Current flowing through a copper ring changes with time according to the graph shown in figure. If self inductance of the coil is 8 mH, the emf induced in the ring is

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16 mV

8 mV

12 mV

4 mV

answer is A.

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Detailed Solution

didt=−42A/s=−2 A/sε=−Ldidt=−8×−2volt=16 mV

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