The number of free electrons per 100 mm of ordinary copper wire is `2xx10^(21)`. The average drift speed of electorn is `0.25 mm//s`. What is the current flowing?

Options

A. 0.8A
B. 8A
C. 80A
D. 5A

Explanation

To solve the problem, we need to find the current flowing through a copper wire given the number of free electrons and the average drift speed of the electrons. We can use the formula for current in terms of drift velocity, number of charge carriers, and the charge of the carriers.

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of free electrons per 100 mm of copper wire, \( n = 2 \times 10^{21} \).
- Drift speed of electrons, \( v_d = 0.25 \, \text{mm/s} = 0.25 \times 10^{-3} \, \text{m/s} \).
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \).
- Length of the wire segment considered, \( L = 100 \, \text{mm} = 0.1 \, \text{m} \).

2. **Calculate the number of free electrons per meter**:
\[ n_{\text{per meter}} = \frac{n}{L} = \frac{2 \times 10^{21}}{0.1} = 2 \times 10^{22} \, \text{electrons/m} \]

3. **Use the formula for current**:
The current \( I \) can be calculated using the formula:
\[ I = n \cdot A \cdot e \cdot v_d \]
where \( A \) is the cross-sectional area of the wire. However, since we don't have the area, we can express the current in terms of the number of electrons per unit length.

4. **Substituting the values**:
Since we are looking for the current without the area, we can express it as:
\[ I = n_{\text{per meter}} \cdot e \cdot v_d \]
Substituting the values:
\[ I = (2 \times 10^{22}) \cdot (1.6 \times 10^{-19}) \cdot (0.25 \times 10^{-3}) \]

5. **Perform the calculations**:
\[ I = 2 \times 10^{22} \cdot 1.6 \times 10^{-19} \cdot 0.25 \times 10^{-3} \]
\[ I = 2 \times 1.6 \times 0.25 \times 10^{22 - 19 - 3} \]
\[ I = 0.8 \times 10^{0} = 0.8 \, \text{A} \]

6. **Final answer**:
The current flowing through the copper wire is \( I = 0.8 \, \text{A} \).