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Magnetic circuit: current in coil B

Short answer

Central flux = 4 mWb. Reluctances: S_c = 2.653×10⁵ AT/Wb, S_o = 1.592×10⁶ AT/Wb. Total mmf = 4×10⁻³×2.653×10⁵ + 2×10⁻³×1.592×10⁶ = 4244 AT. Coil A mmf = 1200 AT. Coil B mmf = 3044 AT, so I = 3044/300 = 10.15 A.

For the magnetic circuit shown in the figure below, find out the current to be passed through coil B so that magnetic flux in CD section is 2 mWb. Given μr = 1000. [Figure: a rectangular magnetic core with corners E (top left), C (top right), F (bottom left), D (bottom right) and a central limb B (top) to A (bottom). Coil B with N1 = 300 turns and current I1 and coil A with N2 = 400 turns and current I2 are wound on the central limb. The central limb has cross-section A1 = 6 cm² and the two outer limbs (E-F side and C-D side) have cross-section A2 = 3 cm² each.] Given: I2 = 3 A, A1 = 6 cm², A2 = 3 cm², AB = CD = EF = 20 cm, BC = AD = BE = AF = 20 cm.

The central limb is common to both outer paths, which are in parallel. We find the mmf needed to drive the flux through the centre and one outer path and subtract the mmf of coil A.

Reading the figure and assumptions

  • Central limb AB: 20 cm, area A1 = 6 cm² = 6×10⁻⁴ m², carries both coils (coil B: N1 = 300; coil A: N2 = 400, I2 = 3 A).
  • Each outer path (B–C–D–A, and B–E–F–A) has length 20 + 20 + 20 = 60 cm and area A2 = 3×10⁻⁴ m².
  • The two outer paths are identical, so each carries half of the centre flux. Φ in CD = 2 mWb, so centre flux = 4 mWb.
  • μr = 1000. Both coils are assumed to magnetise in the same direction (aiding).

Equivalent magnetic circuit

 F (coils) ---[S_centre]---+---[S_outer]---+
                           |               |
                           +---[S_outer]---+

Step 1: reluctances

S_c = l/(μ0 μr A1) = 0.2/(4π×10⁻⁷×1000×6×10⁻⁴)
    = 2.653×10⁵ AT/Wb
S_o = 0.6/(μ0 μr A2) = 0.6/(4π×10⁻⁷×1000×3×10⁻⁴)
    = 1.592×10⁶ AT/Wb

Step 2: total mmf required

F = Φ_c S_c + Φ_o S_o
  = 4×10⁻³ × 2.653×10⁵ + 2×10⁻³ × 1.592×10⁶
  = 1061 + 3183 = 4244 AT

Step 3: coil B

N2 I2 = 400 × 3 = 1200 AT
N1 I1 = 4244 − 1200 = 3044 AT
I1    = 3044/300 = 10.15 A

(If coil A were opposing, I1 would be (4244 + 1200)/300 = 18.1 A.)

Answer: I1 ≈ 10.15 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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