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DC current for iron ring with 4 mm air gap

Short answer

B = 0.1/(5e-3) = 20 T. Air-gap mmf = (20/μ0)×0.004 = 63662 AT, iron mmf = (20/(μ0×2000))×1.496 = 11905 AT. Total mmf = 75567 AT; dc current I = 75567/500 ≈ 151 A.

An iron ring has a mean length of 1.5 m and cross-sectional area of 50 cm². It has a radial air gap of 4 mm. The ring is wound with 500 turns. What dc current would be needed in the coil to produce a flux of 100 mWb in the air gap? Assume that μr = 2000.

Required mmf = ampere-turns for the gap + ampere-turns for the iron, with the same flux density in both (no leakage or fringing). Iron length li=1.5−0.004=1.496l_i = 1.5 - 0.004 = 1.496 m.

Data: A=50A = 50 cm² =5×10−3= 5\times10^{-3} m², lg=4l_g = 4 mm, N=500N = 500, ϕ=100\phi = 100 mWb, μr=2000\mu_r = 2000.

Flux density

B=ϕA=0.15×10−3=20 TB = \frac{\phi}{A} = \frac{0.1}{5\times10^{-3}} = 20\ \text{T}

Air gap

Hg=Bμ0=204π×10−7=1.5915×107 AT/mATg=Hglg=1.5915×107×0.004=63662 AT\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{20}{4\pi\times10^{-7}} = 1.5915\times10^{7}\ \text{AT/m} \\ AT_g &= H_g l_g = 1.5915\times10^{7}\times0.004 = 63662\ \text{AT} \end{aligned}

Iron

Hi=Hgμr=1.5915×1072000=7958 AT/mATi=7958×1.496=11905 AT\begin{aligned} H_i &= \frac{H_g}{\mu_r} = \frac{1.5915\times10^{7}}{2000} = 7958\ \text{AT/m} \\ AT_i &= 7958\times1.496 = 11905\ \text{AT} \end{aligned}

Current

NI=63662+11905=75567 ATI=75567500=151.1 A\begin{aligned} NI &= 63662 + 11905 = 75567\ \text{AT} \\ I &= \frac{75567}{500} = 151.1\ \text{A} \end{aligned}

Answer: I ≈ 151 A (dc). (B = 20 T is far above what iron can carry; the figure assumes μr stays constant at 2000 as given.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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