A 50 turns circular coil has a radius of 3;cm | vedclass

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(B) The induced $e.m.f.$ is given by Faraday's Law: $|e| = N \frac{|\Delta \Phi|}{\Delta t} = N \frac{A \Delta B \cos 	heta}{\Delta t}$.Given: $N = 5

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$17.7$

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(B) The induced $e.m.f.$ is given by Faraday's Law: $|e| = N \frac{|\Delta \Phi|}{\Delta t} = N \frac{A \Delta B \cos 	heta}{\Delta t}$.Given: $N = 50$,$r = 3\;cm = 3 	imes 10^{-2}\;m$,$\Delta B = (0.35 - 0.10)\;T = 0.25\;T$,$\Delta t = 2\;ms = 2 	imes 10^{-3}\;s$,and $	heta = 0^\circ$ (since the field is normal to the area).Area $A = \pi r^2 = \pi (3 	imes 10^{-2})^2 = 9\pi 	imes 10^{-4}\;m^2$.Substituting the values:$|e| = \frac{50 	imes (0.25) 	imes (9\pi 	imes 10^{-4})}{2 	imes 10^{-3}}$$|e| = \frac{50 	imes 0.25 	imes 9 	imes 3.1416 	imes 10^{-4}}{2 	imes 10^{-3}}$$|e| = \frac{12.5 	imes 28.274 	imes 10^{-4}}{2 	imes 10^{-3}} = \frac{353.43 	imes 10^{-4}}{2 	imes 10^{-3}} = 176.715 	imes 10^{-1} \approx 17.7\;V$.

$A$ $50$ turns circular coil has a radius of $3\;cm$. It is kept in a magnetic field acting normal to the area of the coil. The magnetic field $B$ increases from $0.10\;T$ to $0.35\;T$ in $2\;ms$. The average induced $e.m.f.$ in the coil is.......$V$.

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