---
title: "Iron core current for 1 T air-gap flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:iron-core-air-gap-1-tesla"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/iron-core-air-gap-1-tesla
site: BCT Engineering
---

# Iron core current for 1 T air-gap flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2073 Magh, 10 marks; 2069 Bhadra (asked 2 times)
- Video: [Magnetic Circuits With Air Gap](https://www.youtube.com/watch?v=T99_3l0UqB8) — Benchmark Learn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/iron-core-air-gap-1-tesla
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

A magnetic circuit consists of a circular iron core having mean length of 10 cm and cross sectional area of 100 square mm. The air gap is 2 mm and the core has 600 turns of winding. Calculate the magnitude of current to be passed through the winding to produce air gap flux of 1 tesla (permeability of iron = 4000).

## Short answer

Summing ampere-turns for gap (B*lg/μ0=1591 AT) and iron (Bl_i/(μ0 μr)=19.9 AT) gives total 1611.4 AT. Current I=1611.4/600=2.69 A.

## Answer

The current needed is about **2.69 A**. The air gap needs almost all of the mmf because its reluctance is far larger than that of iron.

**Given (SI units):** iron length lᵢ = 10 cm = 0.1 m; area A = 100 mm² = 1×10⁻⁴ m²; air gap lg = 2 mm = 2×10⁻³ m; N = 600; μr = 4000; B = 1 T; μ0 = 4π×10⁻⁷ H/m.

*Assumption:* the 10 cm is the length of the iron path and the gap is added to it. Fringing is neglected, so the flux density is the same in iron and gap.

**Step 1: flux**
Φ = B × A = 1 × 1×10⁻⁴ = 1×10⁻⁴ Wb

**Step 2: reluctances**
```text
Sᵢ = lᵢ/(μ0 μr A) = 0.1/(4π×10⁻⁷ × 4000 × 10⁻⁴)
   = 1.989×10⁵ AT/Wb
Sg = lg/(μ0 A) = 2×10⁻³/(4π×10⁻⁷ × 10⁻⁴)
   = 1.592×10⁷ AT/Wb
S  = Sᵢ + Sg = 1.612×10⁷ AT/Wb
```

**Step 3: mmf and current**
```text
mmf = N I = Φ × S = 10⁻⁴ × 1.612×10⁷ = 1611.4 AT
I   = 1611.4 / 600 = 2.686 A
```
Check by the ampere-turn method: AT(gap) = (B/μ0) lg = 1591.5 AT, AT(iron) = (B/μ0μr) lᵢ = 19.9 AT, total 1611.4 AT.

**Answer: I ≈ 2.69 A**

## Check yourself

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

1. Why does almost all of the required mmf go to the 2 mm air gap?
   A. Iron has zero reluctance
   B. Air carries more flux than iron
   C. Air has a far larger reluctance than iron of similar size
   D. The gap has more turns of winding
   Answer: C. Reluctance is l/(mu0 mur A); air has mur = 1 versus 4000 for iron, so the gap's reluctance is much larger and it needs most of the mmf. (Key point: Air gap dominates mmf)

2. How is the total reluctance of the core with the gap found?
   A. Add the reciprocals of the reluctances
   B. Take the larger of the two
   C. Multiply the two reluctances
   D. Add iron and gap reluctances because they are in series
   Answer: D. The same flux passes through iron and gap, so they form a series magnetic circuit and S = Si + Sg. (Key point: Series circuit method)

3. With B = 1 T and A = 100 mm^2, what flux is used and how is the mmf found?
   A. Phi = 1x10^-2 Wb, and mmf = Phi x S
   B. Phi = 1x10^-4 Wb, and mmf = Phi x S
   C. Phi = 1 Wb, and mmf = B x l
   D. Phi = 1x10^-4 Wb, and mmf = Phi / S
   Answer: B. 100 mm^2 = 1x10^-4 m^2, so Phi = BA = 1x10^-4 Wb, and mmf = Phi x S = N I. (Key point: Flux and mmf)

4. What current results from mmf of about 1611 AT in the 600-turn winding?
   A. About 1611 A
   B. About 9.7 A
   C. About 0.37 A
   D. About 2.69 A
   Answer: D. I = mmf/N = 1611.4/600 = 2.69 A. (Key point: Final current)

## Related in Electrical Machine

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