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Consider a non conducting plate of radius a and mass $m$ which has a charge $q$ distributed uniformly over it. The plate is rotated about its axis with an angular speed $ω$ Show that the magnetic moment $M$ and the angular momentum $L$ of the plate are related as $\frac{M}{L} = \frac{q}{2 m} .$

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$\frac{2 q}{m}$

$\frac{3 q}{2 m}$

$\frac{q}{m}$

$\frac{q}{2 m}$

answer is D.

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Detailed Solution

If $σ$ is the surface charge density, then $q = σ π a^{2}$

Current $i = σ ω r d r$

The magnetic moment of the element ring

$d M = i (d A) = σ ω r d r (π r^{2}) = π σ ω r^{3} d r \Rightarrow M = π σ ω \int_{0}^{a} r^{3} d r = π a^{2} σ \frac{ω a^{2}}{4} = \frac{q ω a^{2}}{4}$

The angular momentum of the disc about its axis $L = \frac{m a^{2}}{2} ω$

$\frac{M}{L} = \frac{\frac{q ω a^{2}}{4}}{\frac{m a^{2}}{2} ω} = \frac{q}{2 m}$

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