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(C) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_{0}$ times the net c

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zero at any point inside the pipe

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(C) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_{0}$ times the net current $I_{	ext{enclosed}}$ passing through the surface bounded by the path: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{	ext{enclosed}}$.For an internal point inside the pipe where the radial distance $r Thus,$I_{	ext{enclosed}} = 0$.Applying Ampere's law: $B(2 \pi r) = \mu_{0}(0) \implies B = 0$.Therefore,the magnetic field is zero at any point inside the pipe.For an external point where $r > R$,the path encloses the total current $I$,so $B(2 \pi r) = \mu_{0} I \implies B = \frac{\mu_{0} I}{2 \pi r}$.Hence,option $(c)$ is correct.

$A$ direct current $I$ flows along the length of an infinitely long straight thin-walled pipe. Then the magnetic field is

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