Inductor Energy Calculator

Find stored energy U = ½LI²

Parameters

Hⓘ
Aⓘ
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Controls

xⓘ

Calculated Values

Stored Energy:
0.23;J0.23;J

Examples

L=0.05 H, I=3 A

Stored energy.

    Visualization

    Energy Stored in an Inductor

    An inductor stores energy in its magnetic field. When current I flows through inductance L, stored energy is U = ½LI² — directly analogous to capacitor energy U = ½CV². Building current from zero requires doing work against the back EMF ε = −L dI/dt that opposes changes in current (Lenz's law for inductors).

    Power delivered to inductor: P = Iε = LI dI/dt. Integrating from I = 0 to I gives U = ∫P dt = ½LI². Energy density in vacuum field is u = B²/(2μ₀); integrating over volume recovers ½LI² for a solenoid.

    Example: L = 0.05 H, I = 3 A gives U = 0.5×0.05×9 = 0.225 J — enough to produce a visible spark if current is interrupted suddenly (large dI/dt → large voltage spike V = L dI/dt).

    In RL circuits, energy flows from battery into magnetic storage and dissipates in resistor. In ideal LC circuits, energy oscillates between magnetic (L) and electric (C) forms at frequency ω = 1/√(LC).

    Practical uses: ignition coils multiply voltage by interrupting primary current; switch-mode power supplies store energy in inductors each cycle; MRI gradient coils are large inductors with significant stored energy (safety concern when quenching).

    Real inductors have resistance and core losses; not all input energy remains stored — some heats the winding. At high frequency, skin effect and core hysteresis matter.

    Key Concepts

    • U = ½LI²
    • ε = −L dI/dt (back EMF)
    • Energy builds as I increases
    • Analogous to ½CV²
    • u = B²/(2μ₀) field density
    • Spark from rapid dI/dt

    Real-World Applications

    • Ignition and spark coils
    • Switch-mode power supplies
    • LC oscillators
    • MRI gradient coil safety
    • Flyback transformers
    • Class 12 electromagnetic energy

    Explore Further

    More magnetism tools

    Physics Equations

    Energy:
    U=12LI2U = \frac{1}{2}LI^2

    Step-by-Step Solution

    See how the main results are calculated.

    1

    Step 1: Energy in Inductor

    Equation:

    U=12LI2U = \frac{1}{2}LI^2

    Explanation:

    Energy stored in the magnetic field of the inductor.

    2

    Step 2: Given

    Result:

    L=0.05 H,I=3 AL = 0.05\ \text{H},\quad I = 3\ \text{A}
    3

    Step 3: Square the Current

    Calculation:

    I2=(3)2=9.0000I^2 = (3)^2 = 9.0000
    4

    Step 4: Energy

    Calculation:

    U=12×0.05×9.0000=0.225000 JU = \frac{1}{2} \times 0.05 \times 9.0000 = 0.225000\ \text{J}

    Result:

    U=0.225000JU = 0.225000 J
    5

    Step 5: Back EMF

    Inductor opposes sudden current changes.

    Equation:

    E=−LdIdt\mathcal{E} = -L\frac{dI}{dt}

    Explanation:

    Opening a circuit quickly can cause large voltage spikes.

    6

    Step 6: Compare to Capacitor

    Electric energy in C; magnetic energy in L.

    Equation:

    UC=12CV2U_C = \frac{1}{2}CV^2

    Explanation:

    LC circuits exchange energy between fields.

    Frequently Asked Questions (FAQ)

    Where is energy stored?

    In magnetic field around inductor; U = ½LI² equivalent form.

    Ideal vs real inductor?

    Real has resistance; some energy dissipates as heat.

    Practice MCQs

    1. Inductor energy:
    2. Double current:
    3. Back EMF when I increases:
    4. SI unit of L:
    5. Opening RL circuit quickly:
    6. Capacitor analog: