A long solenoid has 20 turns per cm. A small | vedclass

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(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Given $n = 20 	ext{ tu

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(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Given $n = 20 	ext{ turns/cm} = 2000 	ext{ turns/m}$.$B = (4\pi 	imes 10^{-7} 	ext{ T m/A}) 	imes (2000 	ext{ m}^{-1}) 	imes I = 8\pi 	imes 10^{-4} I 	ext{ T}$.The magnetic flux $\phi$ through the loop of area $A = \frac{4}{\pi} 	ext{ cm}^2 = \frac{4}{\pi} 	imes 10^{-4} 	ext{ m}^2$ is $\phi = B \cdot A$.The induced emf $\varepsilon$ is given by Faraday's Law: $\varepsilon = \left| \frac{\Delta \phi}{\Delta t} ight| = A \left| \frac{\Delta B}{\Delta t} ight| = A \mu_0 n \left| \frac{\Delta I}{\Delta t} ight|$.Substituting the values: $\Delta I = 3.0 	ext{ A} - 1.0 	ext{ A} = 2.0 	ext{ A}$,$\Delta t = 0.2 	ext{ s}$.$\varepsilon = \left( \frac{4}{\pi} 	imes 10^{-4} 	ext{ m}^2 ight) 	imes (4\pi 	imes 10^{-7} 	ext{ T m/A}) 	imes (2000 	ext{ m}^{-1}) 	imes \left( \frac{2.0 	ext{ A}}{0.2 	ext{ s}} ight)$.$\varepsilon = (4 	imes 10^{-4}) 	imes (4 	imes 10^{-7}) 	imes (2000) 	imes (10) 	ext{ V}$.$\varepsilon = 32 	imes 10^{-7} 	ext{ V} = 3.2 	imes 10^{-6} 	ext{ V} = 3.2 \mu 	ext{V}$.

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