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

Using Ampere's circuital law:

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

For a point just outside the cylindrical tube, the radial distance from the axis is $r = b$.

Considering a circular Amperian loop of radius $r = b$ coaxial with the tube, the total current enclosed by this loop is the entire current $i$ carried by the tube.

$B (2\pi b) = \mu_0 i$

$B = \dfrac{\mu_0 i}{2\pi b}$