Question

In the figure, a conducting rod of length $l=1$ meter and mass $m=1kg$ moves with initial velocity, u $=5m / s$ . 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 = 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 rod becomes v_f = 1 m/s 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= \, -l \, B \, i_{1}dt-\frac{B^{2} l^{2}}{R}v \, dt \, $
$\Rightarrow \, m\displaystyle \int d v=-lB \, \displaystyle \int i_{1}dt-\frac{B^{2} l^{2}}{R} \, \displaystyle \int v \, d t$
$\Rightarrow \, m\left(v_{f} - u\right)=-l \, BQ-\frac{B^{2} l^{2}}{R}x$
( $v_{1}$ = velocity, when it has moved a distance 'x')
$\Rightarrow \, Q=\frac{- \frac{B^{2} \, l^{2}}{R} \, x - m \left(v_{f} - u\right)}{B \, l}=1C$