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

Answer:

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

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;

$L = \frac{\mu_oN^2A}{l}$

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

$L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H$

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