A charge $Q$ is uniformly distributed over the surface of non conducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure

• Correct answer
  A

  
  
• B

  
  
• C

  
  
• D

  
  

Solution

Consider ring like element of disc of radius $r$ and thickness $\mathrm{dr}$. If $\sigma$ is charge per unit area, then charge on the element $d q=\sigma(2 \pi r d r)$ current ' $i$ ' associated with rotating charge $\mathrm{dq}$ is $\mathrm{i}=\frac{(\mathrm{dq}) \mathrm{w}}{2 \pi}=\sigma \mathrm{wrdr}$ Magnetic field $\mathrm{dB}$ at center due to element $\mathrm{dB}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}}=\frac{\mu_0 \sigma \omega \mathrm{dr}}{2}$ $\begin{aligned} & \mathrm{B}_{\text {net }}=\int \mathrm{dB}=\frac{\mu_0 \sigma \omega}{2} \int_0^{\mathrm{R}} \mathrm{dr}=\frac{\mu_0 \sigma \omega \mathrm{R}}{2} \\ & \Rightarrow \quad \mathrm{B}_{\text {net }}=\frac{\mu_0 \mathrm{Q} \omega}{2 \pi \mathrm{R}} \quad\left[\because \mathrm{Q}=\sigma \pi \mathrm{R}^2\right] \end{aligned}$ So if $Q$ and $w$ are unchanged then $\mathrm{B}_{\mathrm{net}} \propto \frac{1}{\mathrm{R}}$ Hence variation of $B_{\text {net }}$ with $R$ should be a rectangular hyperbola as represented in (A).

Hence, the correct answer is option: 'A'.