In a straight coaxial cable, the central conductor and the outer...

By Ampere's circuital law, the line integral of the magnetic field around any closed loop is proportional to the net current enclosed by the loop:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$

For a straight coaxial cable, consider a circular Amperian loop of radius $r$ outside the cable ($r > R_{outer}$). The central conductor carries a current $I$ in one direction, and the outer conductor carries an equal current $I$ in the opposite direction.

The net current enclosed by the Amperian loop outside the cable is:

$I_{enclosed} = I - I = 0$

Substituting this into Ampere's law:

$B (2\pi r) = \mu_0 (0)$

$B = 0$

Thus, the magnetic field vanishes everywhere outside the cable. In all other regions (inside the inner conductor, between the conductors, and inside the outer conductor), the enclosed current is non-zero, so the magnetic field is non-zero.