Magnetic Energy

Suppose that, at $t=0$, a solenoid of inductance $L$, and resistance $R$, is connected across the terminals of a battery of voltage $V$. The circuit equation is

$$V = L\,\frac{d I}{dt} + R\,I.$$ (2.358)

[See Equation (2.326).] The power output of the battery is $V\,I$. [Every charge $q$ that goes around the circuit falls through a potential difference $q\,V$. In order to raise it back to the starting potential, so that it can perform another circuit, the battery must do work $q\,V$. See Section 2.1.5. The work done per unit time (i.e., the power) is $n\,q\,V$, where $n$ is the number of charges per unit time passing a given point on the circuit. But, $I=n\,q$, so the power output is $V\,I$.] Thus, the net work done by the battery in raising the current in the circuit from zero at time $t=0$ to $I_T$ at time $t=T$ is

$$W = \int_0^T V\,I \,dt.$$ (2.359)

Using the circuit equation (2.358), we obtain

$$W = L \int_0^T I\,\frac{dI}{dt} \,dt + R \int_0^T I^2\,dt,$$ (2.360)

giving

$$W = \frac{1}{2}\, L\, I_T^{\,2} + R \int_0^T I^2\,dt.$$ (2.361)

The second term on the right-hand side of the previous equation represents the irreversible conversion of electrical energy into heat energy by the resistor. (See Section 2.1.11.) The first term is the amount of energy stored in the solenoid at time $T$. This energy can be recovered after the solenoid is disconnected from the battery. Suppose that the battery is disconnected at time $T$. The circuit equation is now

$$0 = L\,\frac{dI}{dt} + RI,$$ (2.362)

giving

$$I= I_T \exp \left[ - \frac{R}{L} \,(t-T)\right],$$ (2.363)

where we have made use of the initial condition $I(T) = I_T$. Thus, the current decays away exponentially. The energy stored in the solenoid is dissipated as heat in the resistor. The total heat energy appearing in the resistor after the battery is disconnected is

$$\int_T^\infty I^2 \,R \,dt = \frac{1}{2}\, L\, I_T^{\,2},$$ (2.364)

where use has been made of Equation (2.363). Thus, the heat energy appearing in the resistor is equal to the energy stored in the solenoid. This energy is actually stored in the magnetic field generated inside the solenoid.

Consider, again, our circuit with two solenoids wound on top of one another. (See the previous section.) Suppose that each solenoid is connected to its own battery. The circuit equations are thus

$$V_1 = R_1\, I_1 + L_1\,\frac{d I_1}{dt} +M\,\frac{d I_2}{dt},$$ (2.365)

$$V_2 = R_2 \,I_2 + L_2\, \frac{d I_2}{d t} + M\,\frac{d I_1}{dt},$$ (2.366)

where $V_1$ is the voltage of the battery in the first circuit, et cetera. The net work done by the two batteries in increasing the currents in the two circuits, from zero at time 0, to $I_1$ and $I_2$ at time $T$, respectively, is

$$W = \int_0^T (V_1\, I_1 + V_2 \,I_2 )\,dt$$

$$= \int_0^T (R_1 \,I_1^{\,2} + R_2 \,I_2^{\,2})\,dt +\frac{1}{2}\, L_1 \,I_1^{\,2} + \frac{1}{2}\, L_2 \,I_2^{\,2}$$

$$\phantom{=} + M \int_0^T \left(I_1\, \frac{dI_2}{dt} + I_2\,\frac{d I_1}{dt } \right)dt.$$ (2.367)

Thus,

$$W = \int_0^T (R_1 \,I_1^{\,2} + R_2\, I_2^{\,2} )\,dt + \frac{1}{2} \,L_1 \,I_1^{\,2} + \frac{1}{2}\, L_2 \,I_2^{\,2} + M\, I_1\, I_2.$$ (2.368)

