---
title: "Circular core inductance and flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:circular-core-inductance-flux"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/circular-core-inductance-flux
site: BCT Engineering
---

# Circular core inductance and flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2070 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/circular-core-inductance-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/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

Total reluctance: iron 1.958×10⁵ AT/Wb, gap 6.366×10⁶ AT/Wb, sum 6.562×10⁶ AT/Wb. Inductance L = N²/S = 500²/6.562×10⁶ = 0.0381 H = 38.1 mH. At 10 A, mmf = 5000 AT, flux = 5000/6.562×10⁶ = 7.62×10⁻⁴ Wb (0.762 mWb).

## Answer

Inductance of a coil on a magnetic core is L = N²/S, where S is the total reluctance of the core including the air gap.

**Given:** A = 5 cm² = 5×10⁻⁴ m²; total mean length l = 25 cm (including the gap); gap lg = 4 mm; N = 500; μr = 2000; I = 10 A.

**Step 1: reluctances** (iron length = 0.25 − 0.004 = 0.246 m)
```text
Sᵢ = 0.246/(4π×10⁻⁷ × 2000 × 5×10⁻⁴) = 1.958×10⁵ AT/Wb
Sg = 0.004/(4π×10⁻⁷ × 5×10⁻⁴)        = 6.366×10⁶ AT/Wb
S  = 6.562×10⁶ AT/Wb
```

**Step 2: inductance**
L = N²/S = 500²/6.562×10⁶ = 2.5×10⁵/6.562×10⁶ = 0.0381 H = **38.1 mH**

**Step 3: flux at 10 A**
```text
mmf = N I = 500 × 10 = 5000 AT
Φ   = 5000/6.562×10⁶ = 7.62×10⁻⁴ Wb
```
The flux density is B = Φ/A = 1.52 T. This is acceptable for iron, so a constant μr = 2000 is a fair assumption.

**Answer: L ≈ 38.1 mH; Φ ≈ 0.762 mWb**

## Check yourself

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

1. Which formula gives the inductance of a coil on a magnetic core?
   A. L = N S
   B. L = N/S
   C. L = S/N^2
   D. L = N^2/S
   Answer: D. L = N^2/S where S is the total reluctance including the gap. (Key point: Inductance formula)

2. What iron length is used when the 25 cm includes the 4 mm gap?
   A. 0.246 m
   B. 0.25 m
   C. 0.004 m
   D. 0.254 m
   Answer: A. The gap is part of the 25 cm, so iron length is 0.25 - 0.004 = 0.246 m. (Key point: Iron length)

3. What inductance results with S = 6.562x10^6?
   A. 38.1 mH
   B. 3.81 mH
   C. 13.1 mH
   D. 381 mH
   Answer: A. L = 500^2/6.562x10^6 = 0.0381 H. (Key point: Inductance value)

4. How is the flux at 10 A found?
   A. N I x S
   B. N I / S = 5000/6.562x10^6 = 0.762 mWb
   C. B x l
   D. L x I = 0.38 Wb
   Answer: B. Phi = mmf/S = 7.62x10^-4 Wb. (Key point: Flux)

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