Question

A disc of radius $R$ rotates with a constant angular velocity $\omega$ about its own axis. The surface charge density of this disc varies as $\sigma =a{r}^2$ where $r$ is the distance from the centre of the disc. Determine the magnetic field intensity at the centre of the disc.

• A. ${\mu }_{0}a\omega R^3$
• B. $\frac{{\mu }_{0}a\omega R^3}{6}$
• C. $\frac{{\mu }_{0}a\omega R^3}{8}$
• D. $\frac{{\mu }_{0}a\omega R^3}{3}$

Answer 1

Answer: (B)

$\text{dA}=\left(2\pi \text{r}\right)\cdot \text{dr}$
$\mathrm{dq}=\sigma \cdot \mathrm{dA}=\left(2\pi {\mathrm{ar}}^3\right)\cdot \mathrm{dr}$
$\mathrm{i}=(\mathrm{dq})\mathrm{f}=(\mathrm{dq})\frac{\omega }{2\pi }=\left(\mathrm{a}\omega {\mathrm{r}}^3\right)\cdot \mathrm{dr}$
$\text{dB}=\frac{{\mu }_0\text{i}}{2\text{r}}=\frac{a{\mu }_0\omega {\text{r}}^2\text{dr}}{2}$
$\therefore \text{B}={\int }_0^{\text{R}}\text{dB}=\frac{{\mu }_{0}a\omega R^3}{6}$