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JEE Main202622 January 2026Morning ShiftPhysicsElectromagnetic InductionActual

X P Q Y is a vertical smooth long loop having a total resistance R where P X is parallel to Q Y and separation between them is l . A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod C D of length L(L>l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is _ _ _ _ m / s .( g = acceleration due to gra

Options

  1. A( 8 m g R B^2 l^2 )
  2. B( 2 m g R B^2 L^2 )
  3. C( m ~g R B^2 l^2 )
  4. D( 2 m g R B^2 l^2 )

Correct answer

C. ( m ~g R B^2 l^2 )

Step-by-step solution

As rod CD slides down with velocity v , it cuts magnetic field lines. Induced EMF: = BLv Induced current in the circuit: I = BLv R Magnetic force on the rod (upward, by Lenz's law): F_ mag = BIL = B^2L^2v R At terminal velocity, acceleration is zero. Applying Newton's second law: mg - F_ mag = 0 mg = B^2L^2v_ terminal R v_ terminal = mgR B^2L^2

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