---
title: "Iron ring current for 0.8 Wb air-gap flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:iron-ring-air-gap-0-8-wb"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/iron-ring-air-gap-0-8-wb
site: BCT Engineering
---

# Iron ring current for 0.8 Wb air-gap flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2072 Magh
- 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_machine/iron-ring-air-gap-0-8-wb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

An iron ring has a mean length of 2 m and cross-sectional area of 0.01 m². It has a radial air gap of 4 mm. The ring is wound with 250 turns. What dc current would be needed in the coil to produce a flux of 0.8 Weber in the air gap? Assume that μr = 400.

## Short answer

Iron reluctance = 1.996/(4π×10⁻⁷×400×0.01) = 3.971×10⁵ AT/Wb, air gap reluctance = 0.004/(4π×10⁻⁷×0.01) = 3.183×10⁵ AT/Wb. Total S = 7.154×10⁵ AT/Wb. mmf = 0.8×7.154×10⁵ = 5.723×10⁵ AT. Current I = 5.723×10⁵/250 = 2.29×10³ A.

## Answer

The coil must supply the mmf for the iron and the air gap, both carrying 0.8 Wb.

**Given:** l = 2 m; A = 0.01 m²; gap lg = 4 mm; N = 250; Φ = 0.8 Wb; μr = 400.

*Note:* the figures give B = Φ/A = 80 T, which is physically impossible (iron saturates near 2 T). We follow the data as given, using the constant μr = 400, as the question intends.

**Step 1: reluctances** (iron length = 2 − 0.004 = 1.996 m)
```text
Sᵢ = 1.996/(4π×10⁻⁷ × 400 × 0.01) = 3.971×10⁵ AT/Wb
Sg = 0.004/(4π×10⁻⁷ × 0.01)       = 3.183×10⁵ AT/Wb
S  = Sᵢ + Sg = 7.154×10⁵ AT/Wb
```

**Step 2: mmf**
F = Φ × S = 0.8 × 7.154×10⁵ = 5.723×10⁵ AT

**Step 3: current**
I = F/N = 5.723×10⁵/250 = 2289 A

(If the whole 2 m is used as the iron length, S = 7.162×10⁵ AT/Wb and I = 2292 A, a negligible difference.)

**Answer: I ≈ 2.29×10³ A (about 2290 A)**

## Check yourself

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

1. The coil must supply mmf for which parts?
   A. The iron and the air gap
   B. The iron only
   C. The winding resistance
   D. The air gap only
   Answer: A. Both iron and gap carry the same flux, so their mmf drops add. (Key point: Mmf sum)

2. Which reluctances are used?
   A. Both with mur
   B. Both without mur
   C. Iron l/(mu0 A) and gap lg/(mu0 mur A)
   D. Iron l_i/(mu0 mur A) and gap lg/(mu0 A)
   Answer: D. The gap has mur = 1, while iron uses mur = 400. (Key point: Reluctance formula)

3. B = 0.8/0.01 = 80 T. How is this handled?
   A. mur is changed to 1
   B. It is noted as unphysical but the constant mur = 400 is used as the question intends
   C. B is limited to 2 T
   D. The data is rejected
   Answer: B. The solution follows the given data with constant mur. (Key point: Realism check)

4. Total mmf is about 5.72x10^5 AT with 250 turns. What current is needed?
   A. About 229 A
   B. About 5.7 A
   C. About 22.9 A
   D. About 2289 A
   Answer: D. I = 5.723x10^5/250 = 2289 A. (Key point: Current)

## Close variants set in other exams

- [Iron ring current for 0.82 Wb air-gap flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Airon-ring-air-gap-0-82-wb) — Electrical Machine, chapter 1, asked 2×

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