Question

The magnetic field at distance y from the centre on the axis of a disk of radius r and uniform surface charge density $\sigma$ spinning with angular velocity $\omega$ is,

• A. $\frac{{\mu }_0\sigma \omega }{3}(\frac{r^2-2y^2}{\sqrt{r^2-y^2}}+2y)$
• B. $\frac{{\mu }_0\sigma \omega }{2}\left(\frac{r^2-y^2}{\sqrt{r^2+y^2}}\right)$
• C. $\frac{{\mu }_0\sigma \omega }{3}(\frac{r^2-2y^2}{\sqrt{r^2+y^2}}-2y)$
• D. $\frac{{\mu }_0\sigma \omega }{3}(\frac{r^2+2y^2}{\sqrt{r^2+y^2}}-2y)$

Answer 1

Answer: (D)

$\frac{{\mu }_0\sigma \omega }{3}(\frac{r^2+2y^2}{\sqrt{r^2+y^2}}-2y)$

Charge on a ring of radius x and width dx
$dq=2\left(\pi xdx\right)\sigma$
Current, $dl=\frac{dq}{dt}=\frac{2\pi x\sigma dx}{dt}=\omega \sigma xdx$
$dB=\frac{{\mu }_{0}dl{r}^2}{2(x^2+y^2{)}^{3/2}}$
$B=\frac{{\mu }_0\sigma \omega }{2}(\frac{r^2+2y^2}{\sqrt{r^2+y^2}}-2y)$.