Inductance of a winding on a core

What inductance do these turns have on this core?

The formula

L = \frac{μ_{0} \cdot μ \cdot N^{2} \cdot A}{l}

LaTeX

L = \frac{\mu_0 \cdot μ \cdot N^{2} \cdot A}{l}

Work it out

Fill from a part the library

N Turns

Turns of wire on the core.

μ Relative permeability

The core material's μᵣ from its datasheet. 1 for air; 10 for a powdered-iron toroid like a T106-2; 2000 and up for MnZn ferrite.

A Core cross-section

cm²

The area the flux passes through — the core's effective area Aₑ.

l Magnetic path length

The mean length of the flux path around the core — its effective length lₑ. For an air-cored solenoid, the winding length.

or see a worked example with typical numbers

Convert the area to square metres and the path length to metres.
Multiply μ₀ (4π × 10⁻⁷ H/m) by the relative permeability, by the turns squared, by the area, and divide by the path length. The result is in henries.
The core's A_L value is the same expression with N = 1: the inductance one turn would have. It is how a datasheet describes a core without committing to a winding.

All the flux stays in the core and the core is not gapped. A gapped core has a much lower effective permeability than its material, and a datasheet A_L value already accounts for the gap — prefer it where there is one.
The permeability is the initial, small-signal figure. It falls as the core is driven harder and collapses at saturation; a VIC choke carrying a resonant current may be well past small-signal.
For an air-cored winding this is the long-solenoid formula, and it overstates the inductance of a short fat coil. The Wheeler calculation handles that geometry.