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

The correct answer is: outside the cable.

To understand why, we apply Ampère's Circuital Law, which states that the line integral of the magnetic field $B$ around a closed Amperian loop is proportional to the net current enclosed by that loop:

$\oint B·\mathrm{dl}={\mu }_0{I}_{\text{enc}}$

Let's analyze each region of the coaxial cable systematically.

Setup of the Problem:
- The inner (central) conductor carries current $I$ in one direction (say, out of the page).
- The outer conductor carries equal current $I$ in the opposite direction (into the page).
- We draw circular Amperian loops of radius $r$ centered on the cable axis at different regions.

Region 1 — Inside the inner conductor:
Only a fraction of the inner conductor's current is enclosed. The magnetic field is non-zero and increases with $r$. So $B\ne 0$ here.

Region 2 — In between the two conductors:
The Amperian loop encloses the entire inner conductor current $I$, but none of the outer conductor's current yet. So:

$I_{\text{enc}}=I\ne 0$

Therefore, $B\ne 0$ in the region between the conductors.

Region 3 — Outside the cable (beyond the outer conductor):
Now the Amperian loop encloses both conductors. Since the currents are equal in magnitude but opposite in direction:

$I_{\text{enc}}=I-I=0$

Applying Ampère's Law:

$\oint B·\mathrm{dl}={\mu }_0\times 0=0$

Since by symmetry $B$ is uniform along the circular loop, this gives:

$B\times 2\pi r=0\Rightarrow B=0$

Conclusion:
The net current enclosed by any Amperian loop drawn outside the coaxial cable is zero, because the equal and opposite currents in the two conductors completely cancel each other. As a result, the magnetic field is zero outside the cable. This is precisely why coaxial cables are used in practice — they confine the magnetic field within the cable itself and produce no external electromagnetic interference.