---
title: "Ring with sinusoidal current, inductance and emf"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:em-oscillating-ring-inductance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/em-oscillating-ring-inductance
site: BCT Engineering
---

# Ring with sinusoidal current, inductance and emf

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2071 Magh (old course), 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/em-oscillating-ring-inductance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

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 $I = 5\sin 314t$ is passed through the winding. Calculate inductance and average value of the emf induced in the coil.

## 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.

## Answer

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 $e = -L\,di/dt$.

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

**Inductance**

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\
L &= \frac{\mu_0\mu_r N^2 A}{l} = \frac{4\pi\times10^{-7}\times4000\times600^2\times3.142\times10^{-4}}{0.3} \\
&= 1.895\ \text{H}
\end{aligned}
$$

**Induced emf**

$$
e = -L\frac{di}{dt} = -L\times5\times314\cos314t = -2975\cos314t\ \text{V}
$$

So the peak emf is $E_m = L\omega I_m = 1.895\times314\times5 = 2975$ V.

**Average value**

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

$$
E_{avg} = \frac{2}{\pi}E_m = \frac{2}{\pi}\times2975 = 1894\ \text{V}
$$

This is the same as $4fLI_m$ with $f = 314/2\pi \approx 50$ Hz.

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives the inductance of the coil on the iron ring?
   A. L = mu_0 mu_r N A / l
   B. L = mu_0 mu_r N^2 A / l
   C. L = N / (mu_0 mu_r A l)
   D. L = mu_0 N^2 l / A
   Answer: B. L = N^2 / S with S = l/(mu A), giving mu_0 mu_r N^2 A / l = 1.895 H. (Key point: Inductance formula)

2. For i = 5 sin 314t, what is the induced emf expression?
   A. e = -L x 5 sin 314t
   B. e = -L x 5 x 314 cos 314t
   C. e = L x 5 / 314
   D. e = -L x 314 sin 314t
   Answer: B. e = -L di/dt = -L x 5 x 314 cos 314t = -2975 cos 314t V. (Key point: Induced emf)

3. How is the average emf (half cycle) related to the peak?
   A. 2/pi x peak
   B. peak / sqrt(2)
   C. peak / 2
   D. equal to peak
   Answer: A. The half-cycle average of a sinusoid is 2/pi of the peak, giving about 1894 V. (Key point: Average value)

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