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Circular core, current for 1 T air gap (mean diameter)

Short answer

Total mmf is found by adding iron and air-gap mmf (Bℓ/μ0μr and Bℓ/μ0). For a 10 cm mean diameter core with 600 turns and 1 T flux density, the required current is about 2.76 A.

A magnetic circuit consists of a circular iron core having mean diameter of 10 cm, cross sectional area of 100 mm2^2 and air gap of 2 mm. The core has 600 turns of winding. Calculate the magnitude of current to be passed through the winding to produce air gap flux density of 1 Tesla. Given that relative permeability of the core is 4000.

Given: mean diameter D=10D = 10 cm, A=100A = 100 mm2=10−4^2 = 10^{-4} m2^2, lg=2l_g = 2 mm, N=600N = 600, Bg=1B_g = 1 T, μr=4000\mu_r = 4000.

With no leakage or fringing, the flux density is the same (1 T) in iron and in the gap.

Path lengths

li=πD−lg=0.31416−0.002=0.31216 ml_i = \pi D - l_g = 0.31416 - 0.002 = 0.31216\ \text{m}

Mmf for each part

Fi=Bliμ0μr=0.312164π×10−7×4000=62.10 ATFg=Blgμ0=0.0024π×10−7=1591.5 ATF=62.10+1591.5=1653.7 AT\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.31216}{4\pi\times10^{-7}\times4000} = 62.10\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 62.10 + 1591.5 = 1653.7\ \text{AT} \end{aligned}

Current

I=FN=1653.7600=2.756 AI = \frac{F}{N} = \frac{1653.7}{600} = 2.756\ \text{A}

The air gap needs about 96% of the mmf even though it is only 2 mm long.

Answer: I≈2.76I \approx 2.76 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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