Question

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

Answer 1

Let $i_{1}$ and $i_{2}$ be the current through $L$ and $R$ at any time $t$
$\therefore \, \, i=i_{1}+i_{2} \, \Rightarrow \, \, \frac{B l v}{R}=i_{2}$ and $Blv=L\frac{d i_{1}}{d t}$

Force on conducting rod $=m\frac{d v}{d t}=-ilB=-\left(i_{1} + \frac{B l v}{R}\right)lB$
$\Rightarrow \, mdv= \, -lB \, i_{1}dt-\frac{B^{2} l^{2}}{R}v \, dt \, $
$\Rightarrow \, m\displaystyle \int dv=-lB \, \displaystyle \int i_{1}dt-\frac{B^{2} l^{2}}{R} \, \displaystyle \int v \, dt$
$\Rightarrow \, m\left(v_{f} - u\right)=-lBQ-\frac{B^{2} l^{2}}{R}x$
( $v_{1}$ = velocity, when it has moved a distance 'x')
$\Rightarrow \, Q=\frac{- \frac{B^{2} \, l^{2}}{R} \, x - m \left(v_{f} - u\right)}{B \, l}=1C$