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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:
  1. 1/2
  2. 4
  3. 1/4
  4. 1

A
1/2
B
1/4
C
4
D
1
Solution
Verified by Toppr

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×current inclosedpath

=μ0×(πr21πa2)×I2πr1=μ0×Ir12πa2

Now, magnetic filed at point
(B2)=μ02πI2a=μ0I4πa

Required Ratio =B1B2=μ0Ir12πa2×4πaμ0I
=2r1a=2×a2a=1

1028072_430314_ans_421dfe82a36e4992a8790b1f834738b7.PNG

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