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Q.

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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a

zero

b

5μ0I72 πR

c

7μ0I18πR

d

5μ0I36πR

answer is D.

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Detailed Solution

current per unit area of the conductor is Iπ4R2-R2=I3πR2=σapplying Ampere's circuital law ∫B.dl = μ0i =μ0σ9R2π4-πR2=μ0σ5πR24 B.2π3R2=μ0σ5πR24 B=μ0σ5πR2413πR=μ0I3πR25πR2413πR=5I36πR
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