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JEE Main 2016 · Physics · Magnetic Effects of Current · medium

A long wire carrying 5 A is bent into a hairpin: two long straight parallel segments, 20 cm apart, are joined at their ends by a semicircular arc of…

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A long wire carrying 5 A5\ \mathrm{A} is bent into a hairpin: two long straight parallel segments, 20 cm20\ \mathrm{cm} apart, are joined at their ends by a semicircular arc of radius 10 cm10\ \mathrm{cm}, so the current enters along one straight segment, goes round the arc and leaves along the other. Each straight segment extends to infinity and ends at the arc along a line through the centre of the arc. The magnetic field at the centre of the arc is nearly (μ0=4π×10−7 T m A−1\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}})
  1. A. 10.0 μT10.0\ \mu\mathrm{T}
  2. B. 15.7 μT15.7\ \mu\mathrm{T}
  3. C. 25.7 μT25.7\ \mu\mathrm{T}
  4. D. 31.4 μT31.4\ \mu\mathrm{T}
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Answer: (C) 25.7 μT25.7\ \mu\mathrm{T}

At the centre of the arc all three parts give fields in the same direction. The semicircle gives B1=μ0I4R=4π×10−7×50.4=15.7 μTB_1=\dfrac{\mu_0I}{4R}=\dfrac{4\pi\times10^{-7}\times5}{0.4}=15.7\ \mu\mathrm{T}. Each semi-infinite straight wire, whose end is level with the centre, gives B2=μ0I4πR=5.0 μTB_2=\dfrac{\mu_0I}{4\pi R}=5.0\ \mu\mathrm{T}. Therefore B=B1+2B2=μ0I4R(1+2π)=15.7+10.0=25.7 μTB=B_1+2B_2=\dfrac{\mu_0I}{4R}\left(1+\dfrac2\pi\right)=15.7+10.0=25.7\ \mu\mathrm{T}. Dropping the straight segments gives only the arc value 15.7 μT15.7\ \mu\mathrm{T}, and counting them as infinite wires would give an incorrect μ0I2πR\dfrac{\mu_0 I}{2\pi R} per wire.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake