Question

When a current of \(3\text{ A}\) flows through a uniform copper wire of radius \(0.6\text{ mm}\), the average drift speed of the electrons is \(V\). If \(1.5\text{ A}\) current is flowing in another uniform copper wire of radius \(1.2\text{ mm}\), then the average drift speed of the electrons is:

• A. \(V\)
• B. \(\frac{V}{2}\)
• C. \(\frac{V}{4}\)
• D. \(\frac{V}{8}\)

Hint:

For the same material, drift velocity is directly proportional to current and inversely proportional to the square of the radius of the wire.

Answer 1

The Correct Option is D

Concept: Drift velocity is related to current by \[ I=nAe\,v_d \] where \(n\) is number density of free electrons, \(A\) is cross-sectional area, \(e\) is electronic charge and \(v_d\) is drift velocity. For the same material, \[ v_d \propto \frac{I}{A} \] Since \[ A=\pi r^2 \] we have \[ v_d \propto \frac{I}{r^2} \]

Step 1: Write the ratio of drift velocities. For the first wire, \[ I_1=3\text{ A}, \qquad r_1=0.6\text{ mm} \] For the second wire, \[ I_2=1.5\text{ A}, \qquad r_2=1.2\text{ mm} \] Therefore, \[ \frac{v_2}{v_1} = \frac{I_2}{I_1} \cdot \frac{r_1^2}{r_2^2} \] \[ = \frac{1.5}{3} \cdot \left(\frac{0.6}{1.2}\right)^2 \] \[ = \frac12 \times \frac14 \] \[ = \frac18 \]

Step 2: Obtain the required drift velocity. Since \[ v_1=V \] \[ v_2=\frac{V}{8} \] \[ \boxed{\frac{V}{8}} \]