---
title: "Steel ring air gap total reluctance"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:steel-ring-air-gap-reluctance"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/steel-ring-air-gap-reluctance
site: BCT Engineering
---

# Steel ring air gap total reluctance

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2074 Bhadra, 5 marks
- Video: [Magnetic Circuits - Magnetic Circuit with an Air Gap](https://www.youtube.com/watch?v=Up0E8Rc0i-c) — Energy Conversion Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/steel-ring-air-gap-reluctance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/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 of current. The air gap takes 60% of total magnetomotive force. Find the total reluctance.

## Short answer

Total mmf = 550×3 = 1650 AT. Air gap mmf = 0.6×1650 = 990 AT. Air gap reluctance S_g = 2×10⁻³/(4π×10⁻⁷×3.142×10⁻⁴) = 5.066×10⁶ AT/Wb. Flux = 990/5.066×10⁶ = 1.954×10⁻⁴ Wb. Total reluctance = 1650/1.954×10⁻⁴ = 8.44×10⁶ AT/Wb.

## Answer

Total reluctance = total mmf ÷ total flux. The mmf is known, and the air-gap mmf (60 % of it) lets us find the flux from the air-gap reluctance.

**Given:** mean radius = 12 cm, so mean length l = 2π × 0.12 = 0.754 m; cross-section radius 1 cm, so A = π(0.01)² = 3.142×10⁻⁴ m²; gap lg = 2 mm; N = 550; I = 3 A.

**Step 1: total mmf**
F = N I = 550 × 3 = 1650 AT

**Step 2: air-gap mmf and reluctance**
```text
Fg = 0.6 × 1650 = 990 AT
Sg = lg/(μ0 A)
   = 2×10⁻³/(4π×10⁻⁷ × 3.142×10⁻⁴)
   = 5.066×10⁶ AT/Wb
```

**Step 3: flux**
Φ = Fg/Sg = 990 / 5.066×10⁶ = 1.954×10⁻⁴ Wb

**Step 4: total reluctance**
S = F/Φ = 1650 / 1.954×10⁻⁴ = 8.44×10⁶ AT/Wb

Check: since the gap takes 60 % of the mmf, S = Sg/0.6 = 5.066×10⁶/0.6 = 8.44×10⁶ AT/Wb. The steel part is 3.38×10⁶ AT/Wb (40 %).

**Answer: total reluctance ≈ 8.44×10⁶ AT/Wb (A/Wb)**

## Check yourself

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

1. How is total mmf found?
   A. B x l
   B. N x I x l
   C. N x I = 550 x 3 = 1650 AT
   D. N / I
   Answer: C. mmf = N I = 1650 AT. (Key point: Total mmf)

2. What is the air-gap mmf?
   A. 990 AT
   B. 1500 AT
   C. 660 AT
   D. 1650 AT
   Answer: A. 60% of 1650 = 990 AT. (Key point: Gap mmf)

3. How is the flux found?
   A. Gap mmf divided by gap reluctance
   B. Gap mmf times area
   C. Total mmf divided by length
   D. Total mmf times gap reluctance
   Answer: A. Phi = Fg/Sg = 990/5.066x10^6 = 1.954x10^-4 Wb. (Key point: Flux from the gap)

4. Since the gap takes 60% of the mmf, what is the total reluctance?
   A. Sg + 0.6
   B. Sg / 0.6 = 8.44x10^6 AT/Wb
   C. Sg / 0.4
   D. Sg x 0.6
   Answer: B. Same flux, so S = F/Phi = Sg/0.6 = 8.44x10^6 AT/Wb. (Key point: Shortcut check)

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