Question

1.5*10²⁰ electrons flow through a cross section of a 3.70 mm diameter iron wire in 4.00 s. The electron density of iron is n = 8.5*10²⁸ m^-3.

What is the electron drift speed?

Answer 1

Concepts and reason

The concept required to solve the problem is the drift speed of the electron through a conductor.

First, calculate the area of the cross section of the conductor. Later, compute the current flowing through the conductor. Finally, use known quantities in the drift speed formula to calculate the drift speed of the electrons through the conductor.

Fundamentals

Drift speed is the average speed acquired by a particle such as an electron due to the applied electric field across the conductor.

The drift speed of electrons through a conductor is given by,

${v_d} = \frac{I}{{Ane}}$

Here, $I$ is the current, $e$ is the charge on an electron, $n$ is the number density and$A$ is the cross section of the conductor.

The drift speed of the electrons through a conductor is given by,

${v_d} = \frac{I}{{Ane}}$

Here, $I$ is the current, $e$ is the charge on an electron, $n$ is the number density and$A$ is the cross section of the conductor.

The area of cross section of the conductor will be,

$A = \pi {r^2}$ …… (1)

Here, $r$ is the radius.

The diameter of the cross section of the conductor is,

$d = 3.70{\rm{ mm}}$

The radius will be,

$r = \frac{d}{2}$

Substitute $3.70{\rm{ mm}}$ for $d$ in the above equation.

$\begin{array}{c}\\r = \frac{{3.70{\rm{ mm}}}}{2}\\\\ = 1.85{\rm{ mm}}\\\end{array}$

Substitute $1.85{\rm{ mm}}$ for $r$ in equation (1).

$\begin{array}{c}\\A = \pi {\left( {\left( {1.85{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)} \right)^2}\\\\ = 10.8 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The charge is given by,

$Q = Ne$

Here, $N$ is the number of electrons and $e$ is the charge on an electron.

Substitute $1.5 \times {10^{20}}$ for $N$ and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $e$ in the above equation.

$\begin{array}{c}\\Q = \left( {1.5 \times {{10}^{20}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\\ = 24{\rm{ C}}\\\end{array}$

The current is given by,

$I = \frac{Q}{t}$

Here, $Q$ is the charge and $t$ is the time.

Substitute $24{\rm{ C}}$ for $Q$ and $4.00{\rm{ s}}$ for $t$ in the above equation.

$\begin{array}{c}\\I = \frac{{24{\rm{ C}}}}{{4.00{\rm{ s}}}}\\\\ = 6.00{\rm{ A}}\\\end{array}$

The drift speed of the electrons through the conductor is given by,

${v_d} = \frac{I}{{Ane}}$

Substitute $6.00{\rm{ A}}$ for$I$,$10.8 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^2}$ for$A$, $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $e$ and $8.5 \times {10^{28}}{\rm{ }}{{\rm{m}}^{ - 3}}$ for $n$in the above equation.

$\begin{array}{c}\\{v_d} = \frac{{6.00{\rm{ A}}}}{{\left( {10.8 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}} \right)\left( {8.5 \times {{10}^{28}}{\rm{ }}{{\rm{m}}^{ - 3}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}\\\\ = 40.8 \times {10^{ - 6}}{\rm{ m / s }}\\\end{array}$

Ans:

The drift speed of the electron is$40.8 \times {10^{ - 6}}{\rm{ m / s}}$.