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Question
Obtain an expression for motional emf from Lorentz force.
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Solution
Consider a straight conducting rod AB of length l in a uniform magnetic field `vec"B"` which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity `vec"v"` towards right side.
When the rod moves, the free electrons present in it also move with same velocity `vec"v"` in `vec"B"`. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation
`vec"F"_"B" = - "e"(vec"v" xx vec"B")` ....(1)
The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field E, the coulomb force starts acting on the free electrons along AB and is given by
`vec"F"_"E" = - "e"vec"E"` ....(2)
The magnitude of the electric field `vec"E"` keeps on increasing as long as the accumulation of electrons at the end A continues. The force `vec"F"_"E"` also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force `vec"F"_"B"` and the coulomb force `vec"F"_"E"` balance each other and no further accumulation of free electrons at the end A takes place, i.e.,
`|vec"F"_"B"| = |vec"F"_"E"|`
`|-"e" (vec"v" xx vec"B")| = |-"e" vec"E"|`
vB sin 90° = E
vB = E ……. (3)
The potential difference between two ends of the rod is
(a) 
(b)
Motional emf from Lorentz force
V = El
V = vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf.
ε = Blv ….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.
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