Question

A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.

Answer 1

Let the velocity of the rod at an instant be v and the charge on the capacitor be q.

The emf induced in the rod is given by
e = Blv

The potential difference across the terminals of the capacitor and the ends of the rod must be the same, as they are in parallel.

∴ $\frac{q}{C}=Blv$

And,

q = C × Blv = CBlv

Current in the circuit:
i = $\frac{dq}{dt}=\frac{d\left(CBlv\right)}{dt}$
$\Rightarrow i=CBl\frac{dv}{dt}=CBla$ (a = acceleration of the rod)
The force on the rod due to the magnetic field and its weight are in opposite directions.
∴ mg − ilB = ma
⇒ mg − cBla × lB = ma
⇒ ma + cB²l²a = mg
⇒ a(m + cB²l²) = mg
$\Rightarrow a=\frac{mg}{m+c{B}^2{l}^2}$