Question

A solenoid 30 cm long is wound with 250 turns of wire. The cross-sectional area of the coil is 3.5 cm^2. What is the self-inductance of the solenoid?

Answer 1

The value is
`L = 9.164 *10^(-5) \ H`

Step-by-step explanation:

From the question we are told that

The length of the solenoid is
`l = 30 \ cm = 0.3 \ m`

The number of turns is
`N = 250 \ turns`

The cross-sectional area is
`A = 3.5 \ cm^2 = 3.5 *10^(-4) \ m^2`

Generally the self inductance of the solenoid is

`L = (\mu_o * N^2 * A )/( l )`

Here
`\mu_o` is the permeability of free space with value
`\mu_o = 4\pi *10^(-7) \ T \cdot m/A`

So

`L = ( 4 \pi*0^(-7) * 250^2 * 3.5 *10^(-4) )/( 0.3 )`

=>
`L = 9.164 *10^(-5) \ H`