Inductor Energy Calculator — Joules Stored in a Magnetic Field

Background.

An inductor carrying current stores energy in the magnetic field around and inside it, and the amount is U = ½LI². This page computes it, and also runs the problem backwards: give it an energy and a current and it returns the inductance needed, or give it an energy and an inductance and it returns the current required. All three are the same equation rearranged, so whichever way you go, the full set comes back and the numbers agree.

The square matters. Energy rises with the square of current but only linearly with inductance, so current is always the stronger lever — doubling it quadruples the stored energy while doubling the inductance merely doubles it. That is why a switching converter's inductor is sized by its peak current rather than its henries, and why a 20 percent error in your current estimate becomes a 44 percent error in the energy.

An inductor is the mirror image of a capacitor, and the differences catch people out. A capacitor stores energy in an electric field, holds it after the supply is removed, and is dangerous because it stays charged. An inductor stores energy in a magnetic field, holds it only while the current keeps flowing, and is dangerous for the opposite reason: interrupting that current is what hurts. Voltage across a coil is L times the rate of change of current, so opening a switch — where the rate of change is enormous — makes the coil produce whatever voltage it takes to keep the current going. That is arcing across relay contacts, and destroyed transistors, and it is why every inductive load in a real design has a flyback diode, a snubber or a clamp across it.

The page also reports the flux linkage, Λ = LI, because it is the quantity that actually cannot change instantaneously. Energy is what the coil holds; flux linkage is what it defends. The published example used to verify this page makes the connection concrete: OpenStax describes a 50-turn coil of 4.0 × 10⁻⁴ H carrying 5.0 A with a flux of 4.0 × 10⁻⁵ Wb per turn. Multiplying turns by flux gives 2.0 × 10⁻³ Wb, and LI gives the same 2.0 × 10⁻³ Wb — two routes, one number, which is a genuine check rather than a restatement.

One limit worth stating before you use the result. This assumes an ideal coil: no winding resistance, no core loss and, most importantly, no saturation. A real ferrite or powdered-iron core loses inductance rapidly once the flux density approaches its limit, so ½LI² computed from a datasheet's small-signal inductance badly overstates the energy a saturated coil actually holds. If you are near the rated current, use the inductance at that current, not the headline figure.

What is inductor energy calculator?

The energy stored in an inductor is the work done against the back-emf while establishing its current. As current rises, the changing flux induces a voltage that opposes the change, and the source has to push against it: the instantaneous power is vi = Li(di/dt), and integrating from zero to the final current I gives U = ½LI². OpenStax University Physics prints this as Equation 14.20. Physically the energy is not in the wire but in the magnetic field, distributed with an energy density of B²/2μ₀ — which is why an inductor with a large air gap can store far more energy than its inductance alone suggests, and why gapped cores exist. The behaviour is the exact dual of a capacitor's. A capacitor's energy scales with the square of voltage and persists after disconnection; an inductor's scales with the square of current and disappears the moment the current does. A capacitor resists a sudden change in voltage; an inductor resists a sudden change in current, and it does so by generating whatever voltage that takes. Those dual properties are what make the pair useful together in filters, resonant circuits and switching converters, where energy is shuttled between an electric field and a magnetic one many thousands of times a second.

How to use this calculator.

1. 01Pick what you want to solve for. Energy is the default; inductance and current are the inverse problems that come up when sizing a converter or a storage coil.
2. 02Enter the inductance and choose its unit — nanohenries through henries. In inductance mode this field is ignored.
3. 03Enter the current through the coil. In current mode this field is ignored. Use peak current, not average, if you are checking whether a core will saturate.
4. 04Enter the target energy in joules if you are solving for inductance or current. In energy mode this field is ignored.
5. 05Read the energy in whichever magnitude fits — joules for magnet and converter coils, millijoules for ordinary power electronics, microjoules for RF chokes.
6. 06Check the flux linkage if you are designing the switch-off path: it is what the coil will fight to preserve when the current is interrupted.
7. 07Provide a discharge path — flyback diode, snubber or clamp — before you switch any inductive load. The energy has to go somewhere.

The formula.

