A conducting rod MN of mass m and length 'l' is placed on parallel smooth conducting rails connected to an uncharged capacitor of capacitance C and a battery of emf `epsilon` as shown. A uniform magnetic field B is existing perpendicular to the plane of the rails. The steady state velocity acquired by the conducting rod MN after closing switch S is (neglect the resistance of the parallel rails and the conducting rod)

Options

A. `(2CBl epsilon)/((m+CB^(2)l^(2)))`
B. `(CBl epsilon)/((m+CB^(2)l^(2)))`
C. `(CBl epsilon)/(2(m+CB^(2)l^(2)))`
D. `(CBl epsilon)/(4(m+CB^(2)l^(2)))`

Answer

B

Explanation

`(q)/(C)+Bvl=epsilon`
`q=C(epsilon-Bvl)`
Now, `m(dv)/(dt)=Bil`
`mdv=Blidt`
`m underset(0)overset(v)intdv=Blunderset(0)overset(q)intdq`
`mv=Blq`
`thereforeq=(mv)/(Bl)` ...(i)
From (i) and (ii)
`(mv)/(Bl)=Cepsilon-CBlv`
`impliesv=(CBl epsilon)/((m+CB^(2)l^(2)))`