Figure shows a conducting rod `PQ` in contact with metal rails `RP` and `SQ`, which are `0.25 m` apart in a uniform magnetic field of flux density `0.4T` acting perpendicular to the plane of the paper. Ends `R` and `S` are connected through a `5 Omega` resistance. What is the emf when the rod moves to the right with a velocity of `5ms^(-1)` ? What is the magnitude and direction of the current through the `5 Omega` resistance? If the rod `PQ` moves to the left with the same speed, what will be the new current and its direction?

Explanation

`|e| = Blv = 0.4 xx 0.25 xx 5 = 0.5V`
Current, `I = (|e|)/(R ) = (0.5V)/(5 Omega) = 0.1A`
As the rod `'PQ'` moves to right as shown, the free electrons in it experience a Lorentz force. According to `F.L.H`., the force is towards the end `'Q'` of rod.
`:.` They move from `P` to `Q`, hence the end of the rod `P` becomes deficient of electrons `rArr V_(P) gt V_(Q)`
Applying Flemming's right hand rule, the current in the rod shall flow from `Q` to `P`.
(b) : If the rod `PQ` moves to the left with the same speed, then the current of `0.1A` will flow in the rod `PQ` from `P` to `Q`