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Question

A solenoid of 400 turns per meter and area 0.015 m² has a current drop from 5 A to 2 A in 0.3 s. What
is the self-induced emf? (\( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \))

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Explanation

**Field decreasing to zero** induces emf trying to maintain field, current direction such that its field adds to original. For 150 turns area 0.06 m² B 0.14 T to zero in 0.3 s, e=150×0.06×0.14/0.3=4.2 V, as earlier, showing linear dependence on N, A, ΔB/Δt. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (400)² × 0.015 × 1 = 0.003016 H . ε = L (Δ I/Δ t) = 0.003016 × (2 - 5/0.3) = 0.003016 × (-10) = 0.03016 V ≈ 0.03 V . Using Φ = B A cosθ, e = -N dΦ/dt

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