Electro-mechanics

$\textcolor{#3D99F6}{Fun Problem}$
I was struggling in this question from last 1hr because I was taken $D=1$ instead of $2$ . So it's a lesson for me to read the data carefully.
Now come to the question. The basic equations are $BvD-iR-L\frac{di}{dt}=0$ $F-BiD=ma$ After substituting correct value you will reach $10-10i=\dot{v}$ $10v-i=\dot{i}$
In steady state velocity is constant, Therefore $\dot{v}=0$ and $\dot{i}=0$
Now ,by solving equations you will reach $\textcolor{#20A900}{\large\boxed{v_{steady}=0.1}}$
Edited : I have 2 more methods .
$2$ nd method : At steady state $v=constant$ $a=0$
Net force in rod is zero.
Let current be $I$ (which is constant at steady state) $I =\frac{BDv}{R}$
Magnetic force on rod $F_{m}= BID=\frac{B^{2}D^{2}v}{R}$
At Equilibrium $F_{ext}=F_{m}$ $\frac{B^{2}D^{2}v}{R}=F$ $v= \frac{FR}{B^{2}D^{2}}$ $\textcolor{#20A900}{\large\boxed{v_{steady}=0.1}}$ (3) rd method:
Workdone by Force $F$ = Heat dissipated in resistor $R$
As $\textcolor{fuchsia}{\frac{dI}{dT}=0}$ . Hence inductor offers $0$ resitance and hence energy stored in it becomes constant.
At this moment . Let the velocity of rod is $v$
Power dissipated in resistor $R$ $I^{2}R=\frac{B^{2}D^{2}v^{2}}{R}$ . $\frac{B^{2}D^{2}v^{2}}{R}=Fv\cos0°$ $\textcolor{#20A900}{\large\boxed{v_{steady}=0.1}}$
I hope you enjoyed all these $\textcolor{junglegreen}{3}$ methods.