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Solenoid Inductance Calculator (L = μ₀μrN²A/l, with Wheeler short-coil estimate)

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Understanding the solenoid inductance calculator — concept and principle

This calculator finds the inductance of a solenoid coil from the number of turns, length, diameter and core permeability using the long-solenoid formula, and adds Wheeler's estimate for air-core coils with end effects. It also gives the field inside, the stored energy, the cross-section area and turns per centimetre. Electronics hobbyists and students use it to wind coils.

Solenoid inductanceA 200-turn air-core coilN200Length · diameter10 cm · 2 cmμ_r · I1 · 1 AL = μ₀ N² A ÷ ℓInductance (ideal)157.914 µHWheeler estimate144.477 µHField inside2.513 mT
The long-solenoid formula ignores the ends; Wheeler’s correction brings it down about 9 % for this 5 : 1 coil.

How it is calculated

L = μ₀ μ_r N² A ÷ l, with A = π d²/4 and μ₀ = 4π × 10⁻⁷ H/m

Wheeler (air core): L (µH) = r² N² ÷ (9r + 10l), r and l in inches

Field inside B = μ₀ μ_r N I ÷ l; energy = ½ L I²

The principle

Current in a coil creates a magnetic field inside it; in a long solenoid the field is nearly uniform and proportional to turns per length. The flux through each turn is that field times the area, and inductance is the total flux linkage per ampere, N × flux ÷ I. Because both the field and the number of linked turns scale with N, L grows with N². The simple formula ignores the weaker field near the ends, so it overestimates short coils; Wheeler's empirical formula corrects for this in air-core coils.

Worked example

Default inputs: N = 200, length 10 cm, diameter 2 cm, μ_r = 1, I = 1 A.

  1. A = π × 1² = 3.142 cm²; turns per cm = 20; length ÷ diameter = 5.
  2. L = 4π × 10⁻⁷ × 200² × 3.142 × 10⁻⁴ ÷ 0.1 = 157.914 µH.
  3. Wheeler estimate = 144.477 µH, about 9% lower because of end effects.
  4. B = 4π × 10⁻⁷ × 200 × 1 ÷ 0.1 = 2.513 mT; energy = 0.07896 mJ.
How Inductance, long-solenoid formula (µH) changes with Number of turns NHow Inductance, long-solenoid formula ch…020040060080050100150200250300400Example: 157.914Number of turns N
Each point is the calculator re-run with only “Number of turns N” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Number of turns NInductance, long-solenoid formula (µH)
509.8696
10039.4784
15088.8264
200157.914
250246.74
300355.306
400631.655

How to read the result

For a coil five times longer than its diameter, the two estimates differ by under 10%; for shorter coils the gap grows and Wheeler's value is the better guide. A ferromagnetic core multiplies L, but its effective permeability depends on the core shape, any air gap and the current, so cored results are rough.

Common mistakes

Keep in mind: Measure the finished coil with an LCR meter; wire spacing, layers and self-capacitance change the result.

Frequently asked questions

Why does inductance grow with N²?

More turns make more field and also link more of it.

When is Wheeler's formula accurate?

For single-layer air-core coils with length greater than about 0.8 times the radius, to about 1%.

Does a core always multiply L by μ_r?

Only for a closed core; open rod cores give much less.

How is energy stored?

In the magnetic field, ½LI².

Sources: OpenStax, University Physics Volume 2 (2016), section 14.2 "Self-Inductance and Inductors" · H. A. Wheeler, single-layer coil inductance formula (standard approximation)

Checked 9 Nov 2026. OpenStax, University Physics Volume 2 (2016), §14.2 “Self-Inductance and Inductors”, Eq. 14.13 (L = μ₀N²A/l, long solenoid, end effects ignored). Wheeler’s single-layer air-coil approximation and energy ½LI²: standard formulas.

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