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(A) Given: Radius $r = 10 \, cm = 0.1 \, m$, Resistance $R = 2 \, \Omega$, Time $t = 0.25 \, s$, Induced emf $E = 3.8 	imes 10^{-3} \, V$, Magnetic f

Vedclass · Accepted Answer

$504$

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(A) Given: Radius $r = 10 \, cm = 0.1 \, m$, Resistance $R = 2 \, \Omega$, Time $t = 0.25 \, s$, Induced emf $E = 3.8 	imes 10^{-3} \, V$, Magnetic field $B = 3 	imes 10^{-5} \, T$.Area of the coil $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \, m^2$.The magnetic flux changes as the coil rotates by $180^{\circ}$. The initial flux $\phi_i = B A \cos(0^{\circ}) = B A$ and the final flux $\phi_f = B A \cos(180^{\circ}) = -B A$.The change in flux $\Delta \phi = \phi_f - \phi_i = -B A - B A = -2 B A$.The magnitude of induced emf is $|E| = N \frac{|\Delta \phi|}{t} = N \frac{2 B A}{t}$.Substituting the values: $3.8 	imes 10^{-3} = N \frac{2 	imes (3 	imes 10^{-5}) 	imes (0.01 \pi)}{0.25}$.$3.8 	imes 10^{-3} = N \frac{6 	imes 10^{-7} 	imes 3.14}{0.25}$.$3.8 	imes 10^{-3} = N 	imes 7.536 	imes 10^{-6}$.$N = \frac{3.8 	imes 10^{-3}}{7.536 	imes 10^{-6}} \approx 504$ turns.

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