MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Magnetic field at a point due to a semi-infinite straight wire at perpendicular distance $r$ is $B = \frac{\mu_0 I}{4\pi r}$.Explanation:The vertical wires $BC$ and $ET$ are separated by a distance $a$.Point $O$ is the midpoint, so its perpendicular distance from each vertical wire is $\frac{a}{2}$.The horizontal parts $AB$ and $DE$ lie along the line passing through $O$, so their magnetic field at $O$ is zero.Each vertical part behaves like a semi-infinite straight wire.Field due to $BC$ at $O$:$$B_1 = \frac{\mu_0 I}{4\pi \left(\frac{a}{2}\right)} = \frac{\mu_0 I}{2\pi a}$$Field due to $ET$ at $O$:$$B_2 = \frac{\mu_0 I}{4\pi \left(\frac{a}{2}\right)} = \frac{\mu_0 I}{2\pi a}$$By the right-hand thumb rule, both fields act in the same direction, perpendicular to the plane of the paper, outwards.So the net magnetic induction is:$$B = B_1 + B_2 = \frac{\mu_0 I}{2\pi a} + \frac{\mu_0 I}{2\pi a} = \frac{\mu_0 I}{\pi a}$$Answer:$$\frac{\mu_0 I}{\pi a}$$Hence, the correct option is D.