Here, $$B=0.50T;l=15cm =15*10^{-2}m$$;
$$R=9.0mQ=9.0*10^{3}fl$$
(a) Now, $$e=BVl$$
Here, $$V=12cm$$ $$s^{-1}=12*10^{-2}ms^{-1}$$
$$\therefore e=0.50*12*10^{-2}*15*10^{2}=9*10^{-3}V$$
If $$q$$ is change on as electron, then the elctrons in the rod will experience magnetic Lorentz force $$-q[\underset{v}{\rightarrow}+\underset{B}{\rightarrow}]P.Q$$. Hence, the end $$P$$ of the rod will become positive and the end $$Q$$ will become negative.
(b) When the switch $$K$$ is open, the elctron collect at the end $$Q$$. Therefore, excess change is build up at the end $$Q$$. However, when the switch $$K$$ is closed, the accumulated charge at the end $$Q$$ flows through the circuit.
(c) The magnetic Lorentz force on electron is cancelled by the electronic force acting on it due to the electronic filed set up accross the two ends due to accumulation of positive and negative charges at the end $$P$$ and $$Q$$ respectively.
(d) Retarding force, $$F=BIl=0.50*\frac{e}{R}*15*10^{-2}$$
$$=0.50*\frac{9*10^{-3}}{9*10^{-3}}*15*10^{-2}=7.5*10^{-2}N$$.
(e) When the switch $$K$$ is closed, power required by theb external agent against the retarding force,
$$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$$.
(f) Power dissipated as heat,
$$\frac{e^{2}}{R}=\frac{[9*10^{-3}]^{2}}{9*10^{-3}}=9*10^{-3}W$$/
The sorce of this power is the power of the external agent against the retarding force,
$$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$$.
(g) The motion of rod does not cut field lines, hence no induced e.m.f. is produced.