An inductor of self-inductance L = 2 mH carries time-varying current I = t²e⁻ᵗ (SI units). At what time does its induced emf vanish?
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An inductor of self-inductance L = 2 mH carries time-varying current I = t²e⁻ᵗ (SI units). At what time does its induced emf vanish?

G1 Physics · Accepted Answer

Step 1: An inductor of fixed inductance generates voltage ε = −L(dI/dt). Step 2: Given I(t) = t²e⁻ᵗ, use product rule: dI/dt = e⁻ᵗ(2t − t²) = te⁻ᵗ(2 − t). Step 3: Set dI/dt=0: t=0 or t=2 s. For positive nonzero time, t=2 s. Answer C. Final answer: (C) 2 s.