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
- AI₁=-I₂= B v R , I= 2 B v R
- BI₁=I₂= B v 3 R , I= 2 B v 3 R
- CI₁=I₂=I= B v R
- 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