A loop `ABCD` containing two resistors as shown in figure is placed in a uniform magnetic fielf `B` directed outwards to the plane of page. A sliding conductor `EF` of length `l` and of negligible resistance moves to th right with a uniform velocity `v` as shown in Fig. Determine the current in each branch.

Explanation

The magnetic field induction `B`, length `l` and the velocity `v` of the conductor `EF` are mutually perpendicular, hence the emf induction in it is `e = Blv` (with end `F` of the rod at higher potential)
`:.` The effective electric circuit can be redrawn as shwon Fig.

The resistance `R_(1)` and `R_(2)` are in parallel, so the equilvalent resistance `R` is given by `(1)/(R ) = (1)/(R_(1))+(1)/(R_(2))`
From Ohms law, the total current `i = (e)/(R )`
`i = Blv (1/R_(1)+1/R_(2))`
Current in `AD` is `i_(1) = (BlV)/(R_(1))`, current in `BC` is
`i_(2) = (BlV)/(R_(2))`