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Flux density in circular iron core

Short answer

With no air gap, H = NI/ℓ = 5000 AT/m. Assuming linear μr = 2000, B = μ0 μr H = 12.57 T. This is theoretical; a real core would saturate.

A circular iron core with mean length of 100 cm and cross-sectional area of 50 mm2^2 has 500 turns winding on the core. Calculate the flux density in the core if 10 A current flows in the coil. Take μr=2000\mu_r = 2000 for the iron core and neglect saturation.

With no air gap and a constant μr\mu_r (saturation neglected), the flux density follows directly from the field strength.

Given: l=100l = 100 cm, A=50A = 50 mm2^2 (not needed for BB), N=500N = 500, I=10I = 10 A, μr=2000\mu_r = 2000.

H=NIl=500×101.0=5000 AT/mB=μ0μrH=4π×10−7×2000×5000=12.57 T\begin{aligned} H &= \frac{NI}{l} = \frac{500\times10}{1.0} = 5000\ \text{AT/m} \\ B &= \mu_0\mu_r H = 4\pi\times10^{-7}\times2000\times5000 \\ &= 12.57\ \text{T} \end{aligned}

The corresponding flux is Φ=BA=12.57×50×10−6=6.28×10−4\Phi = BA = 12.57\times50\times10^{-6} = 6.28\times10^{-4} Wb.

A real iron core would saturate near 1.8 to 2 T, so this value is a theoretical result for the given linear assumption.

Answer: B≈12.57B \approx 12.57 T

✓ Answer checked by Claude Opus 5.5 (high effort)

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