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:
1/2
4
1/4
1
A
1/2
B
1/4
C
4
D
1
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Solution
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Here, 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×currentinclosedpath
=μ0×(πr21πa2)×I2πr1=μ0×Ir12πa2
Now, magnetic filed at point (B2)=μ02π⋅I2a=μ0I4πa
∴ Required Ratio =B1B2=μ0Ir12πa2×4πaμ0I =2r1a=2×a2a=1
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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: