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Question

(B) According to Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_{enclosed}$.For a long hollow pipe,the current $I$ flows only through the surface

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Only outside the pipe

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(B) According to Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_{enclosed}$.For a long hollow pipe,the current $I$ flows only through the surface of the pipe.Inside the pipe,any Amperian loop will enclose zero current $(I_{enclosed} = 0)$,so the magnetic field $B = 0$.Outside the pipe,at a distance $r$ from the axis,the Amperian loop encloses the total current $I$,resulting in a magnetic field $B = \frac{\mu_0 I}{2\pi r}$.Therefore,the magnetic field is produced only outside the pipe.

If a long hollow copper pipe carries a current,then the magnetic field is produced:

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