---
title: "Flux density in circular iron core"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:flux-density-1m-core-50mm2"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/flux-density-1m-core-50mm2
site: BCT Engineering
---

# Flux density in circular iron core

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2066 Magh (old course), 8 marks
- Video: [54 - Solved Problems on Magnetic Circuits](https://www.youtube.com/watch?v=iOhQn_EtvIU) — SkanCity Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/flux-density-1m-core-50mm2
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A circular iron core with mean length of 100 cm and cross-sectional area of 50 mm$^2$ has 500 turns winding on the core. Calculate the flux density in the core if 10 A current flows in the coil. Take $\mu_r = 2000$ for the iron core and neglect saturation.

## Short answer

With no air gap, H = NI/ℓ = 5000 AT/m. Assuming linear μr = 2000, B = μ0 μr H = 12.57 T. This is theoretical; a real core would saturate.

## Answer

With no air gap and a constant $\mu_r$ (saturation neglected), the flux density follows directly from the field strength.

**Given:** $l = 100$ cm, $A = 50$ mm$^2$ (not needed for $B$), $N = 500$, $I = 10$ A, $\mu_r = 2000$.

$$
\begin{aligned}
H &= \frac{NI}{l} = \frac{500\times10}{1.0} = 5000\ \text{AT/m} \\
B &= \mu_0\mu_r H = 4\pi\times10^{-7}\times2000\times5000 \\
&= 12.57\ \text{T}
\end{aligned}
$$

The corresponding flux is $\Phi = BA = 12.57\times50\times10^{-6} = 6.28\times10^{-4}$ Wb.

A real iron core would saturate near 1.8 to 2 T, so this value is a theoretical result for the given linear assumption.

**Answer: $B \approx 12.57$ T**

## Check yourself

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

1. How is H found here?
   A. H = NI/l = 5000 AT/m
   B. H = NI x l
   C. H = N/I l
   D. H = I/(N l)
   Answer: A. There is no air gap, so H = 500 x 10 / 1.0. (Key point: Field strength)

2. Which formula gives B?
   A. B = H / (mu_0 mu_r)
   B. B = mu_0 mu_r H
   C. B = mu_0 H / mu_r
   D. B = mu_r / H
   Answer: B. B = 4 pi e-7 x 2000 x 5000 = 12.57 T. (Key point: Flux density)

3. Is the cross-sectional area needed to find B?
   A. Yes, B = H A
   B. Yes, B = NI/A
   C. Only for the flux, not for B
   D. No, and the flux is not found either
   Answer: C. B depends on H and mu; area is used only for Phi = BA. (Key point: Area not needed)

4. Why is 12.57 T not realistic?
   A. mu_r is too low
   B. Real iron saturates near 1.8 to 2 T, but the problem assumes constant mu_r
   C. Area is too small
   D. Current is ac
   Answer: B. The problem neglects saturation, a theoretical assumption. (Key point: Saturation)

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