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JEE Main 2010 (AIEEE) · Motional EMF

Motional EMF question from JEE Main 2010 (AIEEE)

Q55JEE Main 2010 · AIEEEDifficult

Motional EMF

A rectangular loop has a sliding connector PQPQ of length ll and resistance R ΩR\,\Omega and it is moving with a speed vv as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1,I2I_1, I_2 and II are

Figure

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Answer: (b)

  1. PQ acts as a cell of emf BlvBlv and internal resistance RR.
  2. The two side resistors RR are in parallel: R2\dfrac{R}{2}.
  3. I=BlvR+R/2=2Blv3RI = \dfrac{Blv}{R + R/2} = \dfrac{2Blv}{3R}, and it splits equally: I1=I2=Blv3RI_1 = I_2 = \dfrac{Blv}{3R}
  4. So the answer is (b).
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