There is a current of 40 ampere in a wire of `10^(-6)m^(2)` are of cross-section. If the number of free electron per `m^(3)` is `10^(29)` then the drift velocity will be

Options

A. `1.25 xx 10^(3)` m/s
B. `2.5 xx 10^(-3)` m/s
C. `25 xx 10^(-3)` m/s
D. `25 xx 10^(3)` m/s

Explanation

To find the drift velocity (V_d) in the given wire, we can use the formula that relates current (I), electron density (N), charge of an electron (e), cross-sectional area (A), and drift velocity (V_d):

\[ I = N \cdot e \cdot A \cdot V_d \]

Where:
- \( I \) = current in amperes (A)
- \( N \) = number of free electrons per cubic meter (m\(^{-3}\))
- \( e \) = charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs)
- \( A \) = cross-sectional area of the wire in square meters (m\(^{2}\))
- \( V_d \) = drift velocity in meters per second (m/s)

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Current, \( I = 40 \, \text{A} \)
- Cross-sectional area, \( A = 10^{-6} \, \text{m}^2 \)
- Electron density, \( N = 10^{29} \, \text{m}^{-3} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

**Step 2: Rearrange the formula to solve for drift velocity (V_d).**
\[ V_d = \frac{I}{N \cdot e \cdot A} \]

**Step 3: Substitute the known values into the equation.**
\[ V_d = \frac{40}{(10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6})} \]

**Step 4: Calculate the denominator.**
- First, calculate \( N \cdot e \cdot A \):
\[ N \cdot e \cdot A = (10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6}) \]
\[ = 10^{29} \cdot 1.6 \times 10^{-25} \]
\[ = 1.6 \times 10^{4} \]

**Step 5: Now calculate the drift velocity (V_d).**
\[ V_d = \frac{40}{1.6 \times 10^{4}} \]
\[ = 2.5 \times 10^{-3} \, \text{m/s} \]

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

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