Magnetic circuits and electromechanical energy conversion

Why it matters

Every transformer, motor, generator, relay, solenoid valve and LVDT is a magnetic circuit with one or more coils. Sizing the coil current for a required flux, predicting how an air gap changes inductance, and finding the force on a moving iron part are the first calculations in machine design and in electromagnetic sensors. The same energy ideas explain why motors produce torque at all.

Key ideas

Magnetic field quantities

• Magnetic field intensity H (A/m) is set by the current: Ampère's circuital law says the line integral of H around a closed path equals the total current enclosed, N·I for a coil of N turns.
• Flux density B (tesla, T = Wb/m²) is the material's response: B = μ·H = μ0·μr·H, with μ0 = 4π × 10⁻⁷ H/m.
• Flux Φ (weber) is the flux density integrated over an area; for a uniform field Φ = B·A.

Magnetic circuit analogy. When flux is confined to a core of uniform section, Ampère's law becomes Ohm's-law-like:

• MMF F = N·I (ampere-turns, At) plays the role of EMF.
• Flux Φ plays the role of current.
• Reluctance R = l/(μ·A) (At/Wb, i.e. H⁻¹) plays the role of resistance; permeance P = 1/R plays the role of conductance, and permeability μ is the analogue of conductivity.
• Reluctances in series add; parallel paths combine like parallel resistors. A "Kirchhoff's flux law" holds at every junction (flux is continuous because ∇·B = 0).

Why air gaps dominate. Iron has μr of a few hundred to several thousand, so a 1 mm air gap can have more reluctance than 40 cm of iron. Machines need an air gap for rotation, so most of the MMF is spent across it. Gaps also linearise the circuit, which is why inductors for filters are gapped.

Limits of the analogy.

• μ of iron is not constant: the B–H curve saturates (around 1.5–1.8 T for silicon steel), so in real problems R depends on Φ and you read H from the B–H curve.
• Leakage flux takes paths outside the core, and fringing at a gap spreads the flux so the effective gap area is larger than the pole face. Simple problems neglect both — say so.
• Iron losses: hysteresis loss (area of the B–H loop, proportional to f·Bmax^n with n ≈ 1.6–2) and eddy-current loss (proportional to f²·Bmax²·t² for lamination thickness t). Silicon steel and thin insulated laminations reduce them.

Inductance and stored energy. For a linear circuit, flux linkage λ = N·Φ = L·i, so L = N²/R. Energy stored is W = ½·L·i² = ½·λ·i; per unit volume it is B²/(2μ), which is why almost all the energy of a gapped core sits in the gap.

Electromechanical energy conversion. Conversion happens through a coupling magnetic field. Energy balance for a lossless coupling field: electrical energy in = increase in field energy + mechanical work out. For a singly excited system with movable iron:

• Force or torque comes from the change of stored energy (or co-energy) with position: at constant current, F = ½·i²·dL/dx and T = ½·i²·dL/dθ. The part always moves to increase inductance, i.e. to reduce reluctance — this is reluctance torque, used in stepper and reluctance motors.
• On a current-carrying conductor in a field, force F = B·I·l (Lorentz force; motor action). A conductor moving through a field has EMF e = B·l·v (generator action, Faraday's law e = −dλ/dt). Every machine does both at once: the motor generates a back EMF and the generator feels an opposing torque.

Formulas

• F = N·I — MMF in ampere-turns (At); N turns, I in A.
• H = N·I / l — field intensity (A/m) in a uniform core of mean length l (m).
• B = μ0·μr·H — flux density (T); μ0 = 4π × 10⁻⁷ H/m; μr dimensionless. Valid only below saturation.
• Φ = B·A — flux (Wb); A in m², uniform B.
• R = l / (μ0·μr·A) — reluctance (At/Wb or H⁻¹).
• Φ = F / ΣR — series magnetic circuit.
• L = N² / R = N·Φ / I — inductance (H), linear circuit.
• W = ½·L·I² ; w = B² / (2μ0) — stored energy (J) and energy density in air (J/m³).
• F_pull = B²·A / (2μ0) — pull on one pole face of area A (m²) across an air gap (N).
• T = ½·I²·dL/dθ — reluctance torque (N·m), linear circuit, constant current.
• e = B·l·v ; F = B·I·l — motional EMF (V) and force on a conductor (N), with B, l and v mutually perpendicular.

