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

# Mild steel ring: permeability and inductance

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2075 Baisakh (old course), 8 marks
- Video: [Magnetic Circuits - Magnetic Circuit with an Air Gap](https://www.youtube.com/watch?v=Up0E8Rc0i-c) — Energy Conversion Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/mild-steel-ring-permeability-inductance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A mild steel ring of 30 cm mean circumference has a cross-sectional area of 6 cm$^2$ 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; subtracting gap mmf (795.8 AT) leaves 1204.2 AT for iron. Relative permeability μr = B/(μ0 H) ≈ 198, and inductance L = NΦ/I = 75 mH.

## Answer

**Given:** mean circumference $= 30$ cm, $A = 6$ cm$^2 = 6\times10^{-4}$ m$^2$, $N = 500$, $l_g = 1$ mm, $I = 4$ A, $B_g = 1$ T.

#### (i) Relative permeability of the steel

Total mmf: $F = NI = 500\times4 = 2000$ AT.

Mmf used by the gap:

$$
F_g = \frac{B l_g}{\mu_0} = \frac{1\times0.001}{4\pi\times10^{-7}} = 795.8\ \text{AT}
$$

Mmf left for the iron: $F_i = 2000 - 795.8 = 1204.2$ AT. The iron length is $l_i = 0.30 - 0.001 = 0.299$ m.

$$
\begin{aligned}
H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\
\mu_r &= \frac{B}{\mu_0H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6
\end{aligned}
$$

#### (ii) Inductance

$$
\Phi = BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb}
$$

$$
L = \frac{N\Phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H}
$$

**Answer: $\mu_r \approx 198$; $L = 75$ mH**

## Check yourself

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

1. How is the iron mmf found?
   A. Total NI minus gap mmf B l_g/mu_0
   B. Total NI plus gap mmf
   C. Gap mmf only
   D. NI divided by 2
   Answer: A. F_i = 2000 - 795.8 = 1204.2 AT. (Key point: Gap mmf first)

2. How is mu_r of the steel found?
   A. H_i / (mu_0 B)
   B. B / (mu_0 H_i)
   C. B H_i / mu_0
   D. mu_0 / (B H_i)
   Answer: B. With H_i = 1204.2/0.299 = 4027.5 AT/m, mu_r = 1/(4 pi e-7 x 4027.5) = 198. (Key point: Permeability of steel)

3. How is the inductance obtained?
   A. L = N Phi / I
   B. L = Phi / (N I)
   C. L = I / (N Phi)
   D. L = N I Phi
   Answer: A. L = 500 x 6e-4 / 4 = 75 mH. (Key point: Inductance method)

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