A circular coil of radius 8.0 , cm and 20 tur | vedclass

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(C) The maximum induced $emf$ $(\varepsilon_{max})$ in a rotating coil is given by the formula: $\varepsilon_{max} = N \omega A B$,where $N$ is the nu

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$60$

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(C) The maximum induced $emf$ $(\varepsilon_{max})$ in a rotating coil is given by the formula: $\varepsilon_{max} = N \omega A B$,where $N$ is the number of turns,$\omega$ is the angular speed,$A$ is the area of the coil,and $B$ is the magnetic field strength.Given: $N = 20$,$\omega = 50 \, rad \, s^{-1}$,$B = 3.0 	imes 10^{-2} \, T$,and radius $r = 8.0 \, cm = 0.08 \, m$.The area $A = \pi r^2 = \pi 	imes (0.08)^2 = 0.0064 \pi \, m^2$.Substituting the values: $\varepsilon_{max} = 20 	imes 50 	imes (0.0064 \pi) 	imes (3.0 	imes 10^{-2})$.$\varepsilon_{max} = 1000 	imes 0.0064 	imes \pi 	imes 3.0 	imes 10^{-2} = 6.4 	imes \pi 	imes 3.0 	imes 10^{-2} = 19.2 \pi 	imes 10^{-2}$.Using $\pi \approx 3.14159$,$\varepsilon_{max} \approx 19.2 	imes 3.14159 	imes 10^{-2} \approx 60.319 	imes 10^{-2} \, V$.Rounding to the nearest integer,we get $60$.

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