Inductor Energy Calculator
Enter inductance and current to compute the magnetic energy stored in an inductor E = ½·L·I².
Input Data
Results
At a glance:An inductor stores energy in its magnetic field. The energy is E = ½·L·I², where L is inductance and I the current. Equivalently E = ½·L·I² = ½·Φ·I = B²·V/(2μ₀) for a uniform field in volume V. The energy is released when the current falls (e.g. in a flyback diode protecting a switching transistor). Unlike a resistor, an ideal inductor dissipates no average power — it stores and returns energy each cycle. The energy grows with the square of current, so doubling I quadruples stored energy. This tool returns E in J and mJ, plus flux Φ = L·I.
Formula
E = ½·L·I²
Flux: Φ = L·I
$$E = \frac{1}{2} L I^2, \quad \Phi = L I, \quad w = \frac{B^2}{2\mu}$$How to Use
- Enter the inductance L (H).
- Enter the current I (A).
- The calculator returns the stored energy (J and mJ) and flux.
Case Studies
SMPS inductor
L = 10 mH, I = 2 A.
E = ½×0.01×4 = 0.02 J (20 mJ).
Released each switch-off cycle through a diode.
FAQ
Why is energy proportional to I²?
Building current requires work against the back-EMF; integrating P = L·I·dI/dt from 0 to I gives ½LI², the same quadratic form as kinetic or capacitor energy.
Where does the energy go when current stops?
It is returned to the circuit (often via a diode) as the field collapses, inducing a voltage that drives current until the energy is spent.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.