The figure shows a metal rod $PQ$ resting on the smooth rails $AB$ and positioned between the poles of a permanent magnet. The rails,the rod,and the magnetic field are in three mutually perpendicular directions. $A$ galvanometer $G$ connects the rails through a switch $K$. Length of the rod $= 15 \; cm$,$B = 0.50 \; T$,resistance of the closed loop containing the rod $= 9.0 \; m\Omega$. Assume the field to be uniform.
$(a)$ Suppose $K$ is open and the rod is moved with a speed of $12 \; cm \; s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced $emf$.
$(b)$ Is there an excess charge built up at the ends of the rod when $K$ is open? What if $K$ is closed?
$(c)$ With $K$ open and the rod moving uniformly,there is no net force on the electrons in the rod $PQ$ even though they do experience magnetic force due to the motion of the rod. Explain.
$(d)$ What is the retarding force on the rod when $K$ is closed?
$(e)$ How much power is required (by an external agent) to keep the rod moving at the same speed $(= 12 \; cm \; s^{-1})$ when $K$ is closed? How much power is required when $K$ is open?
$(f)$ How much power is dissipated as heat in the closed circuit? What is the source of this power?
$(g)$ What is the induced $emf$ in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

(A) Given: Length of the rod,$l = 15 \; cm = 0.15 \; m$,Magnetic field strength,$B = 0.50 \; T$,Resistance of the closed loop,$R = 9.0 \; m\Omega = 9.0 \times 10^{-3} \; \Omega$,Speed of the rod,$v = 12 \; cm \; s^{-1} = 0.12 \; m \; s^{-1}$.
$(a)$ Induced $emf$ $e = Bvl = 0.50 \times 0.12 \times 0.15 = 9 \times 10^{-3} \; V = 9 \; mV$. By Fleming's Right-Hand Rule,the polarity is such that end $P$ is positive and end $Q$ is negative.
$(b)$ When $K$ is open,an excess charge builds up at the ends of the rod due to the separation of charges by the magnetic Lorentz force. When $K$ is closed,the current flows continuously,so the excess charge is maintained by the current flow.
$(c)$ When $K$ is open,the magnetic force on the electrons is balanced by the electric force due to the accumulated charges at the ends of the rod. Thus,the net force on the electrons is zero.
$(d)$ Retarding force $F = IBl$. Current $I = e/R = (9 \times 10^{-3} \; V) / (9 \times 10^{-3} \; \Omega) = 1 \; A$. Thus,$F = 1 \times 0.50 \times 0.15 = 0.075 \; N = 75 \; mN$.
$(e)$ When $K$ is closed,power $P = Fv = 0.075 \times 0.12 = 9 \times 10^{-3} \; W = 9 \; mW$. When $K$ is open,no current flows,so no power is required to maintain the motion (ignoring friction).
$(f)$ Power dissipated as heat $P = I^2R = (1)^2 \times 9 \times 10^{-3} = 9 \; mW$. The source of this power is the external agent moving the rod.
$(g)$ If the magnetic field is parallel to the rails,the velocity vector is parallel to the magnetic field vector. Thus,$e = Bvl \sin(\theta) = 0$,as $\theta = 0^\circ$.