---
title: "Current for wrought iron ring with and without air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:wrought-iron-bar-ring-600turns"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/wrought-iron-bar-ring-600turns
site: BCT Engineering
---

# Current for wrought iron ring with and without air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2076 Bhadra, 8 marks
- Video: [EE213 - 03 - Analysis of magnetic circuits - example](https://www.youtube.com/watch?v=HRGsgPCZFXg) — MAFarooqi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/wrought-iron-bar-ring-600turns
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A wrought iron bar of 30 cm long and 2 cm in diameter is bent into circular shape. It is wound 600 turns of winding. Calculate the current required to produce a flux of 0.05 mWb in the core in the following two cases: a) with no air gap b) with an air gap of 4 mm. Given that μr = 4000.

## Short answer

For B=0.159 T: (a) no air gap, mmf = 9.50 AT, current = 15.8 mA. (b) with 4 mm gap, iron mmf 9.37 AT plus gap mmf 506.6 AT gives total 516 AT, current required = 0.86 A.

## Answer

The bar length becomes the mean length of the ring. Area is that of a 2 cm diameter circle.

$$
A = \frac{\pi d^2}{4} = \frac{\pi (0.02)^2}{4} = 3.1416\times10^{-4}\ \text{m}^2
$$

$$
B = \frac{\phi}{A} = \frac{0.05\times10^{-3}}{3.1416\times10^{-4}} = 0.1592\ \text{T}
$$

#### a) No air gap ($l_i = 0.30$ m)

$$
\begin{aligned}
H_i &= \frac{B}{\mu_0\mu_r} = \frac{0.1592}{4\pi\times10^{-7}\times4000} = 31.66\ \text{AT/m} \\
NI &= H_i l_i = 31.66\times0.30 = 9.50\ \text{AT} \\
I &= \frac{9.50}{600} = 0.0158\ \text{A}
\end{aligned}
$$

**Answer (a): I ≈ 15.8 mA.**

#### b) With a 4 mm air gap

The gap is cut from the same bar, so iron length $l_i = 0.30 - 0.004 = 0.296$ m.

$$
\begin{aligned}
H_g &= \frac{B}{\mu_0} = \frac{0.1592}{4\pi\times10^{-7}} = 126651\ \text{AT/m} \\
AT_g &= 126651\times0.004 = 506.6\ \text{AT} \\
AT_i &= 31.66\times0.296 = 9.37\ \text{AT} \\
NI &= 506.6 + 9.37 = 516.0\ \text{AT} \\
I &= \frac{516.0}{600} = 0.860\ \text{A}
\end{aligned}
$$

**Answer (b): I ≈ 0.86 A.**

A gap of only 4 mm raises the required current about 54 times, because the reluctance of air is 4000 times that of the same length of iron.

## Check yourself

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

1. For a 2 cm diameter bar and 0.05 mWb, the flux density is about:
   A. 0.0159 T
   B. 0.159 T
   C. 1.59 T
   D. 15.9 T
   Answer: B. A = 3.1416e-4 m^2, so B = 0.05e-3 / 3.1416e-4 = 0.159 T. (Key point: Core area and B)

2. Without a gap, NI = H_i l_i = 9.5 AT. The current is:
   A. 15.8 mA
   B. 0.86 A
   C. 1.58 A
   D. 9.5 A
   Answer: A. I = 9.5/600 = 0.0158 A. (Key point: No-gap current)

3. With a 4 mm gap the ampere-turns are dominated by:
   A. The iron
   B. The air gap (about 507 AT)
   C. The winding resistance
   D. Leakage
   Answer: B. H_g = B/mu0 is 4000 times H_i, so the gap needs about 506.6 AT against 9.4 AT for the iron. (Key point: Gap effect)

4. In part (b) the iron length used is:
   A. 0.304 m
   B. 0.30 m
   C. 0.296 m
   D. 0.004 m
   Answer: C. The gap is cut from the same bar, so l_i = 0.30 - 0.004 = 0.296 m, giving I = 0.86 A. (Key point: Iron length with gap)

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