JEE Physics 2026 Question with Solution

Given: An infinitely long straight wire carrying current $I$ is bent to form a circular loop of radius $r$ and a straight wire segment in the same plane.

Find: The magnetic field at the centre $O$ of the circular loop.

The total magnetic field at $O$ is the vector sum of the fields due to the circular loop and the infinitely long straight wire.

Field due to the circular loop at its centre:

$$B_{\text{loop}} = \frac{\mu_0 I}{2r}$$

Field due to an infinitely long straight wire at distance $r$:

$$B_{\text{wire}} = \frac{\mu_0 I}{2 \pi r}$$

Based on the current direction in the diagram, the current in the loop is counter-clockwise, so by the right-hand thumb rule the magnetic field at $O$ is out of the page, corresponding here to $\hat{i}$.

The straight wire also produces a magnetic field at $O$ in the same direction, so the two fields add.

Therefore,

$$\vec{B}_{\text{net}} = \vec{B}_{\text{loop}} + \vec{B}_{\text{wire}}$$ $$\vec{B}_{\text{net}} = \left( \frac{\mu_0 I}{2r} + \frac{\mu_0 I}{2 \pi r} \right) \hat{i}$$ $$\vec{B}_{\text{net}} = \frac{\mu_0 I}{2 \pi r}(\pi + 1) \hat{i}$$

Therefore, the correct option is B.

Given: The magnetic field at the centre $O$ is produced by both the circular part and the straight part of the current-carrying wire.

Find: Whether the two contributions add or subtract.

Use the right-hand thumb rule separately for each segment:

1. For the circular loop, the current is counter-clockwise, so the field at the centre is directed out of the plane.
2. For the straight wire, at the point $O$ located a distance $r$ above the wire, the field is also out of the plane in the given coordinate convention, that is along $\hat{i}$.

Since both magnetic fields are along the same direction, they reinforce each other.

Hence,

$$\vec{B}_{\text{net}} = \frac{\mu_0 I}{2r}\hat{i} + \frac{\mu_0 I}{2 \pi r}\hat{i}$$ $$\vec{B}_{\text{net}} = \frac{\mu_0 I}{2 \pi r}(\pi + 1)\hat{i}$$

So the magnetic field is $\frac{\mu_0 I}{2 \pi r}(\pi + 1)\hat{i}$.