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A 0.2 H inductor carries current that rises uniformly from 0 to 4 A in 0.1 s. The induced emf across it is
A4 V
B16 V
C2 V
D8 V
Answer & Solution
Correct answer: D. 8 V
1. Self-induced emf ε = L (dI/dt).
2. dI/dt = (4 − 0)/0.1 = 40 A/s.
3. ε = 0.2 × 40 = 8 V (opposing the rise in current).
_Source: Maharashtra Balbharati Std XII Physics, Ch 12 "Electromagnetic Induction" §12.6_
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