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JEE Main 2013 (23 Apr, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2013 (23 Apr, Shift 1)

Q59JEE Main 2013 · 23 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

A rectangular loop of wire, supporting a mass m, hangs with one end in a uniform magnetic field B⃗\vec{B} pointing into the plane of the paper. A clockwise current is set up such that i>mg/Bai > mg/Ba, where a is the width of the loop. Then:

Figure

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Answer: (a)

  1. Clockwise current: in the top side PQ the current flows along +x+x; B⃗\vec{B} is into the page (−k^-\hat{k}).
  2. F⃗=ia i^×(−Bk^)=iaB j^\vec{F} = i a\,\hat{i}\times(-B\hat{k}) = iaB\,\hat{j}: upward. Forces on the vertical sides cancel.
  3. iaB>mgiaB > mg, so the weight rises.
  4. The magnetic force itself does no work; the energy comes from the source driving the current, i.e. work is done on the system.
  5. So the answer is (a).
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