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Magnetic circuit, analogy with electric circuit

Short answer

A magnetic circuit is a closed flux path. Analogies: mmf ↔ emf, flux ↔ current, reluctance ↔ resistance. From B = μH, H = NI/l, Φ = BA = μA NI/l = NI/(l/(μA)), so Ohm's law: Φ = F/S.

Define magnetic circuit and hence list out the similarities between magnetic and electric circuits. Deduce Ohm's law for magnetic circuit.

A magnetic circuit is the closed path followed by magnetic flux, made up of ferromagnetic material (and possibly air gaps), in which the flux is set up by an mmf such as that of a current-carrying coil.

Similarities with an electric circuit

Electric circuitMagnetic circuit
emf EE (volt)mmf F=NIF = NI (ampere-turn)
Current II (A)Flux Φ\Phi (Wb)
Resistance R=ρlAR = \dfrac{\rho l}{A}Reluctance S=lμAS = \dfrac{l}{\mu A}
Conductivity σ\sigmaPermeability μ\mu
Current density JJFlux density BB
Ohm's law I=E/RI = E/RΦ=F/S\Phi = F/S
Series: R=R1+R2R = R_1 + R_2Series: S=S1+S2S = S_1 + S_2
Parallel: G=G1+G2G = G_1 + G_2Parallel: P=P1+P2P = P_1 + P_2
Kirchhoff's laws apply∑Φ=0\sum\Phi = 0 at a junction, ∑F=∑ΦS\sum F = \sum\Phi S around a loop

Differences: current flow dissipates I2RI^2R energy, but a constant flux needs no energy to maintain it. Resistance is nearly constant, but reluctance changes with BB because μ\mu is not constant (saturation). Flux leaks out of the core (leakage, fringing), but current can be kept inside the conductor.

Ohm's law for the magnetic circuit

H=NIlB=μH=μNIlΦ=BA=μAl NI=NIl/μA\begin{aligned} H &= \frac{NI}{l} \\ B &= \mu H = \frac{\mu NI}{l} \\ \Phi &= BA = \frac{\mu A}{l}\,NI = \frac{NI}{l/\mu A} \end{aligned} Φ=FSi.e. flux=mmfreluctance\Phi = \frac{F}{S}\qquad \text{i.e. flux} = \frac{\text{mmf}}{\text{reluctance}}

This is Ohm's law for a magnetic circuit: the flux set up is directly proportional to the mmf and inversely proportional to the reluctance of the path.

✓ Answer checked by Claude Opus 5.5 (high effort)

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