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Calcimator

Inductance Calculator

Calculate the inductance of a solenoid from turns, length, cross-section area, and core permeability using L = μ₀μᵣN²A/l.

About this calculator

This calculator finds the self-inductance of a solenoid coil from its winding and core geometry, using L = μ₀μᵣN²A/l — the standard formula for a long, tightly wound solenoid where μ₀ is the permeability of free space, μᵣ is the core material's relative permeability, N is the turn count, A is the cross-sectional area, and l is the coil's length. Because inductance scales with the square of turns but only linearly with length and area, adding turns is by far the most powerful way to increase inductance — doubling the turns quadruples L, while doubling the length only halves it. The core material matters just as much: an air core (relative permeability of 1) gives a baseline value, while an iron or ferrite core can multiply that by hundreds or thousands, which is why real inductors and transformers use magnetic cores.

The calculator also reports reactance at 60 Hz and 1 kHz (X_L = 2πfL), showing how the same inductor blocks AC current increasingly as frequency rises, and automatically rescales the displayed inductance into nH, μH, mH, or H depending on magnitude so the number stays readable. The underlying formula assumes an idealized long solenoid (length much greater than diameter) with uniform winding and no fringing or leakage flux at the ends — a short, fat coil in practice will have somewhat lower inductance than this ideal formula predicts.

Inputs

ft
sq ft

Results

Inductance

125.66

UnitμH
Reactance at 60 Hz0.05 Ω
Reactance at 1 kHz0.79 Ω
How to Use This Calculator
  1. Enter Number of Turns, Solenoid Length, and Cross-Section Area.
  2. Set Relative Permeability (μᵣ).
  3. Review the Inductance result.
  4. Use Unit and Reactance at 60 Hz (Ω) to inform your decision.
  5. Use the chart to visualize the results and explore different scenarios by adjusting inputs.

How the result changes with Number of Turns

Number of TurnsInductance
5031.42
7570.69
150282.74
250785.4

What each input means

Number of Turns
Total number of wire turns in the solenoid coil.
Solenoid Length
Length of the solenoid (not the wire length).
Cross-Section Area
Cross-sectional area of the solenoid core. For a circular core of radius r: A = πr².
Relative Permeability (μᵣ)
Relative permeability of the core material. Air/vacuum = 1, iron ≈ 200–5000, ferrite ≈ 1000–10000.

How this is calculated

Worked example, using the default values

  1. Identify Input Parameters
    4 parameters
    Number of Turns = 100, Solenoid Length = 0.1, Cross-Section Area = 0.001, Relative Permeability (μᵣ) = 1 = 4 input(s) provided
  2. Calculate Inductance
    Inductance = l
    125.6637 = 125.6637
  3. Calculate Unit
    Unit = displayUnit
    μH = μH
  4. Calculate Reactance at 60 Hz
    Reactance at 60 Hz
    0.0474 = 0.0474

Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.

Frequently Asked Questions

Why does adding turns increase inductance so much more than adding length or area?

In L = μ₀μᵣN²A/l, the number of turns N appears squared, while cross-sectional area A and length l each appear only to the first power (and length is in the denominator). That means doubling the turn count quadruples inductance, while doubling the area only doubles it and doubling the length actually halves it — turns are by far the most powerful lever in this formula.

How much difference does the core material make?

The relative permeability μᵣ multiplies directly into the inductance formula, so it scales the result linearly. An air or vacuum core uses μᵣ = 1 as a baseline, while iron cores (roughly 200-5,000) or ferrite cores (roughly 1,000-10,000) can multiply that same coil's inductance by two to four orders of magnitude — which is exactly why real inductors and transformers use magnetic cores instead of air.

Why is the reactance at 1 kHz so much higher than at 60 Hz for the same coil?

Reactance is computed as X_L = 2πfL, so for a fixed inductance it scales directly with frequency. Since 1 kHz is roughly 16.7 times higher than 60 Hz, the reactance at 1 kHz comes out proportionally higher too — this is the basic reason inductors block high-frequency AC signals far more strongly than low-frequency ones.

Why might a real coil have less inductance than this calculator predicts?

The L = μ₀μᵣN²A/l formula assumes an idealized long solenoid — one whose length is much greater than its diameter, with perfectly uniform winding and no flux leakage at the ends. A short, fat coil doesn't meet that assumption: field lines fringe and leak out near the open ends rather than staying confined along the coil's axis, so its measured inductance will typically come in lower than this ideal formula's result.

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