A long solenoid of radius 2, cm has 100, turn | vedclass

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(A) The magnetic field inside the solenoid is given by $B = \mu_0 n I$,where $n = 100\, turns/cm = 10^4\, turns/m$ and $I = 5\, A$.$B = (4\pi 	imes 1

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$2 	imes 10^{-4}\, C$

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(A) The magnetic field inside the solenoid is given by $B = \mu_0 n I$,where $n = 100\, turns/cm = 10^4\, turns/m$ and $I = 5\, A$.$B = (4\pi 	imes 10^{-7}) 	imes 10^4 	imes 5 = 20\pi 	imes 10^{-3}\, T = 0.02\pi\, T$.The magnetic flux through the coil is $\phi = N B A_{coil}$,where $N = 100$ and $A_{coil} = \pi r^2 = \pi (0.01)^2 = \pi 	imes 10^{-4}\, m^2$.When the current is reversed,the change in flux is $\Delta \phi = \phi_f - \phi_i = (-NBA) - (NBA) = -2NBA$.The induced charge $q$ is given by $q = \frac{|\Delta \phi|}{R} = \frac{2NBA}{R}$.Substituting the values: $q = \frac{2 	imes 100 	imes (0.02\pi) 	imes (\pi 	imes 10^{-4})}{20} = \frac{4 	imes 10^2 	imes 0.01 	imes \pi^2 	imes 10^{-4}}{20} = \frac{4 	imes 10^{-4} 	imes \pi^2}{20}$.Using $\pi^2 \approx 10$,$q = \frac{4 	imes 10^{-4} 	imes 10}{20} = 2 	imes 10^{-4}\, C$.

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