Question

Obtain an expression for motional emf from Lorentz force.

Answer 1

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.