---
title: "Iron ring with 2 mm saw cut, exciting current"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:iron-ring-300turns-gap-flux-density"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/iron-ring-300turns-gap-flux-density
site: BCT Engineering
---

# Iron ring with 2 mm saw cut, exciting current

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2078 Poush, 8 marks
- 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_machines/iron-ring-300turns-gap-flux-density
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

An iron ring of 0.15 meter diameter and 0.001 m$^2$ in cross section with a saw cut 2 mm wide is wound with 300 turns of wire. The gap flux density is 1 Tesla. The relative permeability of the iron is 800. Determine the exciting current and inductance.

## Short answer

Flux Φ = B_g A = 1 mWb. Iron path l_i = 0.4692 m, iron mmf = 466.8 AT, gap mmf = 1591.5 AT, total mmf = 2058.3 AT. Exciting current I = 2058.3/300 = 6.86 A. Inductance L = NΦ/I = 300×10⁻³/6.86 = 0.0437 H = 43.7 mH.

## Answer

**Given:** diameter $D = 0.15$ m (taken as mean diameter), $A = 10^{-3}$ m$^2$, $l_g = 2$ mm, $N = 300$, $B_g = 1$ T, $\mu_r = 800$.

**Flux and lengths**

$$
\begin{aligned}
\Phi &= B_gA = 1\times10^{-3} = 1\ \text{mWb} \\
l_i &= \pi\times0.15 - 0.002 = 0.4692\ \text{m}
\end{aligned}
$$

Since the flux and area are the same in iron and gap (no fringing), $B_i = 1$ T.

**Mmf**

$$
\begin{aligned}
F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{1\times0.4692}{4\pi\times10^{-7}\times800} = 466.8\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\
F &= 466.8 + 1591.5 = 2058.3\ \text{AT}
\end{aligned}
$$

**Exciting current**

$$
I = \frac{F}{N} = \frac{2058.3}{300} = 6.861\ \text{A}
$$

**Inductance**

$$
L = \frac{N\Phi}{I} = \frac{300\times10^{-3}}{6.861} = 0.0437\ \text{H}
$$

**Answer: $I \approx 6.86$ A; $L \approx 43.7$ mH**

## Check yourself

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

1. With no fringing, what is the flux density in the iron if the gap B is 1 T?
   A. 1/800 T
   B. 800 T
   C. 1 T
   D. 0.5 T
   Answer: C. Flux and area are the same in iron and gap, so B is equal. (Key point: Flux density in iron)

2. Which expression gives the gap mmf?
   A. B l_g / mu_0
   B. B l_g mu_0
   C. B mu_0 / l_g
   D. B l_g / (mu_0 mu_r)
   Answer: A. F_g = H l_g = B l_g / mu_0 = 1591.5 AT. (Key point: Mmf of gap)

3. Which part needs the larger mmf, and what is the total?
   A. Iron 1591 AT; total 2058 AT
   B. Gap 1591 AT; total 2058 AT
   C. Gap 467 AT; total 2058 AT
   D. Both equal
   Answer: B. The gap needs 1591.5 AT and the iron 466.8 AT, giving 2058.3 AT. (Key point: Share of mmf)

4. How is the inductance found here?
   A. L = I / (N Phi)
   B. L = N Phi / I
   C. L = Phi / I
   D. L = N^2 I
   Answer: B. L = flux linkage per ampere = 300 x 1e-3 / 6.861 = 43.7 mH. (Key point: Inductance)

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