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

To determine how many electrons flow past a given cross-section of a copper wire in 10.0 minutes, given the current is 80.0 mA, we need to use the formula for current, which relates charge and time:

\( I = \frac{Q}{t} \)

where:

• \( I \) is the current (Amperes)
• \( Q \) is the total charge (Coulombs)
• \( t \) is the time (seconds)

First, convert the current from milliamperes to amperes:

\( 80.0 \text{ mA} = 0.080 \text{ A} \)

Next, convert time from minutes to seconds:

\( 10.0 \text{ minutes} = 10 \times 60 = 600 \text{ seconds} \)

Rearrange the formula to solve for \( Q \):

\( Q = I \times t \)

Substitute the values into the equation:

\( Q = 0.080 \times 600 = 48 \text{ C} \)

Next, calculate the number of electrons using the charge of a single electron, \( e = 1.602 \times 10^{-19} \text{ C} \):

\( n = \frac{Q}{e} = \frac{48}{1.602 \times 10^{-19}} \approx 3.0 \times 10^{20} \text{ electrons} \)

Therefore, the correct answer is \( 3.0 \times 10^{20} \) electrons.

Answer 2

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: \[ \text{Number of electrons} = \frac{48.0}{1.6 \times 10^{-19}} = 3.0 \times 10^{20} \, \text{electrons} \]