MHTCET 18 Apr 2026 Shift 1 Paper

Concept:The magnetic field at the centre of a circular arc carrying current $I$ is $B = \frac{\mu_0 I}{4 \pi r} \theta$, where $\theta$ is the angle subtended in radians.For a semi-infinite straight wire at perpendicular distance $r$, the field is $B = \frac{\mu_0 I}{4 \pi r}$.Explanation:The wire is divided into three parts: the semi-infinite straight wire $AB$, the quarter-circular arc $BC$, and the semi-infinite straight wire $CD$.For the semi-infinite wire $AB$, the field at $O$ is $B_1 = \frac{\mu_0 I}{4 \pi r}$.The quarter-circular arc $BC$ subtends $\theta = \frac{\pi}{2}$ at $O$, so $B_2 = \frac{\mu_0 I}{4 \pi r} \cdot \frac{\pi}{2}$.For the semi-infinite wire $CD$, the field is $B_3 = \frac{\mu_0 I}{4 \pi r}$.Using the right-hand thumb rule, all three fields point out of the plane of the paper.Thus, the net field is their sum: $B_{\text{net}} = B_1 + B_2 + B_3$.$B_{\text{net}} = \frac{\mu_0 I}{4 \pi r}\left(1 + \frac{\pi}{2} + 1\right) = \frac{\mu_0 I}{4 \pi r}\left(2 + \frac{\pi}{2}\right)$.Answer:$\displaystyle B_{\text{net}} = \frac{\mu_0}{4 \pi} \frac{I}{r}\left(2 + \frac{\pi}{2}\right)$Hence, the correct option is C.