AIIMS 2019 Physics Magnetic Force MCQ Question

To find the magnetic force per unit length on the small element P at the center of curvature, we consider the contributions from both the semi-circular arc and the straight wire. The magnetic field at point P due to the semi-circular arc carrying current $i$ is given by $B = \frac{μ₀i}{4r}$, while the straight wire contributes a field of $B = \frac{μ₀i}{2r}$. The net magnetic field at point P is the vector sum of these contributions, leading to a force per unit length on the element due to the wire's field acting on the arc's current, resulting in $F/L = \frac{μ₀i²}{4r}$, which corresponds to option A.

Options B, C, and D are incorrect as they miscalculate the contributions or do not account for the correct vector nature of the magnetic fields from both current-carrying elements.