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

Two long vertical conducting rails 0.4 m apart are joined at the top by a resistor of 0.4 Ω. A horizontal rod of mass 20 g and negligible resistance…

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Two long vertical conducting rails 0.4 m0.4\ \mathrm{m} apart are joined at the top by a resistor of 0.4 Ω0.4\ \Omega. A horizontal rod of mass 20 g20\ \mathrm{g} and negligible resistance slides down the rails without friction while remaining in contact with them. A uniform horizontal magnetic field of 0.5 T0.5\ \mathrm{T} is perpendicular to the plane of the rails. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The terminal speed of the rod is
  1. A. 2 m s−12\ \mathrm{m\,s^{-1}}
  2. B. 4 m s−14\ \mathrm{m\,s^{-1}}
  3. C. 1 m s−11\ \mathrm{m\,s^{-1}}
  4. D. 0.5 m s−10.5\ \mathrm{m\,s^{-1}}
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Answer: (A) 2 m s−12\ \mathrm{m\,s^{-1}}

At speed vv the induced emf is BLvBLv, the current is BLv/RBLv/R, and the magnetic force on the rod is BIL=B2L2v/RBIL=B^2L^2v/R, directed upward. Terminal speed is reached when this balances the weight: mg=B2L2vRmg=\dfrac{B^2L^2v}{R}, so v=mgRB2L2=(0.02)(10)(0.4)(0.25)(0.16)=0.080.04=2 m s−1v=\dfrac{mgR}{B^2L^2}=\dfrac{(0.02)(10)(0.4)}{(0.25)(0.16)}=\dfrac{0.08}{0.04}=2\ \mathrm{m\,s^{-1}}. Note that the mass must be in kilograms; using 2020 in place of 0.020.02 gives a speed a thousand times too large.
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