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DC current for iron rod ring with 4 mm air gap

Short answer

With given 250 mWb, B = 0.25/(2e-4) = 1250 T. Gap mmf ≈ 3.98e6 AT, iron mmf ≈ 1.98e5 AT, total NI = 4.177e6 AT, I = 4.177e6/500 = 8354 A. Unrealistic; if flux is 250 μWb, B = 1.25 T, I ≈ 8.35 A.

An 80 cm long iron rod has cross-sectional area of 200 sq.mm. It is bent into a circular ring with an air gap of 4 mm and the ring is wound with 500 turns of winding. What dc current would be needed in the coil to produce a flux of 250 mWb in the air gap? Assume that μr = 4000.

When the rod is bent into a ring and a 4 mm gap is cut, the iron length is li=0.80−0.004=0.796l_i = 0.80 - 0.004 = 0.796 m. The same flux passes through iron and gap.

Data: A=200A = 200 mm² =2×10−4= 2\times10^{-4} m², lg=4l_g = 4 mm, N=500N = 500, μr=4000\mu_r = 4000, ϕ=250\phi = 250 mWb.

Flux density

B=ϕA=0.252×10−4=1250 TB = \frac{\phi}{A} = \frac{0.25}{2\times10^{-4}} = 1250\ \text{T}

Ampere-turns

ATg=Bμ0lg=12504π×10−7×0.004=3.979×106 ATATi=Bμ0μrli=9.947×1084000×0.796=1.979×105 ATNI=3.979×106+1.979×105=4.177×106 AT\begin{aligned} AT_g &= \frac{B}{\mu_0}l_g = \frac{1250}{4\pi\times10^{-7}}\times0.004 = 3.979\times10^{6}\ \text{AT} \\ AT_i &= \frac{B}{\mu_0\mu_r}l_i = \frac{9.947\times10^{8}}{4000}\times0.796 = 1.979\times10^{5}\ \text{AT} \\ NI &= 3.979\times10^{6} + 1.979\times10^{5} = 4.177\times10^{6}\ \text{AT} \end{aligned}

Current

I=4.177×106500=8354 AI = \frac{4.177\times10^{6}}{500} = 8354\ \text{A}

Answer (data as given): I ≈ 8354 A.

A flux density of 1250 T is physically impossible, so the flux is most likely a misprint for 250 μWb. Then B=1.25B = 1.25 T, ATg=3978.9AT_g = 3978.9 AT, ATi=197.9AT_i = 197.9 AT, NI=4176.8NI = 4176.8 AT and

I=4176.8500=8.35 AI = \frac{4176.8}{500} = 8.35\ \text{A}

which is the realistic answer (the method is identical; the current scales directly with flux).

✓ Answer checked by Claude Opus 5.5 (high effort)

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