A rod of length l is moved horizontally with a uniform velocity 'v' in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod. How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor?

(a) By Faraday's laws of electromagnetic induction, V = dΦ/dt. V = B dA/dt. V = Blv as dA/dt = lv. This is the induced emf across the ends of the rod.
(b) We have the force on the free electrons from Q to P as →Fb = →qv × →B where v is the velocity of the rod as well as of the free electrons inside it, B is the uniform magnetic field. The free electrons will move towards P and positive charge will appear at Q. An electrostatic field is developed from Q to P in the wire which exerts a force →Fe = q→E on each electron. The charge keeps on gathering until →Fb = →Fe and the resultant force on each electron is zero. |q→v × →B| = |q→E| ⇒ vB = E. The potential difference between Q and P is then V = El = vBl which is maintained by the magnetic force of the moving electron producing an emf, e = Bvl