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Relative permeability and inductance of mild steel ring

Short answer

Total mmf 2000 AT; gap mmf 796 AT leaves 1204 AT for iron. Iron H = 4027.5 AT/m, so μr = 198. Flux = 0.6 mWb, inductance L = Nφ/I = 0.075 H.

A mild steel ring of 30 cm mean circumference has a cross-sectional area of 6 cm² and has a winding of 500 turns on it. The ring is cut through at a point so as to provide an air gap of 1 mm in the magnetic circuit. It is found that a current of 4 A in the winding produces a flux density of 1 T in the air gap. Find (i) the relative permeability of the mild steel and (ii) inductance of the winding.

Total mmf = gap ampere-turns + iron ampere-turns. Iron length = 30 − 0.1 = 29.9 cm. Same B in iron and gap (no leakage or fringing).

Data: l=0.30l = 0.30 m, A=6A = 6 cm² =6×10−4= 6\times10^{-4} m², N=500N = 500, lg=1l_g = 1 mm, I=4I = 4 A, B=1B = 1 T.

(i) Relative permeability

NI=500×4=2000 ATATg=Bμ0lg=14π×10−7×10−3=795.8 ATATi=2000−795.8=1204.2 ATHi=1204.20.299=4027.5 AT/mμr=Bμ0Hi=14π×10−7×4027.5=197.6\begin{aligned} NI &= 500\times4 = 2000\ \text{AT} \\ AT_g &= \frac{B}{\mu_0} l_g = \frac{1}{4\pi\times10^{-7}}\times10^{-3} = 795.8\ \text{AT} \\ AT_i &= 2000 - 795.8 = 1204.2\ \text{AT} \\ H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\ \mu_r &= \frac{B}{\mu_0 H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6 \end{aligned}

(ii) Inductance

ϕ=BA=1×6×10−4=6×10−4 WbL=NϕI=500×6×10−44=0.075 H\begin{aligned} \phi &= BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb} \\ L &= \frac{N\phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H} \end{aligned}

Answer: μr ≈ 198, L = 0.075 H (75 mH). (Taking the iron length as the full 30 cm gives μr ≈ 198 as well.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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