Q.(a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{-7}\ \text{m}^2$ carrying a current of $1.5\ \text{A}$. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $9.0 \times 10^{3}\ \text{kg/m}^3$, and its atomic mass is $63.5\ \text{u}$.

Concept: Drift Velocity — the small net velocity electrons acquire under an applied field, linked to current by $I = neAv_d$, so $v_d = \dfrac{I}{neA}$.

(a) Free-electron number density $n$. Each Cu atom gives one conduction electron, so

Drift speed. With $I = 1.5\ \text{A}$, $A = 1.0\times10^{-7}\ \text{m}^2$, $e = 1.6\times10^{-19}\ \text{C}$:

(b) (i) Thermal (rms) speed of Cu atoms at $\sim300\ \text{K}$ is $\approx 3.4\times10^{2}\ \text{m/s}$, about $3\times10^{5}$ times larger than $v_d$. (ii) The field propagates at $\approx 3\times10^{8}\ \text{m/s}$, about $3\times10^{11}$ times larger than $v_d$.

Principle

The current in a metal is carried by free electrons that drift with a tiny average velocity $v_d$ superimposed on their fast random thermal motion. Current and drift speed are related by

where $n$ is the free-electron number density, $e$ the electron charge, and $A$ the cross-sectional area.

(a) Drift speed

Step 1 — number density $n$. Each copper atom donates one conduction electron, so $n$ equals the atomic number density:

Computing: $\dfrac{9.0\times10^{3}}{63.5\times10^{-3}} = 1.417\times10^{5}\ \text{mol/m}^3$, and

Step 2 — substitute. With $I = 1.5\ \text{A}$, $A = 1.0\times10^{-7}\ \text{m}^2$, $e = 1.6\times10^{-19}\ \text{C}$:

(b) Comparisons

1. Thermal speed of copper atoms. From kinetic theory $\tfrac12 m v_{rms}^2 = \tfrac32 k_B T$, with atomic mass $m = \dfrac{63.5\times10^{-3}}{6.022\times10^{23}} \approx 1.05\times10^{-25}\ \text{kg}$ and $T = 300\ \text{K}$:

   $v_{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3(1.38\times10^{-23})(300)}{1.05\times10^{-25}}} \approx 3.4\times10^{2}\ \text{m/s}.$

   Thus $v_{rms}/v_d \approx 343/(1.1\times10^{-3}) \approx 3\times10^{5}$: the thermal speed exceeds the drift speed by about five orders of magnitude.
2. Field-propagation speed. The electric field that drives the drift travels along the conductor at nearly the speed of light, $c \approx 3\times10^{8}\ \text{m/s}$, so

   $\frac{c}{v_d} \approx \frac{3\times10^{8}}{1.1\times10^{-3}} \approx 3\times10^{11}.$

   The field reaches every electron almost instantly, which is why the bulb lights immediately even though each electron only crawls.
   ✓Final answer

   $v_d \approx 1.1\times10^{-3}\ \text{m/s}$; the thermal speed of Cu atoms is $\approx 3.4\times10^{2}\ \text{m/s}$ (about $3\times10^{5}$ times larger) and the field propagates at $\approx 3\times10^{8}\ \text{m/s}$ (about $3\times10^{11}$ times larger).

Method: Drift Velocity Formula from Current–Charge Relation

This method uses the fundamental relation between current, charge carrier density, and drift velocity.

Steps

Step 1: Write the drift velocity formula

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

where:

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

Rearranging for $v_d$:

Step 2: Find $n$, the number density of conduction electrons

Given: each copper atom contributes one conduction electron.

So $n$ = number of copper atoms per cubic metre.

First, find number of atoms per mole: Avogadro’s number $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$.

Mass of one mole of copper = atomic mass = $63.5\ \text{g} = 63.5 \times 10^{-3}\ \text{kg}$.

