---
title: "DC current for iron rod ring with 4 mm air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:iron-rod-200sqmm-4mm-500turns-250mwb"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-rod-200sqmm-4mm-500turns-250mwb
site: BCT Engineering
---

# DC current for iron rod ring with 4 mm air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2069 Poush, 8 marks
- Video: [Iron ring | Electromagnetism | Magnetic Circuit | Problem 1](https://www.youtube.com/watch?v=bjzbJLAnf4M) — Aryabhat Institute
- 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-rod-200sqmm-4mm-500turns-250mwb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

An 80 cm long iron rod has cross-sectional area of 200 sq.mm. It is bent into a circular ring with an air gap of 4 mm and the ring is wound with 500 turns of winding. What dc current would be needed in the coil to produce a flux of 250 mWb in the air gap? Assume that μr = 4000.

## Short answer

With given 250 mWb, B = 0.25/(2e-4) = 1250 T. Gap mmf ≈ 3.98e6 AT, iron mmf ≈ 1.98e5 AT, total NI = 4.177e6 AT, I = 4.177e6/500 = 8354 A. Unrealistic; if flux is 250 μWb, B = 1.25 T, I ≈ 8.35 A.

## Answer

When the rod is bent into a ring and a 4 mm gap is cut, the iron length is $l_i = 0.80 - 0.004 = 0.796$ m. The same flux passes through iron and gap.

**Data:** $A = 200$ mm² $= 2\times10^{-4}$ m², $l_g = 4$ mm, $N = 500$, $\mu_r = 4000$, $\phi = 250$ mWb.

#### Flux density

$$
B = \frac{\phi}{A} = \frac{0.25}{2\times10^{-4}} = 1250\ \text{T}
$$

#### Ampere-turns

$$
\begin{aligned}
AT_g &= \frac{B}{\mu_0}l_g = \frac{1250}{4\pi\times10^{-7}}\times0.004 = 3.979\times10^{6}\ \text{AT} \\
AT_i &= \frac{B}{\mu_0\mu_r}l_i = \frac{9.947\times10^{8}}{4000}\times0.796 = 1.979\times10^{5}\ \text{AT} \\
NI &= 3.979\times10^{6} + 1.979\times10^{5} = 4.177\times10^{6}\ \text{AT}
\end{aligned}
$$

#### Current

$$
I = \frac{4.177\times10^{6}}{500} = 8354\ \text{A}
$$

**Answer (data as given): I ≈ 8354 A.**

A flux density of 1250 T is physically impossible, so the flux is most likely a misprint for **250 μWb**. Then $B = 1.25$ T, $AT_g = 3978.9$ AT, $AT_i = 197.9$ AT, $NI = 4176.8$ AT and

$$
I = \frac{4176.8}{500} = 8.35\ \text{A}
$$

which is the realistic answer (the method is identical; the current scales directly with flux).

## Check yourself

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

1. What iron length is used after the 4 mm gap is cut in the 80 cm rod?
   A. 0.796 m
   B. 0.804 m
   C. 0.80 m, the gap being ignored
   D. 0.4 m, half the rod
   Answer: A. The ring length is 0.80 m of which 4 mm is air, so iron = 0.80 − 0.004 = 0.796 m. (Key point: Iron length)

2. What is the flux density and what does it show?
   A. B = 1.25 T, a normal value
   B. B = 0.25/(2×10^-4) = 1250 T, which is physically impossible, hinting at a misprint
   C. B = 0.0125 T, very low
   D. B = 1250 T, which is normal for iron
   Answer: B. Area 200 mm² = 2×10^-4 m²; 250 mWb gives 1250 T, so the flux is likely 250 μWb. (Key point: Unit conversion)

3. Which part needs most ampere-turns?
   A. The iron (about 3.98×10^6 AT) versus gap (1.98×10^5 AT)
   B. Both need equal ampere-turns
   C. The air gap (about 3.98×10^6 AT) versus iron (about 1.98×10^5 AT)
   D. The gap needs none because μ0 is small
   Answer: C. Iron H is reduced by μr = 4000, but the 4 mm gap has μr = 1. (Key point: Ampere-turn split)

4. If the flux is read as 250 μWb, what is the current?
   A. About 83.5 A
   B. About 8354 A, unchanged
   C. About 0.835 A
   D. About 8.35 A, since current scales directly with flux
   Answer: D. The circuit is linear, so 1000 times less flux needs 1000 times less current: 8354/1000 ≈ 8.35 A. (Key point: Scaling)

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