COMEDK 2025 Afternoon Solved Paper

Concept:Net force on a long straight wire due to a rectangular current-carrying coil is calculated by considering only the two sides of the coil parallel to the wire, using the formula for force between parallel currents.Explanation:Given: Coil current $I_1 = 10\ \mathrm{A}$, breadth $= 9\ \mathrm{cm} = 0.09\ \mathrm{m}$, length $= 10\ \mathrm{cm} = 0.10\ \mathrm{m}$.Straight wire current $I_2 = 20\ \mathrm{A}$, placed $1\ \mathrm{cm} = 0.01\ \mathrm{m}$ from the near side, parallel to the length.Only the two sides of the coil parallel to the wire produce force on the straight wire.Distance to near side, $d_1 = 0.01\ \mathrm{m}$; distance to far side, $d_2 = 0.01 + 0.09 = 0.10\ \mathrm{m}$.Length of each parallel side, $L = 0.10\ \mathrm{m}$.Force between two parallel currents: $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$.$\mu_0 = 4\pi \times 10^{-7}\ \mathrm{N/A^2}$.Force due to near side (attractive if currents in same direction): $F_1 = \frac{4\pi \times 10^{-7} \times 10 \times 20 \times 0.10}{2\pi \times 0.01} = 4.0 \times 10^{-4}\ \mathrm{N}$.Force due to far side (repulsive because currents opposite relative to near side): $F_2 = \frac{4\pi \times 10^{-7} \times 10 \times 20 \times 0.10}{2\pi \times 0.10} = 4.0 \times 10^{-5}\ \mathrm{N}$.These forces act in opposite directions on the straight wire.Net force: $F_{\text{net}} = F_1 - F_2 = 4.0 \times 10^{-4} - 4.0 \times 10^{-5} = 3.6 \times 10^{-4}\ \mathrm{N}$.Direction is towards the coil, but only magnitude is asked.Answer:$3.6 \times 10^{-4}\ \mathrm{N}$ (Option B).