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Electromagnet core with radial gap, DC current

Short answer

Area 7.854×10⁻⁵ m² and flux 0.2 mWb give B = 2.546 T. Iron and gap mmf sum to 6257 AT, so with 2000 turns the required dc current is 3.13 A.

The core of an electromagnet is made of an iron rod 1 cm diameter, bent into a circle of mean diameter 10 cm, a radial air gap of 1 mm being left between the ends of the rod. Calculate the direct current needed in coil of 2000 turns uniformly spaced around the core to produce a magnetic flux of 0.2 mWb in the air gap. Assume that the relative permeability of the iron is 150.

Given: rod diameter =1= 1 cm, mean diameter of ring =10= 10 cm, lg=1l_g = 1 mm, N=2000N = 2000, Φ=0.2\Phi = 0.2 mWb, μr=150\mu_r = 150.

Area, flux density, iron length

A=π4(0.01)2=7.854×10−5 m2B=ΦA=0.2×10−37.854×10−5=2.546 Tli=π×0.10−0.001=0.3132 m\begin{aligned} A &= \frac{\pi}{4}(0.01)^2 = 7.854\times10^{-5}\ \text{m}^2 \\ B &= \frac{\Phi}{A} = \frac{0.2\times10^{-3}}{7.854\times10^{-5}} = 2.546\ \text{T} \\ l_i &= \pi\times0.10 - 0.001 = 0.3132\ \text{m} \end{aligned}

Mmf

Fi=Bliμ0μr=2.546×0.31324π×10−7×150=4230.6 ATFg=Blgμ0=2.546×0.0014π×10−7=2026.4 ATF=6257.0 AT\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{2.546\times0.3132}{4\pi\times10^{-7}\times150} = 4230.6\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{2.546\times0.001}{4\pi\times10^{-7}} = 2026.4\ \text{AT} \\ F &= 6257.0\ \text{AT} \end{aligned}

Current

I=6257.02000=3.129 AI = \frac{6257.0}{2000} = 3.129\ \text{A}

The flux density of 2.55 T is beyond the saturation level of real iron, so the given μr=150\mu_r = 150 is a data assumption of the problem.

Answer: I≈3.13I \approx 3.13 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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