### Magnetic Field

Build an Ampere path with one side inside the cylinder parallel to the axis, two sides perpendicular to the axis and the fourth outside again parallel to the axis (like a coil).
Now $\oint_L \mathbf{B}\cdot \mathbf{r}=Bl$ where $l$ is the length of the side inside.
The current passing through the area is $I=(\omega/2\pi)Q l$, so that $Bl=4\pi K I=4\pi K\omega Ql/2\pi\rightarrow B=2 K\omega Q$.