---
title: "Iron ring current for 0.82 Wb air-gap flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:iron-ring-air-gap-0-82-wb"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/iron-ring-air-gap-0-82-wb
site: BCT Engineering
---

# Iron ring current for 0.82 Wb air-gap flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2069 Poush
- 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-0-82-wb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

An iron ring has a mean length of 2.5 m and A = 0.04 m². It has a radial air gap of 5 mm. The ring is wound with 200 turns, what dc current would be needed in the coil to produce a flux of 0.82 Wb in the air gap. Assume that μr = 5000.

## Short answer

Iron reluctance = 2.495/(4π×10⁻⁷×5000×0.04) = 9.927×10³ AT/Wb, air gap reluctance = 0.005/(4π×10⁻⁷×0.04) = 9.947×10⁴ AT/Wb. Total S = 1.094×10⁵ AT/Wb. mmf = 0.82×1.094×10⁵ = 8.97×10⁴ AT. Current I = 8.97×10⁴/200 = 449 A.

## Answer

The mmf of the coil is the sum of the mmf for the iron part and the air gap, with the same flux through both.

**Given:** l = 2.5 m; A = 0.04 m²; gap lg = 5 mm; N = 200; Φ = 0.82 Wb; μr = 5000.

*Note:* B = 0.82/0.04 = 20.5 T is far beyond real saturation. The calculation follows the constant μr = 5000 given in the question.

**Step 1: reluctances** (iron length = 2.5 − 0.005 = 2.495 m)
```text
Sᵢ = 2.495/(4π×10⁻⁷ × 5000 × 0.04) = 9.927×10³ AT/Wb
Sg = 0.005/(4π×10⁻⁷ × 0.04)        = 9.947×10⁴ AT/Wb
S  = 1.0940×10⁵ AT/Wb
```

**Step 2: mmf**
F = Φ S = 0.82 × 1.0940×10⁵ = 8.971×10⁴ AT

**Step 3: current**
I = F/N = 8.971×10⁴/200 = 448.5 A

The air gap supplies 91 % of the mmf: 0.82 × 9.947×10⁴ = 81.6×10³ AT.

**Answer: I ≈ 449 A**

## Check yourself

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

1. What is the same in iron and gap?
   A. The flux (0.82 Wb)
   B. The mmf drop
   C. The permeability
   D. The reluctance
   Answer: A. Series circuit: same flux, different mmf drops. (Key point: Series flux)

2. Which part supplies most of the mmf?
   A. The 5 mm air gap (about 91%)
   B. Equal split
   C. The iron
   D. The winding
   Answer: A. Sg = 9.947x10^4 against Si = 9.927x10^3, so the gap takes about 91%. (Key point: Dominant part)

3. With total reluctance 1.094x10^5 and 200 turns, what current?
   A. About 449 A
   B. About 90 A
   C. About 4.5 A
   D. About 4490 A
   Answer: A. F = 0.82 x 1.094x10^5 = 8.97x10^4 AT; I = F/200 = 448.5 A. (Key point: Current)

## Close variants set in other exams

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