The coil of area 0.1 m^2 has 500 turns. Afte | vedclass

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(D) The average induced emf is given by Faraday's Law: $e = -N \frac{\Delta \phi}{\Delta t}$.Here,$N = 500$,$A = 0.1 \ m^2$,$B = 4 	imes 10^{-4} \ Wb

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$0.2$

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(D) The average induced emf is given by Faraday's Law: $e = -N \frac{\Delta \phi}{\Delta t}$.Here,$N = 500$,$A = 0.1 \ m^2$,$B = 4 	imes 10^{-4} \ Wb/m^2$,and $\Delta t = 0.1 \ s$.The change in magnetic flux is $\Delta \phi = BA(\cos 	heta_2 - \cos 	heta_1)$.Initially,the coil is perpendicular to the field (or at $0^o$ to the normal),so $	heta_1 = 0^o$. After rotation,$	heta_2 = 90^o$.$\Delta \phi = (4 	imes 10^{-4}) 	imes (0.1) 	imes (\cos 90^o - \cos 0^o) = 4 	imes 10^{-5} 	imes (0 - 1) = -4 	imes 10^{-5} \ Wb$.The induced emf is $e = -500 	imes \frac{-4 	imes 10^{-5}}{0.1} = 500 	imes 4 	imes 10^{-4} = 0.2 \ V$.

The coil of area $0.1 \ m^2$ has $500$ turns. After placing the coil in a magnetic field of strength $4 \times 10^{-4} \ Wb/m^2$,if it is rotated through $90^o$ in $0.1 \ s$,the average emf induced in the coil is......$V$

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