We define drift speed of an electron as the average speed of the electron inside the conductor, which is the distance travelled by the electron between two collisions divided by the average collision time or relaxation time.


Using the laws of motion for uniform acceleration, we may write the drift velocity to be


Now as the average velocity of an electron would be almost zero after any collision, we may say that $\mathbf{u}$ is zero, and hence we may also write


However in Fundamentals of Physics by Resnik and Halliday the drift velocity (which it states is the average velocity of any electron in the conductor) is given as


Meanwhile in Concepts of Physics by prof. H.C.Verma it is given


So which is actually correct?


2 Answers 2


The velocity given in Halliday $$\frac{eE\tau}{m}$$ Is just the average final velocity of the collection of electrons in the wire. While the velocity given in H C Verma is average velocity in the whole path of the collection of electrons. There really are two averages here. One is the average over the velocity of collection of charges in the wire at any instant, and the second is the average of all these averages over the whole path. As shown in Halliday the velocity of the collection at the final instant is as given above. To calculate the the average velocity over path in uniform acceleration one can use the relation $$v_d =\frac{v_H+u}{2}$$ Where $v_H$ is the velocity given by Halliday. Since the average velocity of all charges is $0$ right after collision, the drift velocity becomes $$v_d= \frac{v_H}{2}$$

Edit: After some looking about I found out that the above both derivations are wrong, since statistical motion is not really taken into proper account in the derivations. See this SE post for a much better understanding of the motion of electrons in a conductor. The drift velocity in Halliday is the correct expression in certain approximation. The problem in the above derivations is the definition of $\tau$. It is not exactly the relaxation time but the time required to loose the information of electrons velocity.

  • $\begingroup$ Yes but it has been written in Halliday that the avg velocity of the drifting electron is $\frac{eE\tau}{m}$ $\endgroup$
    – SK Dash
    Dec 8, 2019 at 3:43
  • $\begingroup$ @SKDash I have edited my answer, see the link in the answer for a better understanding. $\endgroup$ Dec 9, 2019 at 0:23

When you consider

$$\mathbf v_\text d=\dfrac 1 2 \frac{e\mathbf E}{m}\tau$$

You're actually first of all finding the total distance traveled by the charge in time $\tau$ during which the velocity wasn't constant. After that you decide to divide the distance by the total time (average collision time,$\tau$) so you end up with

$$\dfrac{\Delta s}{\Delta t}$$

which gives you the average speed during that time interval $\Delta t$. So what you're finding is the average velocity during the travel through the mean free path.

On the other hand, when you do

$$\mathbf v_\text d=\frac{e\mathbf E}{m}\tau$$

you're finding the final velocity of the charge just before the impending collision.

I think Verma sir's approach is a bit better because he talks about the average velocity of the charge.


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