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: Relationship between Drift Velocity and Current
The formula connecting current and drift velocity is: \[ I = n A e v_d \] Where:
\( I = 10 \, \text{A} \) (Current),
\( A = 5 \times 10^{-6} \, \text{m}^2 \) (Cross
sectional Area),
\( e = 1.6 \times 10^{-19} \, \text{C} \) (Elementary Charge),
\( v_d = 2 \times 10^{-3} \, \text{m/s} \) (Drift Velocity),
\( n \) represents the density of free electrons (number of free electrons per unit volume).
Step 2: Calculation of Electron Density (\( n \))
Rearranging the formula to solve for \( n \): \[ n = \frac{I}{A e v_d} \] Substituting the provided values: \[ n = \frac{10}{(5 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) \cdot (2 \times 10^{-3})} \]
Step 3: Simplification of the Expression
Calculating the denominator: \[ (5 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) = 8 \times 10^{-25} \] \[ (8 \times 10^{-25}) \cdot (2 \times 10^{-3}) = 16 \times 10^{-28} \] The expression for \( n \) becomes: \[ n = \frac{10}{16 \times 10^{-28}} \] Performing the division and exponentiation: \[ n = \frac{10}{16} \times 10^{28} = 0.625 \times 10^{28} = 6.25 \times 10^{27} \]Final Result: The number of free electrons per cubic meter of the wire is \( 6.25 \times 10^{27} \).