Start from what establishing a current in a coil costs. As the current rises, the changing flux induces a back-emf v = L di/dt opposing it, and the source must supply power vi = Li(di/dt) to push through. Integrating that from zero current to the final value I gives U = ½LI². The factor of one half comes from the integral, exactly as it does for a capacitor, and dropping it is the standard error.

Working the default: L = 100 mH = 0.1 H and I = 3 A. Squaring the current gives 9, multiplying by 0.1 gives 0.9, and half of that is 0.45 J — which is 450 mJ or 450,000 µJ. The flux linkage is Λ = LI = 0.1 × 3 = 0.3 Wb. A useful consistency check falls out of these two: U should also equal ½ΛI, and ½ × 0.3 × 3 = 0.45 J.

The inverse modes are one line each. Inductance from energy and current is L = 2U/I², and current from energy and inductance is I = √(2U/L). The square root is why a negative energy is refused rather than allowed to produce a NaN. Running the default through both inverses returns 100 mH and 3 A, closing the round trip.

On the exponents, verified against the algebra rather than recalled: U ∝ I² means doubling the current multiplies the energy by four, while U ∝ L¹ means doubling the inductance only doubles it. Flux linkage behaves differently again — Λ = LI is linear in both, so doubling the current doubles the flux linkage while quadrupling the energy. All three relationships are pinned by tests.

The published cross-check is worth following because it uses a second, independent route to the same quantity. OpenStax Example 14.2 describes a 50-turn coil with L = 4.0 × 10⁻⁴ H carrying 5.0 A, and states its flux as 4.0 × 10⁻⁵ Wb per turn. Flux linkage by the definition of inductance is LI = 4.0 × 10⁻⁴ × 5.0 = 2.0 × 10⁻³ Wb. Flux linkage by counting turns is NΦ = 50 × 4.0 × 10⁻⁵ = 2.0 × 10⁻³ Wb. The two agree, and the same coil's stored energy is ½ × 4.0 × 10⁻⁴ × 25 = 5 mJ.

Rounding happens once, at the end, to ten decimal places, with intermediate arithmetic at forty significant digits on a private numeric context. Energy appears in three magnitudes because the practical range spans nanojoules in an RF choke to hundreds of kilojoules in a magnet coil, and no single unit survives that span at a fixed decimal precision.

What is not modelled: winding resistance, core loss, hysteresis, eddy currents, frequency dependence, and — most consequentially — saturation. Once a core saturates its inductance collapses, so the energy computed from a nameplate inductance can be far above what the coil really holds at that current.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), University Physics Volume 2, §14.3 'Energy in a Magnetic Field'. Primary source for the equation implemented here, Equation 14.20: 'U = 1/2 L I^2'. The same section gives the magnetic energy density 'u_m = B^2 / 2μ_0', which is where the energy physically resides — in the field rather than in the conductor. Retrieved 2026-07-29.
2. [2]OpenStax (Rice University), University Physics Volume 2, §14.2 'Self-Inductance and Inductors'. Source of the definition of self-inductance L = NΦ_m/I, from which the flux-linkage output Λ = LI follows, and of the henry as one volt-second per ampere. Its Example 14.2 supplies this page's published cross-check: a 50-turn coil with L = 4.0×10⁻⁴ H at 5.0 A and Φ_m = 4.0×10⁻⁵ Wb per turn, giving NΦ_m = LI = 2.0×10⁻³ Wb by two independent routes. Retrieved 2026-07-29.
3. [3]Engineering LibreTexts / James M. Fiore, Introduction to Circuit Analysis, §6.2.2 'Inductance and Inductors'. Independent second authority, consulted to check the expression rather than to derive it: 'W = ½LI², where W is energy in joules, L is inductance in henries, and I is current in amps', together with the statement that 'the current through an inductor cannot change instantaneously' — the property behind the flyback warning on this page. Retrieved 2026-07-29.
4. [4]MIT OpenCourseWare, 6.002 Circuits and Electronics, Spring 2007 (Prof. Anant Agarwal). Course materials on energy storage in the inductor element and on switching inductive loads. Landing page verified to load and to list the course number, term and instructor. Retrieved 2026-07-29.