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JEE Main 2020 · Physics · Magnetic Effects of Current · medium

A metal rod of length 1.0 m rotates in a horizontal plane about a vertical axis through one end at 50 rad s^-1, in a uniform vertical magnetic field…

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A metal rod of length 1.0 m1.0\ \mathrm{m} rotates in a horizontal plane about a vertical axis through one end at 50 rad s−150\ \mathrm{rad\,s^{-1}}, in a uniform vertical magnetic field of 0.2 T0.2\ \mathrm{T}. The potential difference between the mid-point of the rod and its free end is
  1. A. 3.75 V3.75\ \mathrm{V}
  2. B. 5 V5\ \mathrm{V}
  3. C. 2.5 V2.5\ \mathrm{V}
  4. D. 1.25 V1.25\ \mathrm{V}
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Answer: (A) 3.75 V3.75\ \mathrm{V}

A small element drdr at distance rr moves with speed ωr\omega r perpendicular to both itself and BB, so dε=Bωr drd\varepsilon = B\omega r\,dr. Between radii r1=0.5 mr_1 = 0.5\ \mathrm{m} and r2=1.0 mr_2 = 1.0\ \mathrm{m}: ε=12Bω(r22−r12)=12(0.2)(50)(1−0.25)=3.75 V\varepsilon = \tfrac12B\omega(r_2^2 - r_1^2) = \tfrac12(0.2)(50)(1 - 0.25) = 3.75\ \mathrm{V}. For the full rod it would be 5 V5\ \mathrm{V} and for the inner half 1.25 V1.25\ \mathrm{V}; the emf is not proportional to length.
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