---
title: "Iron ring reluctance and flux with saw cut"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:ring-reluctance-flux-saw-cut"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/ring-reluctance-flux-saw-cut
site: BCT Engineering
---

# Iron ring reluctance and flux with saw cut

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2071 Bhadra, 8 marks
- Video: [Iron ring | Electromagnetism | Magnetic Circuit | Problem 1](https://www.youtube.com/watch?v=bjzbJLAnf4M) — Aryabhat Institute
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/ring-reluctance-flux-saw-cut
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

An iron ring of mean diameter 100 cm and cross sectional area 10 cm² is wound with 1000 turns and has μr = 2000. Compute (i) reluctance (ii) flux produced when the current through the coil is 1 A (iii) flux in the ring if a saw cut of 1 mm length is made, the current through the coil remaining the same.

## Short answer

(i) Reluctance = 3.1416/(4π×10⁻⁷×2000×10⁻³) = 1.25×10⁶ AT/Wb. (ii) Flux = 1000/1.25×10⁶ = 0.8 mWb. (iii) With 1 mm gap, S_g = 7.958×10⁵ AT/Wb, total S = 2.046×10⁶ AT/Wb, flux = 0.489 mWb.

## Answer

**Given:** mean diameter 100 cm, so l = πD = 3.1416 m; A = 10 cm² = 1×10⁻³ m²; N = 1000; μr = 2000; I = 1 A; saw cut 1 mm. (The 1 mm removed from the iron path is ignored in the iron length; the change is 0.03 %.)

#### (i) Reluctance of the ring
```text
S = l/(μ0 μr A)
  = 3.1416/(4π×10⁻⁷ × 2000 × 10⁻³)
  = 1.25×10⁶ AT/Wb
```

#### (ii) Flux with 1 A
```text
mmf = N I = 1000 × 1 = 1000 AT
Φ   = mmf/S = 1000/1.25×10⁶ = 8.0×10⁻⁴ Wb = 0.8 mWb
```

#### (iii) Flux with the saw cut
The saw cut is an air gap in series with the iron.
```text
Sg      = lg/(μ0 A) = 10⁻³/(4π×10⁻⁷ × 10⁻³)
        = 7.958×10⁵ AT/Wb
S_total = 1.25×10⁶ + 7.958×10⁵ = 2.046×10⁶ AT/Wb
Φ       = 1000/2.046×10⁶ = 4.89×10⁻⁴ Wb
```

| Quantity | Without cut | With cut |
|---|---|---|
| Reluctance (AT/Wb) | 1.25×10⁶ | 2.046×10⁶ |
| Flux (mWb) | 0.800 | 0.489 |

**Answer: (i) 1.25×10⁶ AT/Wb; (ii) 0.8 mWb; (iii) 0.489 mWb.** A 1 mm gap, only 0.03 % of the length, cuts the flux by about 39 %, since air has a very high reluctance.

## Check yourself

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

1. Which formula gives the ring's reluctance?
   A. B / H
   B. N^2 / l
   C. mu0 mur A / l
   D. l/(mu0 mur A)
   Answer: D. S = l/(mu0 mur A) = 3.1416/(4pi x10^-7 x 2000 x 10^-3) = 1.25x10^6 AT/Wb. (Key point: Reluctance formula)

2. What flux flows at 1 A without the cut?
   A. 1.25 mWb
   B. 8 mWb
   C. 0.489 mWb
   D. 0.8 mWb
   Answer: D. Phi = N I / S = 1000/1.25x10^6 = 0.8 mWb. (Key point: Flux without cut)

3. How is the saw cut treated?
   A. As zero reluctance
   B. As an air gap in series, adding its reluctance
   C. As a change in turns
   D. As a parallel path
   Answer: B. The gap lies in series with the iron, giving S_total = 1.25x10^6 + 7.958x10^5 = 2.046x10^6. (Key point: Effect of the cut)

4. What happens to the flux with the 1 mm cut?
   A. It doubles
   B. It falls to zero
   C. It drops to about 0.489 mWb (about 39% less)
   D. It barely changes
   Answer: C. Air has high reluctance, so even a 1 mm gap cuts the flux a lot. (Key point: Result of the cut)

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