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.

Options

A. `(Blv)/(R_(1)+R_(2)+R)`
B. `(BlV(R_(1)+R_(2)))/(R+(R_(1)+R_(2)))`
C. `(Blv(R_(1)+R_(2)))/(R R_(1)+R R_(2)+R_(1)R_(2))`
D. `Blv((1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3)))`

Answer

C

Explanation

inductor behaves like a cell of emf. `epsilon'=Blv` and `Blv=epsilon`
Hence `i=(Blv)/(R+(R_(1)R_(2))/(R_(1)+R_(2)))`
`=(Blv(R_(1)+R_(2)))/(R R_(1)+R R_(2)+R_(1)R_(2))`