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 $\sigma =ar^{2}$ where $r$ is the distance from the centre of the disc. Determine the magnetic field intensity at the centre of the disc.

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

Answer 1

Answer: (B)

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