The flux linked with a coil is 0.8Wb when a 2A current is flowing through it. If this current begins to increase at the rate of 400A/s, the induced emf in the coil will be
A
20V
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B
40V
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C
80V
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D
160V
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Solution
The correct option is D160V ϕ=Li 0.8=L×2 L=0.82=0.4 emf=Ldidt =0.4×400 =160V