---
title: "Iron ring: mmf, flux, field strength, permeability"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:iron-ring-1-2m-mmf-flux"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/iron-ring-1-2m-mmf-flux
site: BCT Engineering
---

# Iron ring: mmf, flux, field strength, permeability

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2077 Chaitra, 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/iron-ring-1-2m-mmf-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

An iron ring of mean length 1.2 m and cross sectional area of 0.005 m$^2$ is wound with a coil of 900 turns. If a current of 2 A in the coil produces a flux density of 1.2 T in the iron ring, calculate: (i) the mmf (ii) total flux in the ring (iii) the magnetic field strength (iv) the relative permeability of iron at this flux density.

## Answer

**Given:** $l = 1.2$ m, $A = 0.005$ m$^2$, $N = 900$, $I = 2$ A, $B = 1.2$ T.

#### (i) Mmf

$$
F = NI = 900\times2 = 1800\ \text{AT}
$$

#### (ii) Total flux

$$
\Phi = BA = 1.2\times0.005 = 6\times10^{-3}\ \text{Wb} = 6\ \text{mWb}
$$

#### (iii) Magnetic field strength

$$
H = \frac{F}{l} = \frac{1800}{1.2} = 1500\ \text{AT/m}
$$

#### (iv) Relative permeability

$$
\mu_r = \frac{B}{\mu_0 H} = \frac{1.2}{4\pi\times10^{-7}\times1500} = 636.6
$$

**Answer: mmf = 1800 AT; $\Phi$ = 6 mWb; $H$ = 1500 AT/m; $\mu_r \approx 637$**

## Check yourself

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

1. How is the mmf calculated?
   A. F = NI = 1800 AT
   B. F = BA
   C. F = H / l
   D. F = N/I
   Answer: A. mmf = turns x current = 900 x 2. (Key point: Mmf)

2. What is the total flux?
   A. B/A = 240 Wb
   B. BA = 6 mWb
   C. B l = 1.44 Wb
   D. H A = 7.5 Wb
   Answer: B. Phi = B x A = 1.2 x 0.005 = 6 mWb. (Key point: Total flux)

3. How is H found?
   A. H = F x l
   B. H = B A
   C. H = F / l = 1500 AT/m
   D. H = F / A
   Answer: C. H is mmf per metre of path: 1800/1.2. (Key point: Field strength)

4. How is mu_r obtained?
   A. mu_r = mu_0 H / B
   B. mu_r = B / (mu_0 H)
   C. mu_r = H / B
   D. mu_r = B H
   Answer: B. B = mu_0 mu_r H, so mu_r = 1.2/(4 pi e-7 x 1500) = 637. (Key point: Relative permeability)

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