Suppose you have a two wires. Each has a current of $5 \text{ A}$. One is made of copper (Cu) and has radius $0.5 \text{ mm}$. The other is made of zinc (Zn) and has radius $0.1 \text{ mm}$. What is the drift speed of electrons in each wire? You may want to consult the table below.

• The copper wire has $I=5 \text{ A}$, $r = 0.5 \text{ mm}$.
• The zinc wire has $I=5 \text{ A}$, $r = 0.1 \text{ mm}$.
• The charge of an electron is $q=-1.6\cdot 10^{-19} \text{ C}$.

• Drift speed for both wires.
• Electron charge density for both wires.
• Electron current for both wires.
• Cross-sectional area for both wires.

• The table is accurate for our wires.
• The wires have circular cross-sections.
• The wires do not experience any external electric field.

• We represent electron current as $i=nAv_{avg}$.
• We represent current as $I=qi$. Current is charge per second. Electron current is electrons per second. We multiply by $q$ (the electron charge) to get charge per second.

We can look up electron density $n$ in the table. It is labeled as “density of valency electrons”, which to us just means the density of free electrons, or electrons that are free to drift through the wire. For copper, we get $n_{\text{Cu}}=8.47\cdot 10^{22} \text{ cm}^{-3}$. For zinc, we get $n_{\text{Zn}}=13.2\cdot 10^{22} \text{ cm}^{-3}$.

To find the cross-sectional area of the wire, we just use the area of a circle. We know the radius, so this should be easy: $A=\pi r^2$.

We are given current, and we can solve for electron current using the charge of an electron: $i = \frac{I}{q}$.

We now have enough information to solve for the drift speed of electrons. We use positive numbers below, since we care only about speed for now, not direction.

Current ($I$), radius ($r$), electron density ($n$), and electron charge ($q$) are all things we know for our two wires. When we plug in the numbers, we get the following: