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:
Mark the correct statement about the motion of conductor

Options

A. It is falling down with constant acceleration `g`
B. It is falling down with constant acceleration but not equal to `g`
C. It is falling down with increasing acceleration
D. It is falling down with decreasing acceleration

Answer

B

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`
as `I=B_(0)Clxxa`
As a is constant, `I` is also constant.