---
title: "Core with 4 mm air gap: inductance and flux"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:circular-core-4mm-gap-inductance-flux"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/circular-core-4mm-gap-inductance-flux
site: BCT Engineering
---

# Core with 4 mm air gap: inductance and flux

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2070 Bhadra, 6 marks
- Video: [Magnetic Circuits With Air Gap](https://www.youtube.com/watch?v=T99_3l0UqB8) — Benchmark Learn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/circular-core-4mm-gap-inductance-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A circular iron core has a cross-sectional area of 5 sq.cm. and mean length of 25 cm including an air gap of 4 mm. The core is wound with 500 turns of winding. Calculate the inductance of the coil. If a dc current of 10 Ampere passed through the coil, calculate magnetic flux in the core. Given that relative permeability of the core is 2000.

## Short answer

Iron and air-gap reluctances in series total 6.562×10⁶ AT/Wb. Then L = N²/S = 38.1 mH. With 10 A dc, flux Φ = NI/S = 0.762 mWb.

## Answer

**Given:** $A = 5$ cm$^2 = 5\times10^{-4}$ m$^2$, total mean length $= 25$ cm including the 4 mm gap, $N = 500$, $\mu_r = 2000$, $I = 10$ A.

**Lengths:** $l_g = 4$ mm, $l_i = 0.25 - 0.004 = 0.246$ m.

**Reluctances**

$$
\begin{aligned}
S_i &= \frac{0.246}{4\pi\times10^{-7}\times2000\times5\times10^{-4}} = 1.958\times10^{5}\ \text{AT/Wb} \\
S_g &= \frac{0.004}{4\pi\times10^{-7}\times5\times10^{-4}} = 6.366\times10^{6}\ \text{AT/Wb} \\
S &= 6.562\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

**Inductance**

$$
L = \frac{N^2}{S} = \frac{500^2}{6.562\times10^{6}} = 0.0381\ \text{H}
$$

**Flux for 10 A**

$$
\Phi = \frac{NI}{S} = \frac{500\times10}{6.562\times10^{6}} = 7.62\times10^{-4}\ \text{Wb}
$$

The flux density is $B = \Phi/A = 1.52$ T.

**Answer: $L \approx 38.1$ mH; $\Phi \approx 0.762$ mWb**

## Check yourself

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

1. Which formula gives L from the total reluctance S?
   A. L = S / N^2
   B. L = N / S
   C. L = N^2 / S
   D. L = N S
   Answer: C. L = N^2/S = 250000/6.562e6 = 38.1 mH. (Key point: Inductance from reluctance)

2. Which reluctance dominates?
   A. Iron, 1.96e5
   B. Gap, 6.37e6
   C. Both equal
   D. Neither
   Answer: B. The 4 mm air gap has about 30 times the iron reluctance. (Key point: Dominant reluctance)

3. How is flux at 10 A obtained?
   A. Phi = NI / S = 0.762 mWb
   B. Phi = N I S
   C. Phi = S / NI
   D. Phi = L N
   Answer: A. Phi = 5000/6.562e6 = 7.62e-4 Wb. (Key point: Flux)

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