Question

A uniform wire of area of cross section $1 \times 10^{-7} m^2$ carries a current of $1.6 \text{ A}$. If the number density of electrons is $5 \times 10^{28} m^{-3}$, the drift velocity of electrons (in $\text{mm s}^{-1}$) is

• A. 1
• B. 3
• C. 2
• D. 4

Hint:

Always look for cancellations before computing large powers. Here, the $1.6$ in current and $1.6$ in the electron charge cancel out immediately, simplifying the arithmetic.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field. It relates macroscopic current to microscopic particle movement.
Step 2: Key Formula or Approach:
The formula connecting current and drift velocity is:
\[ I = nAev_d \] Rearrange to solve for drift velocity \(v_d\):
\[ v_d = \frac{I}{nAe} \] Where \(e \approx 1.6 \times 10^{-19} \text{ C}\) is the elementary charge.
Step 3: Detailed Explanation:
Given values:
Current \(I = 1.6 \text{ A}\)
Cross-sectional area \(A = 1 \times 10^{-7} \text{ m}^2\)
Number density \(n = 5 \times 10^{28} \text{ m}^{-3}\)
Elementary charge \(e = 1.6 \times 10^{-19} \text{ C}\)
Substitute into the formula:
\[ v_d = \frac{1.6}{(5 \times 10^{28}) \times (1 \times 10^{-7}) \times (1.6 \times 10^{-19})} \] Cancel the 1.6 from numerator and denominator:
\[ v_d = \frac{1}{5 \times 10^{28} \times 10^{-7} \times 10^{-19}} \] Combine the powers of 10:
\[ 10^{28 - 7 - 19} = 10^2 \] \[ v_d = \frac{1}{5 \times 10^2} = \frac{1}{500} \text{ m/s} \] Convert from meters per second to millimeters per second (multiply by 1000):
\[ v_d = \frac{1}{500} \times 1000 \text{ mm/s} = 2 \text{ mm/s} \] Step 4: Final Answer:
The drift velocity is 2 mm s\(^{-1}\).