A conducting rod `PQ` of mass `m` and length `l` is placed on two long parallel (smooth and conducting) rails connected to a capacitor as shown. The rod `PQ` is connected to a non conducting spring of spring constant `k`, which is initially in relaxed state. The entire arrangement is placed in a magnetic feild perpendicular to the plane of figure.
Neglect the resistance of the rails and rod.
Now, the rod is imparted a velocity `v_(0)` towards right, then acceleration of the rod as a function of its displacement `x` given by

Options

A. `(k x)/(m)`
B. `(k x)/(m+B^(2)l^(2)C)`
C. `(k x)/(m-B^(2)l^(2)C)`
D. `(k x)/(B^(2)l^(2) C)`

Answer

C

Explanation

Let the velocity of rod be `v` when it has been displacement by `x`. Due to motion of rod an emf will be induced in rod given by `e = B v l`, due to this induced emf, charging of the capacitor takes place as a current, flows in the circuit [for very small time] a result of this current, the rod experiences a magnetic force given by `IBl`.

From Newton's second law,
`IBl + kx = ma[a = (dv)/(dt)]`
`I = (d)/(dt)[Q] = (d)/(dt)[C xx Bvl] = CBl xx (dv)/(dt)`
`rArr a = (kx)/(m-B^(2)l^(2)C^(2)) = omega^(2)x`
Which also shows that rod is performing `SHM`