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

Answer

C

Explanation

Circuit can be resuced as

`I=(e)/(3R//2)=(2vlB)/(3R)`
`rArr" "I_(1)=I_(2)=(I)/(2)=(vlB)/(3R)`