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Iron ring: mmf, flux, field strength, permeability

An iron ring of mean length 1.2 m and cross sectional area of 0.005 m2^2 is wound with a coil of 900 turns. If a current of 2 A in the coil produces a flux density of 1.2 T in the iron ring, calculate: (i) the mmf (ii) total flux in the ring (iii) the magnetic field strength (iv) the relative permeability of iron at this flux density.

Given: l=1.2l = 1.2 m, A=0.005A = 0.005 m2^2, N=900N = 900, I=2I = 2 A, B=1.2B = 1.2 T.

(i) Mmf

F=NI=900×2=1800 ATF = NI = 900\times2 = 1800\ \text{AT}

(ii) Total flux

Φ=BA=1.2×0.005=6×10−3 Wb=6 mWb\Phi = BA = 1.2\times0.005 = 6\times10^{-3}\ \text{Wb} = 6\ \text{mWb}

(iii) Magnetic field strength

H=Fl=18001.2=1500 AT/mH = \frac{F}{l} = \frac{1800}{1.2} = 1500\ \text{AT/m}

(iv) Relative permeability

μr=Bμ0H=1.24π×10−7×1500=636.6\mu_r = \frac{B}{\mu_0 H} = \frac{1.2}{4\pi\times10^{-7}\times1500} = 636.6

Answer: mmf = 1800 AT; Φ\Phi = 6 mWb; HH = 1500 AT/m; μr≈637\mu_r \approx 637

✓ Answer checked by Claude Opus 5.5 (high effort)

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