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Three-coil core: air-gap flux

Short answer

Net mmf is the algebraic sum: 335×1.6 + 600×4 − 600×3 = 1136 AT. Iron reluctance is 1.414×10⁶ AT/Wb, air-gap reluctance 9.947×10⁶ AT/Wb, total 1.136×10⁷ AT/Wb. Flux = 1136 / 1.136×10⁷ = 1.0×10⁻⁴ Wb (0.1 mWb).

A rectangular iron core is shown in figure 1. It has a mean length of magnetic path of [?] cm, cross-section of (2 cm × 2 cm), relative permeability of 1400 and an air-gap of 5 mm cut in the core. The three coils carried by the core have number of turns Na = 335, Nb = 600 and Nc = 600; and the respective currents are 1.6 A, 4 A and 3 A. The directions of the currents are as shown in the figure. Find the flux in the air-gap. [Figure 1: a rectangular iron core with an air gap in the top limb; coil a (current Ia) on the upper left, coil b (current Ib) on the right limb, and coil c (current Ic) on the bottom limb; current directions as shown in the figure.]

The flux in the gap is the net mmf of the three coils divided by the total reluctance.

Given and assumptions

  • The mean path length is unreadable in the question. The standard textbook version of this problem uses l = 100 cm (1 m), so that value is used here.
  • Cross-section 2 cm × 2 cm = 4×10⁻⁴ m²; μr = 1400; gap lg = 5 mm.
  • Current directions (from the figure): coils a and b magnetise the core in the same direction, and coil c opposes them.

Step 1: net mmf

Fa = 335 × 1.6 = 536 AT   (+)
Fb = 600 × 4   = 2400 AT  (+)
Fc = 600 × 3   = 1800 AT  (−)
F  = 536 + 2400 − 1800 = 1136 AT

Step 2: reluctances

Sᵢ = (1.0 − 0.005)/(4π×10⁻⁷ × 1400 × 4×10⁻⁴)
   = 1.414×10⁶ AT/Wb
Sg = 0.005/(4π×10⁻⁷ × 4×10⁻⁴)
   = 9.947×10⁶ AT/Wb
S  = 1.136×10⁷ AT/Wb

Step 3: flux Φ = F/S = 1136/1.136×10⁷ = 1.0×10⁻⁴ Wb

Answer: Φ ≈ 0.1 mWb (B = Φ/A = 0.25 T). If instead all three coils aided each other, F = 4736 AT and Φ ≈ 0.42 mWb. Because the air gap dominates the reluctance, the result depends only slightly on the assumed length: with l = 60 cm, Φ ≈ 0.105 mWb.

✓ Answer checked by Claude Opus 5.5 (high effort)

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