An air-cored solenoid with length 20 , cm,are | vedclass

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(B) Given: Length $\ell = 20 \, cm = 0.2 \, m$,Area $A = 20 \, cm^2 = 20 	imes 10^{-4} \, m^2$,Number of turns $N = 400$,Initial current $I_1 = 2 \,

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$4$

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(B) Given: Length $\ell = 20 \, cm = 0.2 \, m$,Area $A = 20 \, cm^2 = 20 	imes 10^{-4} \, m^2$,Number of turns $N = 400$,Initial current $I_1 = 2 \, A$,Final current $I_2 = 0 \, A$,Time interval $dt = 10^{-3} \, s$.The magnetic field inside the solenoid is $B = \frac{\mu_0 N I}{\ell}$.The magnetic flux linked with the solenoid is $\phi = B A N = \frac{\mu_0 N^2 I A}{\ell}$.The induced back $emf$ is given by $\varepsilon = \frac{d\phi}{dt} = \frac{\mu_0 N^2 A}{\ell} \cdot \frac{dI}{dt}$.Substituting the values:$\varepsilon = \frac{4\pi 	imes 10^{-7} 	imes (400)^2 	imes 20 	imes 10^{-4} 	imes (2 - 0)}{0.2 	imes 10^{-3}}$.$\varepsilon = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 160000 	imes 20 	imes 10^{-4} 	imes 2}{0.2 	imes 10^{-3}}$.$\varepsilon = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 1.6 	imes 10^5 	imes 20 	imes 10^{-4} 	imes 2}{2 	imes 10^{-4}} = 4.02 \, V \approx 4 \, V$.

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