A part of an infinitely long current carrying wire is bent into a cir…

The correct answer is Option (4) → \( \displaystyle\frac{\mu_0 I}{2\pi r}(\pi-1) \) normally inward

The field at the centre is the sum of two parts: the straight wire, which is effectively one infinitely long wire at distance \( r \) from the centre, and the circular loop of radius \( r \). The direction of each comes from the right-hand rule.

Straight wire: the current flows to the right and the centre lies above the wire, so the field there points out of the page.

    \( \displaystyle B_1 = \frac{\mu_0 I}{2\pi r} \quad \text{(outward)} \)

Circular loop: the arrows show the current going round the loop clockwise, so its field at the centre points into the page.

    \( \displaystyle B_2 = \frac{\mu_0 I}{2r} = \frac{\mu_0 I}{2\pi r}\,\pi \quad \text{(inward)} \)

The two fields are opposite, and \( B_2 \gt B_1 \), so the resultant is along \( B_2 \):

    \( \displaystyle B = B_2 - B_1 = \frac{\mu_0 I}{2\pi r}(\pi - 1) \quad \text{normally inward} \)