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

CONCEPT:

• The Current-Carrying Conductor produces a magnetic Field Around it. 

Biot-Savarts Law: 

The direction and Magnitude of this magnetic field can be obtained from Biot-Savarts Law. 

\(\overrightarrow{dB} = \frac{μ _{0}\overrightarrow{I}\times \overrightarrow{dl}}{4\pi r^{2}}\)

Where dB = Magnetic Field produced due to small wire of length dl, I = Current in wire, μ0 = Permittivity of free space, dl = Small Current Element

• The magnetic field generated due to the current-carrying circular conductor at its center is given as:

\(\Rightarrow B=\dfrac{\mu_0 i}{2r}\)

• This result has been obtained from Biot Savarts Law.

EXPLANATION:

• The Magnetic Field Produced due to the Semi-Circular wire is half of that of Circular wire.
• So, the magnetic field produced at the center "O" of the wire will be

\(\Rightarrow B = \frac{1}{2}\times \dfrac{\mu_0 i}{2r}=\dfrac{\mu_0 i}{4r}\)

Hence option 2 is the correct option.