Self-inductance: 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) A) 4.19 nH B) 4.19 pH C) 4.19 µH D) 4.19 mH E) 4.19 H

Question

Quizplus · Accepted Answer

The formula for self-inductance of a solenoid is L = μN²Aℓ, where μ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and ℓ is the length of the solenoid. Plugging in the given values, we get:
L = (4π × 10^-7 T ∙ m/A) × (100 turns)^2 × (1.00 × 10^-4 m^2) × (0.300 m)
L = 4.19 × 10^-6 H = 4.19 µH

Therefore, the answer is C) 4.19 µH.

Self-Inductance: What Is the Self-Inductance of a Solenoid 30