(a) A rod of length l is moved horizontal with a uniform velocity `v` in a direction perendicular 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 end of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.

Explanation

The conductor is moving in a direction perpendicular to the field with constant velocity under the influence of some external agent.
The electrons in the conductor experience a force
`vecF=q(vecvxxvecB)`
That is directed along the length `l` perpendicular to both `v` to `B`
Under the influence of this force, the electrons move to the lower end of the conduction and accumulate there leaving a net positive charge at the upper end.As a result of this charge separation an electric field is product inside the conductor.The charges accumulate at both ends until the downward magnetic force `qvB` is balanced by the upward electric force `qE`.At this point, electrons stop moving .The condition of equilibrium requires that
`q [vecE+(vecvxxvecB)]=0`
`vecE=-(vecvxxvecB)`
`V=-int vecE.dvecl`
`DeltaV=(vecvxxvecB).vecl`
`=V_|_l_|_v`
`B_|_` is the component of magnetic field perpendicular to the velocity.
(a)`phi_(B)=Blx`
`epsilon=(-dphi_(B))/(dt)=-d/(dt)(Blx)`
`epsilon=Blv`