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 speed of rod is `v`, the charge on capacitor is :

Options

A. `(Q_(0)-(RM)/(LBC)v)`
B. `(Q_(0)-(M)/(RC^(2)B)v)`
C. `(Q_(0)-(M)/(LB)v)`
D. `(Q_(0)-(M)/(RCB)v)`

Answer

C

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)
`rArr underset(0) overset(v)(int) mdv = - underset(Q_(0))overset(q)(int) BLdq`
solving we get
`q =Q_(0)-(Mv)/(BL)`....(2)
Also `(q)/(C)=BLV+IR =BLv-R(dq)/(dt)`......(3)
from equation (1) and (3)
`(q)/(C)=BLv+(mR)/(BL)(dv)/(dt)`....(4)
from (4) when `(dv)/(dt) =0`
`rArr (q)/(C) =BLv`....(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)`.