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^-7T ∙ m/A)

Question

Quizplus · Accepted Answer

The formula for the self-inductance of a solenoid is L = μn^2Aℓ, where μ is the permeability of free space, n is the number of turns per unit length (n = N/ℓ), A is the cross-sectional area, and ℓ is the length of the solenoid. Substituting the given values, we get:

n = N/ℓ = 100/0.3 = 333.33 turns/m
L = μn^2Aℓ = (4π × 10^-7 T·m/A)(333.33 turns/m)^2(1.00 × 10^-4 m^2)(0.30 m) = 4.19 µH

Therefore, the best choice is C, 4.19 µH.

What Is the Self-Inductance of a Solenoid 30