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(cd) 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 speed of rod 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)
`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)`.