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Magnetic circuits and electromagnetic fields

Electrical and Electronics Engineering (B.E./B.Tech) · Circuits, fields and materials · free study notes for exams, placements and interviews

In short

A magnetic circuit is the closed path that flux follows through iron and air when a current-carrying coil drives it, analysed with an Ohm's-law analogy: MMF (ampere-turns) drives flux against reluctance. Electromagnetic field theory is the underlying physics - Coulomb, Gauss, Biot-Savart, Ampere and Faraday - from which inductance, capacitance, induced EMF and electromagnetic force all follow. Together they explain every transformer, motor, generator, relay, choke, solenoid and current sensor, and they decide core size, exciting current and iron loss. The topic carries high weight in university exams and in GATE EE (Electromagnetic Fields and Magnetic Circuits), and is standard interview ground for transformer and motor design, testing and protection roles at BHEL, Siemens, ABB, Crompton, Kirloskar Electric, Havells, Tata Power, NTPC and the state utilities.

Explained

Field quantities and the magnetisation (B-H) curve

Magnetic circuits: reluctance and the electric analogy

Air gaps, leakage and fringing

AC excitation: hysteresis, eddy currents and core loss

Electrostatic fields

Magnetostatics, Faraday's law and inductance

Key points

Formulas and examples

F = NI (ampere-turns, AT) · H = NI/l (A/m) · B = μ₀μ_r H (T) · Φ = BA (Wb)
S = l/(μ₀μ_r A) (AT/Wb) · Φ = F/S (Hopkinson) · ΣNI = ΣH·l around a loop · ΣΦ = 0 at a node
L = NΦ/I = N²/S = μ₀μ_r N²A/l (H) · M = N₁N₂/S · k = M/√(L₁L₂) · L_series = L₁+L₂±2M
e = −N dΦ/dt · e = Blv · E_rms = 4.44 f N Φ_m · F = BIl · T = NBIA
W = ½LI² = ½ΦNI ; energy density = B²/(2μ) J/m³ ; pull on one pole face F = B²A/(2μ₀) N
P_core = k_h f B_mn V + k_e f² B_m² t² V/ρ (hysteresis + eddy, W)
E = Q/(4πεr²) · E = λ/(2πεr) · V = −∫E·dl · C = εA/d · C_coax = 2πε/ln(b/a) F/m · W = ½CV²
∮D·dA = Q_enc · ∮H·dl = I_enc · ∇×E = −∂B/∂t · ∇×H = J + ∂D/∂t · δ = √(ρ/πfμ)
Symbols: N turns, I A, l m, A m², μ₀ = 4π×10⁻⁷ H/m, ε₀ = 8.854×10⁻¹² F/m, f Hz, t lamination thickness m, ρ resistivity Ω·m, σ conductivity S/m

Example 1 - series circuit with an air gap. A cast-steel ring, mean length 0.40 m, cross-section 2 cm², has a 2 mm radial air gap and a 500-turn coil. Required gap flux density 0.8 T; μ_r of steel = 1,000; neglect leakage and fringing. Find the coil current.

Example 2 - inductance and stored energy of the same circuit.

Example 3 - lifting magnet. A horseshoe electromagnet has two pole faces, each 25 cm², working at B = 1.0 T. Find the load it can hold.

Example 4 - separating core loss. At constant B_m, total core loss is 60 W at 50 Hz and 25 W at 25 Hz. Take P_h = Af and P_e = Bf².

Example 5 - skin depth in copper. ρ = 1.72 × 10⁻⁸ Ω·m, μ = μ₀.

Magnetic circuitElectric circuit
MMF F = NI (AT)EMF E (V)
Flux Φ (Wb)Current I (A)
Reluctance S = l/μAResistance R = ρl/A
Permeance 1/SConductance 1/R
Flux density B (T)Current density J (A/m²)
Permeability μ (varies with B)Conductivity σ (constant)
Φ = F/S (Hopkinson)I = E/R (Ohm)
Flux leaks through surrounding airCurrent confined to the conductor
No continuous energy needed to maintain ΦI²R loss continuous
QuantityElectrostaticMagnetostatic
SourceCharge QCurrent I (no magnetic monopole)
FieldE = Q/4πεr²H = I/2πr (long wire)
Flux law∮D·dA = Q_enc∮B·dA = 0
Circulation∮E·dl = 0 (static)∮H·dl = I_enc
Material constantε = ε₀ε_rμ = μ₀μ_r
Circuit elementCapacitance C = Q/VInductance L = NΦ/I
Energy density½εE²½B²/μ

Where it is used

Interview questions

Common mistakes

Related topics

Network theorems and circuit analysis · AC circuits and three-phase systems · Electrical engineering materials · Transformers · DC machines and induction motors · Electrical measurements and instrumentation

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