---
title: "Flux density in steel ring with air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:steel-ring-12cm-550turns-flux-density"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/steel-ring-12cm-550turns-flux-density
site: BCT Engineering
---

# Flux density in steel ring with air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2073 Magh, 8 marks
- Video: [NEW TOPIC : IRON RING WITH AIRGAP | UNIT : MAGNETIC CIRCUIT IN HINDI (LECTURE 22)  @TIKLESACADEMY](https://www.youtube.com/watch?v=HZb3EGXqjoQ) — TIKLE'S 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/steel-ring-12cm-550turns-flux-density
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A steel ring of 12 cm mean radius and of circular cross section 1 cm in radius has an air gap of 2 mm length. It is wound uniformly with 550 turns of wire carrying 3 A. Neglecting magnetic leakage, calculate the magnetic flux density in the core. Given that relative permeability of the steel is 800.

## Short answer

Mean circumference 0.754 m, iron length 0.752 m, cross-sectional area 3.14×10⁻⁴ m². Applying NI = (B/μ₀)(lᵢ/μᵣ + l_g) gives B = (4π×10⁻⁷·550·3)/(0.752/800 + 0.002) = 0.705 T.

## Answer

Flux density follows from $NI = H_i l_i + H_g l_g$ with the same B in the steel and in the gap.

**Data:** mean radius $r = 0.12$ m, section radius 1 cm, $l_g = 2$ mm, $N = 550$, $I = 3$ A, $\mu_r = 800$.

#### Dimensions

$$
\begin{aligned}
l &= 2\pi r = 2\pi\times0.12 = 0.7540\ \text{m} \\
l_i &= 0.7540 - 0.002 = 0.7520\ \text{m} \\
A &= \pi (0.01)^2 = 3.1416\times10^{-4}\ \text{m}^2
\end{aligned}
$$

#### Flux density

$$
\begin{aligned}
NI &= \frac{B}{\mu_0}\left(\frac{l_i}{\mu_r} + l_g\right) \\
B &= \frac{\mu_0 NI}{\dfrac{l_i}{\mu_r} + l_g} = \frac{4\pi\times10^{-7}\times550\times3}{\dfrac{0.7520}{800} + 0.002} \\
&= \frac{2.0735\times10^{-3}}{9.40\times10^{-4} + 2\times10^{-3}} = \frac{2.0735\times10^{-3}}{2.940\times10^{-3}} \\
&= 0.705\ \text{T}
\end{aligned}
$$

Flux: $\phi = BA = 0.705\times3.1416\times10^{-4} = 2.22\times10^{-4}$ Wb.

**Answer: B ≈ 0.705 T** (flux ≈ 0.222 mWb).

## Check yourself

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

1. The solution uses NI = H_i l_i + H_g l_g with:
   A. Different B in steel and gap
   B. The same B in steel and gap
   C. B = 0 in the gap
   D. H the same in both
   Answer: B. No leakage means the same flux and area, so the same B; H differs by mu_r. (Key point: Governing equation)

2. For a circular section of radius 1 cm, A is:
   A. 3.14e-4 m^2
   B. 1e-4 m^2
   C. 6.28e-4 m^2
   D. 3.14e-2 m^2
   Answer: A. A = pi r^2 = pi x (0.01)^2 = 3.1416e-4 m^2. (Key point: Core area)

3. Solving for B gives:
   A. mu0 NI / (l_i/mu_r + l_g)
   B. mu0 NI (l_i/mu_r + l_g)
   C. NI / (mu0 l_i)
   D. mu0 mu_r NI / l_g
   Answer: A. NI = (B/mu0)(l_i/mu_r + l_g) rearranges to this. (Key point: Formula for B)

4. The flux density is about:
   A. 0.0705 T
   B. 0.705 T
   C. 7.05 T
   D. 1.41 T
   Answer: B. 2.0735e-3 / 2.94e-3 = 0.705 T. (Key point: Result)

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