The arrangement shown is placed in a vertical uniform magnetic field. Two metal rods of length l and masses $m_1$ and $m_2$ are pulled apart from rest by a constant force $F$. Find the current in the resistor $R$ as a function of time.

Let $v_1$ and $v_2$ be the instantaneous velocity force the rods $m_1$ and $m_2$ respectively. The emf’s across the rod will be 

$E_1=Bl{v}_1$ and $E_2=Bl{v}_2$

The instantaneous current is

$I=\frac{E_1+E_2}{R}=\frac{Bl(v_1+v_2)}{R}\dots \dots .\left(1\right)$

The acceleration of each rod is given by

$\frac{d{v}_1}{dt}=\frac{F-BIl}{m_1}=\frac{F}{m_1}-\frac{B^2{l}^2}{m_{1}R}(v_1+v_2)$

And $\frac{d{v}_2}{dt}=\frac{F}{m_2}-\frac{B^2{l}^2}{m_{2}R}(v_1+v_2)$

Adding the above two equations 

$\frac{d}{dt}(v_1+v_2)=F\left[\frac{1}{m_1}+\frac{1}{m_2}\right]-\frac{B^2{l}^2}{R}(v_1+v_2)\left[\frac{1}{m_1}+\frac{1}{m_2}\right]\dots \dots .\left(2\right)$

Using equation $\left(1\right),\frac{R}{Bl}\frac{dI}{dt}=\frac{F}{\mu }-\frac{BlI}{\mu }where\mu =\frac{m_1{m}_2}{m_1+m_2}$

Or, $\frac{dI}{F-BlI}=\frac{Bl}{\mu R}dt$

On integrating $I=\frac{F}{Bl}\left[1e^{-B^2{l}^2{t}^{/}\mu R}\right]$