A solenoid of length 0.30 , m has 2000 turns. | vedclass

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(D) The mutual inductance $M$ of the solenoid-coil system is given by $M = \frac{\mu_0 N_1 N_2 A}{l}$.Given: $l = 0.30 \, m$,$N_1 = 2000$,$N_2 = 300$,

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$48 \, mV$

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(D) The mutual inductance $M$ of the solenoid-coil system is given by $M = \frac{\mu_0 N_1 N_2 A}{l}$.Given: $l = 0.30 \, m$,$N_1 = 2000$,$N_2 = 300$,$A = 1.2 	imes 10^{-3} \, m^2$,$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$.The change in current is $\Delta I = I_f - I_i = -2 \, A - 2 \, A = -4 \, A$. The magnitude of change is $|\Delta I| = 4 \, A$.The time interval is $\Delta t = 0.25 \, s$.The induced $e.m.f.$ is $e = M \frac{|\Delta I|}{\Delta t} = \left( \frac{4\pi 	imes 10^{-7} 	imes 2000 	imes 300 	imes 1.2 	imes 10^{-3}}{0.30} ight) 	imes \frac{4}{0.25}$.Calculating the value: $M = \frac{4\pi 	imes 10^{-7} 	imes 6 	imes 10^5 	imes 1.2 	imes 10^{-3}}{0.3} = \frac{4\pi 	imes 7.2 	imes 10^{-5}}{0.3} = 4\pi 	imes 24 	imes 10^{-5} \approx 3.016 	imes 10^{-3} \, H$.$e = 3.016 	imes 10^{-3} 	imes 16 \approx 48.25 	imes 10^{-3} \, V \approx 48 \, mV$.

$A$ solenoid of length $0.30 \, m$ has $2000$ turns. 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 $0.25 \, s$,then the $e.m.f.$ induced in the coil is:

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