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Circular core inductance and flux

Short answer

Total reluctance: iron 1.958×10⁵ AT/Wb, gap 6.366×10⁶ AT/Wb, sum 6.562×10⁶ AT/Wb. Inductance L = N²/S = 500²/6.562×10⁶ = 0.0381 H = 38.1 mH. At 10 A, mmf = 5000 AT, flux = 5000/6.562×10⁶ = 7.62×10⁻⁴ Wb (0.762 mWb).

A circular iron core has a cross-sectional area of 5 sq.cm. and mean length of 25 cm including an air gap of 4 mm. The core is wound with 500 turns of winding. Calculate the inductance of the coil. If a dc current of 10 Ampere passed through the coil, calculate magnetic flux in the core. Given that relative permeability of the core is 2000.

Inductance of a coil on a magnetic core is L = N²/S, where S is the total reluctance of the core including the air gap.

Given: A = 5 cm² = 5×10⁻⁴ m²; total mean length l = 25 cm (including the gap); gap lg = 4 mm; N = 500; μr = 2000; I = 10 A.

Step 1: reluctances (iron length = 0.25 − 0.004 = 0.246 m)

Sᵢ = 0.246/(4π×10⁻⁷ × 2000 × 5×10⁻⁴) = 1.958×10⁵ AT/Wb
Sg = 0.004/(4π×10⁻⁷ × 5×10⁻⁴)        = 6.366×10⁶ AT/Wb
S  = 6.562×10⁶ AT/Wb

Step 2: inductance L = N²/S = 500²/6.562×10⁶ = 2.5×10⁵/6.562×10⁶ = 0.0381 H = 38.1 mH

Step 3: flux at 10 A

mmf = N I = 500 × 10 = 5000 AT
Φ   = 5000/6.562×10⁶ = 7.62×10⁻⁴ Wb

The flux density is B = Φ/A = 1.52 T. This is acceptable for iron, so a constant μr = 2000 is a fair assumption.

Answer: L ≈ 38.1 mH; Φ ≈ 0.762 mWb

✓ Answer checked by Claude Opus 5.5 (high effort)

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