An air-core solenoid consists of 200 turns of wire wound on a form that is 71 cm long and has a inner diameter of 6 cm. A current of 22 A is established in this solenoid. What is its self-inductance? The permeability of free space is 4 π * 10-7 T · m/A. Answer in units of mH.

To calculate this, you need to use the correct expression which is the following:

L = Uo * N² * I * A / l

Where:

Uo = permeability of free space

N: number of turns

I: current

A: Cross sectional area of wire

l: length

The cross sectional area is calculated with the expression:

A = πr²

Replacing:

A = π * (3) ² = 28.3 cm² or 2.83x10^-3 m²

Therefore the self inductance is:

L = 4πx10^-7 * 2.83x10^-3 * 200² / 0.71

L = 2x10^-4