---
title: "Flux and flux density in iron ring with 2 mm gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:iron-ring-80cm-2mm-gap-1000turns"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-ring-80cm-2mm-gap-1000turns
site: BCT Engineering
---

# Flux and flux density in iron ring with 2 mm gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2077 Chaitra, 8 marks
- Video: [NEW TOPIC : IRON RING WITH AIRGAP | UNIT : MAGNETIC CIRCUIT IN HINDI (LECTURE 22)  @TIKLESACADEMY](https://www.youtube.com/watch?v=HZb3EGXqjoQ) — TIKLE'S ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-ring-80cm-2mm-gap-1000turns
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

An iron ring has mean length of 80 cm and cross sectional area of 16 sq. cm. and has a radial air gap of 2 mm. The core is wound with a coil of 1000 turns. If a current of 6 amp is passed through the coil, calculate the magnetic flux and magnetic flux density in the core. Given that the relative permeability of the core is 2000.

## Short answer

With MMF 6000 AT from 1000 turns at 6 A, iron and gap reluctances summed give total 1.193×10⁶ AT/Wb. Flux = 6000 / 1.193×10⁶ = 5.03 mWb, flux density in core = 3.14 T.

## Answer

Flux is found from $\phi = NI/(S_i + S_g)$. Iron length is taken as $l_i = 80 - 0.2 = 79.8$ cm; gap area = core area (no fringing).

**Data:** $A = 16$ cm² $= 16\times10^{-4}$ m², $l_g = 2$ mm, $N = 1000$, $I = 6$ A, $\mu_r = 2000$.

#### MMF

$$
F = NI = 1000 \times 6 = 6000\ \text{AT}
$$

#### Reluctances

$$
\begin{aligned}
S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.798}{4\pi\times10^{-7}\times2000\times16\times10^{-4}} = 1.984\times10^{5}\ \text{AT/Wb} \\
S_g &= \frac{l_g}{\mu_0 A} = \frac{0.002}{4\pi\times10^{-7}\times16\times10^{-4}} = 9.947\times10^{5}\ \text{AT/Wb} \\
S &= S_i + S_g = 1.193\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

#### Flux and flux density

$$
\begin{aligned}
\phi &= \frac{F}{S} = \frac{6000}{1.193\times10^{6}} = 5.03\times10^{-3}\ \text{Wb} \\
B &= \frac{\phi}{A} = \frac{5.03\times10^{-3}}{16\times10^{-4}} = 3.14\ \text{T}
\end{aligned}
$$

**Answer: φ ≈ 5.03 mWb, B ≈ 3.14 T** in the core (and in the gap). The 2 mm gap has five times the reluctance of the whole iron path. (If the full 80 cm is used as the iron length, φ = 5.027 mWb, which still rounds to 5.03 mWb; the difference is negligible.)

## Check yourself

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

1. For an iron ring with an air gap, the flux is found from:
   A. phi = NI/S_i only
   B. phi = NI/(S_i + S_g)
   C. phi = NI x (S_i + S_g)
   D. phi = NI/S_g only
   Answer: B. Iron and gap are in series so their reluctances add. (Key point: Method)

2. S_g = l_g/(mu0 A) is about:
   A. 1.98e5 AT/Wb
   B. 9.95e5 AT/Wb
   C. 1.19e6 AT/Wb
   D. 2.0e3 AT/Wb
   Answer: B. 0.002/(4 pi e-7 x 16e-4) = 9.947e5 AT/Wb, about five times S_i. (Key point: Gap reluctance)

3. With F = 6000 AT and S = 1.193e6, the flux is:
   A. 5.03 mWb
   B. 50.3 mWb
   C. 0.503 mWb
   D. 5.03 Wb
   Answer: A. 6000/1.193e6 = 5.03e-3 Wb. (Key point: Flux result)

4. The flux density in the core (equal to that in the gap) is:
   A. 0.314 T
   B. 3.14 T
   C. 31.4 T
   D. 1.57 T
   Answer: B. B = 5.03e-3 / 16e-4 = 3.14 T. (Key point: Flux density)

## Close variants set in other exams

- [Number of turns for iron ring with air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Airon-ring-1p25m-turns-0p5wb) — Electrical Machines I, chapter 1, asked 2×

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