The system differs from the one examined in the foregoing problem (Fig ) by a capacitor of capacitance `C` replacing the resistance `R`. Find the accelearation of the connector.

Explanation

From Lenz's law, the current the copper bar is directed from 1 to 2 or in other words, the induced current in the current is in clockwise sense.
Potential difference across the capacitor plates,
`(q)/(C) xi_("in")` or, `q = C xi_("in")`

Hence, the induced current in the loop,
`i = (dq)/(dt) = C (d xi_(i n))/(dt)`
But the variation of magnetic flux through the loop is caused by the movement of the bat.
So, the infuced `e.m.f., xi_(i n) = B lv`
and, `(d xi_(i n))/(dt) = Bl (dv)/(dt) = Blw`
Hence, `i = C (d xi)/(dt) = C B l w`
Now, the forces acting on the bars are the weight and the Ampere's force, where `F_(amp) = i l B (CB l w) B = C l^(2) B^(2) w`.
From Newton's second law, for the rod, `F_(x) = mw_(x)`
or, `mg sin alpha - C l^(2) B^(2) w = mw`
Hence `w = (mg sin alpha)/(C l^(2) B^(2) + m) = (g sin alpha)/(1 + (l^(2) B^(2) C)/(m))`