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(B) The magnetic field at the centre $O$ is the vector sum of the fields due to the straight segments $AB$, $DE$ and the circular arc $BCD$.Using the

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$-\frac{\mu_0}{2 \pi} \frac{ I }{ r }(\pi-1) \hat{ i }$

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(B) The magnetic field at the centre $O$ is the vector sum of the fields due to the straight segments $AB$, $DE$ and the circular arc $BCD$.Using the Biot-Savart law, the field due to a semi-infinite wire at a distance $r$ from its end is $\vec{B} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.For segment $AB$, the field at $O$ is $\vec{B}_{AB} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.For segment $DE$, the field at $O$ is $\vec{B}_{DE} = \frac{\mu_0 I}{4 \pi r} \hat{k}$.For the circular arc $BCD$, the angle subtended is $2\pi - 0 = 2\pi$ (a full circle minus the gap). However, looking at the geometry, the arc is $2\pi - 	heta$ where $	heta$ is the angle of the gap. Assuming the gap is negligible, the arc is a full circle. The field is $\vec{B}_{BCD} = -\frac{\mu_0 I}{2 r} \hat{k}$.Summing these: $\vec{B}_O = \vec{B}_{AB} + \vec{B}_{DE} + \vec{B}_{BCD} = \frac{\mu_0 I}{4 \pi r} \hat{k} + \frac{\mu_0 I}{4 \pi r} \hat{k} - \frac{\mu_0 I}{2 r} \hat{k} = \frac{\mu_0 I}{2 \pi r} \hat{k} - \frac{\mu_0 I}{2 r} \hat{k} = \frac{\mu_0 I}{2 \pi r} (1 - \pi) \hat{k} = -\frac{\mu_0 I}{2 \pi r} (\pi - 1) \hat{k}$.Note: The direction is along the $z$-axis $(\hat{k})$, not $\hat{i}$ as given in the options. Assuming the question intended $\hat{k}$ for the $z$-axis direction.

An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is:

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