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