A non-conducting disc having uniform positive charge Q, is rotating about its axis with uniform angular velocity $\omega$. The magnetic field at the center of the disc is.

Consider a ring of radius x and thickness dx.

Equivalent current in this ring $=\frac{\omega }{2\pi }$ x charge on ring  $=\frac{\omega }{2\pi }\times \left(2\pi \times dx\right)\frac{Q}{\pi R^2}$( anticlockwise which produces outward magnetic field )
dB (due to this ring)   $=\frac{{\mu }_0}{2x}\left(\frac{\omega }{2\pi }\frac{2xQ}{R^2}dx\right)\therefore B=\underset{0}{\overset{R}{\int }}\frac{{\mu }_0\omega }{2\pi }\frac{Q}{R^2}dx.$
$=\frac{{\mu }_0\omega Q}{2\pi R^2}.R=\frac{{\mu }_0\omega Q}{2\pi R}.$