MHTCET 25 Apr 2025 Shift 2 Paper

Concept:The net magnetic field at point $p$ is the vector sum of the fields due to the straight wire segments and the curved arc, using the Biot-Savart law.Explanation:Consider the three portions labelled 1, 2 and 3 in the figure.Portion 1 is a straight wire whose line of action passes through $p$.For such a wire, the angle $\theta = 0^\circ$, so the magnetic field at $p$ is $B_1 = 0$.Portion 2 is a semicircular arc of radius $r$ carrying current $I$.The magnetic field at the centre of a circular arc is$B = \frac{\mu_0 I}{4\pi r} \times \text{angle subtended (in radians)}$.For a semicircle, the angle is $\pi$ radians, so$B_2 = \frac{\mu_0}{4\pi} \cdot \frac{\pi I}{r} = \frac{\mu_0 I}{4 r}$.Using the right-hand rule, its direction is out of the paper.Portion 3 is a straight wire segment carrying current $I$, with its perpendicular distance from $p$ equal to $r$.For a finite straight wire, the field is$B_3 = \frac{\mu_0 I}{4 \pi r}$.By the right-hand rule, this field is also directed out of the paper.Since $B_2$ and $B_3$ act in the same direction, they add directly:$B_{\text{net}} = B_2 + B_3 = \frac{\mu_0 I}{4 r} + \frac{\mu_0 I}{4 \pi r}$.Rearranging gives$B_{\text{net}} = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 r}$.Answer:$B_{\text{net}} = \frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}$Hence, the correct option is C.