A solenoid is 30 cm long, has 10 turns per centimetre and cross-sectional area 5 cm². Its current changes uniformly from 2 A to 4 A in 3.14 s. What is the magnitude of the self-induced emf if it is written as α×10⁻⁵ V?
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Question

A solenoid is 30 cm long, has 10 turns per centimetre and cross-sectional area 5 cm². Its current changes uniformly from 2 A to 4 A in 3.14 s. What is the magnitude of the self-induced emf if it is written as α×10⁻⁵ V?

Options

G1 Physics · Accepted Answer

The total number of turns is N = 10 turns per cm × 30 cm = 300. Convert the dimensions: l = 30 cm = 0.30 m, A = 5 cm² = 5×10⁻⁴ m². For an air-core solenoid, L = μ₀N²A ÷ l. Therefore, L = 4π×10⁻⁷ × 300² × 5×10⁻⁴ ÷ 0.30. This gives L = 6π×10⁻⁵ H. The current-change rate is |dI ÷ dt| = (4-2) ÷ 3.14 = 2 ÷ 3.14 A·s⁻¹. Hence, |ε| = L|dI ÷ dt| = 6π×10⁻⁵ × 2 ÷ 3.14. Using π ≈ 3.14: |ε| ≈ 12×10⁻⁵ V. Therefore, α = 12. Correct option: B.