---
title: "Cast steel ring current from B-H data"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:cast-steel-ring-magnetization-data"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/cast-steel-ring-magnetization-data
site: BCT Engineering
---

# Cast steel ring current from B-H data

Electrical Machine · 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_machine/cast-steel-ring-magnetization-data
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/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

Area A = 7.069×10⁻⁴ m², B = Φ/A = 0.75 mWb / 7.069×10⁻⁴ m² = 1.061 T. From B-H table, interpolated H = 979 AT/m. Iron mmf = 979×0.799 = 782 AT, gap mmf = (1.061/μ₀)×10⁻³ = 844 AT, total 1626 AT. Required current I = 1626/600 = 2.71 A.

## Answer

As the permeability of cast steel is not constant, we use the B–H data: find B in the core, read H from the table, then add the mmf of the iron and the gap.

**Given:** diameter 3 cm; mean length 80 cm; gap lg = 1 mm; N = 600; gap flux Φ = 0.75 mWb.

**Step 1: flux density** (same in iron and gap, no leakage or fringing)
```text
A = π(0.03)²/4 = 7.069×10⁻⁴ m²
B = Φ/A = 0.75×10⁻³ / 7.069×10⁻⁴ = 1.061 T
```

**Step 2: H for steel by interpolation**
B = 1.061 T lies between 0.9 T (800 AT/m) and 1.08 T (1000 AT/m).
```text
H = 800 + 200 × (1.061 − 0.9)/(1.08 − 0.9)
  = 800 + 200 × 0.894 = 979 AT/m
```

**Step 3: ampere-turns**
```text
Iron length = 0.80 − 0.001 = 0.799 m
AT(iron)    = 979 × 0.799 = 782 AT
AT(gap)     = (B/μ0) lg = (1.061/4π×10⁻⁷) × 10⁻³ = 844 AT
AT total    = 1626 AT
```

**Step 4: current**
I = 1626/600 = 2.71 A

**Answer: I ≈ 2.71 A** (iron 782 AT + gap 844 AT)

## Check yourself

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

1. Why is the table used instead of a constant mur?
   A. The ring has no gap
   B. The permeability of cast steel is not constant
   C. Air has varying permeability
   D. The coil has no turns
   Answer: B. Steel saturates, so H must be read from the magnetisation data for the given B. (Key point: Why use B-H data)

2. How is B found?
   A. Gap flux divided by length
   B. N I / l
   C. Gap flux divided by core area, 1.061 T
   D. Flux times area
   Answer: C. B = 0.75x10^-3/7.069x10^-4 = 1.061 T, same in iron and gap. (Key point: Flux density)

3. B = 1.061 T lies between which table points, giving H of about 979 AT/m?
   A. 1.08 T and 1.18 T
   B. 0.32 T and 0.6 T
   C. 0.6 T and 0.9 T
   D. 0.9 T (800) and 1.08 T (1000)
   Answer: D. Linear interpolation between 0.9 T and 1.08 T gives H = 979 AT/m. (Key point: Interpolation)

4. Iron 782 AT plus gap 844 AT with 600 turns gives what current?
   A. 1.30 A
   B. 2.71 A
   C. 27.1 A
   D. 0.27 A
   Answer: B. I = 1626/600 = 2.71 A. (Key point: Total current)

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