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Most Important Selected Qs for JEE AdvancedPhysicsElectromagnetic Induction

A rectangular loop has a sliding connector P Q of length and resistance R and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I ₁, I ₂ and I are

Options

  1. AI₁=-I₂= B v R , I= 2 B v R
  2. BI₁=I₂= B v 3 R , I= 2 B v 3 R
  3. CI₁=I₂=I= B v R
  4. Dl₁=I₂= B v 6 R , I= B v 3 R

Correct answer

B. I₁=I₂= B v 3 R , I= 2 B v 3 R

Step-by-step solution

A moving conductor is equivalent to a battery of emf = vB (motion emf) Equivalent circuit I=I₁+I₂ Applying Kirchoff's law aligned & I ₁ R + IR - vB =0 (1) & I ₂ R + IR - vB =0 (2) aligned Adding (1) and (2) aligned & 2 IR + IR =2 vB & I = 2 vB 3 R ; I ₁= I ₂= vB 3 R aligned

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