Solenoid inductance with a core

What inductance does this coil have, wound on this core?

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

Method

1. Look up the core's relative permeability. Air is 1; a soft ferromagnetic core multiplies the field the same current produces by hundreds or thousands of times.
2. Convert the area to square metres and the length to metres.
3. Square the turns count — inductance grows with the square of N, not linearly, so doubling the winding quadruples the inductance.
4. Multiply μ₀ (4π×10⁻⁷ H/m) by the relative permeability, the turns squared and the area, then divide by the length.

Assumptions

• This is the ideal infinite-solenoid formula: the field is treated as uniform inside the coil and zero outside it. A real, finite coil's inductance is somewhat lower — a long, thin coil (length several times its diameter) is where this approximation is best.
• The core is unsaturated. Relative permeability collapses toward 1 as a ferromagnetic core saturates, and the values used here are the typical low-field figures for each material, not a saturation curve.
• The core fills the coil's cross-section uniformly. A rod that is narrower than the winding, with an air gap around it, has a much lower effective permeability than the bare material figure.