A circular coil of area 100 ,cm^2 and 20 turn | vedclass

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(D) Given that:Area of coil,$A = 100 \,cm^2 = 100 	imes 10^{-4} \,m^2 = 10^{-2} \,m^2$Number of turns,$N = 20$Magnetic field,$B = 2 \,Wb/m^2$Time int

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(D) Given that:Area of coil,$A = 100 \,cm^2 = 100 	imes 10^{-4} \,m^2 = 10^{-2} \,m^2$Number of turns,$N = 20$Magnetic field,$B = 2 \,Wb/m^2$Time interval,$\Delta t = 0.2 \,s$The magnetic flux $\phi$ is given by $\phi = N B A \cos 	heta$,where $	heta$ is the angle between the area vector $\hat{n}$ and the magnetic field $B$.Initial position: The plane of the coil makes an angle of $30^{\circ}$ with the field. Therefore,the angle between the area vector $\hat{n}$ and the field $B$ is $	heta_1 = 90^{\circ} - 30^{\circ} = 60^{\circ}$.Initial flux,$\phi_1 = N B A \cos 60^{\circ} = 20 	imes 2 	imes 10^{-2} 	imes 0.5 = 0.2 \,Wb$.Final position: The coil is perpendicular to the field. Therefore,the area vector $\hat{n}$ is parallel to the field $B$,so $	heta_2 = 0^{\circ}$.Final flux,$\phi_2 = N B A \cos 0^{\circ} = 20 	imes 2 	imes 10^{-2} 	imes 1 = 0.4 \,Wb$.Induced emf,$e = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_2 - \phi_1}{\Delta t} = -\frac{0.4 - 0.2}{0.2} = -\frac{0.2}{0.2} = -1 \,V$.The magnitude of the induced emf is $|e| = 1 \,V$.

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