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(B) Given:$N_1 = 2000$,$L = 0.30 \ m$,$N_2 = 300$,$A = 1.2 	imes 10^{-3} \ m^2$.The rate of change of current is $\frac{di}{dt} = \frac{I_f - I_i}{\D

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$4.8 	imes 10^{-2} \ V$

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(B) Given:$N_1 = 2000$,$L = 0.30 \ m$,$N_2 = 300$,$A = 1.2 	imes 10^{-3} \ m^2$.The rate of change of current is $\frac{di}{dt} = \frac{I_f - I_i}{\Delta t} = \frac{-2 - 2}{0.25} = -16 \ A/s$. The magnitude is $16 \ A/s$.The mutual inductance $M$ of the solenoid-coil system is given by $M = \frac{\mu_0 N_1 N_2 A}{L}$.Substituting the values:$M = \frac{4\pi 	imes 10^{-7} 	imes 2000 	imes 300 	imes 1.2 	imes 10^{-3}}{0.30} = 3.016 	imes 10^{-3} \ H$.The induced emf $e$ is given by $e = M \left| \frac{di}{dt} ight|$.$e = (3.016 	imes 10^{-3}) 	imes 16 = 4.825 	imes 10^{-2} \ V \approx 4.8 	imes 10^{-2} \ V$.

$A$ solenoid has $2000$ turns wound over a length of $0.30 \ m$. The area of its cross-section is $1.2 \times 10^{-3} \ m^2$. Around its central section,a coil of $300$ turns is wound. If an initial current of $2 \ A$ in the solenoid is reversed in $0.25 \ s$,the emf induced in the coil is equal to

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