Energy in Inductors and Magnetic Fields

Building up a current in an inductor takes work, because the back emf pushes against the current the whole time. That work is not lost: it is stored, and it comes back out when the current falls. Just as a capacitor's energy lives in its electric field, an inductor's lives in its magnetic field.

Key idea

An inductor carrying current $I$ stores $U = \tfrac{1}{2}LI^2$. A magnetic field stores energy with density $$u_B = \frac{B^2}{2\mu_0}\quad(\text{J/m}^3).$$

Why it works

While the current rises, the source must push charge against the back emf $L\,dI/dt$. The power it supplies to the inductor is $L\dfrac{dI}{dt}\,I$, so the work to go from $0$ to $I$ is $$W = \int_0^I LI'\,dI' = \tfrac{1}{2}LI^2.$$ For a long solenoid, $L = \mu_0N^2A/\ell$ and $B = \mu_0NI/\ell$, so $\tfrac{1}{2}LI^2 = \dfrac{B^2}{2\mu_0}(A\ell)$: the energy density $B^2/(2\mu_0)$ times the volume holding the field.

As the current builds, energy flows from the battery into the inductor's growing magnetic field.

Worked examples

Worked example: Energy in an inductor

How much energy does a $2.0\ \text{H}$ inductor store at $3.0\ \text{A}$?

$U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$.

Answer: $U = 9.0\ \text{J}$

Worked example: A strong magnet

Find the energy density in a $1.5\ \text{T}$ field, a typical strength for a hospital MRI magnet. Compare with an electric field of $3.0 \times 10^{6}\ \text{V/m}$, near where air breaks down.

$u_B = \dfrac{(1.5)^2}{2(4\pi \times 10^{-7})} = 8.95 \times 10^{5}\ \text{J/m³}$.
$u_E = \tfrac{1}{2}\varepsilon_0E^2 = 39.8\ \text{J/m³}$.
The magnetic field holds about $2.2 \times 10^{4}$ times as much energy per cubic meter.

Answer: $u_B \approx 8.95 \times 10^{5}\ \text{J/m³}$

Worked example: Two routes for a solenoid

A solenoid $50\ \text{cm}$ long with 1000 turns of radius $2.0\ \text{cm}$ carries $2.0\ \text{A}$. Find its stored energy from the field, then from $\tfrac{1}{2}LI^2$.

$B = \mu_0 nI = (4\pi \times 10^{-7})(2000)(2.0) = 5.03\ \text{mT}$, so $u = B^2/(2\mu_0) = 10.1\ \text{J/m³}$.
Volume $\pi r^2\ell = 6.28 \times 10^{-4}\ \text{m³}$, so $U = 6.32\ \text{mJ}$.
$L = \mu_0N^2A/\ell = 3.16\ \text{mH}$ and $\tfrac{1}{2}LI^2 = 6.32\ \text{mJ}$. The same.

Answer: $U \approx 6.32\ \text{mJ}$

The stored energy is also why the current in an RL circuit keeps flowing for a while after the battery is removed: the field's $\tfrac{1}{2}LI^2$ drives the current until the resistor has turned all of it into heat. In an LC circuit, the same energy flows into a capacitor instead, and comes back.

Common mistake

Mixing the inductor and capacitor formulas, writing $\tfrac{1}{2}LV^2$ or $\tfrac{1}{2}CI^2$. Each device's energy depends on what builds its field: current for an inductor ($\tfrac{1}{2}LI^2$), voltage, that is charge, for a capacitor ($\tfrac{1}{2}CV^2$). Check with units: H·A² = (V·s/A)·A² = V·A·s = J, while H·V² is not a joule.

Energy in Inductors and Magnetic Fields

Worked examples

Practice

Builds on

Leads to

Sources and further reading