Question

A thin non conducting disc of mass $M=2\text{kg},$ charge $Q=2\times {10}^{-2}\text{C}$ and radius $R=\frac{1}{6}\text{m}$ is placed on a frictionless horizontal plane with its centre at the origin of the coordinate system. A non uniform, radial magnetic field $\overrightarrow{B}=B_0\hat{r}$ exists in space, where $B_0=10\text{T and}\hat{r}$ is a unit vector in the radially outward direction. The disc is set in motion with an angular velocity $\omega =x\times {10}^2\text{rad/sec},$ about an axis passing through its centre and perpendicular to its plane, as shown in the figure. At what value of $x,$ the disc will lift off from the surface?

• A. 9
• B. 3
• C. 4
• D. 2

Answer 1

The correct option is A 9

The Force an any small part of the disc is in the vertically upward direction.
$dF=\left(\frac{Q}{\pi R^2}2\pi rdr\right)\omega r{B}_0$
$dF=\frac{2Q\omega B_0}{R^2}{r}^{2}dr$
$F=\frac{2}{3}Q\omega B_{0}R=Mg$
$\Rightarrow \omega =\frac{3Mg}{2Q{B}_{0}R}=\frac{3\times 2\times 10}{2\times 2\times {10}^{-2}\times 10\times \frac{1}{6}}=9\times {10}^2\text{rad/s}$