Self-Inductance & Energy in Inductors

You’ll be able to

Define self-inductance through ε = −L di/dt and L = NΦ/i
Derive the inductance of an ideal solenoid from its field and flux linkage
Compute stored energy with U = ½Li² and the field energy density B²/(2μ₀)

A circuit fights changes in its own current

A coil’s current creates a magnetic field, and that field threads the coil’s own turns. If the current changes, so does that self-flux — and by Faraday’s law the coil induces an EMF in itself, directed (Lenz) to oppose the change. The coil’s geometry sets how strongly, captured by the self-inductance L = NΦ_B/i: total flux linkage per ampere. The unit is the henry (1 H = 1 V·s/A). An inductor is a component built to have large L — a coil that gives current the electrical equivalent of inertia.

Self-inductance

ε_L = −L·di/dt, L = NΦ_B/i

The back-EMF depends on how fast the current changes, not on how big it is. A steady current through an ideal inductor drops zero volts.

Worked example

Derive the self-inductance of an ideal solenoid with n turns per unit length, cross-sectional area A, and length ℓ.

1.Field inside from Ampère’s law (Unit 5): B = μ₀ni when the solenoid carries current i.
2.Flux through one turn: Φ = BA = μ₀niA.
3.The solenoid has N = nℓ turns, so the total flux linkage is NΦ = (nℓ)(μ₀niA) = μ₀n²ℓA·i.
4.Divide by the current: L = NΦ/i = μ₀n²Aℓ. Equivalently, substitute n = N/ℓ to get L = μ₀N²A/ℓ.
5.Read the dependence: L grows with the square of the winding density — doubling n doubles the field per ampere and doubles the turns being threaded. Note L is pure geometry times μ₀; the current cancels out, as an inductance must not depend on i.

Answer: L = μ₀n²Aℓ = μ₀N²A/ℓ

Solenoid inductance and stored energy

L = μ₀n²Aℓ = μ₀N²A/ℓ, U = ½Li², u = B²/(2μ₀)

U = ½Li² is the work done against the back-EMF to establish the current; u is that energy per unit volume, stored in the field itself.

The energy lives in the field

Ramping a current up from zero requires work against the back-EMF: dW = ε·i dt = Li·di, and integrating from 0 to i gives U = ½Li² — the magnetic twin of the capacitor’s ½CV². For a solenoid, substitute L = μ₀n²Aℓ and i = B/(μ₀n): U = ½·μ₀n²Aℓ·B²/(μ₀n)² = (B²/2μ₀)·(Aℓ). Since Aℓ is the interior volume, the energy density of a magnetic field is u = B²/(2μ₀) — the exact parallel of u = ½ε₀E² for electric fields. The energy is not “in the wire”; it is stored in the field filling the space.

Worked example

A solenoid has 500 turns wound over 0.25 m with cross-sectional area 1.0 × 10⁻³ m². Find its inductance and the energy it stores at a current of 2.0 A.

1.Use L = μ₀N²A/ℓ with N = 500: N² = 2.5 × 10⁵.
2.Numerator: μ₀N²A = (4π × 10⁻⁷)(2.5 × 10⁵)(1.0 × 10⁻³) = π × 10⁻⁴ ≈ 3.14 × 10⁻⁴ (henry·meters).
3.Divide by ℓ = 0.25 m: L = 1.26 × 10⁻³ H ≈ 1.3 mH.
4.Stored energy: U = ½Li² = ½(1.26 × 10⁻³)(2.0)² = 2.5 × 10⁻³ J = 2.5 mJ.
5.Perspective: modest numbers — which is why power-supply inductors are wound with hundreds of turns on high-permeability cores to boost L.

Answer: L ≈ 1.3 mH, U ≈ 2.5 mJ

Tip

Think of inductance as inertia for current: L plays the role of mass, i the role of velocity, and U = ½Li² mirrors ½mv². Just as a mass resists changes in velocity but not velocity itself, an inductor resists changes in current but lets a steady current sail through with no voltage drop.

Checkpoint

A solenoid is rewound with double the turns per unit length, same length and area. Its self-inductance:

Checkpoint

A constant current I flows through an ideal inductor L. The EMF across the inductor is:

Answer the 2 checkpoints as you read.