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Magnetic circuit, similarities with electric circuit, Ohm's law

Short answer

A magnetic circuit is the closed path of magnetic flux. It is analogous to an electric circuit: flux φ corresponds to current, mmf F = NI to emf, reluctance S = l/(μA) to resistance. Ohm's law for the magnetic circuit states 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. It is usually made of high-permeability iron (sometimes with air gaps) and is excited by a coil carrying current, e.g. the core of a transformer, relay or electrical machine.

     +-----------------+
     |    phi -->      |
   ==|== N turns       |  iron core,
   ==|== I             |  mean length l,
     |                 |  area A
     +-----------------+

Similarities between magnetic and electric circuits

Magnetic circuitElectric circuit
Flux ϕ\phi (Wb)Current II (A)
MMF F=NIF = NI (AT)EMF EE (V)
Reluctance S=l/(μ0μrA)S = l/(\mu_0\mu_r A)Resistance R=ρl/AR = \rho l / A
Permeance =1/S= 1/SConductance =1/R= 1/R
Permeability μ\muConductivity σ\sigma
Flux density B=ϕ/AB = \phi/ACurrent density J=I/AJ = I/A
Magnetic field strength HHElectric field strength EE
ϕ=F/S\phi = F/SI=E/RI = E/R
Reluctances add in series; permeances add in parallelResistances add in series; conductances add in parallel
Kirchhoff's laws: ∑ϕ=0\sum \phi = 0 at a node, ∑F=∑ϕS\sum F = \sum \phi S in a loop∑I=0\sum I = 0, ∑E=∑IR\sum E = \sum IR

Points of difference: flux does not truly flow, no energy is spent to maintain a constant flux (but I2RI^2R loss occurs with current), and μ\mu is not constant because iron saturates, while R is nearly constant.

Ohm's law for a magnetic circuit

Take a uniform ring of mean length ll (m), area AA (m²), relative permeability μr\mu_r, wound with NN turns carrying II.

By Ampere's circuital law, ∮H dl=NI\oint H\,dl = NI, so

H=NIlH = \frac{NI}{l}

Then

B=μ0μrH=μ0μrNIlϕ=BA=μ0μrA NIlϕ=NIlμ0μrA=FS\begin{aligned} B &= \mu_0\mu_r H = \frac{\mu_0\mu_r NI}{l} \\ \phi &= BA = \frac{\mu_0\mu_r A\, NI}{l} \\ \phi &= \frac{NI}{\dfrac{l}{\mu_0\mu_r A}} = \frac{F}{S} \end{aligned}

So flux = mmf ÷ reluctance, which is Ohm's law for the magnetic circuit (I=E/RI = E/R). For a composite circuit (iron + air gap) the reluctances add: ϕ=NI/(Si+Sg)\phi = NI/(S_i + S_g).

✓ Answer checked by Claude Opus 5.5 (high effort)

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