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(A) 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 and $i$ is the current.

Vedclass · Accepted Answer

$9 \pi^2 \ mA$

Vedclass · Answer

(A) 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 and $i$ is the current.Here,$n = 200 \ 	ext{turns/cm} = 20000 \ 	ext{turns/m}$.The magnetic flux $\phi$ linked with the circular coil is $\phi = N B A$,where $N = 100$ is the number of turns in the coil and $A = \pi r^2$ is the area of the coil with $r = 0.03 \ m$.Thus,$\phi = N (\mu_0 n i) (\pi r^2)$.According to Faraday's law,the induced $EMF$ is $\varepsilon = -\frac{\Delta \phi}{\Delta t}$.The induced current $I_{ind} = \frac{|\varepsilon|}{R} = \frac{N \mu_0 n \pi r^2}{R} 	imes \frac{\Delta i}{\Delta t}$.Substituting the values: $I_{ind} = \frac{100 	imes (4 \pi 	imes 10^{-7}) 	imes 20000 	imes \pi 	imes (0.03)^2}{4} 	imes \frac{2 - 0}{0.04}$.$I_{ind} = \frac{100 	imes 4 \pi 	imes 10^{-7} 	imes 20000 	imes \pi 	imes 0.0009}{4} 	imes 50$.$I_{ind} = 9 \pi^2 	imes 10^{-3} \ A = 9 \pi^2 \ mA$.

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