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Iron ring current for 0.82 Wb air-gap flux

Short answer

Iron reluctance = 2.495/(4π×10⁻⁷×5000×0.04) = 9.927×10³ AT/Wb, air gap reluctance = 0.005/(4π×10⁻⁷×0.04) = 9.947×10⁴ AT/Wb. Total S = 1.094×10⁵ AT/Wb. mmf = 0.82×1.094×10⁵ = 8.97×10⁴ AT. Current I = 8.97×10⁴/200 = 449 A.

An iron ring has a mean length of 2.5 m and A = 0.04 m². It has a radial air gap of 5 mm. The ring is wound with 200 turns, what dc current would be needed in the coil to produce a flux of 0.82 Wb in the air gap. Assume that μr = 5000.

The mmf of the coil is the sum of the mmf for the iron part and the air gap, with the same flux through both.

Given: l = 2.5 m; A = 0.04 m²; gap lg = 5 mm; N = 200; Φ = 0.82 Wb; μr = 5000.

Note: B = 0.82/0.04 = 20.5 T is far beyond real saturation. The calculation follows the constant μr = 5000 given in the question.

Step 1: reluctances (iron length = 2.5 − 0.005 = 2.495 m)

Sᵢ = 2.495/(4π×10⁻⁷ × 5000 × 0.04) = 9.927×10³ AT/Wb
Sg = 0.005/(4π×10⁻⁷ × 0.04)        = 9.947×10⁴ AT/Wb
S  = 1.0940×10⁵ AT/Wb

Step 2: mmf F = Φ S = 0.82 × 1.0940×10⁵ = 8.971×10⁴ AT

Step 3: current I = F/N = 8.971×10⁴/200 = 448.5 A

The air gap supplies 91 % of the mmf: 0.82 × 9.947×10⁴ = 81.6×10³ AT.

Answer: I ≈ 449 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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