Coil inductance (Wheeler)

How much inductance will this coil have when I have wound it?

L = \frac{0.8 \cdot \left(N \cdot r\right)^{2}}{6 \cdot r + 9 \cdot l + 10 \cdot d}

Method

1. Work in inches. Wheeler’s formula is dimensional — the constants 0.8, 6, 9 and 10 assume inches and return microhenries, and it gives nonsense in millimetres.
2. Find the mean radius: the former radius plus half the winding depth. Using the former radius alone underestimates a thick winding badly.
3. Multiply turns by mean radius and square the product. Inductance goes as turns squared, so doubling the turns quadruples the inductance — this is the term that dominates.
4. Form the denominator: six times the mean radius, plus nine times the winding length, plus ten times the winding depth. Spreading the same turns over a longer former reduces the inductance.
5. Multiply the numerator by 0.8 and divide. The answer is in microhenries.

Assumptions

• Air core. A ferrite or iron core multiplies this by the core’s effective permeability, which can be hundreds — this figure is then not even the right order of magnitude.
• Wheeler’s fit is accurate to about one per cent for coils whose length and diameter are broadly comparable. A very long thin solenoid or a very short flat pancake falls outside what it was fitted to.
• The winding is even and closely packed. Gaps, uneven layers and random winding all move the real figure.
• This is the low-frequency inductance. Approaching the coil’s self-resonance the apparent inductance rises and then inverts; above it the coil is a capacitor.
• Nothing magnetic is nearby. A steel bench or a clamp within a coil diameter changes the answer.