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Iron ring with saw cut, 4 mWb flux

Short answer

Iron length 0.9415 m, area 4.909e-4 m², B=8.15 T. AT_iron=Bl_i/(μ0 μr)=7631 AT, AT_gap=Blg/μ0=6485 AT, total 14116 AT. I=14116/500=28.2 A.

A ring of 30 cm mean diameter is made up of round iron rod 2.5 cm in diameter. A saw cut of 1 mm is made on the ring. It is uniformly wound with 500 turns of wire. Calculate the current required by the exciting coil to produce a total flux of 4 mWb. Assume a relative permeability of iron at this flux density as 800 (neglect leakage and fringing).

Treat the ring as a series magnetic circuit of iron plus the 1 mm saw-cut air gap. The exciting current is about 28.2 A.

Given: mean diameter D = 0.30 m; rod diameter d = 0.025 m; gap lg = 1 mm; N = 500; Φ = 4 mWb; μr = 800.

Step 1: dimensions

Mean length  l  = πD = π × 0.30 = 0.9425 m
Iron length  lᵢ = 0.9425 − 0.001 = 0.9415 m
Area         A  = πd²/4 = π(0.025)²/4 = 4.909×10⁻⁴ m²

Step 2: flux density B = Φ/A = 4×10⁻³ / 4.909×10⁻⁴ = 8.15 T. This is far above real saturation, but the question tells us to use μr = 800, so we follow it.

Step 3: ampere-turns

AT(iron) = B lᵢ/(μ0 μr)
         = 8.149 × 0.9415 /(4π×10⁻⁷ × 800)
         = 7631 AT
AT(gap)  = B lg/μ0
         = 8.149 × 10⁻³ /(4π×10⁻⁷)
         = 6485 AT
AT total = 14 116 AT

Step 4: current I = 14 116 / 500 = 28.2 A

Answer: I ≈ 28.2 A (iron 7631 AT + gap 6485 AT)

✓ Answer checked by Claude Opus 5.5 (high effort)

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