---
title: "Iron ring: reluctance and flux with saw cut"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:ring-saw-cut-mean-diameter-100cm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/ring-saw-cut-mean-diameter-100cm
site: BCT Engineering
---

# Iron ring: reluctance and flux with saw cut

Electrical Machines · 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_machines/ring-saw-cut-mean-diameter-100cm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

An iron ring of mean diameter 100 cm and cross sectional area 10 cm$^2$ is wound with 1000 turns and has $\mu_r = 2000$. Compute (i) reluctance (ii) flux produced in the ring 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

Mean length l = π m. (i) Reluctance S = l/(μ₀μ_r A) = 1.25×10⁶ AT/Wb. (ii) Flux without gap = NI/S = 1000/1.25e6 = 8×10⁻⁴ Wb = 0.8 mWb. (iii) With 1 mm saw cut, iron path l_i = π–0.001 m, S_total = S_i + S_g = 2.0454×10⁶ AT/Wb, flux = 1000/2.0454e6 = 4.89×10⁻⁴ Wb (0.489 mWb).

## Answer

**Given:** $D = 100$ cm, $A = 10$ cm$^2 = 10^{-3}$ m$^2$, $N = 1000$, $\mu_r = 2000$, $I = 1$ A.

#### (i) Reluctance (no saw cut)

$$
\begin{aligned}
l &= \pi D = \pi\ \text{m} \\
S &= \frac{l}{\mu_0\mu_r A} = \frac{\pi}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.25\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

#### (ii) Flux with 1 A

$$
\Phi = \frac{NI}{S} = \frac{1000\times1}{1.25\times10^{6}} = 8\times10^{-4}\ \text{Wb} = 0.8\ \text{mWb}
$$

#### (iii) Flux with a 1 mm saw cut

The iron path becomes $l_i = \pi - 0.001 = 3.1406$ m, and the gap is $l_g = 1$ mm.

$$
\begin{aligned}
S_i &= \frac{3.1406}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.2496\times10^{6} \\
S_g &= \frac{0.001}{4\pi\times10^{-7}\times10^{-3}} = 7.958\times10^{5} \\
S_{total} &= 2.0454\times10^{6}\ \text{AT/Wb} \\
\Phi &= \frac{1000}{2.0454\times10^{6}} = 4.889\times10^{-4}\ \text{Wb}
\end{aligned}
$$

A gap of only 1 mm cuts the flux from 0.8 mWb to 0.489 mWb, because air has a much higher reluctance than iron.

**Answer: (i) $1.25\times10^6$ AT/Wb; (ii) 0.8 mWb; (iii) 0.489 mWb**

## Check yourself

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

1. How is the length of the iron ring found from its mean diameter?
   A. l = pi D
   B. l = D
   C. l = pi D^2 / 4
   D. l = 2D
   Answer: A. The mean circumference is pi D = pi m. (Key point: Iron path length)

2. What is the flux with 1 A before the saw cut?
   A. 8 mWb
   B. 0.8 mWb
   C. 1.25 mWb
   D. 0.08 mWb
   Answer: B. Phi = NI/S = 1000/1.25e6 = 0.8 mWb. (Key point: Flux without cut)

3. Why does a 1 mm saw cut reduce the flux from 0.8 to 0.489 mWb?
   A. Iron reluctance rises
   B. The mmf falls
   C. The area decreases
   D. Air has much higher reluctance, adding 7.96e5 AT/Wb in series
   Answer: D. The gap adds a large series reluctance while mmf stays the same. (Key point: Effect of saw cut)

4. What is the reluctance of the 1 mm gap?
   A. l_g/(mu_0 mu_r A)
   B. l_g/(mu_0 A)
   C. mu_0 A / l_g
   D. l_g mu_0 / A
   Answer: B. S_g = 0.001/(4 pi e-7 x 1e-3) = 7.958e5 AT/Wb. (Key point: Gap reluctance)

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