Inductance
30.843 µH
Mode
Coil Design
Inductance
30.843 µH
Solenoid formula: L = µ0 × N² × A / l
Inputs
Result
r = 0.0125m, A = π(0.0125)² = 4.91×10⁻⁴ m². L = 4π×10⁻⁷ × 2500 × 4.91×10⁻⁴ / 0.05 = 30.8µH.
Inputs
Result
XL = 2π × 1000 × 0.001 = 6.28Ω.
L = µ0 × N² × A / l, where µ0 = 4π×10⁻⁷ H/m, N = turns, A = cross-section area, l = coil length. A 50-turn coil, 25mm diameter, 50mm long has L ≈ 48.4µH.
XL = 2πfL: the opposition to AC current by an inductor, measured in ohms. It increases with frequency—inductors block high frequencies and pass low frequencies.
E = ½LI². A 1mH inductor carrying 1A stores 0.5mJ. Energy scales with current squared—doubling current quadruples stored energy.
More turns (L ∝ N²), larger diameter, shorter coil, or add a ferrite core (µr = 100–5000). A ferrite core boosts inductance by its relative permeability.
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Last Updated: Mar 20, 2026
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