Question

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm

Answer 1

The self-inductance in henries for the solenoid is 0.0274 H.

Step-by-step explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

`L = (\mu_oN^2A)/(l)`

where;

`\mu_o` is permeability of free space =
`4\pi*10^(-7) \ H/m`

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

`L = (4\pi *10^(-7)(1500)^2*(0.00126))/(0.13) \\\\L = 0.0274 \ H`

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.