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Iron ring reluctance and flux with saw cut

Short answer

(i) Reluctance = 3.1416/(4π×10⁻⁷×2000×10⁻³) = 1.25×10⁶ AT/Wb. (ii) Flux = 1000/1.25×10⁶ = 0.8 mWb. (iii) With 1 mm gap, S_g = 7.958×10⁵ AT/Wb, total S = 2.046×10⁶ AT/Wb, flux = 0.489 mWb.

An iron ring of mean diameter 100 cm and cross sectional area 10 cm² is wound with 1000 turns and has μr = 2000. Compute (i) reluctance (ii) flux produced when the current through the coil is 1 A (iii) flux in the ring if a saw cut of 1 mm length is made, the current through the coil remaining the same.

Given: mean diameter 100 cm, so l = πD = 3.1416 m; A = 10 cm² = 1×10⁻³ m²; N = 1000; μr = 2000; I = 1 A; saw cut 1 mm. (The 1 mm removed from the iron path is ignored in the iron length; the change is 0.03 %.)

(i) Reluctance of the ring

S = l/(μ0 μr A)
  = 3.1416/(4π×10⁻⁷ × 2000 × 10⁻³)
  = 1.25×10⁶ AT/Wb

(ii) Flux with 1 A

mmf = N I = 1000 × 1 = 1000 AT
Φ   = mmf/S = 1000/1.25×10⁶ = 8.0×10⁻⁴ Wb = 0.8 mWb

(iii) Flux with the saw cut

The saw cut is an air gap in series with the iron.

Sg      = lg/(μ0 A) = 10⁻³/(4π×10⁻⁷ × 10⁻³)
        = 7.958×10⁵ AT/Wb
S_total = 1.25×10⁶ + 7.958×10⁵ = 2.046×10⁶ AT/Wb
Φ       = 1000/2.046×10⁶ = 4.89×10⁻⁴ Wb
QuantityWithout cutWith cut
Reluctance (AT/Wb)1.25×10⁶2.046×10⁶
Flux (mWb)0.8000.489

Answer: (i) 1.25×10⁶ AT/Wb; (ii) 0.8 mWb; (iii) 0.489 mWb. A 1 mm gap, only 0.03 % of the length, cuts the flux by about 39 %, since air has a very high reluctance.

✓ Answer checked by Claude Opus 5.5 (high effort)

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