JEE Main 2012 — Magnetic Effects of Current Question with 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 $dq=\sigma (2\pi rdr)$ 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).