There is a current of 40 A in a wire `10^(-6)m^(2)` area of cross-section. If the number of free electrons per cubic metre is `10^(29)`, then the drift velocity is :

Options

A. `250xx10^(-3)m//sec`
B. `25.0xx10^(-3)m//sec`
C. `2.50xx10^(-3)m//sec`
D. `1.25xx10^(-3)m//sec`

Explanation

To find the drift velocity of electrons in a wire, we can use the formula that relates current (I), number of free electrons per unit volume (n), charge of an electron (e), cross-sectional area of the wire (A), and drift velocity (v_d):

\[ I = n \cdot e \cdot A \cdot v_d \]

From this equation, we can rearrange it to solve for drift velocity (v_d):

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

### Step-by-step Solution:

1. **Identify the given values:**
- Current, \( I = 40 \, \text{A} \)
- Cross-sectional area, \( A = 10^{-6} \, \text{m}^2 \)
- Number of free electrons per cubic meter, \( n = 10^{29} \, \text{m}^{-3} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

2. **Substitute the values into the drift velocity formula:**
\[ v_d = \frac{I}{n \cdot e \cdot A} \]
\[ v_d = \frac{40}{(10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6})} \]

3. **Calculate the denominator:**
\[ n \cdot e \cdot A = (10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6}) = 1.6 \times 10^{4} \, \text{C/m}^3 \]

4. **Now substitute this back into the drift velocity equation:**
\[ v_d = \frac{40}{1.6 \times 10^{4}} \]

5. **Perform the division:**
\[ v_d = 2.5 \times 10^{-3} \, \text{m/s} \]

### Final Answer:
The drift velocity \( v_d \) is \( 2.5 \times 10^{-3} \, \text{m/s} \).