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Turns of air-cored coil

Short answer

For an air-core solenoid, L = μ0 N² A/l, so N = √(L l/(μ0 A)). With L=100 µH, l=0.025 m, A=2×10^-4 m²: N ≈ √9947 ≈ 100 turns.

An air cored coil is 2.5cm long and has an average cross-sectional area of 2cm². Determine the number of turns if the coil has an inductance of 100 μH.

For an air-cored coil (solenoid), μr = 1 and

L = μ₀ N² A / l ⟹ N = √( L l / (μ₀ A) )

Given

  • l = 2.5 cm = 0.025 m
  • A = 2 cm² = 2 × 10⁻⁴ m²
  • L = 100 μH = 100 × 10⁻⁶ H
  • μ₀ = 4π × 10⁻⁷ H/m

Calculation

  • Numerator: L l = 100 × 10⁻⁶ × 0.025 = 2.5 × 10⁻⁶
  • Denominator: μ₀ A = 4π × 10⁻⁷ × 2 × 10⁻⁴ = 2.513 × 10⁻¹⁰
  • N² = (2.5 × 10⁻⁶) / (2.513 × 10⁻¹⁰) = 9947.2
  • N = √9947.2 = 99.7

Answer: about 100 turns.

✓ Answer checked by Claude Opus 5.5 (high effort)

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