A long cylindrical tube with inner and outer radii a and b carries...

The current $i$ flows through the cross-section of the cylindrical tube between the inner radius $a$ and outer radius $b$.

For a point just inside the tube (in the hollow region), the radial distance $r$ from the central axis is $r \le a$.

Considering a circular Amperian loop of radius $r \le a$ concentric with the axis of the tube, the current enclosed by this loop is zero, as the current flows only in the region $a < r < b$.

According to Ampere's circuital law:

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

Since $I_{\text{enclosed}} = 0$, we get:

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

Thus, the magnitude of the magnetic field at a location just inside the tube is zero.