---
title: "Number of turns of iron-core inductor"
subject: "Basic Electrical Engineering"
chapter: 4
key: "semester_1/basic_electrical_engineering:inductor-turns-iron-core-500mh"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductor-turns-iron-core-500mh
site: BCT Engineering
---

# Number of turns of iron-core inductor

Basic Electrical Engineering · Chapter 4: Inductance & Capacitance in Electric Circuits

- Asked in: 2071 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductor-turns-iron-core-500mh
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

An inductor is to be made with copper wire wound on a circular iron core having mean length of 40 cm with cross-sectional area of 50 sq mm. If the required value of inductance is 500 mH, calculate the number of turns required given that relative permeability of the core is 1500.

## Short answer

Using L = μ0 μr N² A/l, N = √(L l/(μ0 μr A)). l=0.4 m, A=50×10^-6 m², μr=1500, L=0.5 H yields N ≈ 1457 turns.

## Answer

For a coil on a closed magnetic core of mean length l and area A, the inductance is

L = μ₀ μr N² A / l

so N = √( L l / (μ₀ μr A) )

#### Given
- L = 500 mH = 0.5 H
- l = 40 cm = 0.4 m
- A = 50 mm² = 50 × 10⁻⁶ m²
- μr = 1500, μ₀ = 4π × 10⁻⁷ H/m

#### Calculation
- μ₀ μr = 4π × 10⁻⁷ × 1500 = 1.885 × 10⁻³ H/m
- Denominator: μ₀ μr A = 1.885 × 10⁻³ × 50 × 10⁻⁶ = 9.425 × 10⁻⁸
- Numerator: L l = 0.5 × 0.4 = 0.2
- N² = 0.2 / (9.425 × 10⁻⁸) = 2.122 × 10⁶
- N = √(2.122 × 10⁶) = 1456.7

**Answer: about 1457 turns** (say 1460 turns). Leakage and fringing are neglected.

## Check yourself

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

1. Which rearrangement gives N?
   A. N = L mu A / l
   B. N = sqrt(L A/(mu l))
   C. N = L l/(mu0 mur A)
   D. N = sqrt(L l/(mu0 mur A))
   Answer: D. From L = mu0 mur N^2 A/l. (Key point: Formula)

2. Which unit conversions are used?
   A. 0.5 H, 40 m, 50 x 10^-4 m^2
   B. 0.5 H, 0.4 m, 50 x 10^-6 m^2
   C. 0.5 H, 0.4 m, 50 x 10^-2 m^2
   D. 500 H, 40 m, 50 m^2
   Answer: B. mH, cm and mm^2 convert to SI. (Key point: Units)

3. How many turns?
   A. About 146
   B. About 1457
   C. About 37
   D. About 14570
   Answer: B. N^2 = 0.2/9.425e-8 = 2.122e6, N = 1457. (Key point: Result)

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