Energy Stored in an Inductor

Equivalent forms

U = \frac{B^2}{2\mu_0} V

U = \tfrac{1}{2} \Phi I

Identical algebraic form to (1/2)kx², (1/2)mv², and (1/2)CV² — every linear energy-storage element wears the same 1/2 stamp.

Symbol	Name	SI unit	Dimension	Range
$U$	Stored magnetic energyoutput Energy stored in the inductor's magnetic field	J	M·L²·T⁻²	0 – 100
$L$	Inductance Inductance of the coil	H	M·L²·T⁻²·I⁻²	0.001 – 1
$I$	Current Current flowing through the inductor	A	I	0 – 20

Unit systems

Symbol	SI	CGS	Imperial
$U$	J	erg	ft·lbf
$L$	H	abH	H
$I$	A	statA	A

Where it holds

Valid for linear inductors (constant L) at any current. Fails for ferromagnetic cores near saturation where L drops abruptly and stored energy grows more slowly.

Dimensional analysis

$[U] = [L][I]^{2} = (H)$ $(A^{2}) = (V\cdot s$ /A $)(A^{2}) = V\cdot A\cdot s = J \checkmark$

Discovery

James Clerk Maxwell · 1865

While unifying electricity and magnetism, Maxwell explicitly identified the energy density of the magnetic field as B²/(2μ₀), giving the inductor a definite, locatable energy content.

Try this

Why does opening a switch on an energized inductor cause a bright spark — even from a low-voltage battery?

A 50 mH inductor carries a current of 3 A. How much energy is stored in its magnetic field?

Research status: stable

Real-world applications

Switching power supplies (buck/boost converters) shuttle energy through inductors many times per second
Ignition coils in cars use magnetic energy collapse to fire spark plugs
Pulsed magnets for fusion experiments and high-field labs
Magnetic Energy Storage (SMES) for grid stabilization

Common misconceptions

Inductor energy is stored in the field, not in moving charges — the carriers themselves carry negligible energy
Removing the source doesn't immediately stop the current; energy must go somewhere first
L is a property of geometry, not of how much energy you put in

Experimental verification

Discharge a coil through a known resistance and integrate

I^{2}R dt

; the total dissipated energy matches

(1/2)L I^{2} to

within a few percent. The same is verified by calorimetry in pulsed-power experiments.

Derivation

Instantaneous power into inductor:

P = \varepsilon I = L (dI/dt) I

.

Total work to build up current from 0 to I:

W = \int _{0}

ᴵ

L i di = (1/2) L I^{2}

.

This work is stored as magnetic field energy.

Limiting cases

I \to 0

⟶

U \to 0

No current means no magnetic field, hence no stored energy.

I \to 2I

⟶

U \to 4U

Energy scales quadratically with current — doubling I quadruples U.

Inductor disconnected suddenly⟶ Large voltage spikeU cannot vanish instantly;

\varepsilon = -L dI/dt

becomes huge when dI/dt is forced large, producing the familiar arc.

What if…

What if the current is doubled?

Stored energy quadruples — quadratic dependence on I.

What if the inductor is connected in series with a capacitor?

Energy sloshes back and forth between $(1/2)LI^{2} and (1/2)CV^{2} at$ the LC resonant frequency — an oscillating circuit.

What if you suddenly insert a resistor in parallel?

The stored energy decays exponentially as heat with time constant $\tau = L/R$ .

1

Energy in a 50 mH choke

Given ·

L:: 0.05
I:: 3

Find · U

Steps

Identify: $L = 0.05\,\mathrm{H}$ , $I = 3$ A
Apply $U = (1/2) L I^{2}$
Compute $I^{2} = 9$
$U = 0.5 \times 0.05 \times 9 = 0.225\,\mathrm{J}$

Result ·

U = 0.225\,\mathrm{J}

2

Voltage spike on interruption

Given ·

L:: 0.5
I:: 2
dt:: 0.0001

Find ·

\varepsilon at

interruption

Steps

When switch opens, current must drop from 2 A to 0 in $dt = 100 \mu s$ (limited by arc)
$\varepsilon = L dI/dt = 0.5 \times 2 / 10^{-4} = 10^{4} V$
This 10 kV spike easily ionizes the air gap — visible as a spark

Result ·

\varepsilon \approx 10\,\mathrm{kV}

Self-Inductance of a Solenoid→Energy Stored in a Capacitor→Faraday's Law of Induction→LC Oscillation Frequency→