Question

What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T • m/A)

Answer 1

Answer:

$L=4.19*10^{-6}H$

Explanation:

The self-inductance of a solenoid is defined as:

$L=\frac{\mu N^2A}{l}$

Here $\mu_0$ is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

$L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H$