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Number of turns of iron-core inductor

Short answer

Using L = μ0 μr N² A/l, N = √(L l/(μ0 μr A)). l=0.4 m, A=50×10^-6 m², μr=1500, L=0.5 H yields N ≈ 1457 turns.

An inductor is to be made with copper wire wound on a circular iron core having mean length of 40 cm with cross-sectional area of 50 sq mm. If the required value of inductance is 500 mH, calculate the number of turns required given that relative permeability of the core is 1500.

For a coil on a closed magnetic core of mean length l and area A, the inductance is

L = μ₀ μr N² A / l

so N = √( L l / (μ₀ μr A) )

Given

  • L = 500 mH = 0.5 H
  • l = 40 cm = 0.4 m
  • A = 50 mm² = 50 × 10⁻⁶ m²
  • μr = 1500, μ₀ = 4π × 10⁻⁷ H/m

Calculation

  • μ₀ μr = 4π × 10⁻⁷ × 1500 = 1.885 × 10⁻³ H/m
  • Denominator: μ₀ μr A = 1.885 × 10⁻³ × 50 × 10⁻⁶ = 9.425 × 10⁻⁸
  • Numerator: L l = 0.5 × 0.4 = 0.2
  • N² = 0.2 / (9.425 × 10⁻⁸) = 2.122 × 10⁶
  • N = √(2.122 × 10⁶) = 1456.7

Answer: about 1457 turns (say 1460 turns). Leakage and fringing are neglected.

✓ Answer checked by Claude Opus 5.5 (high effort)

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