutualInd^2. \end{equation*}  We have thus proved the general result that the magnitude of the mutual inductance $\mutualInd$ of any two coils is necessarily less than or equal to the geometric mean of the two self-inductances. ($\mutualInd$ itself may be positive or negative, depending on the sign conventions for the currents $I_1$ and $I_2$.) \begin{equation} \label{Eq:II:17:42} \abs{\mutualInd}<\sqrt{\selfInd_1\selfInd_2}. \end{equation}  The relation between $\mutualInd$ and the self-inductances is usually written as \begin{equation} \label{Eq:II:17:43} \mutualInd=k\sqrt{\selfInd_1\selfInd_2}. \end{equation}  The constant $k$ is called the coefficient of coupling. If most of the flux from one coil links the other coil, the coefficient of coupling is near one; we say the coils are “tightly coupled.” If the coils are far apart or otherwise arranged so that there is very little mutual flux linkage, the coefficient of coupling is near zero and the mutual inductance is very small. For calculating the mutual inductance of two coils, we have given in Eq. (17.30) a formula which is a double line integral around the two circuits. We might think that the same formula could be used to get the self-inductance of a single coil by carrying out both line integrals around the same coil. This, however, will not work, because the denominator $r_{12}$ of the integrand will go to zero when the two line elements $d\FLPs_1$ and $d\FLPs_2$ are at the same point on the coil.  The self-inductance obtained from this formula is infinite. The reason is that this formula is an approximation that is valid only when the cross sections of the wires of the two circuits are small compared with the distance from one circuit to the other. Clearly, this approximation doesn’t hold for a single coil. It is, in fact, true that the inductance of a single coil tends logarithmically to infinity as the diameter of its wire is made smaller and smaller. We must, then, look for a different way of calculating the self-inductance of a single coil. It is necessary to take into account the distribution of the currents within the wires because the size of the wire is an important parameter. We should therefore ask not what is the inductance of a “circuit,” but what is the inductance of a distribution of conductors. Perhaps the easiest way to find this inductance is to make use of the magnetic energy. We found earlier, in Section 15–3, an expression for the magnetic energy of a distribution of stationary currents: \begin{equation} \label{Eq:II:17:44} U=\tfrac{1}{2}\int\FLPj\cdot\FLPA\,dV. \end{equation}  If we know the distribution of current density $\FLPj$, we can compute the vector potential $\FLPA$ and then evaluate the integral of Eq. (17.44) to get the energy. This energy is equal to the magnetic energy of the self-inductance, $\tfrac{1}{2}\selfInd I^2$. Equating the two gives us a formula for the inductance: \begin{equation} \label{Eq:II:17:45} \selfInd=\frac{1}{I^2}\int\FLPj\cdot\FLPA\,dV. \end{equation}  We expect, of course, that the inductance is a number depending only on the geometry of the circuit and not on the current $I$ in the circuit. The formula of Eq. (17.45) will indeed give such a result, because the integral in this equation is proportional to the square of the current—the current appears once through $\FLPj$ and again through the vector potential $\FLPA$. The integral divided by $I^2$ will depend on the geometry of the circuit but not on the current $I$. Equation (17.44) for the energy of a current distribution can be put in a quite different form which is sometimes more convenient for calculation. Also, as we will see later, it is a form that is important because it is more generally valid. In the energy equation, Eq. (17.44), both $\FLPA$ and $\FLPj$ can be related to $\FLPB$, so we can hope to express the energy in terms of the magnetic field—just as we were able to relate the electrostatic energy to the electric field. We begin by replacing $\FLPj$ by $\epsO c^2\FLPcurl{\FLPB}$. We cannot replace $\FLPA$ so easily, since $\FLPB=\FLPcurl{\FLPA}$ cannot be reversed to give $\FLPA$ in terms of $\FLPB$. Anyway, we can write \begin{equation} \label{Eq:II:17:46} U=\frac{\epsO c^2}{2}\int(\FLPcurl{\FLPB})\cdot\FLPA\,dV. \end{equation}  The interesting thing is that—with some restrictions—this integral can be written as \begin{equation} \label{Eq:II:17:47} U=\frac{\epsO c^2}{2}\int\FLPB\cdot(\FLPcurl{\FLPA})\,dV. \end{equation}  To see this, we write out in detail a typical term. Suppose that we take the term $(\FLPcurl{\FLPB})_zA_z$ which occurs in the integral of Eq. (17.46). Writing out the components, we get \begin{equation*} \int\biggl(\ddp{B_y}{x}-\ddp{B_x}{y}\biggr)A_z\,dx\,dy\,dz. \end{equation*}  (There are, of course, two more integrals of the same kind.) We now integrate the first term with respect to $x$—integrating by parts. That is, we can say \begin{equation*} \int\ddp{B_y}{x}\,A_z\,dx=B_yA_z-\int B_y\,\ddp{A_z}{x}\,dx. \end{equation*}  Now suppose that our system—meaning the sources and fields—is finite, so that as we go to large distances all fields go to zero. Then if the integrals are carried out over all space, evaluating the term $B_yA_z$ at the limits will give zero. We have left only the term with $B_y(\ddpl{A_z}{x})$, which is evidently one part of $B_y(\FLPcurl{\FLPA})_y$ and, therefore, of $\FLPB\cdot(\FLPcurl{\FLPA})$. If you work out the other five terms, you will see that Eq. (17.47) is indeed equivalent to Eq. (17.46). But now we can replace $(\FLPcurl{\FLPA})$ by $\FLPB$, to get \begin{equation} \label{Eq:II:17:48} U=\frac{\epsO c^2}{2}\int\FLPB\cdot\FLPB\,dV. \end{equation}  We have expressed the energy of a magnetostatic situation in terms of the magnetic field only. The expression corresponds closely to the formula we found for the electrostatic energy: \begin{equation} \label{Eq:II:17:49} U=\frac{\epsO}{2}\int\FLPE\cdot\FLPE\,dV. \end{equation}  One reason for emphasizing these two energy formulas is that sometimes they are more convenient to use. More important, it turns out that for dynamic fields (when $\FLPE$ and $\FLPB$ are changing with time) the two expressions (17.48) and (17.49) remain true, whereas the other formulas we have given for electric or magnetic energies are no longer correct—they hold only for static fields. If we know the magnetic field $\FLPB$ of a single coil, we can find the self-inductance by equating the energy expression (17.48) to $\tfrac{1}{2}\selfInd I^2$. Let’s see how this works by finding the self-inductance of a long solenoid. We have seen earlier that the magnetic field inside a solenoid is uniform and $\FLPB$ outside is zero. The magnitude of the field inside is $B=nI/\epsO c^2$, where $n$ is the number of turns per unit length in the winding and $I$ is the current. If the radius of the coil is $r$ and its length is $L$ (we take $L$ very long, so that we can neglect end effects, i.e., $L\gg r$), the volume inside is $\pi r^2L$. The magnetic energy is therefore \begin{equation*} U=\frac{\epsO c^2}{2}\,B^2\cdot(\text{Vol})=\frac{n^2I^2}{2\epsO c^2}\, \pi r^2L, \end{equation*}  which is equal to $\tfrac{1}{2}\selfInd I^2$. Or, \begin{equation} \label{Eq:II:17:50} \selfInd=\frac{\pi r^2n^2}{\epsO c^2}\,L. \end{equation} 