1
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Question

# A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is:

A
1/2
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B
1/4
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C
4
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D
1
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Solution

## The correct option is D 1Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the amperean path formed at a distance r1=(a2)Iin=(πr21πa2)×I, where I is total current.∴ Magnetic field at (B1)=μ0×current inclosedpath =μ0×(πr21πa2)×I2πr1=μ0×Ir12πa2Now, magnetic filed at point(B2)=μ02π⋅I2a=μ0I4πa∴ Required Ratio =B1B2=μ0Ir12πa2×4πaμ0I=2r1a=2×a2a=1

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