For a current carrying inductor, emf associated in $20 m V$ . Now, current through it changes from $6 A$ to $2 A$ in $2 s$ . The coefficient of mutual inductance is

20 m H

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10 m H

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1 m H

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2 m H

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Solution

The correct option is C $10 m H$
$| e | = L \frac{d I}{d t}$
Here, $e = 20 m V = 20 \times 10^{- 3} V$
Coefficient of mutual inductance,
$20 \times 10^{- 3} = L \times 2$
$∴ L = 10 \times 10^{- 3} = 10 m H$

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