Question

A wire of mass m and length l can freely slide on a pair of smooth, horizontal rails placed in a vertical magnetic field B. The rails are connected by a capacitor of capacitance C. The electric resistance of the rails and wire is zero. If a constant force F acts on the wire and the resulting acceleration is a, then

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 terminal sof the capacitor and the ends of the must be same, as they are in parallel.

$\therefore \frac{q}{C}=Blv$

And,

$q=C\times Blv$

Current in the circuit:

$i=\frac{dq}{dt}=\frac{d\left(CBlv\right)}{dt}$

$i=CBl\frac{dv}{dt}=CBla$ [a=acceleration of the rod]

The force on the rod due to the magnetic field and its weight are in opposite direction

$\therefore mg-ilB=ma$

$mg-CBla\times lB=ma$

$ma+C{B}^2{l}^{2}a=mg$

$a(m+c{B}^2{l}^2)=mg$

$a=\frac{mg}{m+C{B}^2{l}^2}$