---
title: "DC current for iron ring with 4 mm air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:iron-ring-1p5m-4mm-500turns-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/iron-ring-1p5m-4mm-500turns-current
site: BCT Engineering
---

# DC current for iron ring with 4 mm air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2070 Bhadra, 8 marks
- Video: [Magnetic circuits for Iron Ring with Air Gap | ELECTRONICS ENGINEERING VIEWS](https://www.youtube.com/watch?v=4VVPB7XmpCc) — Electronics Engineering Views 👁️👁️
- 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-1p5m-4mm-500turns-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

An iron ring has a mean length of 1.5 m and cross-sectional area of 50 cm². It has a radial air gap of 4 mm. The ring is wound with 500 turns. What dc current would be needed in the coil to produce a flux of 100 mWb in the air gap? Assume that μr = 2000.

## Short answer

B = 0.1/(5e-3) = 20 T. Air-gap mmf = (20/μ0)×0.004 = 63662 AT, iron mmf = (20/(μ0×2000))×1.496 = 11905 AT. Total mmf = 75567 AT; dc current I = 75567/500 ≈ 151 A.

## Answer

Required mmf = ampere-turns for the gap + ampere-turns for the iron, with the same flux density in both (no leakage or fringing). Iron length $l_i = 1.5 - 0.004 = 1.496$ m.

**Data:** $A = 50$ cm² $= 5\times10^{-3}$ m², $l_g = 4$ mm, $N = 500$, $\phi = 100$ mWb, $\mu_r = 2000$.

#### Flux density

$$
B = \frac{\phi}{A} = \frac{0.1}{5\times10^{-3}} = 20\ \text{T}
$$

#### Air gap

$$
\begin{aligned}
H_g &= \frac{B}{\mu_0} = \frac{20}{4\pi\times10^{-7}} = 1.5915\times10^{7}\ \text{AT/m} \\
AT_g &= H_g l_g = 1.5915\times10^{7}\times0.004 = 63662\ \text{AT}
\end{aligned}
$$

#### Iron

$$
\begin{aligned}
H_i &= \frac{H_g}{\mu_r} = \frac{1.5915\times10^{7}}{2000} = 7958\ \text{AT/m} \\
AT_i &= 7958\times1.496 = 11905\ \text{AT}
\end{aligned}
$$

#### Current

$$
\begin{aligned}
NI &= 63662 + 11905 = 75567\ \text{AT} \\
I &= \frac{75567}{500} = 151.1\ \text{A}
\end{aligned}
$$

**Answer: I ≈ 151 A (dc).** (B = 20 T is far above what iron can carry; the figure assumes μr stays constant at 2000 as given.)

## Check yourself

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

1. How is B found and what is it?
   A. B = φ/A = 0.1/(5×10^-3) = 20 T, the same in iron and gap
   B. B = φ × A = 5×10^-4 T
   C. B = φ/A = 0.2 T
   D. B = φ/A = 20 T only in the gap, less in iron
   Answer: A. With no leakage or fringing the same flux crosses the same area, so B is the same throughout. (Key point: Flux density)

2. How are the air-gap ampere-turns found?
   A. AT_g = B × lg ≈ 0.08 AT
   B. AT_g = (B/μ0) × lg ≈ 63,662 AT
   C. AT_g = (B/(μ0 μr)) × lg ≈ 31.8 AT
   D. AT_g = μ0 × B × lg
   Answer: B. Hg = B/μ0 = 1.59×10^7 AT/m, multiplied by 4 mm. (Key point: Gap ampere-turns)

3. Why are the iron ampere-turns far smaller than the gap's, and what is their value?
   A. Iron has a longer path, so ATi = 63,662 × 1.496
   B. Iron carries less flux than the gap, so ATi ≈ 0
   C. Hi = Hg/μr, so for the same B the iron needs 2000 times less H; ATi ≈ 11,905 AT over 1.496 m
   D. Iron has μr = 1, so ATi equals ATg
   Answer: C. H = B/(μ0 μr), so Hi = 7958 AT/m; times iron length 1.496 m gives 11,905 AT. (Key point: Iron ampere-turns)

4. What current results, and why is it so large?
   A. I = 75,567 × 500 ≈ 3.8×10^7 A
   B. I ≈ 1.51 A, a normal exciting current
   C. I ≈ 127 A, using only the gap ampere-turns
   D. I = 75,567/500 ≈ 151 A, because B = 20 T is unrealistically high
   Answer: D. NI = ATg + ATi = 75,567 AT; dividing by N = 500 gives about 151 A. (Key point: Current)

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