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(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed loop is equal to $\mu_0$ times the net curre

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(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed loop is equal to $\mu_0$ times the net current enclosed by the loop,i.e.,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$.For a point outside the coaxial cable at a distance $r$ from the axis,we consider an Amperian loop of radius $r$.The net current enclosed by this loop is the sum of the current flowing through the inner conductor $(+i)$ and the current returning through the outer conductor $(-i)$.Therefore,$I_{	ext{enclosed}} = i + (-i) = 0$.Since the net current enclosed is zero,the magnetic field $\vec{B}$ at any point outside the cable must be zero.

$A$ current $i$ ampere flows along the inner conductor of a coaxial cable and returns along the outer conductor of the cable. The magnetic induction at any point outside the conductor at a distance $r$ metre from the axis is

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