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:
Current is the flow of charge. In a conductor, it is related to the drift velocity of free electrons, their number density, the cross-sectional area, and the elementary charge of an electron.
Key Formula or Approach:
The relation between current ($I$) and drift velocity ($v_d$) is:
\[ I = n e A v_d \]
Where $e$ is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)).

Step 2: Detailed Explanation:
Given:
\( I = 1.6 \text{ A} \)
\( A = 1 \times 10^{-7} \text{ m}^2 \)
\( n = 5 \times 10^{28} \text{ m}^{-3} \)
\( e = 1.6 \times 10^{-19} \text{ C} \)
Substitute into the formula:
\[ 1.6 = (5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-7}) \times v_d \]
Divide both sides by 1.6:
\[ 1 = 5 \times 10^{28 - 19 - 7} \times v_d \]
\[ 1 = 5 \times 10^2 \times v_d \]
\[ 1 = 500 v_d \]
\[ v_d = \frac{1}{500} \text{ m/s} = 0.002 \text{ m/s} \]
To convert m/s to mm/s, multiply by 1000:
\[ v_d = 0.002 \times 1000 = 2 \text{ mm/s} \]

Step 3: Final Answer:
The drift velocity of electrons is 2 mm/s.