Clearly, the total magnetic energy stored in the two solenoids is

$$W_B = \frac{1}{2}\, L_1\, I_1^{\,2} + \frac{1}{2} \,L_2 \,I_2^{\,2} + M \,I_1 \,I_2.$$ (2.369)

Note that the mutual inductance term increases the stored magnetic energy if $I_1$ and $I_2$ are of the same sign; that is, if the currents in the two solenoids flow in the same direction, so that they generate magnetic fields that reinforce one another. Conversely, the mutual inductance term decreases the stored magnetic energy if $I_1$ and $I_2$ are of the opposite sign. However, the total stored energy can never be negative, otherwise the coils would constitute a power source (a negative stored energy is equivalent to a positive generated energy). Thus,

$$\frac{1}{2}\, L_1 \,I_1^{\,2} + \frac{1}{2}\, L_2 \,I_2^{\,2} + M\, I_1 \,I_2\geq 0,$$ (2.370)

which can be written

$$\frac{1}{2}\left(\!\sqrt{L_1} \,I_1 + \sqrt{L_2}\,I_2\right)^2 - I_1 \,I_2 \left(\!\sqrt{L_1 \,L_2} -M\right) \geq 0,$$ (2.371)

assuming that $I_1\, I_2 <0$. It follows that

$$M \leq \! \sqrt{L_1 \,L_2}.$$ (2.372)

The equality sign corresponds to the situation in which all of the magnetic flux generated by one solenoid passes through the other. If some of the flux misses then the inequality sign is appropriate. In fact, the previous formula is valid for any two inductively coupled circuits, and effectively sets an upper limit on their mutual inductance.

We intimated previously that the energy stored in an solenoid is actually stored in the surrounding magnetic field. Let us now obtain an explicit formula for the energy stored in a magnetic field. Consider an ideal cylindrical solenoid. The energy stored in the solenoid when a current $I$ flows through it is

$$W = \frac{1}{2} \,L \,I^2,$$ (2.373)

where $L$ is the self inductance. We know that

$$L = \mu_0\, N^2\,\pi\, r^2\, l,$$ (2.374)

where $N$ is the number of turns per unit length of the solenoid, $r$ the radius, and $l$ the length. [See Equation (2.324).] The magnetic field inside the solenoid is approximately uniform, with magnitude

$$B = \mu_0 \,N\,I,$$ (2.375)

and is approximately zero outside the solenoid. [See Equation (2.279).] Equation (2.373) can be rewritten

$$W = \frac{B^2}{2\,\mu_0}\,{\cal V},$$ (2.376)

where ${\cal V} = \pi\, r^2 \,l$ is the volume of the solenoid. The previous formula strongly suggests that a magnetic field possesses an energy density

$$U = \frac{B^2}{2\,\mu_0}.$$ (2.377)

Let us now examine a more general proof of the previous formula. Consider a system of $N$ circuits (labeled $i=1$ to $N$), each carrying a current $I_i$. The magnetic flux through the $i$th circuit is written [see Equation (2.315)]

$${\mit\Phi}_i = \int {\bf B} \cdot d{\bf S}_i = \oint {\bf A} \cdot d{\bf r}_i,$$ (2.378)

where ${\bf B} = \nabla\times{\bf A}$, and $d{\bf S}_i$ and $d{\bf r}_i$ denote a surface element and a line element of this circuit, respectively. The back-emf induced in the $i$th circuit follows from Faraday's law:

$${\cal E}_i = - \frac{d {\mit\Phi}_i}{dt}.$$ (2.379)

[See Equation (2.284).] The rate of work of the battery that maintains the current $I_i$ in the $i$th circuit against this back-emf is

$$P_i = -I_i\,{\cal E}_i= I_i\, \frac{d {\mit\Phi}_i}{dt}.$$ (2.380)

Thus, the total work required to raise the currents in the $N$ circuits from zero at time 0, to $I_{0\,i}$ at time $T$, is

$$W = \sum_{i=1,N} \int_0^T I_i \,\frac{d{\mit\Phi}_i}{dt}\,dt.$$ (2.381)

