Question

A flat disc of radius $R$ carries an excess charge on its surface. The surface charge density is $\sigma$. The disc rotated about an axis perpendicular to the plane passing through its center with angular velocity $\omega$. Find the torque on the disc , if it is placed in a uniform magnetic field $B$ directed perpendicular to the rotation axis:

• A. $\frac{\sigma \omega \pi B{R}^4}{4}$
• B. $\frac{\sigma \omega \pi B{R}^2}{2}$
• C. $\frac{3\sigma \omega \pi B{R}^4}{4}$
• D. $\frac{2\sigma \omega \pi B{R}^4}{3}$

Answer 1

The correct option is A $\frac{\sigma \omega \pi B{R}^4}{4}$

Conside an infinitesimal elemental ring of radius $r$ and thickness $dr$ on this disc as shown in the figure.


As the ring is rotating with angular speed $\omega$, due to its the equivalent current is

$di=\left(dq\right)f[\because dq=\sigma (2\pi rdr\left)\right]$

$di=\sigma \left(2\pi rdr\right)\left(\frac{\omega }{2\pi }\right)=\sigma \omega rdr$

Magnetic moment of the elemental ring,

$d\mu =di\left(A\right)=\sigma \omega rdr\left(\pi r^2\right)$

Torque experienced by this ring:

$d\tau =d\mu B\sin {90}^{\circ }$

$d\tau =B\sigma \omega \pi r^{3}dr$

Total torque on the disc will be summation of torque experienced by all such elemental rings on the disc.

${\int }_0^{\tau }d\tau ={\int }_0^{R}B\sigma \omega \pi r^{3}dr$

$\tau =\frac{B\sigma \omega \pi R^4}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.