# Faraday’s law and motional emf

Can someone please explain me in layman’s language the difference between the emf induced as per faraday and emf induced as per Lorentz? How does a wire moving through a magnetic field produce emf when there is no open surface ( pls explain me this in terms of faraday’s law )

I assume you're thinking of something like a straight wire of length $$l$$ moving at speed $$v$$ in a uniform magnetic field of magnitude $$B$$, with the wire, motion and field all at right angles to each other. It would be a good idea to draw a diagram and to add to it as you read on. [Let the ends of the wire slide on fixed rails at right angles to the wire, and have the rails joined together at one end by a resistor. In that way the moving wire is part of a circuit whose area changes as the wire moves.]
The Lorentz treatment is that each electron is carried along by the wire and has a velocity component $$v$$ at right angles to the wire, and to the field. It therefore experiences a force component of magnitude $$Bev$$ directed along the wire, so if a free electron could traverse the length of the wire, the work done on it would be $$Bevl.$$ The emf is the work done per unit charge, so$$\mathscr{E}=\frac{Bevl}{e}\ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \mathscr{E}=Bvl$$
I assume that by the Faraday approach you mean one based on flux-cutting. It goes like this… In time $$\Delta t$$ the wire sweeps out an area $$lv\Delta t$$, at right angles to the magnetic field, so it cuts flux of amount $$\Delta \Phi=Blv\Delta t.$$ Alternatively we could say that the flux passing through the circuit (see first paragraph) changes by amount $$\Delta \Phi=Blv\Delta t.$$ The emf is the rate of cutting of flux or rate of change of flux linking the circuit, so$$\mathscr{E}=(-)\frac{d \Phi}{dt} =(-)\frac{Blv\Delta t}{\Delta t}\ \ \ \ \ \ \ \text{so}\ \ \ \ \ \ \ \mathscr{E}=(-)Blv.$$