Question

In the figure, a conducting rod of length $l=1$ meter and mass $m=1kg$ moves with an initial velocity, u $=5m{s}^{-1}$ . On a fixed horizontal frame containing inductor $L=2H$ and resistance $R=1\Omega$ . PQ and MN are smooth, conducting wires. There is a uniform magnetic field of strength $B=1T$ . 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 $v_f=1m{s}^{-1}$ and the rod has travelled a distance $x=3$ meter.

Answer 1

Answer: 1

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 $=m\frac{dv}{dt}=-ilB=-\left(i_1+\frac{Blv}{R}\right)lB$
$\Rightarrow mdv=-lB{i}_{1}dt-\frac{B^2{l}^2}{R}vdt$
$\Rightarrow m\int dv=-lB\int i_{1}dt-\frac{B^2{l}^2}{R}\int vdt$
$\Rightarrow m\left(v_f-u\right)=-lBQ-\frac{B^2{l}^2}{R}x$
( $v_1$ = velocity, when it has moved a distance 'x')
$\Rightarrow Q=\frac{-\frac{B^2{l}^2}{R}x-m\left(v_f-u\right)}{Bl}=1C$