## Section42.4Energy in an Inductor

In Subsection 42.3.1, we have discussed energy dissipation in a RL-circuit. Current in an RL-circuit without an EMF source decays exponentially with time constant equal to $\tau=\mathcal{L}/R\text{.}$

\begin{equation*} I(t) = I(0)\; e^{-R\, t/L}. \end{equation*}

With time all the energy in the circuit dissipates in the resitor. Quantifying the total energy dissipated in the resistor, we found that we can assign energy to the circuit at $t=0\text{,}$ when current in the circuit is $I_0\text{,}$ is

\begin{equation*} U = \frac{1}{2}\mathcal{L} I_0^2. \end{equation*}

Where in the circuit is this energy stored at $t=0\text{.}$ If you look at the circuit, you find that the circuit has magnetic field at $t=0\text{,}$ especially concentrated in the inductor. That is, magnetic energy stored in the inductor, when current $I_0$ is flowing through the inductor is
$$U_B = \frac{1}{2}\mathcal{L} I_0^2.\tag{42.4.1}$$
In the section below, we will write this explicitly in terms of magnetic field.

### Subsection42.4.1Energy in Magnetic Field

We found the an inductor carrying current $I$ has energy $\frac{1}{2} \mathcal{L} I^2\text{.}$ We can now show that this energy is actually in the magnetic field of current in the solenoid.

Let's suppose out inductor is a long solenoid with $n$ turns per unit length. That would mean, inside of the solenoid has magnetic field of magnitude

\begin{equation*} B_\text{in} = \mu_0 n I. \end{equation*}

Using $I$ from this in the energy of solenoid we get

\begin{equation*} U_\text{solenoid} = \frac{1}{2} \mathcal{L} I^2 = \frac{1}{2} \mathcal{L} \dfrac{1}{\mu_0^2 n^2}\, B^2. \end{equation*}

Now, recall that self-inductance of a solenoid of radius $a$ and length $l$ is

\begin{equation*} \mathcal{L} = \pi\mu_0 n^2 la^2. \end{equation*}

Therefore, the energy $U$ of solenoid is

\begin{equation*} U_\text{solenoid} = \frac{1}{2\mu_0} B_{\text{in}}^2 \left( l \pi a^2 \right), \end{equation*}

which can be written as an energy density times volume.

\begin{equation*} U = \frac{1}{2\mu_0} B_{\text{in}}^2 \times \text{Volume of solenoid} \end{equation*}

Therefore, we say that magnetic field contains energy given by the following formula for energy per unit volume, $u_B\text{.}$ (Note: we use a small letter for energy density and capital letter for energy.)

\begin{equation*} u_B = \frac{1}{2\mu_0} B^2\ \ \ \text{(Energy density)} \end{equation*}

(Calculus) Electromagnetic Energy

If magnetic field is not uniform, then the energy in the magnetic field in a particular space volume will be given by an integral over the volume.

\begin{equation*} U_B = \frac{1}{2\mu_0} \int_{\text{Volume}} B^2 dV\ \ \ \text{(Energy in magnetic field)} \end{equation*}

The formula for the energy in the magnetic field is similar to that of energy in the electric field.

\begin{equation*} U_E = \frac{1}{2} \epsilon_0\int_{\text{Volume}} E^2 dV. \ \ \ \text{(Energy in electric field)} \end{equation*}

The sum of energy in electric and magnetic field is called the electromagnetic energy, or energy in the electromagnetic field, $U_{em}\text{.}$

\begin{equation*} U_{em} = U_E + U_B. \end{equation*}

### Subsection42.4.2Energy in Interacting Inductors

The arguments for the energy in an inductor carrying a current can be extended to obtain energy in two coupled circuits.

Suppose one circuit of self-inductance $\mathcal{L}_{1}$ carries a current $I_1$ and another circuit of self-inductance $\mathcal{L}_{2}$ carries a current $I_2\text{.}$ Suppose their mutual inductance is $\mathcal{M}\text{.}$

Then, the energy stored in the two circuits will be

$$U = \frac{1}{2} \mathcal{L}_1 I_1^2 + \frac{1}{2} \mathcal{L}_2 I_2^2 + \mathcal{M} I_1 I_2. \label{eq-energy-mutual-ind}\tag{42.4.2}$$

The reference value of this energy is zero when both currents are zero. At any other value of the currents we require that $U$ be positive.

\begin{equation*} U\ge 0. \end{equation*}

Now, we write the energy expression in Eq. (42.4.2) as follows.

$$U = \frac{1}{2} \mathcal{L}_1 \left( I_1 + \frac{\mathcal{M}}{\mathcal{L}_1} I_2 \right)^2 + \frac{1}{2}\left( \mathcal{L}_2 - \frac{\mathcal{M}^2}{\mathcal{L}_1} \right) I_2^2. \label{eq-energy-mutual-ind-2}\tag{42.4.3}$$

We observe that if the first term is zero, then the second term will place restrictions on $\mathcal{L}_1\text{,}$ $\mathcal{L}_2\text{,}$ and $\mathcal{M}^2\text{.}$ The first term will be zero if currents have the following relation.

\begin{equation*} I_1 + \frac{\mathcal{M}}{\mathcal{L}_1} I_2= 0. \end{equation*}

In this case, the positivity of $U$ requires that the second term in Eq. (42.4.3) be positive. Therefore,

\begin{equation*} \mathcal{L}_2 - \frac{\mathcal{M}^2}{\mathcal{L}_1} \ge 0. \end{equation*}

Rearranging, we find that

\begin{equation*} \mathcal{M}^2 \le \mathcal{L}_1 \mathcal{L}_2 . \end{equation*}

Since the mutual inductance can be positive or negative depending upon the sign convention for the currents, we write the absolute value of the mutual inductance as

\begin{equation*} |\mathcal{M}| \le \sqrt{ \mathcal{L}_1 \mathcal{L}_2 }. \end{equation*}

Therefore, the magnitude of the mutual inductance must always be less than the geometric mean of the self-inductances. We often write the mutual inductance in terms of the self-inductances by introducing a constant $k$ called the coefficient of coupling.

\begin{equation*} |\mathcal{M}| = k \sqrt{ \mathcal{L}_1 \mathcal{L}_2 }. \end{equation*}

In experimental situations, the absolute value of the coefficient of coupling is near 1, $|k|\sim 1\text{,}$ if magnetic flux from one circuit goes through the other circuit and vice-versa, as in an ideal transformer. If the two circuits are so arranged that the magnetic flux from one does not completely go through the other circuit, then $|k|\lt 1\text{.}$