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In the shown figure, there are two long ...

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 :

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))`

Text Solution

Verified by Experts

The correct Answer is:
A
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)`.
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