U-frame $ADCB$ and a sliding rod $PQ$ of resistance $R$, start moving with velocities $v$ and $2v$ respectively, parallel to along wire carrying current $i_0$. When the distance $AP=l$ at $t=0$, determine the current through the inductor of inductance $L$ just before connecting rod $PQ$ loses contact with the U-frame.

Since, $PQ$ and $DC$ both cut the lines of field. 

$\therefore$Motional emf will be induced across both of them. 

Integrating, potential difference across $dx$

$\Rightarrow \int de={\int }_a^{2a}v\left(\frac{{\mu }_0{i}_0}{2\mathrm{\pi x}}\right)dx$

$e_{DC}=\frac{v{\mu }_0{i}_0}{2\mathrm{\pi }}\ln 2$ with $D$ at higher potential

$e_{PQ}=\frac{2v{\mu }_0{i}_0}{2\mathrm{\pi }}\ln 2$ with $P$ at higher potential

The relative velocity of the rod $PQ$ w.r.t. $U$ frame

$v_{rel}=2v-v=v$

Now, time taken by it to loose the contact $t=\frac{l}{v}$

From equivalent electric network

Net emf in the closed loop $QPDC$. 

$e=e_{PQ}-e_{DC}=\frac{v{\mu }_0{i}_0}{2\mathrm{\pi }}\ln 2$

Growth of current in the L-R circuit is given by

$i=i_0\left(1-e^{-tR/L}\right)=\left(\frac{e}{R}\right)\left(1-e^{-tR/L}\right)$

At time $t=\frac{l}{v}$

$\Rightarrow i=\left(\frac{e}{R}\right)\left(1-e^{-\frac{Rl}{vL}}\right)$