Question

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 900 loops of wire wound on a rod 6 cm long with radius 1 cm?

Answer 1

The self-inductance is
`L = 0.0053 \ H`

Step-by-step explanation:

From the question we are told that

The number of loops is
`N = 900`

The length of the rod is
`l =6 \ cm = 0.06 \ m`

The radius of the rod is
`r = 1 \ cm = 0.01 \ m`

The self-inductance for the solenoid is mathematically represented as

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

Now the cross-sectional of the solenoid is mathematically evaluated as

`A = \pi r^2`

substituting values

`A =3.142 * 0.01 ^2`

`A = 3.142 *10^(-4) \ m^2`

and
`\mu_o` is the permeability of free space with a value
`\mu_o = 4\pi * 10^(-7) N/A^2`

substituting values into above equation

`L = ( 4\pi * 10^(-7) ^2* 3.142*10^(-4) * 900^2 )/(0.06)`

`L = 0.0053 \ H`