---
title: "Inductance and average induced emf of iron ring coil"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:iron-rod-ring-inductance-average-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-rod-ring-inductance-average-emf
site: BCT Engineering
---

# Inductance and average induced emf of iron ring coil

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2069 Bhadra, 8 marks
- Video: [Self Inductance and Mutual Inductance Explained](https://www.youtube.com/watch?v=hoTInTKij0o) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-rod-ring-inductance-average-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A 50 cm long iron rod is bent into circular ring and 1000 turns of windings are wound on it. The diameter of the rod is 40 mm and relative permeability of the iron is 5000. Calculate the inductance of the coil. If a time varying current is passed through the coil whose magnitude changes from 2 amp to 10 amp in 5 ms, calculate the average value of emf induced in the coil.

## Short answer

Cross-sectional area = π(0.04)²/4 = 1.2566e-3 m². Inductance L = μ0×5000×1000²×1.2566e-3/0.5 = 15.79 H. Average emf = L×(10-2)/5e-3 = 15.79×1600 = 25.3 kV.

## Answer

Inductance of a toroid: $L = N^2/S = \mu_0\mu_r N^2 A/l$. Average emf = $L \times$ (change of current)/(time).

**Data:** $l = 0.5$ m, $N = 1000$, $d = 40$ mm, $\mu_r = 5000$.

#### Cross-sectional area

$$
A = \frac{\pi d^2}{4} = \frac{\pi(0.04)^2}{4} = 1.2566\times10^{-3}\ \text{m}^2
$$

#### Inductance

$$
\begin{aligned}
L &= \frac{\mu_0\mu_r N^2 A}{l} \\
&= \frac{4\pi\times10^{-7}\times5000\times(1000)^2\times1.2566\times10^{-3}}{0.5} \\
&= 15.79\ \text{H}
\end{aligned}
$$

#### Average induced emf

$$
\begin{aligned}
e_{avg} &= L\frac{\Delta I}{\Delta t} = 15.79\times\frac{10 - 2}{5\times10^{-3}} \\
&= 15.79\times1600 = 25266\ \text{V}
\end{aligned}
$$

**Answer: L ≈ 15.79 H; average emf ≈ 25.3 kV** (it opposes the rise of current, by Lenz's law). The very large value comes from the high μr and fast change of current; in practice saturation would reduce L.

## 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 ring coil?
   A. L = μ0 μr N A / l^2
   B. L = N / S
   C. L = μ0 μr N² l / A
   D. L = μ0 μr N² A / l
   Answer: D. L = N²/S with reluctance S = l/(μ0 μr A). (Key point: Inductance formula)

2. How is the cross-section found from the 40 mm diameter?
   A. A = π d = 0.1257 m²
   B. A = π d²/4 = 1.2566×10^-3 m²
   C. A = d² = 1.6×10^-3 m²
   D. A = π d²/4 with d in mm, = 1256.6 m²
   Answer: B. d = 0.04 m gives A = π(0.04)²/4. (Key point: Area)

3. What is the inductance?
   A. About 0.158 H
   B. About 1579 H
   C. About 15.79 H
   D. About 7.9 H
   Answer: C. L = 4π×10^-7 × 5000 × 10^6 × 1.2566×10^-3 / 0.5 ≈ 15.79 H. (Key point: Inductance value)

4. How is the average induced emf found and what is it?
   A. e = L × I = 15.79 × 10 ≈ 158 V
   B. e = L ΔI × Δt ≈ 0.63 V
   C. e = L × 2/0.005 ≈ 6.3 kV
   D. e = L ΔI/Δt = 15.79 × 8/0.005 ≈ 25.3 kV, opposing the rise of current
   Answer: D. ΔI = 10 − 2 = 8 A in 5 ms gives 1600 A/s; the emf opposes the change by Lenz's law. (Key point: Average emf)

## Related in Electrical Machines I

- [DC current for iron rod ring with 4 mm air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-rod-200sqmm-4mm-500turns-250mwb) — Electrical Machines I, chapter 1
- [Relative permeability and inductance of mild steel ring](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Amild-steel-ring-relative-permeability-inductance) — Electrical Machines I, chapter 1
- [Number of turns for iron ring with air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-ring-1p25m-turns-0p5wb) — Electrical Machines I, chapter 1, asked 2×
- [Flux and flux density in iron ring with 2 mm gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-ring-80cm-2mm-gap-1000turns) — Electrical Machines I, chapter 1, asked 2×
- [DC current for iron ring with 4 mm air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-ring-1p5m-4mm-500turns-current) — Electrical Machines I, chapter 1
- [DC current for ring core with 2 mm air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aring-5p5cm-3p5cm-airgap-0p3mwb) — Electrical Machines I, chapter 1

## Related in other subjects

- [Ring with sinusoidal current, inductance and emf](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aem-oscillating-ring-inductance) — Electrical Machines, chapter 1
- [Iron ring with air gap: current, inductance, energy](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Airon-ring-15cm-stored-energy) — Electrical Machines, chapter 1
- [Iron ring: mmf, flux, field strength, permeability](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Airon-ring-1-2m-mmf-flux) — Electrical Machines, chapter 1
- [Iron ring with 2 mm saw cut, exciting current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Airon-ring-300turns-gap-flux-density) — Electrical Machines, chapter 1