JEE Main2010PhysicsElectromagnetic InductionActual
A rectangular loop has a sliding connector PQ 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 I₁, I₂ and I are
Options
- AI ₁=- I ₂= B v R , I = 2 ~B v R
- BI ₁= I ₂= B v 3 R , I = 2 ~B v 3 R
- Cl₁=I₂=I= B v R
- DI₁=I₂= B v 6 R , I= B v 3 R
Correct answer
B. I ₁= I ₂= B v 3 R , I = 2 ~B v 3 R
Step-by-step solution
A moving conductor is equivalent to a battery of emf = v B (motion emf) Equivalent circuit I = l ₁+ l ₂ applying Kirchoff's law aligned & I ₁ R+I R-v B =0 & I ₂ R+I R-v B =0 aligned adding (1) & (2) aligned & 2 IR + IR =2 vB & I = 2 vB 3 R & I ₁= I ₂= vB 3 R aligned