Question

A non-conducting thin disc of radius $R$ rotates about its axis with an angular velocity $\omega$. The surface charge density on the disc varies with the distance $r$ from the centre as $\sigma (r)={\sigma }_0\left[1+{\left(\frac{r}{R}\right)}^{\beta }\right]$, where ${\sigma }_0$ and $\beta$ are
constants. If the magnetic induction at the center is $B=\left(\frac{9}{10}\right){\mu }_0{\sigma }_0\omega R$, the value of $\beta$ is

• A. $\frac{1}{4}$
• B. $4$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

Answer: (A)

Total charge on the disc, $Q=\sigma \left(\pi R^2\right)$
Period of motion, $T=2\pi /\omega$

So, current in the ring, $dI=\frac{dQ}{T}$
$\because Q=\sigma \pi R^2\Rightarrow dQ=2\sigma \pi rdr$
$\therefore dI=\frac{2\sigma \pi rdr}{2\pi }\times \omega =\sigma \omega rdr$
Magnetic field due to ring is
$dB=\frac{{\mu }_{0}dI}{2r}$
So, magnetic field due to disk is
$B=\int dB={\int }_0^R\frac{{\mu }_0}{2r}\cdot \sigma \omega rdr$
On putting value of $\sigma$, we get
$B=\frac{{\mu }_0{\sigma }_0\omega }{2}{\int }_0^R\left[{}^1+{\left(\frac{r}{R}\right)}^{\beta }\right]dr$
$=\frac{{\mu }_0{\sigma }_0\omega }{2}\left[R+\frac{R^{\beta +1}}{R^{\beta }(1+\beta )}\right]=\frac{{\mu }_0{\sigma }_0\omega }{2}\times \left(\frac{2+\beta }{1+\beta }\right)R$
Now,
$B=\frac{9}{10}{\mu }_0{\sigma }_0\omega R=\frac{{\mu }_0{\sigma }_0\omega R}{2}\times \frac{(2+\beta )}{(1+\beta )}$
$\Rightarrow 5\beta +10=9\beta +9$
$\Rightarrow \beta =\frac{1}{4}$