Practice question
Question
A solenoid of 700 turns per meter and area 0.015 m² has a current drop from 8 A to 5 A in 0.3 s. What
is the self-induced emf? (\( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \))
Explanation
**Magnetic energy density** u = B²/(2μ₀), for B=0.5 T, u=0.25/(2×4π×10⁻⁷)=0.25/(2.513×10⁻⁶)=99471 J/m³, large, but volume small, total energy moderate. Inductor stores energy in field, released when current interrupted causing spark. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (700)² × 0.015 × 1 = 0.00923 H . ε = L (Δ I/Δ t) = 0.00923 × (5 - 8/0.3) = 0.00923 × (-10) = 0.0923 V ≈ 0.092 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L
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