A part of an infinitely long current carrying wire is bent into a circular shape as shown in the diagram. The magnetic field at the centre of the circle is

$\frac{μ_0I}{2πr}(π-1)$ normally inward

The correct answer is Option (4) → $\frac{μ_0I}{2πr}(π-1)$ normally inward

Magnetic field due to circular loop of radius $r$ at centre:

$B_{circle} = \frac{\mu_0 I}{2r}$

Magnetic field due to infinitely long straight wire at perpendicular distance $r$:

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

Both fields act in the opposite direction. Hence, net magnetic field is:

$B = \frac{\mu_0 I}{2r} - \frac{\mu_0 I}{2 \pi r}$

Final Answer: $B = \frac{\mu_0 I}{2r}\left(1 - \frac{1}{\pi}\right)$