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 $I = 5\, A$.The magnetic flux t

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $I = 5\, A$.The magnetic flux through the coil is $\phi = N A B$,where $N = 100$,$A = \pi r^2 = \pi (10^{-2})^2\, m^2$,and $B = 4\pi 	imes 10^{-7} 	imes 10^4 	imes 5 = 0.02\pi\, T$.When the current is reversed,the change in flux is $\Delta \phi = \phi_f - \phi_i = (-NBA) - (NBA) = -2NBA$.The magnitude of the induced charge is $q = \frac{|\Delta \phi|}{R} = \frac{2NBA}{R}$.Substituting the values: $q = \frac{2 	imes 100 	imes \pi 	imes (10^{-2})^2 	imes 0.02\pi}{20} = \frac{2 	imes 100 	imes \pi 	imes 10^{-4} 	imes 0.02\pi}{20} = 2 	imes 10^{-4}\, C$.

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