A thin circular disk of radius R is uniformly | vedclass

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(C) Consider a thin elemental ring of radius $x$ and width $dx$ on the disk.The area of this elemental ring is $dA = 2 \pi x dx$.The charge on this el

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$\frac{1}{4} \pi R^{4} \sigma \omega$

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(C) Consider a thin elemental ring of radius $x$ and width $dx$ on the disk.The area of this elemental ring is $dA = 2 \pi x dx$.The charge on this elemental ring is $dq = \sigma dA = 2 \pi \sigma x dx$.Since the disk rotates with angular speed $\omega$,the time period of rotation is $T = \frac{2 \pi}{\omega}$.The equivalent current $dI$ due to this rotating charge is $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2 \pi}$.Substituting $dq$,we get $dI = \frac{(2 \pi \sigma x dx) \omega}{2 \pi} = \sigma \omega x dx$.The magnetic moment $dM$ of this elemental ring is $dM = dI \cdot A_{ring} = (\sigma \omega x dx) \cdot (\pi x^2) = \pi \sigma \omega x^3 dx$.To find the total magnetic moment $M$,integrate $dM$ from $x = 0$ to $x = R$:$M = \int_{0}^{R} \pi \sigma \omega x^3 dx = \pi \sigma \omega \left[ \frac{x^4}{4} \right]_{0}^{R} = \frac{1}{4} \pi \sigma \omega R^4$.

$A$ thin circular disk of radius $R$ is uniformly charged with surface charge density $\sigma > 0$. The disk rotates about its central axis with a uniform angular speed $\omega$. The magnetic moment of the disk is:

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