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UY1: Magnetic-Field Energy In Inductor

Derive energy stored in an inductor and connect it to magnetic-field energy density in space.

  • University Physics Year 1
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Learning objectives

  • Analyse magnetic forces, induction, inductance, and alternating-current systems with consistent signs.
Why this matters + quick links

This page gives the UY1 working model/result for Magnetic-Field Energy In Inductor. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • Building current in an inductor requires work against back e.m.f.
  • Stored energy:
U = 1/2 LI².
  • Magnetic energy density in a linear medium:
u = B²/2μ.
  • Modelling context: U = (1/2)LI² assumes constant inductance (linear regime). If L depends on i, the safe form is U = ∫₀ⁱ L(i') i' di'.

Prerequisites: Self-Inductance & Inductors
Next uses: L-C Circuit, L-R-C Series Circuit

2) Setup

Consider an inductor of inductance L with current increasing from 0 to I.

  • Instantaneous inductor voltage (passive sign convention):
v_L = Ldi/dt.
  • Source power into inductor:
p = i v_L.
Common traps (factor of 1/2 + what stores energy)
  • The energy is (1/2)LI², not LI².
  • Inductor energy is stored in the magnetic field, not as “charge piled up” (that’s the capacitor picture).
  • If a problem hints at saturation or changing permeability, treat L as not strictly constant.

3) Core derivation/explanation

Start from power:

dU/dt = p = iLdi/dt.

Hence:

dU = Li di.

Integrate from 0 to I:

U = ∫₀^I Li di = 1/2 LI².

So when current falls, this stored energy is released back to the external circuit.

Field viewpoint for a uniform-field region:

U = ∫ u dV, u = B²/2μ

(in vacuum, μ = μ₀).

Units check:

  • (1/2)LI² → (H)(A²) = J.
  • B²/(2μ) → J m⁻³.

Checks (sanity)

  • U → 0 as I → 0.
  • Doubling current multiplies stored energy by 4 (quadratic scaling).

4) Worked example(s)

An inductor has L = 0.40 H carrying I = 3.0 A.

U = (1/2)LI² = (1/2)(0.40)(3.0)² = 1.8 J.

If the field is approximately uniform at B = 0.25 T in vacuum, energy density is:

u = B²/2μ₀ = (0.25)²/(2(4π × 10⁻⁷)) ≈ 2.49 × 10⁴ J m⁻³.

5) Practice set (with hints + answers)

  1. If current doubles in the same inductor, by what factor does stored energy change? Hint: U ∝ I². Answer: Factor of 4.

  2. A 1.2 H inductor carries 0.50 A. Find U. Hint: direct substitution. Answer: U = 0.15 J.

  3. Compare energy density at B = 0.10 T and 0.20 T in same medium. Hint: u ∝ B². Answer: second is 4 times larger.

6) Summary + next steps

  • Inductors store energy in magnetic fields, not in “charge separation” like capacitors.
  • Circuit formula and field-density formula are consistent descriptions of the same energy.
  • This energy viewpoint is central for LC oscillations and AC power flow.

Next: L-R-C Series Circuit Previous: L-C Circuit Back To Electromagnetism (UY1)