---
title: "Cast steel ring with air gap, B-H data"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:cast-steel-ring-bh-table"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/cast-steel-ring-bh-table
site: BCT Engineering
---

# Cast steel ring with air gap, B-H data

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2068 Magh, 8 marks
- Video: [16. Calculation of Current in Cast Steel Ring Using Magnetization Curve With Or Without Air gap](https://www.youtube.com/watch?v=QCb3gnnuxmw) — Electrical Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/cast-steel-ring-bh-table
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A cast steel ring has a circular cross section of 3 cm in diameter and mean circumference of 80 cm. A 1 mm air-gap is cut in the ring which is wound with a coil of 600 turns. Estimate the current required to establish a flux of 0.75 mWb in the air-gap.
Magnetization data:

| H (AT/m) | 200 | 400 | 600 | 800 | 1000 | 1200 | 1400 | 1600 |
|---|---|---|---|---|---|---|---|---|
| B (T) | 0.1 | 0.32 | 0.6 | 0.9 | 1.08 | 1.18 | 1.27 | 1.32 |

## Short answer

Flux density B = 0.75 mWb / 7.069e-4 m2 = 1.061 T. H in steel from given table at this B is 978.9 AT/m. MMF for iron path (0.799 m) = 782.2 AT, for air gap (1 mm) = 844.3 AT, total 1626.5 AT. With 600 turns, current I = 1626.5/600 = 2.71 A.

## Answer

For a material given by a $B$–$H$ table, $H$ for the working $B$ is read (interpolated) from the table instead of using $\mu_r$.

**Given:** section diameter $= 3$ cm, mean circumference $= 80$ cm, $l_g = 1$ mm, $N = 600$, $\Phi_g = 0.75$ mWb.

**Flux density** (the same in steel and gap, leakage and fringing neglected):

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.03)^2 = 7.069\times10^{-4}\ \text{m}^2 \\
B &= \frac{0.75\times10^{-3}}{7.069\times10^{-4}} = 1.061\ \text{T}
\end{aligned}
$$

**$H$ for steel at 1.061 T:** this lies between (800 AT/m, 0.9 T) and (1000 AT/m, 1.08 T).

$$
H = 800 + 200\times\frac{1.061 - 0.9}{1.08 - 0.9} = 978.9\ \text{AT/m}
$$

**Mmf**

$$
\begin{aligned}
l_i &= 0.80 - 0.001 = 0.799\ \text{m} \\
F_i &= H l_i = 978.9\times0.799 = 782.2\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{1.061\times0.001}{4\pi\times10^{-7}} = 844.3\ \text{AT} \\
F &= 1626.5\ \text{AT}
\end{aligned}
$$

**Current**

$$
I = \frac{1626.5}{600} = 2.711\ \text{A}
$$

**Answer: $I \approx 2.71$ A**

## Check yourself

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

1. For steel given by a B-H table, how is H obtained?
   A. H = B / (mu_0 mu_r)
   B. By interpolating H from the table at the working B
   C. H = NI/l
   D. H is taken as 1000 AT/m
   Answer: B. mu_r is not constant, so H is read from the table: about 978.9 AT/m at 1.061 T. (Key point: Using B-H table)

2. How is B obtained?
   A. B = Phi/A = 1.061 T, same in steel and gap
   B. B = Phi A
   C. B = 0.75 T
   D. B different in steel and gap
   Answer: A. A = 7.069e-4 m^2, and with no leakage B is equal in steel and gap. (Key point: Flux density)

3. What is the gap mmf?
   A. 978.9 AT
   B. B l_g / mu_0 = 844.3 AT
   C. 782.2 AT
   D. 1626.5 AT
   Answer: B. F_g = 1.061 x 0.001 / (4 pi e-7). (Key point: Gap mmf)

4. What is the current?
   A. 2.71 A
   B. 1.30 A
   C. 0.78 A
   D. 5.4 A
   Answer: A. (782.2 + 844.3)/600 = 2.71 A. (Key point: Total current)

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