Practice question
Question
A coil of self-inductance 0.8 H has its current decreased from 6 A to 3 A in 0.2 s. What is the
magnitude of the induced emf?
Explanation
**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. ε = L (Δ I/Δ t) . Δ I = 3 - 6 = -3 A , Δ t = 0.2 s . ε = 0.8 × (-3/0.2) = 0.8 × (-15) = -12 V , magnitude = 12 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B
Discussion
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