When a current \(I\) is set up in a wire of radius \(r\), the drift velocity is \(v_d\). If the same current is set up through a wire of radius \(2r\), the new drift speed will be:

Drift Velocity in Wires – Radius Variation

Question: When a current \(I\) is set up in a wire of radius \(r\), the drift velocity is \(v_d\). If the same current is set up through a wire of radius \(2r\), the new drift speed will be:

a) \(0.25\,v_d\)

b) \(0.5\,v_d\)

c) \(2\,v_d\)

d) \(4\,v_d\)

Detailed Solution:

The drift velocity in a conductor is given by:

\[ v_d = \frac{I}{n\,e\,A}, \]

where \(I\) is current, \(n\) is charge carrier density, \(e\) is charge, and \(A\) is cross-sectional area.

For a wire of radius \(r\):

\[ A = \pi r^2, \quad v_d = \frac{I}{n e \pi r^2}. \]

For a wire of radius \(2r\):

\[ A' = \pi (2r)^2 = 4\pi r^2, \quad v_d' = \frac{I}{n e\,4\pi r^2} = \frac{1}{4}\,\frac{I}{n e \pi r^2} = \frac{v_d}{4}. \]

Thus, the drift speed becomes \(0.25\,v_d\).

Correct answer: (a) \(0.25\,v_d\)