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JEE Main 2019 · Physics · Magnetic Effects of Current · hard

A long wire carrying a current of 10 A is bent into a hairpin: it runs up along a straight line, goes round a semicircle of radius 5 cm and returns…

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A long wire carrying a current of 10 A10\ \mathrm{A} is bent into a hairpin: it runs up along a straight line, goes round a semicircle of radius 5 cm5\ \mathrm{cm} and returns along a second straight line parallel to the first. Both straight parts are semi-infinite, each starts at an end of the diameter of the semicircle and is perpendicular to that diameter, lying on the side opposite to the arc. The magnitude of the magnetic field at the centre of the semicircle is nearly (μ0=4π×10−7 T m A−1\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}})
  1. A. 6.3×10−5 T6.3\times10^{-5}\ \mathrm{T}
  2. B. 1.03×10−4 T1.03\times10^{-4}\ \mathrm{T}
  3. C. 4.0×10−5 T4.0\times10^{-5}\ \mathrm{T}
  4. D. 1.43×10−4 T1.43\times10^{-4}\ \mathrm{T}
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Answer: (B) 1.03×10−4 T1.03\times10^{-4}\ \mathrm{T}

Three segments contribute, and by the right-hand rule all three fields point the same way (into the plane for the current sense up on the left, clockwise round the arc, down on the right). The semicircle gives B1=μ0I4R=4π×10−7×104×0.05=6.28×10−5 TB_1=\dfrac{\mu_0I}{4R}=\dfrac{4\pi\times10^{-7}\times10}{4\times0.05}=6.28\times10^{-5}\ \mathrm{T}. Each semi-infinite wire, whose end is at perpendicular distance RR from the centre, gives μ0I4πR=2×10−5 T\dfrac{\mu_0I}{4\pi R}=2\times10^{-5}\ \mathrm{T}. The total is 6.28×10−5+4×10−5≈1.03×10−4 T6.28\times10^{-5}+4\times10^{-5}\approx1.03\times10^{-4}\ \mathrm{T}. Treating the wires as infinite gives 1.43×10−4 T1.43\times10^{-4}\ \mathrm{T}.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake