Practice question
Question
A solenoid of 550 turns per meter and area 0.016 m² has a current drop from 7 A to 4 A in 0.3 s. What
is the self-induced emf? (\( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \))
Explanation
**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (550)² × 0.016 × 1 = 0.00608 H . ε = L (Δ I/Δ t) = 0.00608 × (4 - 7/0.3) = 0.00608 × (-10) = 0.0608 V ≈ 0.061 V . Using
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