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

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will remain the same

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(C) The magnetic field at the center $O$ due to a cylindrical conductor with a hole can be calculated using the principle of superposition. The field is equivalent to the field of a solid conductor minus the field of a cylinder of radius $r$ carrying current in the opposite direction.The magnetic field due to a hole at distance $s$ from the center $O$ is given by $B = \frac{\mu_0 J r^2}{2s}$,where $J$ is the current density. Since $J$ and $r$ are constant,the magnitude of the magnetic field $B_1$ due to the hole at $A$ is $B_1 = \frac{\mu_0 J r^2}{2s}$.When a second identical hole is drilled at $B$,the magnetic field at $O$ is the vector sum of the fields due to the two holes. Let $\vec{B}_A$ and $\vec{B}_B$ be the fields due to holes at $A$ and $B$ respectively. Both have magnitude $B_1$. The angle between $\vec{OA}$ and $\vec{OB}$ is $120^{\circ}$.The resultant magnetic field $B_{net}$ is given by:$B_{net} = \sqrt{B_1^2 + B_1^2 + 2 B_1^2 \cos(120^{\circ})}$$B_{net} = \sqrt{2 B_1^2 + 2 B_1^2 (-0.5)} = \sqrt{2 B_1^2 - B_1^2} = B_1$.Thus,the magnitude of the magnetic field at $O$ remains the same.

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