Question

What will be the effect on the drift velocity of electrons, when the current in a copper wire is passed in a wire, twice of its radius (the same current being maintained)?

Hint:

Use \(v_d = I/(neA)\) with \(A=\pi r^2\); at constant current \(v_d \propto 1/r^2\), so doubling the radius quarters the drift velocity.

Answer 1

Step 1: Start from the microscopic current formula.
For a wire carrying current \(I\), the drift speed of the conduction electrons is
\[ v_d = \frac{I}{n e A} \]
Step 2: Identify what stays fixed and what changes.
The wire is still copper, so \(n\) and \(e\) do not change, and the problem keeps the current \(I\) the same. Only the cross-sectional area \(A=\pi r^2\) changes because the radius changes.

Step 3: Take a ratio of the two situations.
Let the original radius be \(r\) and the new radius be \(2r\). Writing the ratio of drift speeds:
\[ \frac{v_d'}{v_d} = \frac{A}{A'} = \frac{\pi r^2}{\pi (2r)^2} = \frac{r^2}{4r^2} = \frac{1}{4} \]
Step 4: Interpret the result.
Because the same number of electrons per second (same current) now has four times the area to spread across, each electron needs to move only a quarter as fast.
\[ v_d' = \frac{v_d}{4} \]
\[\boxed{v_d' = \tfrac{1}{4}\,v_d}\]