A coil has an area of 0.05,m^2 and it has 800 | vedclass

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(D) Given:Area $A = 0.05\,m^2$Number of turns $N = 800$Magnetic field $B = 4 	imes 10^{-5}\,Wb/m^2$Time interval $\Delta t = 0.1\,s$Initial magnetic

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$0.016$

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(D) Given:Area $A = 0.05\,m^2$Number of turns $N = 800$Magnetic field $B = 4 	imes 10^{-5}\,Wb/m^2$Time interval $\Delta t = 0.1\,s$Initial magnetic flux $\phi_1 = N \cdot B \cdot A \cdot \cos(0^o) = N \cdot B \cdot A$Final magnetic flux $\phi_2 = N \cdot B \cdot A \cdot \cos(90^o) = 0$The average induced emf is given by Faraday's law:$e = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_2 - \phi_1}{\Delta t} = -\frac{0 - NAB}{\Delta t} = \frac{NAB}{\Delta t}$$e = \frac{800 	imes 4 	imes 10^{-5} 	imes 0.05}{0.1}$$e = \frac{1600 	imes 10^{-5}}{0.1} = 16000 	imes 10^{-5} = 0.016\,V$

$A$ coil has an area of $0.05\,m^2$ and it has $800\,turns$. It is placed perpendicularly in a magnetic field of strength $4 \times 10^{-5}\,Wb/m^2$. It is rotated through $90^o$ in $0.1\,s$. The average emf induced in the coil is ........... $V.$

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