---
title: "Relative permeability and inductance of mild steel ring"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:mild-steel-ring-relative-permeability-inductance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/mild-steel-ring-relative-permeability-inductance
site: BCT Engineering
---

# Relative permeability and inductance of mild steel ring

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2075 Baisakh, 8 marks
- Video: [54 - Solved Problems on Magnetic Circuits](https://www.youtube.com/watch?v=iOhQn_EtvIU) — SkanCity Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/mild-steel-ring-relative-permeability-inductance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A mild steel ring of 30 cm mean circumference has a cross-sectional area of 6 cm² and has a winding of 500 turns on it. The ring is cut through at a point so as to provide an air gap of 1 mm in the magnetic circuit. It is found that a current of 4 A in the winding produces a flux density of 1 T in the air gap. Find (i) the relative permeability of the mild steel and (ii) inductance of the winding.

## Short answer

Total mmf 2000 AT; gap mmf 796 AT leaves 1204 AT for iron. Iron H = 4027.5 AT/m, so μr = 198. Flux = 0.6 mWb, inductance L = Nφ/I = 0.075 H.

## Answer

Total mmf = gap ampere-turns + iron ampere-turns. Iron length = 30 − 0.1 = 29.9 cm. Same B in iron and gap (no leakage or fringing).

**Data:** $l = 0.30$ m, $A = 6$ cm² $= 6\times10^{-4}$ m², $N = 500$, $l_g = 1$ mm, $I = 4$ A, $B = 1$ T.

#### (i) Relative permeability

$$
\begin{aligned}
NI &= 500\times4 = 2000\ \text{AT} \\
AT_g &= \frac{B}{\mu_0} l_g = \frac{1}{4\pi\times10^{-7}}\times10^{-3} = 795.8\ \text{AT} \\
AT_i &= 2000 - 795.8 = 1204.2\ \text{AT} \\
H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\
\mu_r &= \frac{B}{\mu_0 H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6
\end{aligned}
$$

#### (ii) Inductance

$$
\begin{aligned}
\phi &= BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb} \\
L &= \frac{N\phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H}
\end{aligned}
$$

**Answer: μr ≈ 198, L = 0.075 H (75 mH).** (Taking the iron length as the full 30 cm gives μr ≈ 198 as well.)

## Check yourself

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

1. To find the iron's H, first:
   A. Divide NI by 2
   B. Subtract the gap ampere-turns (795.8) from the total NI (2000)
   C. Add them
   D. Use mu_r = 1
   Answer: B. AT_i = 2000 - 795.8 = 1204.2 AT is left for the iron. (Key point: Subtracting gap mmf)

2. With iron length 0.299 m, H_i is about:
   A. 403 AT/m
   B. 4027 AT/m
   C. 40 275 AT/m
   D. 1204 AT/m
   Answer: B. 1204.2 / 0.299 = 4027.5 AT/m. (Key point: Iron H)

3. mu_r = B/(mu0 H_i) is about:
   A. 19.8
   B. 198
   C. 1976
   D. 19 760
   Answer: B. 1/(4 pi e-7 x 4027.5) = 197.6. (Key point: Relative permeability)

4. L = N phi / I with phi = BA = 6e-4 Wb gives:
   A. 0.075 H
   B. 0.75 H
   C. 7.5 H
   D. 0.0075 H
   Answer: A. L = 500 x 6e-4 / 4 = 0.075 H. (Key point: Inductance)

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