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What is drift velocity? And why in some books it is expressed as drift speed and not drift velocity?Are these different ?

Does it mean that the electron will have extra velocity opposite to the direction of electric field i.e thermal(random) velocity + drift velocity=Total velocity of electron,in the conductor with a specific direction (opposite to electric field)

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closed as off-topic by John Rennie, Waffle's Crazy Peanut, Dimensio1n0, Kyle Kanos, Brandon Enright Feb 26 '14 at 16:09

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en.wikipedia.org/wiki/Drift_velocity. You might wish to read through this and edit your question to address any specific points arising from it. –  John Rennie Feb 26 '14 at 9:53
I would guess there's no significance to the use of the word speed rather than velocity. Many of us are casual about this when the direction is obvious (i.e. along the wire). –  John Rennie Feb 26 '14 at 9:54
Drift speed = \abs(Drift velocity) –  Dilaton Feb 26 '14 at 17:57

2 Answers 2

Imagine identical charges moving through a conductor. Said conductor has a cross-sectional area $A$. The volume of an element of length $\Delta x$ of the conductor is simply $A \Delta x$. If $n$ represents the number of mobile charge carriers per unit volume, the mobile charge $\Delta Q$ is given $\Delta Q = (nA \Delta x) q$, or the number of carriers times the charge per carrier. If the carriers move at a constant AVERAGE speed (drift speed, $v_d$), the distance they move in $\Delta t$ is given by $\Delta x = v_t \Delta t$. Therefore, $\Delta Q = (nAv_d \Delta t)q$. Dividing by $\Delta Q$ both sides yields:$I = \frac {\Delta Q}{\Delta t} = nqv_dA$, which is the equation for drift speed.

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I read in my 12th book the difference b/w these two but i don't have it now so i can't give you the reference but what i remember is as follows.
The drift speed is defined as: $v_d=\dfrac{l}{\tau}$
Here we define $l$ as $l=\dfrac{1}{2}a(\tau)^2$ here $\tau$ is the mean-time b/w any two sucsessive collisions. So $v_d=\dfrac{1}{2}a\tau$ where $a$ is $\dfrac{eE}{m}$.
Now what about drift velocity? Well some contexts define it as the magnitude of velocity(just before the next collision) averaged out as $\dfrac{v_1+v_2+v_3+.....v_n}{n}$. Now $v_n=u_n+at_n$ Also averaging $u_n$ will give null because avg thermal velocity is assumed to be zero. So we got drift velocity as $v_d=a\tau$. Clearly this is different from previous expression for drift speed. this is in an unusual manner to over simplify things because $v_n$ is the velocity acquired by the nth electron just before the next collision.
Some contexts use drift velocity in place of drift speed i.e. synonymously making no distinction between them.
Actually drift speed is the total distance covered by an electron in n sucsessive collisions divided by total time taken to make n collions where n is very large. A notable thing is we declared mean distance as $l=\dfrac{1}{2}a(\tau)^2$ but actually mean distance should have been $\dfrac{l_1+l_2+l_3+....l_n}{n}$ where $l_n=u_nt_n+\dfrac{1}{2}at_n^2$ so avg speed of electrons should be $v_{avg}=\dfrac{\Sigma l_n}{\Sigma t_n}$
Also $\Sigma l_n=\Sigma u_nt_n+\dfrac{1}{2}a{\Sigma t_n^2}$ here you see we cannot replace the summation with $\tau$ because $\tau$ is $\dfrac{\Sigma_0^nt_n}{n}$ and also cannot compute $\Sigma u_n t_n$ so we cannot find $v_{avg}$ in this way.
To find some reasonable expression for $v_{avg}$ what we do is we find avg speed i.e. we find $\dfrac{\Sigma \dfrac{l_n}{t_n}}{n}$ which again gives us $v_{avg}=\dfrac{1}{2}a\tau$.
For some good related knowledge read this answer:Which derivation of drift velocity is correct?

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