A thin disc of radius R and mass M has charge | vedclass

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(A) Consider a small elemental ring of radius $r$ and thickness $dr$ on the disc.The area of this ring is $dA = 2\pi r dr$.The charge on this ring is

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$\frac{q}{2 M}$

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(A) Consider a small elemental ring of radius $r$ and thickness $dr$ on the disc.The area of this ring is $dA = 2\pi r dr$.The charge on this ring is $dq = \sigma dA = \frac{q}{\pi R^2} (2\pi r dr) = \frac{2q}{R^2} r dr$,where $\sigma$ is the surface charge density.The current $dI$ produced by this rotating ring is $dI = \frac{dq}{T} = \frac{dq \omega}{2\pi} = \frac{q \omega}{\pi R^2} r dr$.The magnetic moment of this ring is $d\mu = dI \cdot A = (\frac{q \omega}{\pi R^2} r dr) (\pi r^2) = \frac{q \omega}{R^2} r^3 dr$.Integrating from $0$ to $R$,the total magnetic moment is $\mu = \int_0^R \frac{q \omega}{R^2} r^3 dr = \frac{q \omega}{R^2} [\frac{r^4}{4}]_0^R = \frac{1}{4} q \omega R^2$.The angular momentum of the disc is $L = I \omega = (\frac{1}{2} M R^2) \omega$.The ratio of magnetic moment to angular momentum is $\frac{\mu}{L} = \frac{\frac{1}{4} q \omega R^2}{\frac{1}{2} M R^2 \omega} = \frac{q}{2M}$.

$A$ thin disc of radius $R$ and mass $M$ has charge $q$ uniformly distributed on it. It rotates with angular velocity $\omega$. The ratio of magnetic moment and angular momentum for the disc is

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