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JEE Main 2024 (29 Jan, Shift 2) · Motional EMF

Motional EMF question from JEE Main 2024 (29 Jan, Shift 2)

Q18JEE Main 2024 · 29 JAN · SHIFT 2Easy

Motional EMF

A horizontal straight wire 5 m5\text{ m} long extending from east to west is falling freely at right angle to horizontal component of earths magnetic field 0.60×10−4 Wb m−20.60 \times 10^{-4}\text{ Wb m}^{-2}. The instantaneous value of emf induced in the wire when its velocity is 10 ms−110\text{ ms}^{-1} is _________ ×10−3 V\times 10^{-3}\text{ V}.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Wire (E–W), velocity (down) and BHB_H (N) are mutually perpendicular, so ε=BHlv\varepsilon = B_H l v.
  2. ε=0.60×10−4×5×10=3×10−3\varepsilon = 0.60\times10^{-4} \times 5 \times 10 = 3\times10^{-3} V
  3. Answer: 33 (i.e. 3×10−33\times10^{-3} V).
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