Volume of one mole of copper:

Number of atoms per cubic metre:

Step 3: Substitute into drift velocity formula

Given:

• $I = 1.5\ \text{A}$
• $A = 1.0 \times 10^{-7}\ \text{m}^2$
• $e = 1.6 \times 10^{-19}\ \text{C}$
• $n = 8.53 \times 10^{28}\ \text{m}^{-3}$

First compute denominator:

Thus:

Answer (a): $\boxed{1.1 \times 10^{-3}\ \text{m/s}}$

(b) Comparisons

1. Thermal speed of copper atoms at ordinary temperatures At room temperature ($T \approx 300\ \text{K}$), the root-mean-square speed of copper atoms is:

   $v_{th} = \sqrt{\frac{3kT}{m}}$

   where $k = 1.38 \times 10^{-23}\ \text{J/K}$ and mass of one copper atom $m = \frac{63.5 \times 10^{-3}}{6.02 \times 10^{23}} = 1.05 \times 10^{-25}\ \text{kg}$.

   $v_{th} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{1.05 \times 10^{-25}}} \approx \sqrt{1.18 \times 10^{5}} \approx 3.4 \times 10^{2}\ \text{m/s}$

   Comparison: Drift speed ($\sim 10^{-3}\ \text{m/s}$) is about $10^5$ times smaller than thermal speed ($\sim 10^2\ \text{m/s}$).
2. Speed of propagation of electric field The electric field propagates at nearly the speed of light: $c \approx 3 \times 10^8\ \text{m/s}$. Comparison: Drift speed is about $10^{11}$ times smaller than the field propagation speed.

Key Insight

The drift velocity is extremely slow — electrons move at millimetres per second — yet the electric signal travels near light speed. This is like a long pipe full of marbles: push one end, and the pulse reaches the other end almost instantly, even though each marble moves only a tiny distance.

Common Mistakes & How to Avoid Them — Drift Velocity

Mistake 1: Forgetting to convert atomic mass unit (u) to kg

The error: Students use $63.5\ \text{u}$ directly in calculations without converting to kg. Since $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$, the mass of one copper atom is:

How to avoid: Always check units — density is in $\text{kg/m}^3$, so atomic mass must be in kg for consistency. Write the conversion step explicitly.

Mistake 2: Confusing number density ($n$) with mass density ($\rho$)

The error: Using $\rho$ (density of copper) directly as $n$ (number of conduction electrons per unit volume).

Correct approach: Number density $n$ is found by:

Since each atom contributes 1 electron:

where:

• $\rho = 9.0 \times 10^3\ \text{kg/m}^3$
• $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$
• $M = 63.5 \times 10^{-3}\ \text{kg/mol}$ (molar mass in kg)

How to avoid: Remember: $n$ is number per volume, not mass per volume. Use Avogadro's number to bridge mass → number.

Mistake 3: Using wrong formula for drift velocity

The error: Writing $v_d = \frac{I}{nA}$ instead of the correct:

where $e = 1.6 \times 10^{-19}\ \text{C}$ is the electron charge.

How to avoid: Drift velocity comes from $I = n e A v_d$. Always check dimensions — current is charge per time, so charge $e$ must appear.

Mistake 4: Arithmetic errors in powers of 10

The error: Mismanaging exponents when calculating $n$ or $v_d$, especially with $10^{23}$ and $10^{-19}$.

How to avoid: Write all numbers in scientific notation before multiplying/dividing. Group powers of 10 separately:

Mistake 5: Not comparing magnitudes correctly in part (b)

The error: Giving numerical values without meaningful comparison.

Correct comparison:

• Drift speed $v_d \approx 10^{-4}\ \text{m/s}$ (very slow — like a snail)
• Thermal speed of copper atoms at 300 K: $v_{th} \approx \sqrt{\frac{3kT}{m}} \approx 10^2\ \text{m/s}$ — $10^6$ times larger
• Electric field propagation speed ≈ speed of light $3 \times 10^8\ \text{m/s}$ — $10^{12}$ times larger

How to avoid: Always express comparisons as ratios (e.g., "thermal speed is $10^6$ times drift speed"). This shows conceptual understanding.

Mistake 6: Thinking drift speed is the same as signal speed

The error: Assuming electrons move at near light speed because the bulb lights instantly.

The truth: Individual electrons drift at $\text{mm/s}$, but the electric field signal propagates at nearly $c$. It's like a hose already full of water — turning on the tap sends a pressure wave instantly, but the water itself moves slowly.

How to avoid: Distinguish clearly between:

• Drift velocity — actual motion of electrons
• Drift velocity — actual motion of electrons
• Signal velocity — speed of energy/information transfer

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