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NEET 2026 · Motional EMF

Motional EMF question from NEET 2026

Q2NEET 2026Medium

Motional EMF

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s−12\text{ cm s}^{-1}, in a direction normal to the shorter side of the loop, will be:

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Answer: (d)

  1. Only the side perpendicular to the velocity cuts the field boundary: ε=Bℓv\varepsilon = B\ell v.
  2. Velocity is normal to the shorter side, so ℓ=3\ell = 3 cm =0.03= 0.03 m, v=0.02v = 0.02 m/s.
  3. ε=0.3×0.03×0.02=1.8×10−4\varepsilon = 0.3 \times 0.03 \times 0.02 = 1.8\times10^{-4} V
  4. So the answer is (d) 1.8×10−41.8\times10^{-4} volt.
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