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Question

Physics Question on Electromagnetic induction

In a coil of self-inductance 55 henry, the rate of change of current is 22 ampere per second. The e.m.f. induced in the coil, is

A

10 V

B

- 10 V

C

5 V

D

- 5 V

Answer

- 10 V

Explanation

Solution

Given : Inductance of the coil (L)=5H(L) = 5 \,H and rate of change of current (dIdt)=2\left(\frac{dI}{dt}\right)=2 A/sec. The induced e.m.f. =L(dIdt)=5×2=10V=-L\left(\frac{dI}{dt}\right)=-5\times2=-10\, V