Question

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.

What is the density of free electrons in the metal?

Express your answer numerically in m−3 to two significant figures.

Answer 1

Concepts and reason

The concept required to solve this problem is drift velocity and the density of free electrons.

Initially, calculate the area of cross section of the metallic wire. Finally, calculate the density of the free electrons in the metal.

Fundamentals

Therefore, the cross-sectional area in terms of the diameter is written as follows:

$A = \frac{{\pi {d^2}}}{4}$

The expression for the density of the free electrons is as follows:

$n = \frac{i}{{{v_{\rm{d}}}eA}}$

Here, i is the current, e is the charge on an electron, A is the area of cross section, and ${v_{\rm{d}}}$ is the drift velocity.

The relation between the radius and the diameter is,

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

Substitute 4.12 mm for d.

$\begin{array}{c}\\r = \frac{{{\rm{4}}{\rm{.22 mm}}}}{2}\\\\ = \left( {\frac{{{\rm{4}}{\rm{.2 mm}}}}{2}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1.0{\rm{ mm}}}}} \right)\\\\ = 0.0021{\rm{ m}}\\\end{array}$

Therefore, the radius of the circular track is equal to 0.0021 m.

The area can be expressed in terms of radius as follows:

$A = \pi {r^2}$

Substitute 0.0021 m for r and $3.14$ for$\pi$.

$\begin{array}{c}\\A = \left( {3.14} \right){\left( {0.0021{\rm{ m}}} \right)^2}\\\\ = 0.000014{\rm{ }}{{\rm{m}}^2}\\\end{array}$

Therefore, the cross-sectional area is equal to$0.000014{\rm{ }}{{\rm{m}}^2}$.

Substitute 8.00 A for i, $5.40 \times {10^{ - 5}}{\rm{ m/s}}$ for ${v_{\rm{d}}}$, $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e, and $0.000014{\rm{ }}{{\rm{m}}^2}$ for A in the equation$n = \frac{i}{{{v_{\rm{d}}}eA}}$.

$\begin{array}{c}\\n = \frac{{8.00{\rm{ A }}}}{{\left( {5.40 \times {{10}^{ - 5}}{\rm{ m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {0.000014{\rm{ }}{{\rm{m}}^2}} \right)}}\\\\ = 6.95 \times {10^{28}}{\rm{ /}}{{\rm{m}}^3}\\\end{array}$

Ans:

The density of the free electrons in the metal is equal to$6.95 \times {10^{28}}{\rm{ /}}{{\rm{m}}^3}$.