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(B) According to Ampere's Circuital Law, the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net cu

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the magnetic field at any point inside the pipe is zero

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(B) According to Ampere's Circuital Law, the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $i_{	ext{enclosed}}$ passing through the surface bounded by the loop.$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{	ext{enclosed}}$For any point inside a thin-walled pipe, we can choose an Amperian loop (a circle) that lies entirely inside the pipe.Since the current $i$ flows only through the walls of the pipe, the current enclosed by this loop is $i_{	ext{enclosed}} = 0$.Therefore, $\oint \vec{B} \cdot d\vec{l} = 0$, which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

$A$ current $i$ flows in an infinitely long, straight and thin-walled pipe. Then,

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