A non-conducting thin disc of radius R rotate | vedclass

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(A) Consider a thin ring of thickness $dr$ at a distance $r$ from the center of the disc.The magnetic field $dB$ at the center due to this ring is giv

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$\frac{1}{4}$

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(A) Consider a thin ring of thickness $dr$ at a distance $r$ from the center of the disc.The magnetic field $dB$ at the center due to this ring is given by $dB = \frac{\mu_0 dI}{2r}$.The current $dI$ due to the rotating ring is $dI = \frac{dQ}{T} = \frac{dQ}{2\pi} \omega = \frac{\sigma(r) (2\pi r dr)}{2\pi} \omega = \sigma(r) \omega r dr$.Substituting $\sigma(r) = \sigma_0 \left[1 + (\frac{r}{R})^\beta \right]$, we get $dI = \sigma_0 \omega \left[1 + (\frac{r}{R})^\beta \right] r dr$.Integrating to find the total magnetic field $B$ at the center:$B = \int_0^R \frac{\mu_0}{2r} \sigma_0 \omega \left[1 + (\frac{r}{R})^\beta \right] r dr = \frac{\mu_0 \sigma_0 \omega}{2} \int_0^R \left[1 + (\frac{r}{R})^\beta \right] dr$.$B = \frac{\mu_0 \sigma_0 \omega}{2} \left[ r + \frac{r^{\beta+1}}{R^\beta (\beta+1)} \right]_0^R = \frac{\mu_0 \sigma_0 \omega}{2} \left[ R + \frac{R}{\beta+1} \right] = \frac{\mu_0 \sigma_0 \omega R}{2} \left( \frac{\beta+2}{\beta+1} \right)$.Given $B = \frac{9}{10} \mu_0 \sigma_0 \omega R$, we equate:$\frac{1}{2} \left( \frac{\beta+2}{\beta+1} \right) = \frac{9}{10} \Rightarrow \frac{\beta+2}{\beta+1} = \frac{9}{5}$.$5\beta + 10 = 9\beta + 9 \Rightarrow 4\beta = 1 \Rightarrow \beta = \frac{1}{4}$.

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