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.

Explanation:

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

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

where is the circuit diagram ?