---
title: "Iron ring with air gap: current, inductance, energy"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:iron-ring-15cm-stored-energy"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/iron-ring-15cm-stored-energy
site: BCT Engineering
---

# Iron ring with air gap: current, inductance, energy

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2067 Mangsir (old course), 8 marks
- 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/iron-ring-15cm-stored-energy
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

An iron ring of mean diameter 15 cm and 10 sq-cm cross sectional area is wound with 200 turns of wire. There is an air gap of 2 mm cut in the ring. For a flux density of 1 Wb/m$^2$ and relative permeability of 500, find the exciting current, the inductance and stored energy.

## Short answer

Total mmf = 2338.4 AT gives current 11.69 A. Inductance from flux linkage L = NΦ/I = 17.1 mH, and stored energy W = ½ LI² = 1.17 J.

## Answer

**Given:** mean diameter $= 15$ cm, $A = 10$ cm$^2 = 10^{-3}$ m$^2$, $N = 200$, $l_g = 2$ mm, $B = 1$ T, $\mu_r = 500$.

**Lengths and flux**

$$
l_i = \pi\times0.15 - 0.002 = 0.4692\ \text{m},\qquad \Phi = BA = 10^{-3}\ \text{Wb}
$$

**Mmf**

$$
\begin{aligned}
F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.4692}{4\pi\times10^{-7}\times500} = 746.8\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\
F &= 2338.4\ \text{AT}
\end{aligned}
$$

**Exciting current**

$$
I = \frac{F}{N} = \frac{2338.4}{200} = 11.69\ \text{A}
$$

**Inductance**

$$
L = \frac{N\Phi}{I} = \frac{200\times10^{-3}}{11.69} = 0.0171\ \text{H}
$$

**Stored energy**

$$
W = \tfrac{1}{2}LI^2 = \tfrac{1}{2}\times0.0171\times11.69^2 = 1.169\ \text{J}
$$

Check: $W = \tfrac12 F\Phi = \tfrac12\times2338.4\times10^{-3} = 1.169$ J. About 68% of this energy is stored in the 2 mm air gap.

**Answer: $I \approx 11.69$ A; $L \approx 17.1$ mH; $W \approx 1.17$ J**

## Check yourself

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

1. Which is correct for the iron and gap mmfs?
   A. Iron 746.8 AT, gap 1591.5 AT
   B. Iron 1591.5 AT, gap 746.8 AT
   C. Both 1591.5 AT
   D. Both 746.8 AT
   Answer: A. F_i = Bl_i/(mu_0 mu_r) = 746.8 AT and F_g = B l_g/mu_0 = 1591.5 AT. (Key point: Mmf split)

2. Which formula gives the stored energy?
   A. W = L I^2
   B. W = 1/2 L I^2 = 1/2 F Phi
   C. W = N Phi I
   D. W = Phi/I
   Answer: B. W = 1/2 x 2338.4 x 1e-3 = 1.17 J. (Key point: Stored energy)

3. Where is most of the energy stored?
   A. In the winding
   B. Equally everywhere
   C. In the iron
   D. In the 2 mm air gap, about 68%
   Answer: D. Energy share follows mmf share, and the gap needs 1591.5 of 2338.4 AT. (Key point: Where energy is stored)

4. How is the exciting current found?
   A. I = F/N = 11.69 A
   B. I = N/F
   C. I = F N
   D. I = Phi/N
   Answer: A. I = 2338.4/200. (Key point: Current)

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