Question

A flat circular disc of radius R has uniform charge distribution of -$\sigma$ upto $r$ and of $+\sigma$ from $r$ upto $R$, such that net charge on the disc is zero. lf the disc is rotating uniformly at the rate of $\omega$ rad$s^{-1}$, its magnetic moment then is given by $\frac{\pi \sigma \omega R^4}{n}$. Find the value of n.

• A. 8
• B. 2
• C. 4
• D. 16

Answer 1

The correct option is A 8

Charge on the portion of the disc having radius r: $q_1=\pi r^2(-\sigma )$

Charge on the portion of disc having inner radius r and outer radius R: $q_2=\pi (R^2-r^2)\sigma$

Given total charge on the disc is zero.

$\therefore q_1+q_2=0$

$\Rightarrow \pi r^2(-\sigma )+\pi (R^2-r^2)\sigma =0$

$\Rightarrow 2r^2=R^2$

$\Rightarrow R=\sqrt{2}r$

For a ring of radius r having charge q uniformly distributed on the ring, current $I=\frac{q\omega }{2\pi }$

magnetic moment $\mu =IA$ where I is the current and A is the area.

For the circular disc we need to integrate to find the magnetic moment

Consider a thick ring of width dx at a distance x from the center of the disc.

Area of the ring width dx: $A=\pi x^2$

Magnetic moment of this annular disc: $dM=\frac{q\omega }{2\pi }\pi x^2=\frac{2\pi xdx\sigma \omega }{2\pi }{x}^2=\pi \omega {\sigma }^{\prime }{x}^{3}dx$

$\therefore M=\omega \left({\int }_0^r\pi \right(-\sigma )x^{3}dx+{\int }_r^R\pi (\sigma \left)x^{3}dx\right)=\omega (-\pi \sigma \frac{r^4}{4}+\pi \sigma (\frac{R^4-r^4}{4}\left)\right)=\pi \sigma \omega \frac{R^4-2r^4}{4}=\pi \sigma \omega \frac{R^4}{8}$

$\Rightarrow n=8$