In the shown figure, there are two long fixed parallel conducting rails (having negligible resistance) and are separated by distance `L`. A uniform rod of resistance `R` and mass `M` is placed at rest on frictionless rails. Now at time `t = 0`, a capacitor having charge `Q_(0)` and capacitance `C` is connected across rails at ends `a` and `b` such that current in rod `(c d)` is from `c` towards `d` and the rod is released. A uniform and constant magnetic field having magnitude `B` exists normal to plane of paper as shown. (Neglect acceleration due to gravity)

When the acceleration of rod is zero, the charge on capacitor is:

Options

A. `(B^(2)L^(2)CQ_(0))/(M+B^(2)L^(2)C)`
B. `(B^(2)R^(4)C^(3)Q_(0))/(M+B^(2)L^(2)C^(2))`
C. `(B^(2)R^(4)C^(3)Q_(0))/(M+B^(2)R^(4)C^(4))`
D. `(B^(2)L^(2)CQ_(0))/(M+B^(2)L^(2)C^(2))`

Answer

A

Explanation

At any instant `t`, the charge on capacitor `q`, velocity of rod `v` and the current `I` through rod are as shown.
`m(dv)/(dt) = BIL = B(-(dq)/(dt))L` ...(1)
`Ϸ underset(0)overset(v) (int) m d v = -underset(Q_(0))overset(q)(int)BL dq`

solving we get `q = Q_(0) - (Mv)/(BL)` ...(2)
Also, `(q)/(C ) = BLV + IR = BLv - R (d q)/(dt)` ...(3)
form equaiton (1) and (3)
`(q)/(C ) = BLv + (mR)/(BL)(dv)/(dt)` .....(4)
from (4) when `(d v)/(dt) = 0 Ϸ= (q)/(C ) = BL v` ......(5)
From (2) and (5). At instant acceleration is zero
`v=(Q_(0)LB)/(M+B^(2)L^(2)C)` and `q=(B^(2)L^(2)CQ_(0))/(M+B^(2)L^(2)C)`