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JEE Advanced 2019 Paper 1 · Physics · Magnetic Effects of Current · hard

A long air-core solenoid of radius 10 cm carries a current that makes the uniform axial magnetic field inside it increase at the steady rate 0.50 T…

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A long air-core solenoid of radius 10 cm10\ \mathrm{cm} carries a current that makes the uniform axial magnetic field inside it increase at the steady rate 0.50 T s−10.50\ \mathrm{T\,s^{-1}} (the field outside is negligible). Consider circles centred on the axis, in a plane perpendicular to it. Which of the following statements is/are correct?
  1. A. The induced electric field at a distance 5.0 cm5.0\ \mathrm{cm} from the axis has magnitude 1.25×10−2 V m−11.25\times10^{-2}\ \mathrm{V\,m^{-1}}.
  2. B. The induced electric field at 20 cm20\ \mathrm{cm} from the axis has magnitude 2.5×10−2 V m−12.5\times10^{-2}\ \mathrm{V\,m^{-1}}.
  3. C. The emf induced in a square loop of side 40 cm40\ \mathrm{cm} enclosing the solenoid axis symmetrically is 80 mV80\ \mathrm{mV}.
  4. D. The work done by the induced field on a 2 μC2\ \mu\mathrm{C} charge taken once round the circle of radius 20 cm20\ \mathrm{cm} is zero.
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Answer: A

By symmetry the induced field is tangential, and Faraday's law gives E (2πr)=∣dΦdt∣E\,(2\pi r)=\left|\dfrac{d\Phi}{dt}\right|. Inside (r<ar<a) the flux is πr2B\pi r^2B, so E=r2dBdt=0.052×0.5=1.25×10−2 V m−1E=\dfrac r2\dfrac{dB}{dt}=\dfrac{0.05}{2}\times0.5=1.25\times10^{-2}\ \mathrm{V\,m^{-1}} at 5 cm5\ \mathrm{cm}; (A) is true. Outside only the solenoid cross-section πa2\pi a^2 carries flux, so E=a22rdBdt=0.010.4×0.5=1.25×10−2 V m−1E=\dfrac{a^2}{2r}\dfrac{dB}{dt}=\dfrac{0.01}{0.4}\times0.5=1.25\times10^{-2}\ \mathrm{V\,m^{-1}} at 20 cm20\ \mathrm{cm}, not 2.5×10−22.5\times10^{-2}; (B) is false. For the square loop the flux is again limited to the solenoid area, so the emf is π(0.1)2×0.5=15.7 mV\pi(0.1)^2\times0.5=15.7\ \mathrm{mV}, not 80 mV80\ \mathrm{mV} (which would use the loop area 0.16 m20.16\ \mathrm{m^2}); (C) is false. The induced field is non-conservative, so the work is qE=2×10−6×0.0157=3.14×10−8 J≠0q\mathcal E=2\times10^{-6}\times0.0157=3.14\times10^{-8}\ \mathrm{J}\neq0; (D) is false. Only (A) is correct.
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Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake