---
title: "Iron ring with saw cut, 4 mWb flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:ring-saw-cut-flux-4mwb"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/ring-saw-cut-flux-4mwb
site: BCT Engineering
---

# Iron ring with saw cut, 4 mWb flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2072 Asoj; 2070 Magh, 8 marks (asked 2 times)
- Video: [A cast iron ring of 40cm mean length......](https://www.youtube.com/watch?v=P6CYy9mH7Qc) — Vishwakumar Sheelavant
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/ring-saw-cut-flux-4mwb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

A ring of 30 cm mean diameter is made up of round iron rod 2.5 cm in diameter. A saw cut of 1 mm is made on the ring. It is uniformly wound with 500 turns of wire. Calculate the current required by the exciting coil to produce a total flux of 4 mWb. Assume a relative permeability of iron at this flux density as 800 (neglect leakage and fringing).

## Short answer

Iron length 0.9415 m, area 4.909e-4 m², B=8.15 T. AT_iron=Bl_i/(μ0 μr)=7631 AT, AT_gap=Blg/μ0=6485 AT, total 14116 AT. I=14116/500=28.2 A.

## Answer

Treat the ring as a series magnetic circuit of iron plus the 1 mm saw-cut air gap. The exciting current is about **28.2 A**.

**Given:** mean diameter D = 0.30 m; rod diameter d = 0.025 m; gap lg = 1 mm; N = 500; Φ = 4 mWb; μr = 800.

**Step 1: dimensions**
```text
Mean length  l  = πD = π × 0.30 = 0.9425 m
Iron length  lᵢ = 0.9425 − 0.001 = 0.9415 m
Area         A  = πd²/4 = π(0.025)²/4 = 4.909×10⁻⁴ m²
```

**Step 2: flux density**
B = Φ/A = 4×10⁻³ / 4.909×10⁻⁴ = 8.15 T. This is far above real saturation, but the question tells us to use μr = 800, so we follow it.

**Step 3: ampere-turns**
```text
AT(iron) = B lᵢ/(μ0 μr)
         = 8.149 × 0.9415 /(4π×10⁻⁷ × 800)
         = 7631 AT
AT(gap)  = B lg/μ0
         = 8.149 × 10⁻³ /(4π×10⁻⁷)
         = 6485 AT
AT total = 14 116 AT
```

**Step 4: current**
I = 14 116 / 500 = 28.2 A

**Answer: I ≈ 28.2 A** (iron 7631 AT + gap 6485 AT)

## Check yourself

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

1. How is the iron length found for the ring with a 1 mm saw cut?
   A. Cross-section perimeter
   B. Mean circumference pi x D minus the 1 mm gap
   C. Diameter D plus the gap
   D. Mean circumference plus the gap
   Answer: B. The ring's mean length is pi x 0.30 = 0.9425 m and the gap takes 1 mm of it, leaving 0.9415 m of iron. (Key point: Iron path length)

2. How is B obtained and what is its value?
   A. B = Phi/A = 0.8 T using A = pi D^2/4
   B. B = N I / l
   C. B = Phi/A = 8.15 T, using A = pi d^2/4
   D. B = Phi x A = 1.96 T
   Answer: C. Area of the 2.5 cm rod is 4.909x10^-4 m^2, so B = 4x10^-3/4.909x10^-4 = 8.15 T. (Key point: Flux density)

3. How is the total mmf of the ring obtained?
   A. Use only the gap AT
   B. Use only the iron AT
   C. Multiply the iron AT by the gap AT
   D. Add AT for iron (B l/(mu0 mur)) and for the gap (B lg/mu0)
   Answer: D. The coil supplies the sum of the mmf drops in the series iron and air-gap parts: 7631 + 6485 = 14 116 AT. (Key point: Ampere-turn sum)

4. With total 14 116 AT and 500 turns, what current is needed?
   A. 14.1 A
   B. 8.2 A
   C. 28.2 A
   D. 56.5 A
   Answer: C. I = 14 116/500 = 28.2 A. (Key point: Current)

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