A coil of effective area $3 \mathrm{~m}^2$ is placed at right angles to a magnetic field of induction $0.05…

Faraday’s Law gives the induced e.m.f. as $ε = -\frac{d\Phi_B}{dt}$, where magnetic flux is $\Phi_B = B A \cos\theta$. The coil is perpendicular to the field so $\theta = 0^\circ$ and $\cos\theta = 1$. Area $A = 3~\mathrm{m}^2$ remains constant.

The magnetic field decreases from $B_1 = 0.05~\mathrm{Wb}/\mathrm{m}^2$ to $B_2 = 0.01~\mathrm{Wb}/\mathrm{m}^2$ over $\Delta t = 10~\mathrm{s}$, giving $\Delta B = -0.04~\mathrm{Wb}/\mathrm{m}^2$.

The change in flux is $\Delta\Phi_B = A \cdot \Delta B = 3 \times (-0.04) = -0.12~\mathrm{Wb}$.

The induced e.m.f. is $\varepsilon = -\frac{\Delta\Phi_B}{\Delta t} = -\frac{-0.12}{10} = 0.012~\mathrm{V}$.

Converting to millivolts: $0.012~\mathrm{V} = 12~\mathrm{mV}$.

Final answer: $12~\mathrm{mV}$