---
title: "Electromagnet core with radial gap, DC current"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:electromagnet-core-radial-gap-0-2mwb"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/electromagnet-core-radial-gap-0-2mwb
site: BCT Engineering
---

# Electromagnet core with radial gap, DC current

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2065 Chaitra (old course), 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/electromagnet-core-radial-gap-0-2mwb
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

The core of an electromagnet is made of an iron rod 1 cm diameter, bent into a circle of mean diameter 10 cm, a radial air gap of 1 mm being left between the ends of the rod. Calculate the direct current needed in coil of 2000 turns uniformly spaced around the core to produce a magnetic flux of 0.2 mWb in the air gap. Assume that the relative permeability of the iron is 150.

## Short answer

Area 7.854×10⁻⁵ m² and flux 0.2 mWb give B = 2.546 T. Iron and gap mmf sum to 6257 AT, so with 2000 turns the required dc current is 3.13 A.

## Answer

**Given:** rod diameter $= 1$ cm, mean diameter of ring $= 10$ cm, $l_g = 1$ mm, $N = 2000$, $\Phi = 0.2$ mWb, $\mu_r = 150$.

**Area, flux density, iron length**

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.01)^2 = 7.854\times10^{-5}\ \text{m}^2 \\
B &= \frac{\Phi}{A} = \frac{0.2\times10^{-3}}{7.854\times10^{-5}} = 2.546\ \text{T} \\
l_i &= \pi\times0.10 - 0.001 = 0.3132\ \text{m}
\end{aligned}
$$

**Mmf**

$$
\begin{aligned}
F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{2.546\times0.3132}{4\pi\times10^{-7}\times150} = 4230.6\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{2.546\times0.001}{4\pi\times10^{-7}} = 2026.4\ \text{AT} \\
F &= 6257.0\ \text{AT}
\end{aligned}
$$

**Current**

$$
I = \frac{6257.0}{2000} = 3.129\ \text{A}
$$

The flux density of 2.55 T is beyond the saturation level of real iron, so the given $\mu_r = 150$ is a data assumption of the problem.

**Answer: $I \approx 3.13$ A**

## Check yourself

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

1. How is B found from the flux 0.2 mWb?
   A. B = Phi A
   B. B = Phi / A = 2.546 T
   C. B = Phi l
   D. B = A / Phi
   Answer: B. A = pi/4 x 0.01^2 = 7.854e-5 m^2, so B = 2.546 T. (Key point: Flux density)

2. Which formula gives the iron mmf?
   A. B l_i mu_0 mu_r
   B. B l_i / mu_0
   C. B l_i / (mu_0 mu_r)
   D. B / (l_i mu_0 mu_r)
   Answer: C. F_i = H l_i = B l_i/(mu_0 mu_r) = 4230.6 AT. (Key point: Iron mmf)

3. How do the mmfs combine and give current?
   A. Subtract; I = 1.0 A
   B. Add; I = F/N = 3.13 A
   C. Multiply
   D. Add; I = F x N
   Answer: B. (4230.6 + 2026.4)/2000 = 3.13 A. (Key point: Total)

4. Why is the 2.55 T figure noteworthy?
   A. It is too low
   B. It exceeds iron saturation, so mu_r = 150 is only a data assumption
   C. It proves the gap is wrong
   D. It shows no leakage
   Answer: B. Real iron saturates well below 2.55 T. (Key point: Realism)

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