In the arrangement shown in the figure, uniform magnetic field B is upward to the plane of paper. Connecter AB is smooth and conducting, gaving mass m and length l. Initially spring has extension `X_(0)`. Spring in non-coducting (neglect induction in spring). Connector is released at t=0. Maximum charge on the capacitor is

Options

A. `X_(0)sqrt((KC)/(1+(B^(2)l^(2)C)/(m)))`
B. `X_(0)sqrt((KC)/(1+(m)/(B^(2)l^(2)C)))`
C. `sqrt((KC)/(2+(B^(2)l^(2)C)/(m)))`
D. `sqrt((4KC)/(1+(B^(2)l^(2)C)/(2m)))`

Answer

B

Explanation

At any tie
`Kx-I//B=ma` …(1)
`(q)/(C)=Bvl`
`I=B//Ca`
From (1) and (2)
`kx=a(B^(2)l^(2)C+m)` …(2)
From (1) and (2)
`kx=a(B^(2)l^(2)C+m)`
`a=((K)/(B^(2)l^(2)C+m))x` ..(3)
since motion of connector will be opposite to the displacement from equilibrium therefore equation (3) represents S.H.M
`omega=sqrt((K)/(B^(2)l^(2)C+m))`
`V_(max)=X_(0)omega`
`thereforeq_(max)=BlCV_(max)`
`q_(max)=BlCX_(0)omega`
`X_(0)sqrt((KC)/(1+(m)/(K^(2)l^(2)C)))`