---
title: "Circular ring with radial air gap, DC current"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:ring-outer-inner-diameter-radial-gap"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/ring-outer-inner-diameter-radial-gap
site: BCT Engineering
---

# Circular ring with radial air gap, DC current

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2074 Bhadra (old course)
- Video: [Magnetic Circuits - Magnetic Circuit with an Air Gap](https://www.youtube.com/watch?v=Up0E8Rc0i-c) — Energy Conversion Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ring-outer-inner-diameter-radial-gap
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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

Mean diameter 4.5 cm; area 3.142×10⁻⁴ m² gives B = 0.9549 T. MMF for iron (52.95 AT) and gap (1519.8 AT) sum to 1572.8 AT, so I = 1.57 A for 1000 turns.

## Answer

The ring is a toroid with a circular cross-section. Its mean diameter and section size follow from the outer and inner diameters.

**Given:** outer diameter $= 5.5$ cm, inner diameter $= 3.5$ cm, $\mu_r = 2000$, $l_g = 2$ mm, $N = 1000$, $\Phi = 0.3$ mWb.

**Geometry**

The radial thickness is $5.5 - 3.5 = 2$ cm, which is the diameter of the circular section. So

$$
\begin{aligned}
D_{mean} &= \frac{5.5 + 3.5}{2} = 4.5\ \text{cm} \\
A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\
l_i &= \pi\times0.045 - 0.002 = 0.13937\ \text{m}
\end{aligned}
$$

**Flux density**

$$
B = \frac{\Phi}{A} = \frac{0.3\times10^{-3}}{3.142\times10^{-4}} = 0.9549\ \text{T}
$$

**Mmf**

$$
\begin{aligned}
F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.9549\times0.13937}{4\pi\times10^{-7}\times2000} = 52.95\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{0.9549\times0.002}{4\pi\times10^{-7}} = 1519.8\ \text{AT} \\
F &= 1572.8\ \text{AT}
\end{aligned}
$$

**Current**

$$
I = \frac{F}{N} = \frac{1572.8}{1000} = 1.573\ \text{A}
$$

**Answer: $I \approx 1.57$ A**

## Check yourself

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

1. How is the mean diameter obtained from outer 5.5 cm and inner 3.5 cm?
   A. Their difference, 2 cm
   B. Their average, 4.5 cm
   C. Their sum, 9 cm
   D. The outer, 5.5 cm
   Answer: B. Mean diameter = (5.5 + 3.5)/2 = 4.5 cm. (Key point: Mean diameter)

2. What gives the circular section diameter?
   A. Sum of the diameters
   B. Inner diameter
   C. Outer diameter
   D. Radial thickness (outer - inner) = 2 cm
   Answer: D. The radial thickness of the ring is the diameter of its circular section. (Key point: Section area)

3. How is B found from the gap flux 0.3 mWb?
   A. B = Phi A
   B. B = Phi/A = 0.955 T
   C. B = A/Phi
   D. B = mu_0 Phi
   Answer: B. B = 0.3e-3 / 3.142e-4 = 0.955 T. (Key point: Flux density)

4. Which is the total mmf requirement?
   A. Only gap mmf
   B. Iron mmf + gap mmf = 1572.8 AT
   C. Iron mmf only
   D. Gap mmf - iron mmf
   Answer: B. The two are in series, so mmfs add; I = 1572.8/1000 = 1.57 A. (Key point: Total mmf)

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