A uniform rod of mas `m` is moving with constant velocity `v_(0)` in a perpendicular uniform magnetic field `B` as shown. The resistance of rod is `r` . The current flowing through rod, `R_(I)` and `R_(2)` will be

Explanation

The rod can be replaced by a battery of emf `e=Bv_(0)L` and internal resistance `r` .
`3r` and `6r` in parallel, `R_(eq)'=(3r.6r)/(3r+6r)=2r`
`i=(Bv_(0)L)/(r+2r)=(Bv_(0)L)/(3r)`
`V_(A)-V_(B)=i.2r=(2Bv_(0)L)/(3)`
`i=(2Bv_(0)L//3)/(6r)=(2Bv_(0)L)/(9r)`
`i_(2)=(2Bv_(0)L//3)/(6r)=(Bv_(0)L)/(9r)`