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: Write the relation between current and drift velocity.
The current in a conductor is related to the drift velocity of its free electrons by
\[ I = n\,e\,A\,v_d \]where \(n\) = free-electron number density, \(e\) = electronic charge, \(A\) = cross-sectional area and \(v_d\) = drift velocity.

Step 2: Solve for drift velocity.
\[ v_d = \frac{I}{n\,e\,A} \]For the same wire material and the same current \(I\), the quantities \(n\), \(e\) and \(I\) are unchanged, so \(v_d \propto \dfrac{1}{A}\).

Step 3: Bring in the area of a circular cross-section.
\(A = \pi r^2\), therefore \(v_d \propto \dfrac{1}{r^2}\).

Step 4: Double the radius.
If the radius becomes \(r' = 2r\), the new area is
\[ A' = \pi (2r)^2 = 4\pi r^2 = 4A \]so the new drift velocity is
\[ v_d' = \frac{I}{n\,e\,(4A)} = \frac{v_d}{4} \]
Conclusion: The drift velocity of the electrons falls to one-fourth of its original value.
\[\boxed{v_d' = \tfrac{1}{4}\,v_d}\]