Question

To metal bars are fixed vertically and are connected on the top by a capacitor $C$. A sliding conductor of length $l$ and mass $m$ slides with its ends in contact with bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement $x\left(t\right)$ of the conductor as a function of time $t$.

Answer 1

Let $v$ be the velocity at some instant.

Then, motional emf, $V=Bvl$
Charge stored in capacitor, $q=CV=\left(CBl\right)v$
Current in the wire, $=\frac{dq}{dt}=\left(CBl\right)\cdot \frac{dv}{dt}$
Magnetic force $F_m=ilB=\left(C{B}^2{l}^2\right)\cdot \frac{dv}{dt}$ (upwards)
$\therefore$ Net force, $F_{net}=mg-F_m$
$m\cdot \frac{dv}{dt}=mg-\left(C{B}^2{l}^2\right)\cdot \frac{dv}{dt}$
$\therefore$ $\frac{dv}{dt}=$ acceleration, $a=\frac{mg}{m+C{B}^2{l}^2}$
Since, $a=$ constant
$\therefore$ $x=\frac{1}{2}a{t}^2=\frac{mg{t}^2}{2(m+C{B}^2{l}^2)}$