Inductance and magnetic energy

Why it matters

Inductive transducers — LVDTs, variable-reluctance pick-ups, proximity sensors — work by changing a self or mutual inductance with displacement. Current transformers and every transformer rely on mutual inductance, and the energy stored in a magnetic field sets the kickback voltage when an inductive load is switched off. Calculating L, M and the stored energy from geometry is the bridge between field theory and circuit models.

Key ideas

Flux linkage. A coil of N turns, each linked by flux Φ, has flux linkage λ = NΦ (weber-turns). For a linear medium, λ is proportional to the current that produces it.

Self-inductance. L = λ/I = NΦ/I (henry, H = Wb/A). It depends only on geometry and the permeability of the surrounding medium, not on the current — as long as the material is linear (no saturation). Its circuit meaning is v = L di/dt: an inductor opposes changes of current (Lenz's law).

Recipe for L (mirror image of the capacitance recipe): assume a current I; find H by Ampère's law; find B = μH and the flux Φ = ∫B·dS; form λ = NΦ; divide by I.

Mutual inductance. M12 = N2Φ12/I1, the flux linkage of coil 2 per ampere in coil 1. Reciprocity gives M12 = M21 = M. The coupling coefficient k = M/√(L1L2) lies between 0 and 1; tightly wound transformers approach k ≈ 1, while an LVDT varies M with core position. Mutual inductance is a property of geometry — it exists in DC circuits too, but induces an EMF only while a current changes.

Magnetic circuits. Ampère's law around a core gives NI = Φ × ℛ, where the reluctance ℛ = ℓ/(μA) plays the role of resistance. Then L = N²/ℛ. An air gap usually dominates the reluctance because μ₀ is thousands of times smaller than μ of iron — a small gap makes L stable against changes of μr and saturation, which is why gapped cores are used in inductors.

Energy. Building up current I in an inductor needs work W = ½LI², stored in the field with density w = ½B·H = ½μH² = B²/(2μ). In a gapped core most of the energy sits in the gap, where H is largest. For coupled coils, W = ½L1I1² + ½L2I2² ± MI1I2.

Internal and external inductance. For a round conductor, flux inside the wire adds an internal inductance μ/(8π) per metre (uniform current, low frequency). At high frequency the skin effect removes it.

Duality check. For any two-conductor transmission line in a uniform medium, (L per metre) × (C per metre) = με — a useful check on calculations.

Formulas

• L = NΦ / I = λ / I — self-inductance (H); Φ in Wb.
• M = N2 Φ12 / I1, k = M / √(L1 L2) — mutual inductance (H), coupling coefficient (0 ≤ k ≤ 1).
• L = μ₀μr N² A / ℓ — long solenoid; A = cross-section (m²), ℓ = length (m).
• L = μ N² h ln(b/a) / (2π) — toroid of rectangular cross-section, height h, radii a and b.
• L/ℓ = (μ / 2π) ln(b/a) — coaxial line, external inductance per metre (H/m).
• L/ℓ = μ / (8π) — internal inductance of a round wire per metre (uniform current).
• ℛ = ℓ / (μA) (A/Wb); NI = Φ Σℛ; L = N² / Σℛ — magnetic circuit.
• W = ½ L I², w = ½ μ H² = B² / (2μ) (J/m³).
• v = L di/dt, v2 = M di1/dt — circuit relations.

Worked examples

Example 1 (standard, solenoid). An air-core solenoid has N = 200 turns, A = 0.01 m², ℓ = 0.5 m. Find L and the energy at 5 A.

1. L = μ₀N²A/ℓ = 4π × 10⁻⁷ × 200² × 0.01 / 0.5 = 4π × 10⁻⁷ × 40 000 × 0.02 = 1.005 mH.
2. W = ½LI² = 0.5 × 1.005 × 10⁻³ × 25 = 12.6 mJ.
3. With a linear iron core of μr = 1000 (below saturation), L would be 1000 times larger, 1.005 H.

Example 2 (GATE level, gapped core). An iron core has mean length 0.4 m, cross-section 4 cm², μr = 2000, and a 1 mm air gap. It carries a 500-turn coil. Neglect fringing and leakage. Find L, the flux at 1 A, the stored energy and the share in the gap.

1. Core reluctance: ℛc = ℓ/(μ₀μr A) = 0.4 / (2000 × 4π × 10⁻⁷ × 4 × 10⁻⁴) = 3.98 × 10⁵ A/Wb.
2. Gap reluctance: ℛg = g/(μ₀A) = 0.001 / (4π × 10⁻⁷ × 4 × 10⁻⁴) = 1.99 × 10⁶ A/Wb.
3. Total ℛ = 2.39 × 10⁶ A/Wb. L = N²/ℛ = 250 000 / 2.387 × 10⁶ = 0.105 H.
4. At 1 A: Φ = NI/ℛ = 500 / 2.387 × 10⁶ = 0.209 mWb (B = 0.524 T); W = ½LI² = 52.4 mJ.
5. Energy divides like the reluctances (same flux through both): gap share = ℛg/ℛ = 83%. Without the gap, L would be 0.628 H — but it would change whenever μr changes.

Example 3 (coaxial line). For the coax with a = 0.5 mm, b = 1.75 mm (polyethylene, μr = 1), find the external inductance per metre.

1. L/ℓ = (μ₀/2π) ln(b/a) = 2 × 10⁻⁷ × ln 3.5 = 2 × 10⁻⁷ × 1.2528 = 0.251 μH/m.
2. Check with C/ℓ = 99.9 pF/m (εr = 2.25): L × C = 0.2506 × 10⁻⁶ × 99.9 × 10⁻¹² = 2.503 × 10⁻¹⁷ s²/m², and μ₀ε₀εr = 4π × 10⁻⁷ × 8.854 × 10⁻¹² × 2.25 = 2.503 × 10⁻¹⁷ ✓.

Common mistakes

• Using N instead of N² in L = μN²A/ℓ (one N makes the flux, the other links it).
• Forgetting to convert cm² to m² (×10⁻⁴) and mm to m.
• Treating μr of iron as constant into saturation — the inductance then falls.
• Adding reluctances in parallel when the core and gap are in series (same flux).
• Thinking mutual inductance exists only with AC; it exists always, but induces EMF only when current changes.
• Taking k > 1 or forgetting the sign (±M) depends on the dot convention.

For GATE IN

Typical items: L of solenoids, toroids and coaxial lines; magnetic circuits with an air gap (L, flux, energy, share of energy in the gap); mutual inductance and coupling coefficient; energy stored and energy density; series-aiding and series-opposing coils (L1 + L2 ± 2M), which is also how M is measured. Practise the reluctance method — it is fast and less error-prone than integrating fields.

Quick check

1. A coil's turns are doubled with the same geometry. How does L change?
2. What is the energy in a 10 mH inductor at 5 A?
3. L1 = 4 mH, L2 = 9 mH, M = 3 mH. What is k?
4. In a gapped core, where is most of the energy stored?
5. What is the SI unit of reluctance?

Answers: 1. It becomes four times larger. 2. 0.125 J. 3. 0.5. 4. In the air gap. 5. A/Wb (ampere-turns per weber, = H⁻¹).