In the figure shown `PQRS` is a fixed resistanceless conducting frame in a uniform and constant magnetic field of strength `B`. A rod `EF` of mass `m` length `l` and resistance `R` can smoothly move on this frame.A capacitor charged to a potential difference `V_(0)` initially is connected as shown in the figure.Find the velocity of the rod as function of time `t` if it is released at `t=0` from rest.

Answer

A, B, C

Explanation

Due to enarged capacitor current will flow in the rod in the rod in downward direction.Hence the rod will experience magnetic force towards right.Then an `emf` (motional) will be induced in the rod.Let the charge on capacitor and speed of rod at any time `t` be `q` and `v` respectively.
Applying loop law we get `q/C-iR=Blv=0` ..(1)
The force on rod is `F=m(dv)/(dt)=Bil`..(2)
differentiating equation (1) and (2) we get `-i/C-R(di)/(dt)-Bl(dv)/(dt)=0`...(3)
From equation (2) and (3) we get `(di)/(dt)=-[(B^(2)l^(2))/(mR)+1/(RC)]i`
`rArr (di)/(dt)=-Kdt`..(4)
where `K=(B^(2)l^(2)C+m)/(mRC)`
at `t=0 sec, q=CV_(0)` and `v=0`
`therefore` from equation (1) the current at `t=0` is `i_(0)=V_(0)/R`
integrating equation `underset(V_(0)//R)overset(i)int(di)/l=-K underset(0)overset(t)int dt`
we get `i=V_(0)/R e^(-Kt)`
from eqation (2) `(dv)=(Bl)/mi dt` or `dv=(Blv_(0))/(mR) e^(-Kt) dt`
integarating the equation `underset(0)overset(v)int dv=underset(0)overset(t)int (Bl)/m V_(0)/R e^(-Kt) dt`
`rArr v=(BlV_(0))/(mR)(-1/K)[e^(-Kt-e^(0))]=(BlV_(0))/(mRK)[1-e^(-Kt)]`
By substituting `K` we get `v=(BlCB_(0))/(m+B^(2)l^(2)C)(1-e^(((B^(2)l^(2))/(mR) 1/(RC))t))`