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. `2.50 xx 10^(-3)m//s`
C. `25.0 xx 10^(-3)m//s`
D. `250 xx 10^(-3)m//s`

Answer

To find the drift velocity of the electrons in the wire, we can use the formula for current (I) in terms of drift velocity (v_d):

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

Where:
- \( I \) is the current (in Amperes),
- \( n \) is the number of free electrons per cubic meter (in \( m^{-3} \)),
- \( A \) is the cross-sectional area of the wire (in \( m^{2} \)),
- \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) Coulombs),
- \( v_d \) is the drift velocity (in \( m/s \)).

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

### Step 2: Rearrange the formula to solve for drift velocity
We need to isolate \( v_d \) in the formula:

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

### Step 3: Substitute the known values into the equation
Now, we can substitute the values into the equation:

\[ v_d = \frac{40}{(10^{29}) \cdot (10^{-6}) \cdot (1.6 \times 10^{-19})} \]

### Step 4: Calculate the denominator
First, calculate the denominator:

\[ n \cdot A \cdot e = (10^{29}) \cdot (10^{-6}) \cdot (1.6 \times 10^{-19}) \]
\[ = 10^{29 - 6} \cdot 1.6 \times 10^{-19} = 10^{23} \cdot 1.6 \times 10^{-19} \]
\[ = 1.6 \times 10^{4} \, C \]

### Step 5: Calculate the drift velocity
Now substitute back into the drift velocity formula:

\[ v_d = \frac{40}{1.6 \times 10^{4}} = 2.5 \times 10^{-3} \, m/s \]

### Final Answer
The drift velocity \( v_d \) is:

\[ v_d = 2.5 \times 10^{-3} \, m/s \]