A solenoid of 1000 turns, length 0.8 m, and area 0.02 m² carries 3 A. The self-inductance of the solenoid is (μ₀ = 4π × 10⁻⁷ T·m/A):
Correct answer: B. 0.038 H
- A. 0.019 H
- B. 0.038 H
- C. 0.057 H
- D. 0.076 H
Explanation
Self-inductance L = μ₀n²Aℓ, where n = N/ℓ = 1000/0.8 = 1250 turns/m, A = 0.02 m², ℓ = 0.8 m. Thus, L = (4π × 10⁻⁷) (1250)²(0.02)(0.8) ≈ 0.038 H.
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