Inductance of a toroid numerical

For a toroid of rectangular cross-section (inner radius a, outer radius b, height h), the field at radius r is B = μ₀NI/(2πr). The flux through one turn is Φ = ∫ B h dr from a to b = (μ₀NIh/2π) ln(b/a). Hence

L = NΦ/I = μ₀N²h ln(b/a)/(2π)

Given: N = 1250, a = 52 mm, b = 95 mm, h = 13 mm (ring thickness), μ₀ = 4π×10⁻⁷ H/m.

μ₀/(2π)    = 2×10⁻⁷ H/m
N²         = 1.5625×10⁶
ln(b/a)    = ln(95/52) = ln(1.827) = 0.6025
L = (2×10⁻⁷)(1.5625×10⁶)(0.013)(0.6025)
  = 4.0625×10⁻³ × 0.6025
  = 2.45×10⁻³ H

Answer: L ≈ 2.45 mH