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(C) The magnetic field at point $O$ due to a cylindrical conductor with a hole can be calculated by the principle of superposition. The field is the v

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will remain the same

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(C) The magnetic field at point $O$ due to a cylindrical conductor with a hole can be calculated by the principle of superposition. The field is the vector sum of the field due to the solid conductor and the field due to a cylinder of current flowing in the opposite direction through the hole.Let the magnetic field produced by the hole at point $O$ be $\vec{B}_{hole}$. The magnitude of this field is $B = \frac{\mu_0 J r^2}{2d}$,where $J$ is the current density,$r$ is the radius of the hole,and $d$ is the distance from the center of the hole to $O$.In the first case,the hole is at $A$,so the field at $O$ is $\vec{B}_A$. In the second case,the hole is at $B$,so the field at $O$ is $\vec{B}_B$. Since the holes are identical and at the same distance from $O$ (let this distance be $s$),the magnitudes are equal: $|\vec{B}_A| = |\vec{B}_B| = B_0$.The angle between the vectors $\vec{OA}$ and $\vec{OB}$ is $120^{\circ}$. The resultant magnetic field magnitude is given by the vector sum: $B_{net} = \sqrt{B_0^2 + B_0^2 + 2 B_0^2 \cos(120^{\circ})}$.Since $\cos(120^{\circ}) = -0.5$,we have $B_{net} = \sqrt{2 B_0^2 - B_0^2} = \sqrt{B_0^2} = B_0$.Thus,the magnitude of the magnetic field at $O$ remains the same.

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