According to Faraday's Law of Induction, the magnitude of the induced emf ($\varepsilon$) is equal to the rate of change of magnetic flux ($\Phi_B$) linked with the coil .

1. Calculate Initial Flux ($\Phi_i$): The coil is perpendicular to the magnetic field, meaning the angle $\theta$ between the area vector and the magnetic field is $0^\circ$ (or $180^\circ$). $\Phi_i = N B A \cos(0^\circ) = N B A$

2. Calculate Final Flux ($\Phi_f$): The plane of the coil is rotated by $90^\circ$, so the angle between the area vector and the magnetic field becomes $90^\circ$. $\Phi_f = N B A \cos(90^\circ) = 0$

3. Calculate Change in Flux ($\Delta \Phi$): $|\Delta \Phi| = |\Phi_f - \Phi_i| = |0 - N B A| = N B A$

4. Calculate Induced EMF ($\varepsilon$): $|\varepsilon| = \frac{|\Delta \Phi|}{\Delta t} = \frac{N B A}{\Delta t}$

Given Values: $N = 800$ $A = 0.05 \text{ m}^2$ $B = 5 \times 10^{-5} \text{ T}$ $\Delta t = 0.1 \text{ s}$

Substitution: $|\varepsilon| = \frac{800 \times (5 \times 10^{-5}) \times 0.05}{0.1}$ $|\varepsilon| = \frac{800 \times 25 \times 10^{-7}}{0.1}$ $|\varepsilon| = \frac{20000 \times 10^{-7}}{0.1}$ $|\varepsilon| = \frac{2 \times 10^{-3}}{0.1}$ $|\varepsilon| = 2 \times 10^{-2} \text{ V} = 0.02 \text{ V}$