Magnetic field due to the current carrying wire as shown in the figure at point "O" will be:

$\frac{\mu_0 I}{4 R}$

The correct answer is Option (2) → $\frac{\mu_0 I}{4 R}$

The magnetic field generated due to the current carrying circular, conductor at it's centre is,

$B = \frac{\mu_0 I}{2 r}$ [Using Biot Savart's law]

∴ Magnetic field produced due to semi-circular coil is,

$B=\frac{1}{2}×\frac{\mu_0 I}{2 r}$

$=\frac{\mu_0 I}{4 r}$