Question

You build a solenoid containing 500 windings over a 0.20-m length, with a loop radius of 0.025 m for each winding. the current in the unit is 3.0

a. what is the inductance of the solenoid?

Answer 1

The inductance of a solenoid is given by:

`L= (\mu_0 N^2 A)/(l)`
where

`\mu_0` is the vacuum permeability
N is the number of turns of the solenoid
A is the area of one loop of the solenoid
l is its length

For the solenoid in our problem, N=500, and the radius of one loop is r=0.025 m, so the area of one loop is

`A=\pi r^2 = \pi (0.025 m)^2 = 1.96 \cdot 10^(-3) m^2`
and its length is l=0.20 m

So the inductance of the solenoid is

`L= ((4 \pi \cdot 10^(-7) NA^(-2))(500)^2(1.96 \cdot 10^(-3) m^2))/( 0.20 m)=3.08 \cdot 10^(-3) H`