Question

An electron takes \(40\times10^3\;s\) to drift from one end of a metal wire of length \(2\;m\) to its other end. The area of cross-section of the wire is \(4\;mm^2\) and it is carrying a current of \(1.6\;A\). The number density of free electrons in the metal wire is

• A. \(8\times10^{28}\;m^{-3}\)
• B. \(6\times10^{28}\;m^{-3}\)
• C. \(4\times10^{28}\;m^{-3}\)
• D. \(5\times10^{28}\;m^{-3}\)

Hint:

For current through a conductor, use \[ I=neAv_d \] where \(n\) is number density, \(e\) is electronic charge, \(A\) is area, and \(v_d\) is drift velocity.

Answer 1

The Correct Option is D

Step 1: Find the drift speed from the transit time.
An electron crosses the length $L = 2\ \text{m}$ in time $t = 40 \times 10^{3}\ \text{s}$, so \[ v_d = \frac{L}{t} = \frac{2}{40 \times 10^{3}} = 5 \times 10^{-5}\ \text{m s}^{-1} \]
Step 2: Convert the cross-section area to SI units.
Given $A = 4\ \text{mm}^2$, we have \[ A = 4 \times 10^{-6}\ \text{m}^2 \]
Step 3: Recall the current relation.
The current carried by drifting electrons is \[ I = n e A v_d \] where $n$ is the free-electron number density.
Step 4: Rearrange for the number density.
Solving for $n$, \[ n = \frac{I}{e A v_d} \]
Step 5: Substitute the values.
Using $I = 1.6\ \text{A}$ and $e = 1.6 \times 10^{-19}\ \text{C}$, \[ n = \frac{1.6}{(1.6 \times 10^{-19})(4 \times 10^{-6})(5 \times 10^{-5})} \]
Step 6: Evaluate the result.
The denominator equals $1.6 \times 10^{-19} \times 2 \times 10^{-10} = 3.2 \times 10^{-29}$, so \[ n = \frac{1.6}{3.2 \times 10^{-29}} = 5 \times 10^{28}\ \text{m}^{-3} \] giving \[ \boxed{5 \times 10^{28}\ \text{m}^{-3}} \]