Coil Inductance Calculator
From the number of turns, the coil radius, and the coil length, get the inductance of a long single-layer air-core solenoid in microhenries and millihenries.
Three inputs, one inductance
Enter the number of turns, the coil radius, and the coil length and the calculator returns the inductance L = µ₀ × N² × A / l in both µH and mH.
Long-coil approximation
This formula assumes a long, thin solenoid (length much greater than the radius). Short, fat coils read higher than the true value.
What is a coil inductance calculator?
Turns, radius, and length in, inductance out
A solenoid is a coil of wire wound in a single layer around a cylinder. A coil inductance calculator turns three measurements into the one number that matters: the inductance, or how strongly the coil opposes a change in the current flowing through it. More turns, a wider coil, or a shorter winding all push the inductance up. That makes these three inputs all you need for winding an air-core RF coil, sizing the inductor in an LC tuned circuit, building a homemade choke, or checking a textbook solenoid problem.
Enter the number of turns, the coil radius, and the coil length to get the inductance in microhenries and millihenries instantly.
One formula built from the turns, the geometry, and the constant µ₀ (the permeability of free space, 4π × 10⁻⁷ H/m).
L = µ₀ × N² × A / l, with A = π × r²N is the number of turns, A = π × r² is the cross-sectional area from the coil radius r (the calculator converts your centimetres to metres), l is the coil length in metres, and µ₀ is the fixed permeability of free space. Inductance grows with the square of the turns and with the cross-sectional area, and falls as the coil gets longer — so a tightly packed, wide coil with many turns gives the most inductance.
Suppose you wind 100 turns on a former of radius 1 cm over a length of 5 cm.
Convert and find the area
r = 1 cm = 0.01 m, so A = π × 0.01² = 3.1416 × 10⁻⁴ m². Length l = 5 cm = 0.05 m.
Apply the formula
L = (4π × 10⁻⁷) × 100² × 3.1416 × 10⁻⁴ / 0.05 = 7.8957 × 10⁻⁵ H.
Scale the units
7.8957 × 10⁻⁵ H = 78.956835 µH = 0.078957 mH.
The inductance tells you how strongly the coil resists a change in current — a higher number means it stores more magnetic energy and reacts more sluggishly when the current rises or falls. The example value (about 78.96 µH for 100 turns, a 1 cm radius, and a 5 cm length) sits in the microhenry range typical of small air-core RF coils. The single most useful insight is that the turns enter as a square: keep the same coil but double the turns and the inductance quadruples, which is why adding a few turns changes the value far more than you might expect. Widen the coil and the inductance climbs with the area (the square of the radius); stretch the same number of turns over a longer former and it drops. The microhenry and millihenry readouts are the same value at two scales — 1000 µH = 1 mH — so use whichever matches the part you are comparing against. To hit a target inductance, the turns are usually your finest control because of that squared relationship.
The formula is the long-solenoid approximation, so a couple of practical points are worth keeping in mind.
Long air-core coils and consistent units
This formula assumes a long, single-layer, air-core solenoid where the length is much greater than the radius. For a short, fat coil it overestimates the inductance — a more accurate result needs Wheeler's or Nagaoka's correction. It also assumes air (or vacuum) inside; a ferrite or iron core multiplies the inductance by the core's relative permeability, which this air-core model does not include. Keep the radius and length in centimetres as labelled; metres or millimetres there will throw the answer off by powers of ten.