Question

Explain the term ‘drift velocity’ of electrons in conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’. 

Answer 1

‘Drift velocity’ of electrons in a conductor - Metals contain a large number of free electrons. These electrons are in continuous random motion. Due to the random motion, the free electrons collide with positive metal ions with high frequency and undergo change in direction at each collision. So the average velocity for the electrons in a conductor is zero.

Now, when this conductor is connected to a source of emf, an electric field is established in the conductor, such that E = `"V"/"L"`

Where V= potential difference across the conductor and
L = length of the conductor.

The electric field exerts an electrostatic force ‘-Ee’ on each free electron in the conductor. The acceleration of each electron is given by

`vec"a" = ("e"vec"E")/"m"`

Where e = electric charge on electron and
m = mass of electron

The negative sign indicates that the force and hence the acceleration is in a direction opposite to the direction of the electric field. Due to this acceleration, the electrons attain a velocity in addition to thermal velocity in the direction opposite to that of electric field.

The average velocity of all the free electrons in the conductor is called the drift velocity of free electrons of the conductor.

`vec"v"_"d" = - ("e"vec"E")/"m" tau`      ....... (1)

Thus, the expression for the drift velocity is 

Electric field, `"E" = - "V"/"L"`       .....(2)

where `tau` = relaxation time between two successive collision.

Let n = number density of electrons in the conductor.

No. of free electrons in the conductor = nAL

Total charge on the conductor, q = nALe

Time taken by this charge to cover the length L of the conductor, `"t" = "L"/"V"_"d"`

current `"I" = "q"/"t"`

`= ("nALe")/"L" xx "v"_"d"`

`= "nAev"_"d"`

Using eq (1) and (2) , we get that

`"I" = "nAe" xx (-("e"vec"E")/"m" tau)`

`= "nAe" xx (- ("e"(-"V"))/("mL") tau)`

`= (("n" "e"^"2A")/("mL")tau)"V"`