Worked examples

Example 1 (standard): core with an air gap. Given: iron core mean length 0.4 m, μr = 2000, cross-section 4 cm² = 4 × 10⁻⁴ m², air gap 1 mm, coil N = 500 turns. Required gap flux density 1.0 T. Neglect leakage and fringing.

1. Flux: Φ = B·A = 1.0 × 4 × 10⁻⁴ = 4 × 10⁻⁴ Wb.
2. Core reluctance: R_c = l/(μ0·μr·A) = 0.4 / (4π × 10⁻⁷ × 2000 × 4 × 10⁻⁴) = 3.98 × 10⁵ At/Wb.
3. Gap reluctance: R_g = l_g/(μ0·A) = 1 × 10⁻³ / (4π × 10⁻⁷ × 4 × 10⁻⁴) = 1.99 × 10⁶ At/Wb.
4. Total R = 2.39 × 10⁶ At/Wb; MMF = Φ·R = 4 × 10⁻⁴ × 2.39 × 10⁶ = 955 At.
5. Current: I = 955 / 500 = 1.91 A. Inductance L = N²/R = 500² / 2.39 × 10⁶ = 0.105 H. Note that the 1 mm gap takes about 83 % of the MMF.

Example 2 (GATE level): lifting electromagnet. Given: a U-shaped electromagnet lifts an iron armature across two air gaps, each 2 mm long with pole-face area 10 cm² = 1 × 10⁻³ m². The gap flux density must be 0.8 T. Neglect iron reluctance, leakage and fringing. Find the MMF and the lifting force.

1. MMF for two gaps in series: F = 2·(B/μ0)·l_g = 2 × (0.8 / 4π × 10⁻⁷) × 2 × 10⁻³ = 2546 At (e.g. 2.55 A in a 1000-turn coil).
2. Force per pole face: F_pull = B²·A/(2μ0) = 0.8² × 1 × 10⁻³ / (2 × 4π × 10⁻⁷) = 254.6 N.
3. Two pole faces: total force = 509 N (about 52 kg).

Common mistakes

• Using cm² or mm in the reluctance formula without converting to m² and m — the answer is off by 10⁴ or 10³.
• Forgetting μr for iron or, conversely, multiplying the air-gap reluctance by μr.
• Arithmetic with powers of ten: 0.5/(4π × 10⁻⁷ × 0.01) is 3.98 × 10⁷ At/Wb, not 3.98 × 10⁶.
• Counting only one air gap in an electromagnet that has two (flux crosses the armature twice).
• Using constant μ for iron in saturation; above the knee you must use the B–H curve.
• Confusing flux Φ (Wb) with flux density B (T), and permeability (material property) with permeance (circuit property).
• Mixing up force sign: a magnetic system always pulls to reduce reluctance (increase inductance).

For GATE IN

Expect short numericals on series and parallel magnetic circuits with an air gap, inductance of a gapped core (L = N²/R) and how it changes with gap length, energy stored in the gap, and force on a plunger or armature from F = B²A/(2μ0) or ½i²dL/dx. The variable-reluctance and variable-gap inductive transducers in the Measurements paper use exactly these relations, so practise sensitivity dL/dx too.

Quick check

1. Doubling the air gap (iron reluctance negligible) changes the inductance by what factor?
2. What is the unit of reluctance?
3. A conductor 0.2 m long moves at 10 m/s perpendicular to a 0.5 T field. What EMF is induced?
4. In which direction does a movable iron plunger move when the coil is energised?

Answers: 1. It halves (L = N²/R and R doubles). 2. Ampere-turns per weber (At/Wb, equivalently H⁻¹). 3. e = B·l·v = 1 V. 4. Towards the position of lower reluctance (higher inductance).