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Ring with sinusoidal current, inductance and emf

Short answer

Compute area A = π(0.02)²/4 = 3.142×10⁻⁴ m², inductance L = μ₀μ_r N² A / l = 1.895 H. Induced emf e = –L di/dt = –2975 cos 314t V; average value over half cycle = (2/π)×2975 = 1894 V.

A 30 cm long circular iron is bent into circular ring and 600 turns of windings are wound on it. The diameter of the rod is 20 mm and relative permeability of the iron is 4000. A time varying current I=5sin⁡314tI = 5\sin 314t is passed through the winding. Calculate inductance and average value of the emf induced in the coil.

A coil on an iron ring has a self-inductance fixed by its turns and the reluctance of the core. The emf induced then follows e=−L di/dte = -L\,di/dt.

Given: l=30l = 30 cm, N=600N = 600, d=20d = 20 mm, μr=4000\mu_r = 4000, i=5sin⁡314ti = 5\sin 314t A.

Inductance

A=π4(0.02)2=3.142×10−4 m2L=μ0μrN2Al=4π×10−7×4000×6002×3.142×10−40.3=1.895 H\begin{aligned} A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\ L &= \frac{\mu_0\mu_r N^2 A}{l} = \frac{4\pi\times10^{-7}\times4000\times600^2\times3.142\times10^{-4}}{0.3} \\ &= 1.895\ \text{H} \end{aligned}

Induced emf

e=−Ldidt=−L×5×314cos⁡314t=−2975cos⁡314t Ve = -L\frac{di}{dt} = -L\times5\times314\cos314t = -2975\cos314t\ \text{V}

So the peak emf is Em=LωIm=1.895×314×5=2975E_m = L\omega I_m = 1.895\times314\times5 = 2975 V.

Average value

The average of a sinusoid over a half cycle is 2π\dfrac{2}{\pi} of its peak (over a full cycle it is zero):

Eavg=2πEm=2π×2975=1894 VE_{avg} = \frac{2}{\pi}E_m = \frac{2}{\pi}\times2975 = 1894\ \text{V}

This is the same as 4fLIm4fLI_m with f=314/2π≈50f = 314/2\pi \approx 50 Hz.

Answer: L≈1.895L \approx 1.895 H; average emf ≈1894\approx 1894 V (peak 2975 V)

✓ Answer checked by Claude Opus 5.5 (high effort)

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