Figure shows a wire `ab` of length `L` and resistance `R` which can slide on a smooth pair of rails. `I_(g)` is current generator which supplies a constant current `i` in the circuit. If the wire `ab` slide at a speed `v` towards right, find the potential difference below `a` and `b`

(b) The current generator `I_(g)` show in the figure sends a constant current `i` through the circiut. The wire `ab` has a length `L` and mass `m` and can slide on the smooth horizontal rails connect to `I_(g)` . The entire system lies in a vertical magnetic field `B` . Find the velocity of the wire as a function of time.

(c) The system containing the rails and the wire of the previous part is kept vertically in a uniform horizontal magnetic field `B` that is perpendicular to the plane of the rails (figure). If is found that the wire stays in equilibrium. If th ewire `ab` is replaced by another wire of double its mass, how long will lit take in falling through a distance equal to its length?
The current generator `I_(g)` shown in figure sends a constant current `i` through the circuit. The wire `cd` is fixed and wire `ab` is made to slide on the smooth, thick rails with a constant velocity `v` toward right. Each of these wires has resistance `r` . Find the current through the wire `cd` .

Explanation

The rod can be replaced by battery `e=Bvl`
`V_(a)-V_(b)=e-iR=BvL-iR`
(b) Force on rod `F=BiL` toward right
Acceleration `a=(F)/(m)=(BiL)/(m)` (constant)
`v=u+at=0+(BiL)/(m)t`
`v=(BiL)/(m)t`
(c) In equilibrium, `Fm=mg`
`BiL=mg`
When mass of rod: `2m`
`2mg` -BiL=2ma
`2mg-mg=2ma implies a=(g)/(2)`
`s=ut+(1)/(2)at^(2)`
`L=0+(1)/(2)(g)/(2)t^(2)`
`t=2sqrt((L)/(g))`
(d) `c` to `c` : path `cbada V_(c)-i_(1)r+Bvl+(i-i_(1))r=V_(c)`
`i_(1)=(Bvl+ir)/(2r)=(Bvl)/(2r)+(i)/(2)`
current in `cd : i-i_(1)=(i)/(2)-(Bvl)/(2r)`