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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

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 B′B' 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 B′B' 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.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMstraightradiuscarriessteadycurrent

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