To solve this problem, we will use Faraday's law of electromagnetic induction.$\\$The induced emf $(\varepsilon)$ in a coil is given by:$\\ \varepsilon = -N \frac{d\Phi}{dt} \\$where:$\\$• $N = 800$ is the number of turns in the coil.$\\$• $\Phi$ is the magnetic flux through one turn of the coil.$\\$• $\frac{d\Phi}{dt}$ is the rate of change of magnetic flux.$\\$The magnetic flux $\Phi$ through the coil is given by:$\\ \Phi = B \cdot A \cdot \cos \theta \\$where:$\\$• $B = 5 \times 10^{-5}$ T is the magnetic field strength.$\\$• $A = 0.05$ m$^2$ is the effective area of the coil.$\\$• $\theta$ is the angle between the magnetic field and the normal to the coil.$\\$Initially, the coil is perpendicular to the magnetic field, so $\theta = 0^\circ. \\$Thus, the initial magnetic flux $\Phi_1$ is:$\\ \Phi_1 = B \cdot A \cdot \cos 0^\circ = B \cdot A \\$When the coil is rotated by $90^\circ,$ the angle $\theta$ becomes $90^\circ. \\$Thus, the final magnetic flux $\Phi_2$ is:$\\ \Phi_2 = B \cdot A \cdot \cos 90^\circ = 0 \\$The change in magnetic flux $(\Delta \Phi)$ is:$\\ \Delta \Phi = \Phi_2 - \Phi_1 = 0 - B \cdot A = -B \cdot A \\$The rate of change of magnetic flux $\left(\frac{d\Phi}{dt}\right)$ is:$\\ \frac{d\Phi}{dt} = \frac{\Delta \Phi}{\Delta t} = \frac{-B \cdot A}{0.1} \\$Substitute the values:$\\ \frac{d\Phi}{dt} = \frac{-(5 \times 10^{-5}) \cdot 0.05}{0.1} \\ \frac{d\Phi}{dt} = -2.5 \times 10^{-5}$ Wb/s$\\$The induced emf $(\varepsilon)$ is:$\\ \varepsilon = -N \frac{d\Phi}{dt} = -800 \times (-2.5 \times 10^{-5}) \\ \varepsilon = 2 \times 10^{-2}$ V$\\$Therefore, the emf induced in the coil is $0.02$ V.$\\$This corresponds to Option 4.