---
title: "Number of turns for iron ring with air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:iron-ring-1p25m-turns-0p5wb"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-ring-1p25m-turns-0p5wb
site: BCT Engineering
---

# Number of turns for iron ring with air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2071 Magh, 8 marks
- Video: [Magnetic circuits for Iron Ring with Air Gap | ELECTRONICS ENGINEERING VIEWS](https://www.youtube.com/watch?v=4VVPB7XmpCc) — Electronics Engineering Views 👁️👁️
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-ring-1p25m-turns-0p5wb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

An iron ring has a mean length of 1.25 m and cross-sectional area of 0.02 m². It has a radial air gap of 5 mm. Calculate the number of turns required to be wound on the core to produce a magnetic flux of 0.5 weber in the air gap with current of 2 amp in the coil. Assume that relative permeability of the core is 1000.

## Short answer

Iron length = 1.25−0.005 = 1.245 m, area 0.02 m². Reluctances: iron 4.95×10⁴ AT/Wb, gap 1.99×10⁵ AT/Wb, total 2.485×10⁵ AT/Wb. MMF = φ × total reluctance = 1.242×10⁵ AT; number of turns N = 1.242×10⁵ / 2 ≈ 62,120.

## Answer

Required mmf = flux × total reluctance; then $N = F/I$. Iron length is taken as $l_i = 1.25 - 0.005 = 1.245$ m; gap area = core area.

**Data:** $A = 0.02$ m², $l_g = 5$ mm, $\phi = 0.5$ Wb, $I = 2$ A, $\mu_r = 1000$.

#### Reluctances

$$
\begin{aligned}
S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{1.245}{4\pi\times10^{-7}\times1000\times0.02} = 4.954\times10^{4}\ \text{AT/Wb} \\
S_g &= \frac{l_g}{\mu_0 A} = \frac{0.005}{4\pi\times10^{-7}\times0.02} = 1.989\times10^{5}\ \text{AT/Wb} \\
S &= S_i + S_g = 2.485\times10^{5}\ \text{AT/Wb}
\end{aligned}
$$

#### MMF and turns

$$
\begin{aligned}
F &= \phi S = 0.5\times2.485\times10^{5} = 1.2424\times10^{5}\ \text{AT} \\
N &= \frac{F}{I} = \frac{1.2424\times10^{5}}{2} = 62120\ \text{turns}
\end{aligned}
$$

**Answer: N ≈ 62,120 turns.** (Using the full 1.25 m as iron length gives N ≈ 62,170.) Note: B = 0.5/0.02 = 25 T, which no iron can reach; the figure follows only from the constant μr given.

## Check yourself

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

1. What is the basic relation used to find the turns N for a given flux?
   A. N = φ / (I × total reluctance)
   B. F = φ × total reluctance, then N = F / I
   C. N = I × φ × total reluctance
   D. F = φ / total reluctance, then N = F × I
   Answer: B. The mmf needed equals flux times total reluctance of the series circuit, and NI = F. (Key point: Method)

2. What iron length is used, and why?
   A. 1.25 m plus 5 mm, because the gap adds to the length
   B. 1.245 m, because the 5 mm gap is part of the 1.25 m mean length
   C. 0.005 m, because only the gap matters
   D. 1.25 m × μr, to allow for permeability
   Answer: B. The ring's mean length includes the gap, so iron length = 1.25 − 0.005 = 1.245 m. (Key point: Iron length)

3. Which reluctance is larger, and what is the total?
   A. Iron is larger because μr = 1000; total ≈ 2.485×10^5 AT/Wb
   B. Air gap is larger; total is their difference, 1.49×10^5 AT/Wb
   C. Air gap 1.989×10^5 is larger than iron 4.954×10^4; total ≈ 2.485×10^5 AT/Wb
   D. They are equal; total is 9.9×10^4 AT/Wb
   Answer: C. The tiny gap with μr = 1 outweighs the iron path; reluctances in series add. (Key point: Dominant reluctance)

4. What number of turns results, and what caution is noted?
   A. About 1,242 turns; the answer is realistic for iron
   B. About 62,120 turns; B = 0.25 T is well within limits
   C. About 124,240 turns, since the current is halved
   D. About 62,120 turns; B = 25 T is unrealistic, so it holds only for constant μr
   Answer: D. N = 1.2424×10^5 / 2 ≈ 62,120, but 0.5 Wb in 0.02 m² gives 25 T, beyond any real iron. (Key point: Result)

## Close variants set in other exams

- [Flux and flux density in iron ring with 2 mm gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-ring-80cm-2mm-gap-1000turns) — Electrical Machines I, chapter 1, asked 2×

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