How to use it
- Choose what to find: the energy, the current, or the inductance of a solenoid.
- Energy: enter the inductance and the current.
- Current: enter the inductance and the energy you want to store.
- Solenoid: enter the number of turns, the coil diameter and length in millimetres, μr of the core (1 for air) and the current.
Formulas
- Stored energy: E = ½ L I²
- Current for an energy: I = √(2E ÷ L)
- Flux linkage: Ψ = L × I (Wb)
- Long solenoid: L = μ₀ μr N² A ÷ ℓ, A = π d² ÷ 4, μ₀ = 1.25663706212 × 10⁻⁶ H/m
Worked example
A 10 mH inductor carries 2 A: E = ½ × 0.01 × 2² = 0.02 J (20 mJ), Ψ = 0.01 × 2 = 0.02 Wb.
An air coil of 100 turns, 20 mm in diameter and 50 mm long: A = π × 0.02² ÷ 4 = 3.1416 × 10⁻⁴ m², L = 1.25663706 × 10⁻⁶ × 100² × 3.1416 × 10⁻⁴ ÷ 0.05 = 78.957 µH. At 2 A it stores 0.000157914 J.
Tips
- Current counts squared. Most of the energy in a power inductor appears at the peak current.
- Check the saturation current. Above it a cored inductor loses inductance and the formula no longer holds.
- The solenoid formula overestimates short coils. For a coil only 2.5 times longer than wide, like the example, the real value is lower; use it as an upper estimate.