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Question: When the current in a coil changes from 8 ampere to 2 ampere in \(3 \times 10^{- 2}\)second, the e.m...

When the current in a coil changes from 8 ampere to 2 ampere in 3×10−23 \times 10^{- 2}second, the e.m.f. induced in the coil is 2 6muvolt2\mspace{6mu} volt. The self inductance of the coil (in millihenry) is

A

1

B

5

C

20

D

10

Answer

10

Explanation

Solution

e=−Ldidt⇒2=−L(8−23×10−2)⇒L=0.01 6muH=10 6mumHe = - L\frac{di}{dt} \Rightarrow 2 = - L\left( \frac{8 - 2}{3 \times 10^{- 2}} \right) \Rightarrow L = 0.01\mspace{6mu} H = 10\mspace{6mu} mH