A uniform magnetic field vec{B} is perpendicu | vedclass

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(A) The induced current $i$ is given by $i = \frac{e}{R} = \frac{1}{R} \frac{d\phi}{dt} = \frac{A}{R} \frac{dB}{dt}$.Here, $A$ is the area of the loop

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$2.8$

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(A) The induced current $i$ is given by $i = \frac{e}{R} = \frac{1}{R} \frac{d\phi}{dt} = \frac{A}{R} \frac{dB}{dt}$.Here, $A$ is the area of the loop: $A = \pi r^2 = \pi 	imes (0.05 	ext{ m})^2 = 25\pi 	imes 10^{-4} 	ext{ m}^2$.The resistance $R$ of the wire is $R = \frac{ho l}{a}$, where $l = 2\pi r$ and $a = \pi (r_{wire})^2$.$l = 2 	imes \pi 	imes 0.05 = 0.1\pi 	ext{ m}$.$a = \pi 	imes (10^{-3} 	ext{ m})^2 = \pi 	imes 10^{-6} 	ext{ m}^2$.$R = \frac{2 	imes 10^{-8} 	imes 0.1\pi}{\pi 	imes 10^{-6}} = 2 	imes 10^{-3} \Omega$.Now, $\frac{dB}{dt} = \frac{iR}{A} = \frac{11 	imes 2 	imes 10^{-3}}{25\pi 	imes 10^{-4}} = \frac{22 	imes 10^{-3}}{25 	imes 3.14 	imes 10^{-4}} \approx 2.8 	ext{ T s}^{-1}$.

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