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(B) Consider a thin ring-like element of the disc with radius $r$ and thickness $dr$.If $\sigma$ is the surface charge density, the charge $dq$ on thi

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$\frac{\mu_0 \sigma \omega R}{2}$

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(B) Consider a thin ring-like element of the disc with radius $r$ and thickness $dr$.If $\sigma$ is the surface charge density, the charge $dq$ on this element is:$dq = \sigma (2 \pi r) dr$The current $di$ associated with the rotating charge $dq$ is:$di = \frac{dq}{T} = \frac{dq \omega}{2 \pi} \quad (\because T = \frac{2 \pi}{\omega})$Substituting $dq$ into the expression for $di$:$di = \frac{(\sigma 2 \pi r dr) \omega}{2 \pi} = \sigma \omega r dr$The magnetic field $dB$ at the center due to this current-carrying ring is:$dB = \frac{\mu_0 di}{2r} = \frac{\mu_0 (\sigma \omega r dr)}{2r} = \frac{\mu_0 \sigma \omega}{2} dr$To find the net magnetic field $B_{	ext{net}}$ at the center, integrate from $r = 0$ to $r = R$:$B_{	ext{net}} = \int_0^R \frac{\mu_0 \sigma \omega}{2} dr = \frac{\mu_0 \sigma \omega}{2} [r]_0^R = \frac{\mu_0 \sigma \omega R}{2}$Thus, the correct option is $B$.

$A$ dielectric circular disc of radius $R$ carries a uniform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$, the magnetic field at the center of the disc is:

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