A coil has 600 turns which produces $5 \times 10^{- 3} W b / t u r n$ of flux when $3 A$ current flows in the wire. This produced $6 \times 10^{- 3} W b / t u r n$ in 1000 turns secondary coil. When the switch is opened, the current drops to zero in $0.2 s$ in primary If the
mutual inductance is M , the induced emf in the secondary is $e_{2}, t h e n$

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M = $2 H$

$M = 1 H$

$e_{2} = 30 V$

$e_{2} = 20 V$

answer is A.

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Detailed Solution

(a) $M = \frac{N_{2} ϕ_{2}}{i_{1}} = \frac{(1000) (6 \times 10^{- 3})}{3} = 2 H$

(b) $|e_{2}| = M (\frac{∆ i_{1}}{∆ t}) = \frac{(2) (3)}{0.2} = 30 V$

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