---
title: "Turns of air-cored coil"
subject: "Basic Electrical Engineering"
chapter: 4
key: "semester_1/basic_electrical_engineering:air-cored-coil-turns-100uh"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/air-cored-coil-turns-100uh
site: BCT Engineering
---

# Turns of air-cored coil

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

- Asked in: 2068 Chaitra, 4 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/air-cored-coil-turns-100uh
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

An air cored coil is 2.5cm long and has an average cross-sectional area of 2cm². Determine the number of turns if the coil has an inductance of 100 μH.

## Short answer

For an air-core solenoid, L = μ0 N² A/l, so N = √(L l/(μ0 A)). With L=100 µH, l=0.025 m, A=2×10^-4 m²: N ≈ √9947 ≈ 100 turns.

## Answer

For an air-cored coil (solenoid), μr = 1 and

L = μ₀ N² A / l  ⟹  N = √( L l / (μ₀ A) )

#### Given
- l = 2.5 cm = 0.025 m
- A = 2 cm² = 2 × 10⁻⁴ m²
- L = 100 μH = 100 × 10⁻⁶ H
- μ₀ = 4π × 10⁻⁷ H/m

#### Calculation
- Numerator: L l = 100 × 10⁻⁶ × 0.025 = 2.5 × 10⁻⁶
- Denominator: μ₀ A = 4π × 10⁻⁷ × 2 × 10⁻⁴ = 2.513 × 10⁻¹⁰
- N² = (2.5 × 10⁻⁶) / (2.513 × 10⁻¹⁰) = 9947.2
- N = √9947.2 = 99.7

**Answer: about 100 turns.**

## Check yourself

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

1. Air-core turns formula?
   A. N = L l/(mu0 A)
   B. N = sqrt(L mu0 l/A)
   C. N = sqrt(L A/(mu0 l))
   D. N = sqrt(L l/(mu0 A))
   Answer: D. mur = 1. (Key point: Formula)

2. Area 2 cm^2 in SI?
   A. 2 m^2
   B. 2 x 10^-4 m^2
   C. 2 x 10^-2 m^2
   D. 2 x 10^-6 m^2
   Answer: B. 1 cm^2 = 10^-4 m^2. (Key point: Unit conversion)

3. Turns?
   A. About 100
   B. About 9947
   C. About 10
   D. About 1000
   Answer: A. N^2 = 9947, N = 99.7. (Key point: Result)

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