Q.The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5 \times 10^{28}\ \text{m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0\ \text{m}$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6}\ \text{m}^2$ and it is carrying a current of $3.0\ \text{A}$.

Concept: Drift Velocity — relates current to the average velocity of charge carriers: $v_d = \dfrac{I}{neA}$.

Given $I=3.0\,\text{A}$, $n=8.5\times10^{28}\,\text{m}^{-3}$, $e=1.6\times10^{-19}\,\text{C}$, $A=2.0\times10^{-6}\,\text{m}^2$, $L=3.0\,\text{m}$:

$v_d = \frac{3.0}{(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})} = \frac{3.0}{2.72\times10^{4}} \approx 1.10\times10^{-4}\,\text{m/s}$ …

Why drift velocity is the key

A current is carried by the net drift of free electrons superimposed on their much faster random thermal motion. The relation linking current to that drift speed is

Once $v_d$ is known, the time to cross the wire's length $L$ is simply $t = L/v_d$.

Step-by-step solution

1. Known quantities

• $n = 8.5\times10^{28}\,\text{m}^{-3}$
• $L = 3.0\,\text{m}$
• $A = 2.0\times10^{-6}\,\text{m}^2$
• $I = 3.0\,\text{A}$
• $e = 1.6\times10^{-19}\,\text{C}$

2. Drift velocity

Computing the denominator: $ne = (8.5\times10^{28})(1.6\times10^{-19}) = 1.36\times10^{10}\,\text{C/m}^3$, and $neA = (1.36\times10^{10})(2.0\times10^{-6}) = 2.72\times10^{4}\,\text{C/(m\cdot s)}$. So

3. Drift time …

Method: Drift Velocity Formula

This problem uses the drift velocity relation that connects current, charge carrier density, cross-sectional area, and drift speed.

Step-by-step solution

Step 1: Recall the formula for current in terms of drift velocity

The current $I$ in a conductor is given by:

where:

• $n$ = number density of free electrons ($8.5 \times 10^{28}\ \text{m}^{-3}$)
• $e$ = charge of an electron ($1.6 \times 10^{-19}\ \text{C}$)
• $A$ = cross-sectional area ($2.0 \times 10^{-6}\ \text{m}^2$)
• $v_d$ = drift velocity of electrons

Step 2: Solve for drift velocity $v_d$

Rearranging:

Substitute the values:

Step 3: Find the time to drift the given length

Time $t$ is distance divided by drift velocity:

$t = 2.72 \times 10^{4}\ \text{s}$ …

Common Mistakes Students Make on Drift Velocity Problems

Mistake 1: Confusing Drift Speed with Actual Electron Speed

The error: Students often think electrons zoom through wires at near light speed. They calculate a tiny drift velocity and panic, thinking something is wrong.

Why it happens: The signal speed (≈ $3 \times 10^8\ \text{m/s}$) is confused with drift speed (≈ $10^{-4}\ \text{m/s}$). Electrons actually drift very slowly — like a snail's pace — but the electric field propagates almost instantly.

How to avoid: Remember:

• Drift velocity ($v_d$) = net average velocity of electrons under an electric field
• Signal speed = speed at which current starts flowing (near light speed)
• A slow drift velocity is correct — expect answers in $\text{mm/s}$ or $\mu\text{m/s}$

Mistake 2: Using Wrong Formula or Misplacing Variables

The error: Students write $I = n e A v_d$ but solve for the wrong quantity, or forget that $n$ is number density (not number of electrons).

Correct formula:

where:

• $I$ = current (A)
• $n$ = number density ($\text{m}^{-3}$)
• $e$ = charge of electron ($1.6 \times 10^{-19}\ \text{C}$)
• $A$ = cross-sectional area ($\text{m}^2$)
• $v_d$ = drift velocity ($\text{m/s}$)

How to avoid: Write the formula before plugging numbers. Solve for $v_d$ explicitly:

Mistake 3: Forgetting to Convert Units

The error: Using area in $\text{cm}^2$ or length in km without converting to SI units.

Example: $2.0 \times 10^{-6}\ \text{m}^2$ is already in SI — but if given as $2.0\ \text{mm}^2$, students forget $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$.

How to avoid: Always convert to metres, seconds, amperes before calculation. Write units beside every number.

Mistake 4: Stopping at Drift Velocity Instead of Finding Time

The error: The question asks: "How long does an electron take to drift from one end to the other?" Students calculate $v_d$ and stop.

What's needed: After finding $v_d$, use:

where $L = 3.0\ \text{m}$.

How to avoid: Read the question twice. Underline what is being asked — here it's time, not velocity.

Mistake 5: Arithmetic Errors with Powers of 10

The error: Mismanaging exponents when dividing:

Students often add/subtract exponents incorrectly. …