Question

Derive the expression of current density in term of drift velocity

Answer 1

Consider a conductor of length l and area of cross section A. Let the number of free electrons per unit volume be n. When a pd V is applied, let the drift velocity of electrons be v.

The total no. of free electrons in the piece of conductor = nAl (no. of free electrons per unit volume x volume)

Total charge due to free electrons; q  = nAle

The time taken by this charge to completely traverse the conductor, t =l/v

Therefore, the current in the conductor, I=q/t = nAle/(l/v) = nAev

i.e; I=nAev

Answer 2

Answer:

J = l/A = nevd

Explanation:

• The average velocity attained by charged particles, (eg. electrons) in a material due to an electric field.
• In a conducting solid, an electron will suffer collisions with fixed heavy ions. After collisions,
• electron will emerge with same speed, but direction changes randomly. If we consider
• $\mathrm{N}$ number of electrons in a given volume, since directons are changed randomly due to collisions,

average velocity of $\mathrm{N}$ electrons will be zero.

This is expressed as $\frac{1}{N} \sum_{i=1}^{N} v_{i}=0$.

If electrons are accelerated by electric field $\mathrm{E}$, then acceleration is given by, $\mathrm{a}=-\mathrm{eE} / \mathrm{m}$

Let us consider an ith electron in a group of $\mathrm{N}$electrons at a given time t.

Let us assume after a previous collision, speed of this ith electron is $\mathrm{v}_{\mathrm{i}}$ and there is an elapsed time $\mathrm{t}_{\mathrm{i}}$ after collision.

Speed $V_{i}$ of this ith electron at time $t$ is given by, $V_{i}=v_{i}-(e E / m) t_{i} \ldots \ldots \ldots \ldots .(3)$

Average velocity of electrons at time $t$ is average of all $V_{i}$ of each electron in the group we have considered.

In eqn.(3), average of $\mathrm{v}_{\mathrm{i}}$ appearing on left side is zero as mentioned in eqn.(1).

Collisions of elecrons do not occur at regular intervals but at random time. Let us denote the average time between successive collisions as $\mathrm{T}$.

Then averaging eqn.(3) over $\mathrm{N}$ electrons at any given time $\mathrm{t}$ gives us average velocity $\mathrm{v}_{\mathrm{d}}$, as given by

$\begin{aligned}&v_{d}=\left(v_{i}\right)_{\text {average }}=\left(v_{i}\right)_{\text {average }}-\frac{e E}{m}\left(t_{i}\right)_{\text {average }} \\&v_{d}=-\frac{e E}{m} T \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots(4)\end{aligned}$

$v_{d}$ is called drift velocity. Due to drift, there will be net transfer of charges across any area perpendicular to Electric field $E$.

Consider a planar area A, located inside the conductor such that normal to area is parallel to Electric filed $\mathrm{E}$.

Then because of drift, in an infinitesimal amount of time $\Delta t$, all electrons to the left of the area at distances upto $\left|v_{d}\right| \Delta t$ would have crossed the area. If $n$ is number of free electrons per unit volume in the metal,

then there are $n \Delta t\left|v_{d}\right| A$such electron. Since each electron carry a charge -e, the total charge transported across this area $A$ to the right in time $\Delta \mathrm{t} is -\mathrm{neA}\left|\mathrm{v}_{\mathrm{d}}\right| \Delta \mathrm{t} .$

Flow of charge per unit time across an area $\mathrm{A}$ is the magnitude of current $\mathrm{I}$.

Then we have, $I=n e A\left|v_{d}\right|$

by substituting $\mathrm{v}_{\mathrm{d}} from eqn.(4) in eqn.(5), I=\frac{n \mathrm{e}^{2}}{m} T \times A|E|$

Current density $\mathrm{J} is$ defined as, $\mathrm{J}=\mathrm{I} / \mathrm{A} , where \mathrm{I}$ is the current flowing in a cross section area $\mathrm{A}$.

hence eqn.(6) is written as,$J=\sigma E$, where conductivity $\sigma$ is expressed as $\sigma=\frac{n e^{2}}{m} T$

mobility $\mu$ is the drift velocity per unit electric field, hence we write, $\mu=\frac{\left|v_{d}\right|}{E}=\left|\frac{e T}{m}\right| \ldots \ldots \ldots . .(7)$

In eqn.(7) we used the relation for drift velocity from Eqn.(4).

Potenital difference changes the Electric field, but this change will not affect mobility as per eqn.(7)

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