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JEE Main 2023 · Physics · Magnetic Effects of Current · medium

A metal rod of length 0.5 m rotates uniformly at 40 rad s^-1 about one end in a plane perpendicular to a uniform magnetic field of 0.4 T. Its free…

Single correct option

A metal rod of length 0.5 m0.5\ \mathrm{m} rotates uniformly at 40 rad s−140\ \mathrm{rad\,s^{-1}} about one end in a plane perpendicular to a uniform magnetic field of 0.4 T0.4\ \mathrm{T}. Its free end slides on a conducting circular ring centred at the pivot, and a resistor of 0.2 Ω0.2\ \Omega joins the pivot to the ring (all other resistances negligible). The external torque needed to keep the rotation uniform is
  1. A. 0.25 N m0.25\ \mathrm{N\,m}
  2. B. 0.50 N m0.50\ \mathrm{N\,m}
  3. C. 1.00 N m1.00\ \mathrm{N\,m}
  4. D. 2.00 N m2.00\ \mathrm{N\,m}
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Answer: (B) 0.50 N m0.50\ \mathrm{N\,m}

Motional emf: E=12Bωl2=12(0.4)(40)(0.25)=2 V\mathcal E=\tfrac12B\omega l^2=\tfrac12(0.4)(40)(0.25)=2\ \mathrm V, so I=2/0.2=10 AI=2/0.2=10\ \mathrm A. The magnetic force on the rod F=BIlF=BIl acts at its midpoint, giving a retarding torque τ=BIl⋅l2=12BIl2=12(0.4)(10)(0.25)=0.50 N m\tau=BIl\cdot\dfrac l2=\tfrac12BIl^2=\tfrac12(0.4)(10)(0.25)=0.50\ \mathrm{N\,m}. Check by power: P=EI=20 W=τωP=\mathcal E I=20\ \mathrm W=\tau\omega, so τ=20/40=0.5 N m\tau=20/40=0.5\ \mathrm{N\,m}.
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