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NEET 2016
Exam
NEET
Year
2016
Subject
Physics
Chapter
Magnetic Effects of Current

NEET 2016 · Physics · Magnetic Effects of Current

A long straight wire of radius a carries a steady current I . the...

A long straight wire of radius a\text{a} carries a steady current I\text{I}. the current is uniformly distributed over its cross-section. The ratio of the magnetic fields B\text{B} and B’\text{B'}, at radial distances a2\dfrac{a}{2} and 2a2a respectively, from the axis of the wire is:

Options

  1. A14\dfrac{1}{4}
  2. B12\dfrac{1}{2}
  3. C1
  4. D44

Correct answer

C. 1

Step-by-step solution

InsidethewireByampere′slaw\dfrac{Insidethewire}{Byamper{e}^{′}slaw}

∫B→.dl→=μ0(ienclosed)∫ \overset{\to}{B} . \overset{\to}{dl} = {\mu}_{0}\left({i}_{enclosed}\right)

∫B.dlcos0=μ0(Iπa2.π(a2)2)∫ B . dlcos 0 = {\mu}_{0}\left(\dfrac{I}{\pi{a}^{2}} . \pi{\left(\dfrac{a}{2}\right)}^{2}\right)

B∫dl=μ0I4B∫ dl = {\mu}_{0}\dfrac{I}{4}

B(2π(a2))=μ0I4B\left(2\pi\left(\dfrac{a}{2}\right)\right) = \dfrac{{\mu}_{0}I}{4}

B=μ0I4\piaB = \dfrac{{\mu}_{0}I}{4\pia}

Outside the wire,

B′=μ0I2\pir=μ0I2π(2a)=μ0I4\pia{B}^{′} = \dfrac{{\mu}_{0}I}{2\pir} = \dfrac{{\mu}_{0}I}{2\pi\left(2a\right)} = \dfrac{{\mu}_{0}I}{4\pia}

So, BB′=1\dfrac{B}{{B}^{′}} = 1.
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