COMEDK 2025 Afternoon Solved Paper

Concept:Induced emf is generated due to a change in magnetic flux linked with the coil, and the induced current is found using Ohm's law.Explanation:The area of the coil is $2\sqrt{2}\;\text{cm}^2$. Convert to SI: $A = 2\sqrt{2} \times 10^{-4}\;\text{m}^2$. The coil is vertical in the east–west direction, so its normal points north–south. The magnetic field of $0.2\;\text{T}$ is applied in the north–east direction, which is at $45^\circ$ to the north. Only the component of $\mathbf{B}$ perpendicular to the coil's plane contributes to the flux. Thus, $B_\perp = B \cos 45^\circ = 0.2 \times \frac{1}{\sqrt{2}}\;\text{T}$. Initial magnetic flux: $\Phi = B_\perp A = \left(0.2 \times \frac{1}{\sqrt{2}}\right) \times \left(2\sqrt{2} \times 10^{-4}\right) = 4 \times 10^{-5}\;\text{Wb}$. The field is removed steadily in $\Delta t = 0.4\;\text{s}$, so the induced emf is $\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{4\times10^{-5}}{0.4} = 1\times10^{-4}\;\text{V}$. Coil resistance $R = 2\;\Omega$, therefore induced current $I = \frac{\mathcal{E}}{R} = \frac{1\times10^{-4}}{2} = 0.5\times10^{-4}\;\text{A}$.Answer:$0.5 \times 10^{-4}\;\text{A}$ (Option B).