The previous expression for the work done is, of course, equivalent to the total energy stored in the magnetic field surrounding the various circuits. This energy is independent of the manner in which the currents are set up. Suppose, for the sake of simplicity, that the currents are ramped up linearly, so that

$$I_i = I_{0\,i}\, \frac{t}{T}.$$ (2.382)

The fluxes are proportional to the currents, so they must also ramp up linearly; that is,

$${\mit\Phi}_i = {\mit\Phi}_{0\,i} \,\frac{t}{T}.$$ (2.383)

It follows that

$$W = \sum_{i=1,N} \int_0^T I_{0\,i} \,{\mit\Phi}_{0\,i}\, \frac{t}{T^2}\,dt,$$ (2.384)

giving

$$W = \frac{1}{2} \sum_{i=1,N} I_{0\,i} \,{\mit\Phi}_{0\,i}.$$ (2.385)

So, if instantaneous currents $I_i$ flow in the $N$ circuits, which link instantaneous fluxes ${\mit\Phi}_i$, then the instantaneous stored energy is

$$W= \frac{1}{2} \sum_{i=1,N} I_i \,{\mit\Phi}_i.$$ (2.386)

Equations (2.378) and (2.386) imply that

$$W = \frac{1}{2} \sum_{i=1,N} I_i \oint {\bf A} \cdot d{\bf r}_i.$$ (2.387)

It is convenient, at this stage, to replace our $N$ line currents by $N$ current distributions of small, but finite, cross-sectional area. Equation (2.387) transforms to give

$$W = \frac{1}{2} \int_V {\bf A} \cdot {\bf j} \, dV,$$ (2.388)

where $V$ is a volume that contains all of the circuits. Note that for an element of the $i$th circuit, ${\bf j} =I_i\, d{\bf r}_i /(dr_i \,A_i)$ and $dV = dr_i \,A_i$, where $A_i$ is the cross-sectional area of the circuit. Now, $\mu_0\, {\bf j} = \nabla\times {\bf B}$ [see Equation (2.271)], so

$$W = \frac{1}{2\,\mu_0} \int_V {\bf A} \cdot \nabla\times{\bf B} \, dV.$$ (2.389)

However,

$$\nabla\cdot({\bf A} \times{\bf B}) \equiv {\bf B} \cdot \nabla\times{\bf A} - {\bf A} \cdot \nabla\times{\bf B}$$ (2.390)

(see Section A.24), which implies that

$$W = \frac{1}{2\,\mu_0} \int_V \left[- \nabla\cdot ({\bf A} \times{\bf B}) +{\bf B}\cdot \nabla \times {\bf A} \right] \,dV.$$ (2.391)

Using the divergence theorem (see Section A.20), and ${\bf B} = \nabla\times{\bf A}$, we obtain

$$W = -\frac{1}{2\,\mu_0} \oint_S {\bf A}\times{\bf B} \cdot d{\bf S} + \frac{1}{2\,\mu_0} \int_V B^2\,dV,$$ (2.392)

where $S$ is the bounding surface of some volume $V$. Let us take this surface to infinity. It is easily demonstrated that the magnetic field generated by a current loop falls of like $r^{-3}$ at large distances. (See Section 2.2.7.) The vector potential falls off like $r^{-2}$. However, the area of surface $S$ only increases like $r^2$. It follows that the surface integral is negligible in the limit $r\rightarrow \infty$. Thus, the previous expression reduces to

$$W = \int \frac{B^2}{2\,\mu_0} \, dV,$$ (2.393)

where the integral is over all space. Because this expression is valid for any magnetic field whatsoever, we can safely conclude that the energy density of a general magnetic field generated by a system of electrical circuits is given by

$$U = \frac{B^2}{2\,\mu_0}.$$ (2.394)

Note, finally, that the fact that a magnetic field possesses an energy density demonstrates that it has a real physical existence, and is not merely an aid to calculating the forces that current-carrying wires exert on one another.