Figure shows the cross-sectional view of the hollow, cylindrical conductor with inner radius 'R' and outer radius '2R'. Cylinder is carrying uniformly distributed current along it's axis. The magnetic induction at a distance 3R/2: from the axis of the cylinder will be

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$\frac{7 μ_{0} I}{18 π R}$

$\frac{5 μ_{0} I}{36 π R}$

$\frac{5 μ_{0} I}{72 π R}$

answer is D.

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Detailed Solution

current per unit area of the conductor is $\frac{I}{π (4 R^{2} - R^{2})} = \frac{I}{3 π R^{2}} = σ$

applying Ampere's circuital law

$\int B . d l = μ_{0} i = μ_{0} σ (\frac{9 R^{2} π}{4} - π R^{2}) = μ_{0} σ (\frac{5 π R^{2}}{4}) B . 2 π \frac{3 R}{2} = μ_{0} σ (\frac{5 π R^{2}}{4}) B = μ_{0} σ (\frac{5 π R^{2}}{4}) \frac{1}{3 π R} = μ_{0} \frac{I}{3 π R^{2}} (\frac{5 π R^{2}}{4}) \frac{1}{3 π R} = \frac{5 I}{36 π R}$

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