Question

If a copper wire carries a current of 80.0 mA, how many electrons flow past a given cross-section of the wire in 10.0 minutes?

• A. \( 0.3 \times 10^{20} \) electrons
• B. \( 3.0 \times 10^{16} \) electrons
• C. \( 9.0 \times 10^{18} \) electrons
• D. \( 3.0 \times 10^{20} \) electrons

Hint:

The number of electrons flowing through a wire can be calculated using the total charge and the charge of one electron.

Answer 1

The Correct Option is D

The charge \( Q \) passed by the current is given by: \[ Q = I \times t \] Where:
- \( I = 80.0 \, \text{mA} = 0.080 \, \text{A} \),
- \( t = 10.0 \, \text{minutes} = 600 \, \text{seconds} \). The charge is: \[ Q = 0.080 \times 600 = 48.0 \, \text{C} \] Since the charge of one electron is \( e = 1.6 \times 10^{-19} \, \text{C} \), the number of electrons is: \[ \text{Number of electrons} = \frac{48.0}{1.6 \times 10^{-19}} = 3.0 \times 10^{20} \, \text{electrons} \]