Two 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 the 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 of the conductor (in metre) after 2s. Assume that  $C{B}^2{\mathcal{l}}^2=4m$ (the mass of the conductor) and  $g=10m/s^2$.

 

Due to the motion of the conductor in magnetic field, an e.m.f. is induced in it. As a result, a current flows through the conductor. According to Lenz’s law, a force $Bi\mathcal{l}$ (due to induced current) opposes the motion of the conductor. Let at some instant t, velocity of the conductor be v.
The net accelerating force on conductor is  $F=mg-Bi\mathcal{l}$ …(1)
Here, induced e.m.f.    $=B\mathcal{l}v$ 
Charge on the capacitor,
  $q=Ce=C\left(B\mathcal{l}v\right)$
Since, v is increasing, the charge and hence the current through the capacitor is also increasing. The current through capacitor is given by
$i_c=\frac{dq}{dt}=CB\mathcal{l}f\frac{dv}{dt}$   …(2)
From equations (1) and (2), we get
  $m\frac{dv}{dt}=mg-B\left(CB\mathcal{l}\frac{dv}{dt}\right)\mathcal{l}$
  $\Rightarrow m\frac{dv}{dt}=mg-B^2{\mathcal{l}}^{2}C\frac{dv}{dt}$
  $\Rightarrow \frac{dv}{dt}[m+B^2{\mathcal{l}}^{2}C]=mg$
  $\Rightarrow a=\frac{dv}{dt}=\frac{mg}{(m+B^2{\mathcal{l}}^{2}C)}$
  $\Rightarrow x\left(t\right)=\frac{1}{2}a{t}^2=\frac{mg{t}^2}{2(m+B^2{\mathcal{l}}^{2}C)}=4$