There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is `1 mm^(2)`. If the number of free electrons per `cm^(2)` is `8.4 xx 10^(22)` then the drift velocity would be

Options

A. `1.0 mm//sec`
B. `1.0 m//sec`
C. `0.1 mm//sec`
D. `0.01 mm//sec`

Explanation

To find the drift velocity of electrons in a copper wire, we can use the formula for drift velocity:

\[ v_d = \frac{I}{n \cdot A \cdot e} \]

where:
- \( v_d \) = drift velocity (in cm/s)
- \( I \) = current (in amperes)
- \( n \) = number of free electrons per unit volume (in electrons/cm³)
- \( A \) = area of cross-section of the wire (in cm²)
- \( e \) = charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs)

### Step-by-Step Solution:

1. **Identify the given values:**
- Current, \( I = 1.344 \, \text{A} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{cm}^2 \)
- Number of free electrons per cm³, \( n = 8.4 \times 10^{22} \, \text{electrons/cm}^3 \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

2. **Convert the area from mm² to cm²:**
\[ A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{cm}^2 \]

3. **Substitute the values into the drift velocity formula:**
\[ v_d = \frac{1.344}{(8.4 \times 10^{22}) \cdot (1 \times 10^{-6}) \cdot (1.6 \times 10^{-19})} \]

4. **Calculate the denominator:**
- First, calculate \( n \cdot A \cdot e \):
\[ n \cdot A \cdot e = (8.4 \times 10^{22}) \cdot (1 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) = 1.344 \times 10^{7} \]

5. **Now calculate the drift velocity:**
\[ v_d = \frac{1.344}{1.344 \times 10^{7}} = 0.01 \, \text{cm/s} \]

6. **Convert the drift velocity to mm/s:**
\[ v_d = 0.01 \, \text{cm/s} = 0.1 \, \text{mm/s} \]

### Final Answer:
The drift velocity is \( 0.1 \, \text{mm/s} \).