In the figure, a conducting rod of length $l=1$ meter and mass $m=1 \mathrm{kg}$ moves with an initial velocity, u $=5 \mathrm{m} {\mathrm{s}}^{-1}$ . On a fixed horizontal frame containing inductor $L=2 \mathrm{H}$ and resistance $R=1 \mathrm{\Omega }$ . PQ and MN are smooth, conducting wires. There is a uniform magnetic field of strength $B=1 \mathrm{T}$. Initially, there is no current in the inductor. Find the total charge in coulomb, flown through the inductor by the time velocity of the rod becomes $v_f= 1 \mathrm{m} {\mathrm{s}}^{-1}$and the rod has travelled a distance $x=3$ meter.

Solution

Let $i_1$ and $i_2$ be the current through $L$ and $R$ at any time $t$

$\therefore i=i_1+i_2 \Rightarrow \frac{Blv}{R}=i_2$ and $Blv=L\frac{d{i}_1}{dt}$



Force on conducting rod $=m\frac{dv}{dt}=-ilB=-\left(i_1+\frac{Blv}{R}\right)lB$

$\Rightarrow mdv=\mathcal{ }-lB i_{1}dt-\frac{B^2{l}^2}{R}v dt$

$\Rightarrow m\int dv=-lB \int i_{1}dt-\frac{B^2{l}^2}{R} \int v dt$

$\Rightarrow m\left(v_f-u\right)=-lBQ-\frac{B^2{l}^2}{R}x$

( $v_1$ = velocity, when it has moved a distance 'x')

$\Rightarrow Q=\frac{-\frac{B^2 l^2}{R} x-m\left(v_f-u\right)}{B\mathcal{ }l}=1C$

Hence, the correct answer is 1.