Drift Velocity - Meaning, Formula, FAQs

Drift Velocity

In a conductor, free electrons are in a state of continuous random motion due to thermal energy. In the absence of any external electric field, these random motions cancel out, resulting in zero net flow of charge.

When an external electric field is applied across the conductor, the electrons continue their random motion, but they also experience a force due to the electric field. As a result, they acquire a small net velocity in a particular direction. This net average velocity of charge carriers is called drift velocity.

Definition:
Drift velocity is the average velocity attained by free charge carriers in a conductor under the influence of an external electric field.

Electrons, being negatively charged, drift opposite to the direction of the electric field, while the direction of conventional current is taken along the field.

Drift Velocity Formula

The drift velocity is given by:

$
v_d=\frac{e E \tau}{m}
$

where $e$ is the charge of an electron, $E$ is the applied electric field, $\tau$ is the relaxation time, and $m$ is the mass of the electron.

Derivation of Drift Velocity

When an electric field $E$ is applied to a conductor, an electron of charge $-e$ experiences a force:

$
F=-e E
$

Using Newton's second law $F=m a$, the acceleration is:

$
a=\frac{-e E}{m}
$

Let $\tau$ be the relaxation time (average time between collisions).
The average velocity gained by the electron is:

$
v_d=a \tau
$

Substituting $a$ :

$
v_d=-\frac{e E \tau}{m}
$

Thus, drift velocity is directly proportional to the electric field and relaxation time, and is opposite to the direction of the electric field.

Drift Velocity in Terms of Electric Field ($E$) and Mobility of Electrons ($\mu$)

$
V_d=\mu E
$

$V_d$ is directly proportional to $E$:

$V_d \propto E$ when the temperature is constant, the greater the electric field, the larger the drift velocity.

Drift velocity varies inversely with the area of the cross-section

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SI Unit of Drift Velocity

The SI unit of drift velocity is meters per second ( $\mathrm{m} / \mathrm{s}$ ).

$V_d=\frac{I}{n A e}$

substituting unit of each term

$V_d=\frac{\mathrm{A}}{\left(\mathrm{m}^{-3}\right)\left(\mathrm{m}^2\right) \mathrm{C}}$

$V_d=\frac{\mathrm{A}}{\mathrm{m}^{-1} \cdot \mathrm{C}}$

$V_d=\frac{\mathrm{C} / \mathrm{s}}{\mathrm{m}^{-1} \cdot \mathrm{C}}$

Thus simplifying we get

$V_d=\frac{\mathrm{m}}{\mathrm{s}}$

Relation Between Current Density and Drift Velocity

The drift velocity is directly proportional to the current density.

$
J=n e V_d
$

where,

$J$ is the Current Density
$n$ is the number density of electrons in a conductor
$e:$ is the charge of an electron
$V_d$ is the drift velocity

Relation Between Drift Velocity and Electric Current

The drift velocity is directly proportional to the electric current.

$
I=n A e v_d
$

hence,

$
I \propto v_d
$

Factors Affecting Drift Velocity

1. Electric field
2. Charge of the electron
3. The number density of electrons
4. Temperature
5. Mobility of electron

Important Terms Related to Drift Velocity

Relaxation Time (τ):

The time interval between two successive collisions of electrons with the Positive ions.

Mean Free Path:

The path between two consecutive collisions is called the free path. The average length of these free paths is called the “Mean Free Path”.

Mobility Of An Electron:

It is the drift velocity of an electron per unit electric field. It indicates how easily an electron can pass through the material (conductor or semiconductor) in the presence of an electric field.

$\mu=\frac{v_d}{E}$

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Solved Examples Based on Drift Velocity

Example 1: Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-section is v . If the electron density in copper is $9 \times 1028 / \mathrm{m} 3$ the value of $v$ in $\mathrm{mm} / \mathrm{s}$ is close to (Take charge of an electron to be $=1.6 \times 10-19 \mathrm{C}$ )

1) 0.02

2) 3

3) 2

4) 0.2

Solution:

$\begin{gathered}I=n e A V_d \\ V_d=\frac{I}{n e A}=\frac{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}{1}=0.02 \times 10^{-3} \mathrm{~m} / \mathrm{s}=0.02 \mathrm{~mm} / \mathrm{sec}\end{gathered}$

Hence, the answer is the option (1).

Example 2: A current of 10 A exists in a wire of a cross-sectional area of 5 mm 2 with a drift velocity of $2 \times 10-3 \mathrm{~ms}-1$. The number of free electrons in each cubic meter of the wire is:

1) 1×1023
2) 625×1025
3) 2×1025
4) 2×106

Solution:

\begin{aligned}
&i=10 \mathrm{~A}, A=5 \mathrm{~mm}^2=5 \times 10^{-6} \mathrm{~m}^2 \text { and } v_d=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}\\
&\text { We know, }\\
&\begin{gathered}
i=n e A v_d \\
10=n \times 1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3} \Rightarrow n=0.625 \times 10^{28}=625 \times 10^{25}
\end{gathered}
\end{aligned}

Hence, the answer is the Option(2).

Example 3: When a 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is $2.5 \times 10^{-4} \mathrm{~ms}^{-1}$. If the electron density in the wire is $8 \times 10^{28} \mathrm{~m}^{-3}$, the resistivity of the material is close to:

1) $1.6 \times 10-8 \Omega \mathrm{~m}$
2) $1.6 \times 10-7 \Omega \mathrm{~m}$
3) $1.6 \times 10-6 \Omega \mathrm{~m}$
4) $1.6 \times 10-5 \Omega \mathrm{~m}$

Solution:

\begin{aligned}
& I=A n e v_{-} d \\
& \Rightarrow V=R A n e v_{-} d \\
& \Rightarrow V(\rho I / A)=A n e v_{-} d \\
& \Rightarrow V A / \rho I=A n e v_{-} d \\
& \Rightarrow \rho=V / I^*\left(1 / n v_{-} d\right) \\
& \Rightarrow \rho=V I /\left(n e v_{-} d\right) \\
& =50.1 \times\left(8 \times 10^{\wedge} 28\right) \times\left(1.6 \times 10^{\wedge}(-19)\right) \times\left(2.5 \times 10^{\wedge}(-4)\right) \\
& \Rightarrow \rho=1.5625 \times 10^{\wedge}(-5) \Omega \mathrm{m} \approx 1.6 \times 10^{\wedge}(-5) \Omega \mathrm{m}
\end{aligned}

Hence, the answer is the option (4).

Example 4: An electron moving in a zigzag path travels a displaces by 0.2 mm in 10 seconds. Its drift speed is (in $\mathrm{m} / \mathrm{sec}$ )
1) $2 \times 10^{-5}$
2) $10^{-5}$
3) $2 \times 10^{-4}$
4) $10^{-4}$

Solution:

Drift velocity

Drift velocity is the average velocity that a particle such as an electron attains in a material due to an electric field.

Drift Velocity = Displacement/time

Displacement $=.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}$

Time = 10 sec

Drift Velocity $V_d=2 \times 10^{-5} \mathrm{~m} / \mathrm{sec}$

Hence, the answer is option (1).

Example 5: Which of the following is correct regarding relaxation time?

1) Relaxation time increases with increase in temperature

2) Relaxation time decreases with increase in temperature

3) A decrease in relaxation time causes a decrease in resistivity

4) Conductivity is independent of Relaxation time

Solution:

Relaxation time ($\tau$)⟶ The time interval between two successive collisions of electrons with the ions/ atoms.

As with an increase in temperature drift velocity increases which will lead to an increase in the rate of collision and hence relaxation time decreases.

Hence, the answer is the option (2).