Solenoid Inductance Calculator

Enter the number of turns, coil diameter, coil length, and core permeability to find a solenoid's self-inductance using L = μ0 μr N² A / l.

Quick Facts

Inductance formula
L = μ₀ μr N² A / l
μ₀ = 4π×10⁻⁷ H/m is the permeability of free space; μr is the core's relative permeability (1 for air).
Turns dominate
L ∝ N²
Doubling the number of turns on a fixed geometry quadruples the inductance.
Validity range
Best for l/d ≳ 4
The long-solenoid formula assumes a coil much longer than it is wide; short coils need Nagaoka's correction.
Typical μr
Air ≈ 1, iron ≈ 200-5,000
Ferromagnetic cores multiply inductance directly through μr.

Your Results

Calculated
Self-Inductance
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L = μ₀ μr N² A / l
Coil Cross-Sectional Area
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A = π × (diameter / 2)²
Turns per Unit Length
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n = N / l
Length-to-Diameter Ratio
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l / d — indicates long-solenoid validity

Ready

Enter coil turns, diameter, length, and core permeability, then press Calculate.

How Solenoid Inductance Is Calculated

A solenoid is a coil of wire wound in tightly spaced turns, typically around a cylindrical form. When a current I flows through it, the coil produces a nearly uniform magnetic field inside its core. For an ideal solenoid — one whose length is much greater than its diameter, with turns wound evenly and closely together — Ampère's law gives an interior field of B = μ₀μrnI, where n = N/l is the number of turns per unit length. Multiplying by the cross-sectional area A gives the flux through one turn, and multiplying by the total number of turns N gives the flux linkage. Dividing that flux linkage by the current defines the self-inductance: L = μ₀μrN²A / l, where A = π(d/2)² is the coil's cross-sectional area from its diameter d.

Deriving the formula from Ampère's and Faraday's laws

Starting from Ampère's law for a long solenoid, the interior magnetic field is B = μ₀μrNI/l. The flux through a single turn of area A is Φ = BA, and because all N turns are linked by essentially the same flux, the total flux linkage is λ = NΦ = μ₀μrN²AI/l. Since inductance is defined by λ = LI, the I terms cancel and you're left with L = μ₀μrN²A/l — an inductance that depends only on geometry and core material, not on the current itself.

Turns, geometry, and core material

Three factors control the result: the number of turns N (inductance scales with N², so doubling the turns quadruples L), the coil's cross-sectional area A and length l, and the core's relative permeability μr. An air or vacuum core has μr ≈ 1; inserting a ferromagnetic core (iron, ferrite, permalloy) can raise μr into the hundreds or thousands, multiplying the inductance by that same factor — which is why transformer and inductor cores use magnetic materials rather than air.

When the long-solenoid approximation breaks down

This formula assumes the field lines close entirely inside a long, thin coil, which holds well when the length is at least about four times the diameter. For short, fat coils (length comparable to or smaller than the diameter), fringing fields at the ends become significant and the ideal formula overestimates inductance — sometimes by 10% or more. Precision work on short coils applies Nagaoka's coefficient, a correction factor derived from elliptic integrals that accounts for the coil's finite length-to-diameter ratio.

Frequently Asked Questions

What is the formula for a solenoid's inductance?
For an ideal (long) solenoid, self-inductance is L = μ₀μrN²A/l, where N is the number of turns, A = π(d/2)² is the cross-sectional area from the coil diameter d, l is the coil's length, μ₀ = 4π×10⁻⁷ H/m is the permeability of free space, and μr is the core material's relative permeability (1 for air or vacuum).
Why does the number of turns affect inductance so strongly?
Inductance scales with the square of the turns count (L ∝ N²), because each additional turn both adds its own flux linkage and links to the flux produced by every other turn. Doubling the number of turns on the same core quadruples the inductance, while doubling only the length halves it — turns matter far more than length.
How much does a core material change the inductance?
Relative permeability (μr) multiplies the result directly. An air-core coil has μr ≈ 1. Inserting a ferrite or laminated iron core typically raises μr into the range of a few hundred to several thousand, boosting inductance by that same factor for the same winding geometry — which is why practical inductors and transformers use magnetic cores instead of air.
When is this formula not accurate?
The formula assumes an idealized long solenoid where the length is at least about 4 times the diameter and the winding is tightly and evenly spaced. Short, wide coils have significant fringing fields at the ends, and the ideal formula can overestimate inductance by 10% or more; for those cases, use Nagaoka's coefficient or a numerical field solver for an accurate value.