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JEE Main 2026 (21 Jan, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2026 (21 Jan, Shift 1)

Q8JEE Main 2026 · 21 JAN · SHIFT 1Medium

Motional EMF

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 Ω2\ \Omega then the force needed to move the rod towards right with constant speed (v)(v) of 1.5 m/s1.5\text{ m/s} is ______ N.

Figure

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Answer: (b)

  1. e=Blv=0.10×1×1.5=0.15 Ve = Blv = 0.10 \times 1 \times 1.5 = 0.15\text{ V}
  2. I=eR=0.152=0.075 AI = \dfrac{e}{R} = \dfrac{0.15}{2} = 0.075\text{ A}
  3. Applied force balances the magnetic drag: F=BIl=0.10×0.075×1F = BIl = 0.10 \times 0.075 \times 1
  4. F=7.5×10−3 NF = 7.5 \times 10^{-3}\text{ N}, so the answer is (b).
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