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Question

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

Vedclass · Accepted Answer

$\frac{\mu_0 q \omega}{2 \pi R}$

Vedclass · Answer

(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.The area of the disc is $A = \pi R^2$. The surface charge density is $\sigma = \frac{q}{\pi R^2}$.The charge on the elemental ring is $dq = \sigma \cdot (2\pi x dx) = \frac{q}{\pi R^2} \cdot 2\pi x dx = \frac{2q x dx}{R^2}$.When the disc rotates with angular frequency $\omega$,this ring acts as a current loop with current $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2\pi} = \frac{2q x dx}{R^2} \cdot \frac{\omega}{2\pi} = \frac{q \omega x dx}{\pi R^2}$.The magnetic field at the center 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 $x = R$:$B = \int_0^R \frac{\mu_0 q \omega}{2 \pi R^2} dx = \frac{\mu_0 q \omega}{2 \pi R^2} [x]_0^R = \frac{\mu_0 q \omega}{2 \pi R}$.

$A$ plastic disc of radius $R$ has a charge $q$ uniformly distributed over its surface. If the disc is rotated at an angular frequency $\omega$ about its axis,the magnetic induction at the center of the disc is:

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