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JEE Main 2025 (4 Apr, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2025 (4 Apr, Shift 1)

Q14JEE Main 2025 · 4 APR · SHIFT 1Medium

Motional EMF

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 12\frac{1}{\sqrt{2}} Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s10\text{ cm/s}, induced emf between points A and E is ________ mV.

Figure

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

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Answer: 10

AEBCD10√2 cmv
  1. Emf between A and E equals that of the straight line AE, using its length perpendicular to v⃗\vec v.
  2. AB and DE are along v⃗\vec v; BC and CD are at 45∘45^\circ, each giving 10sin⁡45∘=10210\sin45^\circ = \frac{10}{\sqrt2} cm across.
  3. Effective length =2×102=102= 2\times\frac{10}{\sqrt2} = 10\sqrt2 cm =0.12= 0.1\sqrt2 m.
  4. ε=12×0.12×0.1=0.01\varepsilon = \frac{1}{\sqrt2}\times0.1\sqrt2\times0.1 = 0.01 V
  5. Answer: 1010 mV.
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