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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’.
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Solution
‘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"`
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RELATED QUESTIONS
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10−7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 × 1028 m−3
(a) drift speed
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The drift velocity of electrons in a conductor connected to a battery is given by vd = `(−"eE" τ)/"m"`. Here, e is the charge of the electron, E is the electric field, τ is the average time between collisions and m is the mass of the electron.
Based on this, answer the following:
- How does the drift velocity change with a change in the potential difference across the conductor?
- A copper wire of length 'l' is connected to a source. If the copper wire is replaced by another copper wire of the same area of cross-section but of length '4l', how will the drift velocity change? Explain your answer.
