---
title: "Magnetizing current for ring with 5 mm air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:ring-airgap-magnetizing-current-1mwb"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ring-airgap-magnetizing-current-1mwb
site: BCT Engineering
---

# Magnetizing current for ring with 5 mm air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2081 Chaitra (new course), 6 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_i/ring-airgap-magnetizing-current-1mwb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A circular ring of magnetic material has a mean length of 0.8 m, cross sectional area of 120 cm² and an air gap of 5 mm. Calculate the magnetizing current required to produce a flux of 1 mWb in the air gap if the ring is wound uniformly with a coil of 220 turns. Take relative permeability of the ring material = 500 and neglect leakage and fringing.

## Short answer

Flux density B = 0.0833 T; air‑gap AT = 331.6 AT, iron AT = 105.4 AT; total AT = 437. With 220 turns, magnetizing current I ≈ 1.99 A.

## Answer

With leakage and fringing neglected, the same flux passes through the iron and the gap, and the area is the same. Iron path length is taken as $l_i = 0.8 - 0.005 = 0.795$ m.

**Data:** $\phi = 1$ mWb, $A = 120$ cm² $= 0.012$ m², $l_g = 5$ mm, $N = 220$, $\mu_r = 500$.

#### Flux density

$$
B = \frac{\phi}{A} = \frac{1\times10^{-3}}{0.012} = 0.0833\ \text{T}
$$

#### Ampere-turns for the air gap

$$
\begin{aligned}
H_g &= \frac{B}{\mu_0} = \frac{0.0833}{4\pi\times10^{-7}} = 66315\ \text{AT/m} \\
AT_g &= H_g l_g = 66315 \times 0.005 = 331.6\ \text{AT}
\end{aligned}
$$

#### Ampere-turns for the iron

$$
\begin{aligned}
H_i &= \frac{B}{\mu_0\mu_r} = \frac{66315}{500} = 132.6\ \text{AT/m} \\
AT_i &= H_i l_i = 132.6 \times 0.795 = 105.4\ \text{AT}
\end{aligned}
$$

#### Magnetising current

$$
\begin{aligned}
NI &= 331.6 + 105.4 = 437.0\ \text{AT} \\
I &= \frac{437.0}{220} = 1.986\ \text{A}
\end{aligned}
$$

**Answer: magnetising current ≈ 1.99 A.** (Taking the iron length as the full 0.8 m gives 1.989 A, practically the same.) Note that the 5 mm gap needs about three times the ampere-turns of the 0.795 m of iron.

## Check yourself

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

1. Neglecting leakage and fringing, the flux density in the iron and gap is:
   A. Different, since the gap is air
   B. Same, B = phi/A = 0.0833 T
   C. Zero in the gap
   D. Larger in the gap by mu_r
   Answer: B. Same flux through the same area gives the same B in iron and gap. (Key point: Flux density)

2. Which expression gives the gap ampere-turns?
   A. (B/mu0) x l_g
   B. (B/(mu0 mu_r)) x l_g
   C. B x l_g
   D. mu0 B l_g
   Answer: A. In air H = B/mu0, so AT_g = H_g l_g = 331.6 AT. (Key point: Air-gap ampere-turns)

3. What are the approximate total ampere-turns and the current?
   A. 105 AT, 0.48 A
   B. 331.6 AT, 1.5 A
   C. 437 AT, 1.99 A
   D. 437 AT, 0.437 A
   Answer: C. NI = 331.6 + 105.4 = 437 AT, and I = 437/220 = 1.99 A. (Key point: Total current)

4. Compared with the 0.795 m of iron, the 5 mm gap needs:
   A. Much fewer ampere-turns
   B. About the same
   C. About three times as many ampere-turns
   D. About 500 times as many
   Answer: C. 331.6 AT for the gap against 105.4 AT for the iron, about three times. (Key point: Gap versus iron)

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