COMEDK 2025 Afternoon Solved Paper

Concept:The induced emf in the secondary coil is given by the rate of change of magnetic flux linked with it, which depends on the rate of change of current in the primary coil.Explanation:The total flux linked with the secondary coil, $\Lambda$, is directly proportional to the primary current $I$.Thus, $\Lambda = k I$, where $k$ is a constant.The induced emf magnitude is $|\mathcal{E}| = \frac{d\Lambda}{dt} = k \frac{dI}{dt}$.From the given data, when $I = 2\ \mathrm{A}$, $\Lambda = 2000 \times 10^{-6}\ \mathrm{Wb}$, so $k = \frac{\Lambda}{I} = \frac{2000 \times 10^{-6}}{2} = 1000 \times 10^{-6}\ \mathrm{Wb/A}$.The rate of increase of current is $\frac{dI}{dt} = 0.2\ \mathrm{A/s}$.Therefore, the induced emf is $|\mathcal{E}| = (1000 \times 10^{-6}) \times 0.2 = 200 \times 10^{-6}\ \mathrm{V} = 2 \times 10^{-4}\ \mathrm{V}$.Answer:Option A: $2 \times 10^{-4}\ \mathrm{V}$