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Cast steel ring current from B-H data

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.

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 |

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)

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).

H = 800 + 200 × (1.061 − 0.9)/(1.08 − 0.9)
  = 800 + 200 × 0.894 = 979 AT/m

Step 3: ampere-turns

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)

✓ Answer checked by Claude Opus 5.5 (high effort)

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