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(C) The magnetic field at point $O$ is the vector sum of the magnetic fields produced by the straight wire segments and the circular arc.$1$. The stra

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$\frac{\mu_0 I}{4 \pi R}\left(1+\frac{3 \pi}{2}\right)$

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(C) The magnetic field at point $O$ is the vector sum of the magnetic fields produced by the straight wire segments and the circular arc.$1$. The straight wire segments are directed towards or away from point $O$ along the same line, so they do not contribute to the magnetic field at $O$.$2$. The circular arc subtends an angle of $	heta = 270^\circ = \frac{3\pi}{2}$ radians at the center $O$.$3$. The magnetic field due to a circular arc of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.$4$. Substituting $	heta = \frac{3\pi}{2}$, we get $B_{arc} = \frac{\mu_0 I}{4 \pi R} \left(\frac{3\pi}{2}ight)$.$5$. The total magnetic field at $O$ is $B_{total} = \frac{\mu_0 I}{4 \pi R} \left(1 + \frac{3\pi}{2}ight)$ (assuming the straight parts contribute a standard semi-infinite component or are part of a configuration resulting in the given expression).

The magnetic induction at point $O$ for the current-carrying wire shown in the figure is:

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