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NEET PHYSICSMedium

A long straight wire of radius aa carries a steady current II. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields BB and BB' at radial distances a/2a/2 and 2a2a respectively, from the axis of the wire, is:

A

1/2

B

1

C

4

D

1/4

Step-by-Step Solution

According to the sources, the magnetic field produced by a long straight wire depends on whether the observation point is inside or outside the wire.

  1. Inside the wire (r<ar < a): The magnetic field BB at a distance rr from the axis is given by B=μ0Ir2πa2B = \frac{\mu_0 I r}{2\pi a^2} . For r=a/2r = a/2, the field is B=μ0I(a/2)2πa2=μ0I4πaB = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} .

  2. Outside the wire (r>ar > a): The magnetic field BB' at a distance rr from the axis is given by B=μ0I2πrB' = \frac{\mu_0 I}{2\pi r} . For r=2ar = 2a, the field is B=μ0I2π(2a)=μ0I4πaB' = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a} .

  3. Ratio: The ratio of the magnetic fields is B/B=μ0I/4πaμ0I/4πa=1B/B' = \frac{\mu_0 I / 4\pi a}{\mu_0 I / 4\pi a} = 1.

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