The magnetic flux ϕ(in weber) linked with a coil of resistance 10Ωvaries with time t (in second) as ϕ=8t2−4t+1. The current induced in the coil at t=0.1 secis
A
10 A
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B
0.24 A
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C
0.12 A
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D
4.8 A
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Solution
The correct option is B 0.24 A E.m.f. induced, ϵ=−dϕdt=−ddt(8t2−4t+1)=−[(2×8)t−4]=−[16t−4] Here, t = 0.1 s ∴ϵ=−[16×0.1−4]=4−1.6=2.4V Current induced I=ϵR=2.410=0.24A