A rectangular loop has a sliding connector PQ of length l and resistance `R (Omega)` 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

A. `I_(1)=-I_(2)=(Blv)/R, I=(2Blv)/R`
B. `I_(1)=I_(2)=(Blv)/(3R),I+(2Blv)/(3R)`
C. `I_(1)=I_(2)=(Blv)/R`
D. `I_(1)=I_(2)=(Blv)/(6R),I=(2Blv)/(3R)`

Answer

B

Explanation

Current `I=(vBl)/(R//2+R)=(2vBl)/(3R)`
`I_(1)=I_(2)=(vBl)/(3R)`