---
title: "Circular core, current for 1 T air gap (mean diameter)"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:core-airgap-current-4000-diameter"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/core-airgap-current-4000-diameter
site: BCT Engineering
---

# Circular core, current for 1 T air gap (mean diameter)

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2078 Baisakh, 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/core-airgap-current-4000-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A magnetic circuit consists of a circular iron core having mean diameter of 10 cm, cross sectional area of 100 mm$^2$ and air gap of 2 mm. The core has 600 turns of winding. Calculate the magnitude of current to be passed through the winding to produce air gap flux density of 1 Tesla. Given that relative permeability of the core is 4000.

## Short answer

Total mmf is found by adding iron and air-gap mmf (Bℓ/μ0μr and Bℓ/μ0). For a 10 cm mean diameter core with 600 turns and 1 T flux density, the required current is about 2.76 A.

## Answer

**Given:** mean diameter $D = 10$ cm, $A = 100$ mm$^2 = 10^{-4}$ m$^2$, $l_g = 2$ mm, $N = 600$, $B_g = 1$ T, $\mu_r = 4000$.

With no leakage or fringing, the flux density is the same (1 T) in iron and in the gap.

**Path lengths**

$$
l_i = \pi D - l_g = 0.31416 - 0.002 = 0.31216\ \text{m}
$$

**Mmf for each part**

$$
\begin{aligned}
F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.31216}{4\pi\times10^{-7}\times4000} = 62.10\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\
F &= 62.10 + 1591.5 = 1653.7\ \text{AT}
\end{aligned}
$$

**Current**

$$
I = \frac{F}{N} = \frac{1653.7}{600} = 2.756\ \text{A}
$$

The air gap needs about 96% of the mmf even though it is only 2 mm long.

**Answer: $I \approx 2.76$ A**

## Check yourself

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

1. What length of iron is used for the mean diameter 10 cm and 2 mm gap?
   A. pi D = 0.314 m
   B. D - l_g
   C. pi D - l_g = 0.312 m
   D. pi D + l_g
   Answer: C. The gap is cut from the ring, so l_i = 0.31416 - 0.002. (Key point: Iron path length)

2. Why can B = 1 T be used for both iron and gap?
   A. Iron has mu_r = 1
   B. Gap has larger area
   C. B is the same by definition
   D. Same flux and same area, with no leakage or fringing
   Answer: D. Equal flux through equal area gives equal B. (Key point: Same B)

3. About what fraction of the total mmf does the gap take?
   A. About 96%
   B. About 4%
   C. About 50%
   D. About 10%
   Answer: A. F_g = 1591.5 of 1653.7 AT, about 96%, since mu_r of iron is 4000. (Key point: Share of mmf)

4. How is the current found from the total mmf?
   A. I = N F
   B. I = F/N = 2.76 A
   C. I = F B
   D. I = N/F
   Answer: B. I = (62.1 + 1591.5)/600 = 2.76 A. (Key point: Current)

## Close variants set in other exams

- [Circular core, current for 1 T air gap (mean length)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Acore-airgap-current-4000-length) — Electrical Machines, chapter 1, asked 2×

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