A conductor of mass `m` and length `l` is sliding smoothly an two vertical conducting rails `AB` and `CD` as shown in figure. The top ends of two conducting rails are joined by a capacitor of capacitance `C`. The conductor is relased from rest when it is very close to `AC` i.e., `y = 0`. A uniform magnetic field `B_(0)` perpendicular to plane of figure existing. Neglect the resistance of rails and connecting wires. Take acceleration due to gravity to be `g`.

Based on above information answer the following questions:
Charge on the capacitor as a function of distance travelled `y` is given by

Options

A. `CB_(0)lsqrt((2m gy)/(m+CB_(0)^(2)l^(2)))`
B. `CB_(0)lsqrt((m+CB_(0)^(2)l^(2))/(2mgy))`
C. `CB_(0)l e^((mgy)/(m+CB_(0)^(2)l^(2)))`
D. `CB_(0)lsqrt((2mgy)/(m+CB_(0)^(2)l^(2)) ) xx e^((mgy)/(m+CB_(0)^(2)l^(2)))`

Answer

A

Explanation

Let us say that in time `t` the conductor falls down by `y` and acquires a velocity `v`, then at this instant induced emf is
`e = B_(0) vl` [with `Q` at higher potential]

Charge on the capacitor at this instant is,
`q = Ce = B_(0) Cl xx v`
the current in the circuit at this instant is,
`I = (d q)/(dt) = B_(0) Cl(dv)/(dt) = B_(0) Cl xx a`
where a is the acceleration of the conductor at this instant.
Writing Newton's `2^(nd)` law equation for conductor,
`mg - IB_(0)l = m a`
` rArr a = (m g)/(m + CB_(0)^(2)l^(2))`,
`v = (m g t)/(m + CB_(0)^(2)l^(2))` and `y = (m g t^(2))/(2(m + CB_(0)^(2) l^(2)))`
After solving above equations, we could have `q` as a function of `y`
`E=r/2 (dB)/(dt) (dB)/(dt) =B_(0) rArr E=a/2 B_(0)`

`d tau=dF.a=(dqE).a tau int d tau=Ea int dq =Eaq`
`tau=I alpha" "alpha=tau/I=(Eaq)/((ma^(2)))=(Eq)/(ma)`
`alpha=(Eq)/(ma)` anticlockwise
`P=tau.w=(Eaq) (alpha t)=Eaq (Eq)/(ma) t`