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(D) To find the magnetic field at a point $P$ at a distance $r = 3R/2$ from the axis,we use Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{	ext

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

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(D) To find the magnetic field at a point $P$ at a distance $r = 3R/2$ from the axis,we use Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{	ext{enclosed}}$.Since the current $i$ is uniformly distributed over the cross-sectional area $A = \pi(2R)^2 - \pi R^2 = 3\pi R^2$,the current density $J = \frac{i}{3\pi R^2}$.The current enclosed by the Amperian loop of radius $r = 3R/2$ is $I_{	ext{enclosed}} = J 	imes 	ext{Area}_{	ext{enclosed}} = J 	imes \pi(r^2 - R^2) = \frac{i}{3\pi R^2} 	imes \pi \left( \left(\frac{3R}{2}ight)^2 - R^2 ight) = \frac{i}{3R^2} 	imes \left( \frac{9R^2}{4} - R^2 ight) = \frac{i}{3R^2} 	imes \frac{5R^2}{4} = \frac{5i}{12}$.Applying Ampere's law: $B(2\pi r) = \mu_0 I_{	ext{enclosed}}$.$B 	imes 2\pi \left(\frac{3R}{2}ight) = \mu_0 \left(\frac{5i}{12}ight)$.$B(3\pi R) = \frac{5\mu_0 i}{12}$.$B = \frac{5\mu_0 i}{36\pi R}$.

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

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