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JEE Main 2022 · Physics · Magnetic Effects of Current · hard

Two parallel, smooth conducting rails inclined at 37^° to the horizontal are a distance 0.5 m apart and are joined at the lower end by a 0.2 Ω…

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Two parallel, smooth conducting rails inclined at 37∘37^\circ to the horizontal are a distance 0.5 m0.5\ \mathrm{m} apart and are joined at the lower end by a 0.2 Ω0.2\ \Omega resistor. A uniform vertical magnetic field of 1 T1\ \mathrm{T} acts in the region. A rod of mass 0.2 kg0.2\ \mathrm{kg} slides down the rails without friction, remaining perpendicular to them. Neglecting other resistances, and with g=10 m s−2g=10\ \mathrm{m\,s^{-2}} and sin⁡37∘=0.6\sin37^\circ=0.6, its terminal speed is
  1. A. 0.96 m s−10.96\ \mathrm{m\,s^{-1}}
  2. B. 1.2 m s−11.2\ \mathrm{m\,s^{-1}}
  3. C. 1.5 m s−11.5\ \mathrm{m\,s^{-1}}
  4. D. 2.4 m s−12.4\ \mathrm{m\,s^{-1}}
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Answer: (C) 1.5 m s−11.5\ \mathrm{m\,s^{-1}}

With a vertical field, only the component of B⃗\vec B normal to the incline, Bcos⁡θB\cos\theta, links the circuit. Emf =BLvcos⁡θ=BLv\cos\theta, current I=BLvcos⁡θRI=\dfrac{BLv\cos\theta}{R}. The magnetic force BILBIL is horizontal; its component along the incline is BILcos⁡θBIL\cos\theta. At terminal speed mgsin⁡θ=B2L2vcos⁡2θRmg\sin\theta=\dfrac{B^{2}L^{2}v\cos^{2}\theta}{R}, so v=(0.2)(10)(0.2)(0.6)(0.25)(0.64)=0.240.16=1.5 m s−1v=\dfrac{(0.2)(10)(0.2)(0.6)}{(0.25)(0.64)}=\dfrac{0.24}{0.16}=1.5\ \mathrm{m\,s^{-1}}.
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