Inductance Calculator
This calculator finds the inductance of a solenoid from its physical dimensions and core material. It also returns the energy stored in the inductor's magnetic field at a given current, and the inductive reactance at a given frequency. Inductance is the property of an electrical conductor that opposes changes in current. When the current through an inductor changes, the changing magnetic flux induces a voltage that opposes the change, which is the principle behind Faraday's Law. The formula for a solenoid is L = mu_r x mu_0 x N squared x A / l, where mu_0 is the permeability of free space (4 x pi x 10 to the minus 7 henries per metre), mu_r is the relative permeability of the core material (1 for air or vacuum, around 200 to 5000 for iron, much higher for ferrite and mumetal), N is the number of turns, A is the cross-sectional area of the core in square metres, and l is the length of the winding in metres. Inductors are used in filters, transformers, switched-mode power supplies, chokes, and oscillator circuits. The inductive reactance is XL = 2 x pi x f x L in ohms, which is the effective resistance the inductor presents to alternating current at frequency f. At 50 Hz (NZ mains frequency), a 12.57 uH inductor has a reactance of about 0.00395 ohms. The energy stored in the magnetic field when carrying current I is E = 0.5 x L x I squared joules. Enter the solenoid dimensions and optional current and frequency values below.
mu_0 = 4 x pi x 10^-7 H/m. Formula assumes an ideal solenoid with uniform field and no fringing effects. Real inductors differ; add 10-20% for practical designs.
How it works
The inductance formula is L = mu_r x mu_0 x N squared x A / l, where mu_0 = 4 x pi x 10^-7 H/m. If a current I is provided, stored energy is E = 0.5 x L x I squared joules. If a frequency f is provided, inductive reactance is XL = 2 x pi x f x L ohms. The formula assumes the solenoid is long compared with its radius (l much greater than the radius), so that the magnetic field inside is approximately uniform and fringing effects at the ends are negligible.
Worked example
A solenoid has 100 turns (N = 100), a length of 0.1 m, a cross-sectional area of 0.0001 m squared (1 cm squared), and an air core (mu_r = 1). Inductance is L = 1 x 4 x pi x 10^-7 x 100 squared x 0.0001 / 0.1 = 4 x pi x 10^-7 x 10000 x 0.0001 / 0.1 = 4 x pi x 10^-7 x 10 = 1.2566 x 10^-5 H = 0.01257 mH (12.57 uH). If carrying 1 A, the stored energy would be E = 0.5 x 1.2566 x 10^-5 x 1 = 6.28 x 10^-6 J = 0.00628 mJ. These match the default values above.
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