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(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.The surface charge density $\sigma = \frac{Q}{\pi R^2}$.The charge on

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(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.The surface charge density $\sigma = \frac{Q}{\pi R^2}$.The charge on the elemental ring is $dq = \sigma (2 \pi x dx) = \frac{Q}{\pi R^2} (2 \pi x dx) = \frac{2Qx dx}{R^2}$.The current $di$ due to the rotation of this ring with angular velocity $\omega$ is $di = \frac{dq}{T} = \frac{dq \omega}{2 \pi} = \frac{2Qx dx}{R^2} \cdot \frac{\omega}{2 \pi} = \frac{Q \omega x dx}{\pi R^2}$.The magnetic field $dB$ at the centre due to this ring is $dB = \frac{\mu_0 di}{2x} = \frac{\mu_0}{2x} \cdot \frac{Q \omega x dx}{\pi R^2} = \frac{\mu_0 Q \omega}{2 \pi R^2} dx$.Integrating from $x = 0$ to $x = R$:$B = \int_0^R \frac{\mu_0 Q \omega}{2 \pi R^2} dx = \frac{\mu_0 Q \omega}{2 \pi R^2} [x]_0^R = \frac{\mu_0 Q \omega}{2 \pi R^2} \cdot R = \frac{\mu_0 Q \omega}{2 \pi R}$.Thus,$B \propto \frac{1}{R}$.This relationship is represented by a rectangular hyperbola,which corresponds to Figure $A$.

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