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(C) Consider an elemental ring of radius $r$ and thickness $dr$.The charge on the ring is $dq = 	ext{Area} 	imes \sigma = (2\pi r dr) \sigma$.The cu

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$\frac{5}{6}$

Vedclass · Answer

(C) Consider an elemental ring of radius $r$ and thickness $dr$.The charge on the ring is $dq = 	ext{Area} 	imes \sigma = (2\pi r dr) \sigma$.The current $di$ through the ring is given by $di = \frac{dq}{T}$, where $T = \frac{2\pi}{\omega}$ is the time period.Thus, $di = \frac{\omega dq}{2\pi} = \frac{\omega (2\pi r dr) \sigma}{2\pi} = \omega r \sigma dr$.The magnetic induction $dB$ at the centre due to this elemental ring is $dB = \frac{\mu_0 di}{2r} = \frac{\mu_0 (\omega r \sigma dr)}{2r} = \frac{\mu_0 \omega \sigma dr}{2}$.Substituting $\sigma = \sigma_0 \left(1 + \sqrt{\frac{r}{R}}ight)$, we get $dB = \frac{\mu_0 \omega \sigma_0}{2} \left(1 + \sqrt{\frac{r}{R}}ight) dr$.To find the total magnetic induction $B$ at the centre, we integrate from $r = 0$ to $r = R$:$B = \int_0^R \frac{\mu_0 \omega \sigma_0}{2} \left(1 + \sqrt{\frac{r}{R}}ight) dr = \frac{\mu_0 \omega \sigma_0}{2} \left[ r + \frac{r^{3/2}}{\sqrt{R} \cdot (3/2)} ight]_0^R = \frac{\mu_0 \omega \sigma_0}{2} \left[ r + \frac{2r^{3/2}}{3\sqrt{R}} ight]_0^R$.Evaluating the limits:$B = \frac{\mu_0 \omega \sigma_0}{2} \left( R + \frac{2R^{3/2}}{3\sqrt{R}} ight) = \frac{\mu_0 \omega \sigma_0}{2} \left( R + \frac{2}{3}R ight) = \frac{\mu_0 \omega \sigma_0}{2} \left( \frac{5}{3}R ight) = \frac{5}{6} \mu_0 \omega \sigma_0 R$.Therefore, $\frac{B}{\mu_0 \sigma_0 \omega R} = \frac{5}{6}$.

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