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Core with 4 mm air gap: inductance and flux

Short answer

Iron and air-gap reluctances in series total 6.562×10⁶ AT/Wb. Then L = N²/S = 38.1 mH. With 10 A dc, flux Φ = NI/S = 0.762 mWb.

A circular iron core has a cross-sectional area of 5 sq.cm. and mean length of 25 cm including an air gap of 4 mm. The core is wound with 500 turns of winding. Calculate the inductance of the coil. If a dc current of 10 Ampere passed through the coil, calculate magnetic flux in the core. Given that relative permeability of the core is 2000.

Given: A=5A = 5 cm2=5×10−4^2 = 5\times10^{-4} m2^2, total mean length =25= 25 cm including the 4 mm gap, N=500N = 500, μr=2000\mu_r = 2000, I=10I = 10 A.

Lengths: lg=4l_g = 4 mm, li=0.25−0.004=0.246l_i = 0.25 - 0.004 = 0.246 m.

Reluctances

Si=0.2464π×10−7×2000×5×10−4=1.958×105 AT/WbSg=0.0044π×10−7×5×10−4=6.366×106 AT/WbS=6.562×106 AT/Wb\begin{aligned} S_i &= \frac{0.246}{4\pi\times10^{-7}\times2000\times5\times10^{-4}} = 1.958\times10^{5}\ \text{AT/Wb} \\ S_g &= \frac{0.004}{4\pi\times10^{-7}\times5\times10^{-4}} = 6.366\times10^{6}\ \text{AT/Wb} \\ S &= 6.562\times10^{6}\ \text{AT/Wb} \end{aligned}

Inductance

L=N2S=50026.562×106=0.0381 HL = \frac{N^2}{S} = \frac{500^2}{6.562\times10^{6}} = 0.0381\ \text{H}

Flux for 10 A

Φ=NIS=500×106.562×106=7.62×10−4 Wb\Phi = \frac{NI}{S} = \frac{500\times10}{6.562\times10^{6}} = 7.62\times10^{-4}\ \text{Wb}

The flux density is B=Φ/A=1.52B = \Phi/A = 1.52 T.

Answer: L≈38.1L \approx 38.1 mH; Φ≈0.762\Phi \approx 0.762 mWb

✓ Answer checked by Claude Opus 5.5 (high effort)

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