A coil of resistance 5.0 Ω has its flux linkage changed uniformly by 0.60 Wb-turns in 0.30 s. What current is induced in it?
Correct answer A. 0.40 A
Explanation
ε = Δ(NΦ)/Δt = 0.60 ÷ 0.30 = 2.0 V. Current I = ε/R = 2.0 V ÷ 5.0 Ω = 0.40 A. The value 2.0 A confuses the emf with the current.
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