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

Answer 3

The number of electrons passing a cross-section equals the total charge that flows divided by the charge on a single electron. Checking the scale of each option against the actual charge delivered in 10 minutes tells us which one is right.

1. \( 0.3 \times 10^{20} \) electrons: The total charge over 600 seconds is \( Q = It = 0.080 \times 600 = 48 \, \text{C} \). Dividing by the electron charge, \( \frac{48}{1.602 \times 10^{-19}} \approx 3.0 \times 10^{20} \), which is ten times larger than this option, so this option is too small by a factor of ten.
2. \( 3.0 \times 10^{16} \) electrons: This is smaller than the correct value by a factor of ten thousand. A current of \( 80 \, \text{mA} \) sustained for a full ten minutes moves far more charge than this tiny electron count would represent, so this option is far too small.
3. \( 9.0 \times 10^{18} \) electrons: This would correspond to a charge of only \( 9.0 \times 10^{18} \times 1.602 \times 10^{-19} \approx 1.44 \, \text{C} \), which is much less than the \( 48 \, \text{C} \) that actually flows in 10 minutes at \( 80 \, \text{mA} \), so this option is too small.
4. \( 3.0 \times 10^{20} \) electrons: This corresponds to a charge of \( 3.0 \times 10^{20} \times 1.602 \times 10^{-19} \approx 48 \, \text{C} \), which matches the charge computed directly from \( Q = It \).

Only the count of \( 3.0 \times 10^{20} \) electrons corresponds to the \( 48 \, \text{C} \) of charge that a steady \( 80.0 \, \text{mA} \) current delivers over 10 minutes.

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