A coil of 200 turns and area 0.20 m^2 is rot | vedclass

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(C) The magnetic flux linked with the coil is given by $\phi = NAB \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\

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$5$

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(C) The magnetic flux linked with the coil is given by $\phi = NAB \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = NAB\omega \sin(\omega t)$.The maximum voltage generated is $\varepsilon_{\max} = NAB\omega$.Given:Number of turns $N = 200$Area $A = 0.20 \ m^2$Magnetic field $B = 0.01 \ T$Frequency $f = 0.5 \ rev/s$,so angular frequency $\omega = 2\pi f = 2\pi(0.5) = \pi \ rad/s$.Substituting the values:$\varepsilon_{\max} = 200 	imes 0.20 	imes 0.01 	imes \pi$$\varepsilon_{\max} = 40 	imes 0.01 	imes \pi = 0.4\pi = \frac{4\pi}{10} = \frac{2\pi}{5} \ V$.Comparing this with $\frac{2\pi}{\beta}$,we get $\beta = 5$.

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