MHTCET 26 Apr 2025 Shift 1 Paper

Concept:The net magnetic field at the centre is the vector sum of the fields due to the long straight wire and the circular loop.Explanation:The magnetic field at $O$ due to the infinitely long straight wire is:$$B_{\text{wire}} = \frac{\mu_0 I}{2 \pi r}$$This field acts in one direction, say outside the plane of the paper.The magnetic field at the centre of a circular loop of radius $r$ is:$$B_{\text{loop}} = \frac{\mu_0 I}{2r}$$This field acts in the opposite direction, into the plane of the paper.Since the two fields are opposite, the net magnetic induction at $O$ is:$$B_O = B_{\text{loop}} - B_{\text{wire}} = \frac{\mu_0 I}{2r} - \frac{\mu_0 I}{2 \pi r}$$$$B_O = \frac{\mu_0 I}{2r} \left(1 - \frac{1}{\pi}\right) = \frac{\mu_0 I}{2 \pi r} (\pi - 1)$$Answer:$$\frac{\mu_0 I}{2 \pi r} (\pi - 1)$$Therefore, the correct option is D.