An annular disc of inner radius $$R$$ and outer radius $$2R$$ has uniform charge density $$\sigma$$. It is rotating about the axis passing through the center and perpendicular to its plane with angular velocity $$\omega$$. If the magnetic field at the center of the ring is $$\frac{\mu_0\sigma\omega R}{n},$$ then find $$n$$.  

Consider a small circular ring of radius $$r$$ and thickness $$dr$$.

Its charge is  $$dq=\sigma(2\pi r\,dr)$$

Since the disc rotates with angular velocity $$\omega$$, the time taken for one revolution is  $$T=\frac{2\pi}{\omega}$$

Hence, the current due to this small ring is  $$dI=\frac{dq}{T}$$

$$dI=\frac{\sigma(2\pi r\,dr)\omega}{2\pi}$$

$$dI=\sigma\omega r\,dr$$

The magnetic field at the centre due to a circular ring is  $$dB=\frac{\mu_0dI}{2r}$$

Therefore,  $$dB=\frac{\mu_0\sigma\omega r\,dr}{2r}$$

$$dB=\frac{\mu_0\sigma\omega}{2}\,dr$$

The annular disc extends from $$r=R$$ to $$r=2R$$.

Hence,  $$B=\int_R^{2R}dB$$

$$B=\frac{\mu_0\sigma\omega}{2}\int_R^{2R}dr$$

$$B=\frac{\mu_0\sigma\omega}{2}(2R-R)$$

$$B=\frac{\mu_0\sigma\omega R}{2}$$

Given,  $$B=\frac{\mu_0\sigma\omega R}{n}$$

Comparing,  $$\frac{1}{n}=\frac{1}{2}$$

Therefore,  $${n=2}$$

Hence, the correct answer is 2.