---
title: "DC current for ring core with 2 mm air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:ring-5p5cm-3p5cm-airgap-0p3mwb"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ring-5p5cm-3p5cm-airgap-0p3mwb
site: BCT Engineering
---

# DC current for ring core with 2 mm air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2074 Bhadra, 8 marks
- Video: [Magnetic Circuits With Air Gap](https://www.youtube.com/watch?v=T99_3l0UqB8) — Benchmark Learn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ring-5p5cm-3p5cm-airgap-0p3mwb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A magnetic core consists of circular ring with outer diameter 5.5 cm and inner diameter 3.5 cm. The relative permeability of the iron is 2000. A radial airgap of 2 mm is cut in this core. Calculate the direct current that will be required in a coil of 1000 turns uniformly distributed around the core to produce a magnetic flux of 0.3 mWb in the airgap. Assume the magnetic leakage is negligible.

## Short answer

Using mean diameter 4.5 cm, iron length 0.1394 m, area 7.854×10⁻⁵ m², flux density B = 0.3 mWb/area = 3.82 T. MMF for air gap 6079 AT, iron 212 AT, total 6291 AT. Required direct current I = 6291/1000 = 6.29 A.

## Answer

**Assumption:** the ring has a circular cross-section whose diameter equals the radial width, $(5.5 - 3.5)/2 = 1$ cm.

#### Dimensions

$$
\begin{aligned}
D_{mean} &= \frac{5.5 + 3.5}{2} = 4.5\ \text{cm} \\
l_{mean} &= \pi D_{mean} = \pi\times0.045 = 0.1414\ \text{m} \\
l_i &= 0.1414 - 0.002 = 0.1394\ \text{m} \\
A &= \frac{\pi(0.01)^2}{4} = 7.854\times10^{-5}\ \text{m}^2
\end{aligned}
$$

#### Flux density

$$
B = \frac{\phi}{A} = \frac{0.3\times10^{-3}}{7.854\times10^{-5}} = 3.82\ \text{T}
$$

#### Ampere-turns

$$
\begin{aligned}
AT_g &= \frac{B}{\mu_0} l_g = \frac{3.82}{4\pi\times10^{-7}}\times0.002 = 3.0396\times10^{6}\times0.002 = 6079\ \text{AT} \\
AT_i &= \frac{B}{\mu_0\mu_r} l_i = \frac{3.0396\times10^{6}}{2000}\times0.1394 = 1519.8\times0.1394 = 211.8\ \text{AT} \\
NI &= 6079 + 211.8 = 6291\ \text{AT}
\end{aligned}
$$

#### Current

$$
I = \frac{6291}{1000} = 6.29\ \text{A}
$$

**Answer: I ≈ 6.29 A (dc).** (B = 3.82 T is beyond the saturation of real iron; the answer assumes μr stays 2000. If a square 1 cm × 1 cm section is assumed instead, A = 1 cm², B = 3 T and I ≈ 4.94 A.)

## Check yourself

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

1. For 0.3 mWb in A = 7.854e-5 m^2, B is:
   A. 0.38 T
   B. 3.82 T
   C. 38.2 T
   D. 1.91 T
   Answer: B. 0.3e-3 / 7.854e-5 = 3.82 T. (Key point: Flux density)

2. The larger share of NI is used by:
   A. The air gap (about 6079 AT)
   B. The iron (about 6079 AT)
   C. Both equally
   D. Neither
   Answer: A. AT_g = 6079 AT versus only 211.8 AT for the iron. (Key point: Dominant mmf)

3. The dc required is:
   A. 0.63 A
   B. 6.29 A
   C. 62.9 A
   D. 4.94 A
   Answer: B. NI = 6291 AT with 1000 turns gives 6.29 A. (4.94 A applies only to the alternative square-section assumption.) (Key point: Current)

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