Skip to main content

Iron ring current for 0.8 Wb air-gap flux

Short answer

Iron reluctance = 1.996/(4π×10⁻⁷×400×0.01) = 3.971×10⁵ AT/Wb, air gap reluctance = 0.004/(4π×10⁻⁷×0.01) = 3.183×10⁵ AT/Wb. Total S = 7.154×10⁵ AT/Wb. mmf = 0.8×7.154×10⁵ = 5.723×10⁵ AT. Current I = 5.723×10⁵/250 = 2.29×10³ A.

An iron ring has a mean length of 2 m and cross-sectional area of 0.01 m². It has a radial air gap of 4 mm. The ring is wound with 250 turns. What dc current would be needed in the coil to produce a flux of 0.8 Weber in the air gap? Assume that μr = 400.

The coil must supply the mmf for the iron and the air gap, both carrying 0.8 Wb.

Given: l = 2 m; A = 0.01 m²; gap lg = 4 mm; N = 250; Φ = 0.8 Wb; μr = 400.

Note: the figures give B = Φ/A = 80 T, which is physically impossible (iron saturates near 2 T). We follow the data as given, using the constant μr = 400, as the question intends.

Step 1: reluctances (iron length = 2 − 0.004 = 1.996 m)

Sᵢ = 1.996/(4π×10⁻⁷ × 400 × 0.01) = 3.971×10⁵ AT/Wb
Sg = 0.004/(4π×10⁻⁷ × 0.01)       = 3.183×10⁵ AT/Wb
S  = Sᵢ + Sg = 7.154×10⁵ AT/Wb

Step 2: mmf F = Φ × S = 0.8 × 7.154×10⁵ = 5.723×10⁵ AT

Step 3: current I = F/N = 5.723×10⁵/250 = 2289 A

(If the whole 2 m is used as the iron length, S = 7.162×10⁵ AT/Wb and I = 2292 A, a negligible difference.)

Answer: I ≈ 2.29×10³ A (about 2290 A)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machine chapter 1 questions →