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.

#### Solution

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.

\[\therefore\frac{q}{C} = Blv\]

And,

q = C × Blv = CBlv

Current in the circuit:-

\[i = \frac{dq}{dt} = \frac{d(CBlv)}{dt}\]

\[\Rightarrow i = CBl\frac{dv}{dt} = CBla\] ........(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 + cB^{2}l^{2}a = mg

⇒ a(m + cB^{2}l^{2}) = mg

\[\Rightarrow a = \frac{mg}{m + c B^2 l^2}\]