(a) Consider a rod PQ of length $l$ moving in a magnetic field $\overset{\rightarrow}{B}$ with a constant velocity $\overset{\rightarrow}{v}$. The length of the rod is perpendicular to the magnetic field and also the velocity is perpendicular to both the rod and field. The free electrons of the rod also move at this velocity $\overset{\rightarrow}{v}$ because of which it experiences a magnetic force. This force is towards $Q$ to $P$.Thus, the free electrons will move towards $P$ and positive charge will appears at $Q$. An electrostatic field $E$ is developed within the wire from $Q$ to $P$. This field exerts a force.$\vec{F}_e = q \overset{\rightarrow}{E}$on each free electron. The charge keeps on gathering until$\vec{F}_b = \vec{F}_e$$\Rightarrow \;\; q \overset{\rightarrow}{v} \times \overset{\rightarrow}{B} | \; = | q \overset{\rightarrow}{E} |$$v B = E$After this, resultant force on the free electrons of the wire $P Q$ becomes zero. The potential difference between the ends $Q$ and $P$ is given by,$V = E l = v B l$ Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf. $e = B v l$(b) Lorentz force acting on a charge $q$ which is moving with a speed $v$ in a (normal) uniform magnetic field $B$, is $B q v$.All the charges will experience the same force. Work done to move the charge from $P$ to $Q$.$W = B q v \times l$$e \; = \; \frac{W}{q} = \; \frac{B q v l}{q} = B l v$