In a co-axial,straight cable,the central cond | vedclass

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(A) According to Ampere's Circuital Law,the magnetic field $\vec{B}$ at a distance $r$ from the axis is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0

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outside the cable

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(A) According to Ampere's Circuital Law,the magnetic field $\vec{B}$ at a distance $r$ from the axis is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$.For a point outside the cable $(r > R_{	ext{outer}})$,the total current enclosed is $I_{	ext{enclosed}} = I_{	ext{inner}} + I_{	ext{outer}}$.Since the currents are equal and in opposite directions,$I_{	ext{inner}} = I$ and $I_{	ext{outer}} = -I$.Therefore,$I_{	ext{enclosed}} = I + (-I) = 0$.Since the enclosed current is zero,the magnetic field $\vec{B}$ outside the cable is zero.

In a co-axial,straight cable,the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:

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