Question

A current of 10 A exists in a wire of cross-sectional area of \( 5 \times 10^{-6} \, \text{m}^2 \) with a drift velocity of \( 2 \times 10^{-3} \, \text{m/s} \). The number of free electrons in each cubic meter of the wire is:

• A. \( 2 \times 10^{25} \)
• B. \( 2 \times 10^{23} \)
• C. \( 625 \times 10^{25} \)
• D. \( 2 \times 10^{6} \)

Hint:

Drift velocity is extremely small, and the number of free electrons in a conductor is very large. Use the equation \( I = n A e v_d \) to determine electron density.

Answer 1

The Correct Option is C

Step 1: Drift Velocity and Current Relation
The formula for current is: \[ I = n A e v_d \] where:
\( I = 10 \, \text{A} \) (current),
\( A = 5 \times 10^{-6} \, \text{m}^2 \) (cross
sectional area),
\( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron),
\( v_d = 2 \times 10^{-3} \, \text{m/s} \) (drift velocity),
\( n \) is the number of free electrons per unit volume.
Step 2: Solving for \( n \)
Rearranging the equation for \( n \): \[ n = \frac{I}{A e v_d} \] Substituting the given values: \[ n = \frac{10}{(5 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) \cdot (2 \times 10^{-3})} \]
Step 3: Simplifying
Evaluating the denominator: \[ (5 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) = 8 \times 10^{-25} \] \[ (8 \times 10^{-25}) \cdot (2 \times 10^{-3}) = 16 \times 10^{-28} \] \[ n = \frac{10}{16 \times 10^{-28}} \] \[ n = \frac{10}{16} \times 10^{28} = 0.625 \times 10^{28} = 625 \times 10^{25} \] Final Answer: The number of free electrons in each cubic meter of the wire is \( 625 \times 10^{25} \).