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

Let us assume, `vec(B)` is directed into the plane of the loop. Then the motinal `e.m.f`
`xi_(In) |int -(vec(v) xx vec(B)).d.vec(l)| = v Bl`
and directed in the same of `(vec(v) xx vec(B))` (Fig)
So, `i = (xi_(I n))/(R + (R_(1) R_(2))/(R_(1) + R_(2))) = (B v l)/(R + R_(u))`
As `R_(1)` and `R_(2)` are in parallel connections.