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(D) For a hollow cylindrical conductor carrying a uniformly distributed current $i$,the magnetic field $B$ at a distance $r$ (where $R < r < 2R$) from

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$\frac{5\mu_0 i}{36\pi R}$

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(D) For a hollow cylindrical conductor carrying a uniformly distributed current $i$,the magnetic field $B$ at a distance $r$ (where $R $B = \frac{\mu_0 I_{enclosed}}{2\pi r}$Here,the current $I_{enclosed}$ within the radius $r$ is the fraction of the total current $i$ proportional to the cross-sectional area:$I_{enclosed} = i \left( \frac{\pi r^2 - \pi R^2}{\pi (2R)^2 - \pi R^2} \right) = i \left( \frac{r^2 - R^2}{4R^2 - R^2} \right) = i \left( \frac{r^2 - R^2}{3R^2} \right)$Substituting $r = \frac{3R}{2}$:$I_{enclosed} = i \left( \frac{(\frac{3R}{2})^2 - R^2}{3R^2} \right) = i \left( \frac{\frac{9R^2}{4} - R^2}{3R^2} \right) = i \left( \frac{\frac{5R^2}{4}}{3R^2} \right) = \frac{5i}{12}$Now,calculate the magnetic field $B$:$B = \frac{\mu_0}{2\pi r} \left( \frac{5i}{12} \right) = \frac{\mu_0}{2\pi (\frac{3R}{2})} \left( \frac{5i}{12} \right) = \frac{\mu_0}{3\pi R} \left( \frac{5i}{12} \right) = \frac{5\mu_0 i}{36\pi R}$

The figure shows the cross-sectional view of a hollow cylindrical conductor with inner radius $R$ and outer radius $2R$. The cylinder carries a uniformly distributed current $i$ along its axis. The magnetic induction at point $P$ at a distance $\frac{3R}{2}$ from the axis of the cylinder will be:

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