A rectangular loop with a sliding connector of length `l` is located in a uniform magnetic field perpendicular to the loop plane. The magnetic induction is equal to B. The connector has an electric resistance `R`, the sides `ab` and `cd` have resistances `R_1` and `R_2`. Neglecting the self-inductance of the loop, find the current flowing in the connector during its motion with a constant velocity v.

Explanation

The equilvalent circuit is

The equilvalent resistance of the cirduit is
`R = (R+(R_(1)R_(2))/(R_(1)+R_(2)))`
Hence the current in the connector is `i = (e)/(R )`
`:. i = (Blv(R_(1)+R_(2)))/((R R_(1)+R R_(2)+R_(1) R_(2)))`