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(D) The magnetic field produced inside a long solenoid is given by $B = \mu_0 n I$.Here, $n$ is the number of turns per unit length and $I$ is the cur

Vedclass · Accepted Answer

$7.54 	imes 10^{-6} \,V$

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(D) The magnetic field produced inside a long solenoid is given by $B = \mu_0 n I$.Here, $n$ is the number of turns per unit length and $I$ is the current.The magnetic flux $\phi$ linked with the loop of area $A$ placed normal to the axis is $\phi = B A = \mu_0 n I A$.The induced emf $e$ is given by Faraday's law: $e = -\frac{d\phi}{dt} = -\mu_0 n A \frac{dI}{dt}$.Given values:$n = 15 	ext{ turns/cm} = 1500 	ext{ turns/m}$.$A = 2.0 	ext{ cm}^2 = 2.0 	imes 10^{-4} 	ext{ m}^2$.$\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20 	ext{ A/s}$.$\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Substituting these values:$|e| = (4 	imes 3.14 	imes 10^{-7}) 	imes 1500 	imes (2.0 	imes 10^{-4}) 	imes 20$.$|e| = 12.56 	imes 10^{-7} 	imes 1500 	imes 2.0 	imes 10^{-4} 	imes 20$.$|e| = 7.536 	imes 10^{-6} 	ext{ V} \approx 7.54 	imes 10^{-6} 	ext{ V}$.

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