Inductance
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
A coil carrying current links its own magnetic flux. If the current changes, so does that flux, and the coil induces an emf in itself that fights the change: εL = −L dI/dt. The inductance L depends only on the coil’s geometry, and the coil stores energy ½LI² in its field.
1. Self-inductance
2. What it means
An inductor is “electrical inertia”: it resists changes in current the way mass resists changes in velocity. Current through an inductor can’t jump instantly.
When current increases, εL opposes it (like a battery facing backward); when current decreases, εL tries to keep it going.
Why N²: more turns make more field (one N) and each turn links that field (another N).
Mutual inductance: a changing current in one coil induces emf in a nearby coil: ε2 = −M dI1/dt. That’s how transformers work.
3. Example, then try it
The current in a 0.15 H inductor rises steadily from 0 to 4.0 A in 0.20 s. Find the emf across it and the energy stored at the end.
|ε| = L ΔI/Δt = (0.15)(4.0) ÷ 0.20 = 3.0 V, opposing the rise.
U = ½LI² = ½(0.15)(4.0)² = 1.2 J.
A solenoid 0.40 m long has 800 turns and radius 1.0 cm, and carries 2.0 A. Find its inductance, then find the stored energy two ways: from ½LI² and from the field’s energy density.
Inductance: A = π(0.010)² = 3.14 × 10⁻⁴ m², so L = μ0N²A/ℓ = (4π × 10⁻⁷)(800)²(3.14 × 10⁻⁴) ÷ 0.40 ≈ 6.3 × 10⁻⁴ H.
From L: U = ½(6.3 × 10⁻⁴)(2.0)² ≈ 1.26 × 10⁻³ J.
Field: B = μ0nI = (4π × 10⁻⁷)(2000)(2.0) ≈ 5.0 × 10⁻³ T, so u = B²/(2μ0) ≈ 10 J/m³.
From u: volume = Aℓ = 1.26 × 10⁻⁴ m³, so U = u × volume ≈ 1.26 × 10⁻³ J. They agree.
Answer: L ≈ 0.63 mH; U ≈ 1.3 mJ either way.
4. Common mistakes
A steady current flows through an ideal inductor (no resistance). What is the voltage across it?
εL = −L dI/dt, and dI/dt = 0 for a steady current.
Ready to practice?
5 questions on Inductance, each with a hint and an explanation for every wrong answer.