Question

A non-conducting disc, having uniformly distributed positive charge $Q$, is rotating about its axis with uniform angular velocity $\omega$. Choose the correct alternative(s).

• A. The magnetic field at the centre of the disc is directed outward.
• B. The magnetic field at the centre of the disc have magnitude $\frac{{\mu }_{0}Q\omega }{2\pi R}$
• C. The magnetic field at the centre of the disc is directed inward.
• D. The magnetic field at the centre of the disc have magnitude $\frac{{\mu }_{0}Q\omega }{4\pi R}$

Answer 1

Answer: (A), (D)

The magnetic field at the centre of the disc have magnitude $\frac{{\mu }_{0}Q\omega }{4\pi R}$

Let us consider an elementary ring of radius $x$ and thickness $dx$ as shown in the figure.

Equivalent current in the elementary ring,

$dI=dq\times f=dq\times \frac{\omega }{2\pi }$

$=\frac{\omega }{2\pi }\times \left(2\pi xdx\right)\frac{Q}{\pi R^2}=\frac{Q\omega xdx}{\pi R^2}$

Magnetic field at the centre of the disc due to this elementary ring,

$dB=\frac{{\mu }_{0}dI}{2R}$

$\Rightarrow {\int }_0^{B}dB={\int }_0^R\frac{{\mu }_{0}Q\omega xdx}{2\pi R^3}$

$\Rightarrow B=\frac{{\mu }_{0}Q\omega }{2\pi R^3}\times [\frac{x^2}{2}{]}_0^R$

$\Rightarrow B=\frac{{\mu }_{0}Q\omega }{4\pi R}\left[\text{Outward}\right]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, options $\left(A\right)$ and $\left(D\right)$ are correct.