A wire carrying a current of 16A and its cross-sectional area `10^(-5 )m^2`. If the free electron density in wire is `4 xx 10^(28) m^(-3)`, the drift velocity is ----x`10^-4`(in m/s)

Options

A. `1.6`
B. `2.5`
C. `6.4`
D. `3.2`

Answer

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

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

Where:
- \( I \) is the current (in Amperes),
- \( n \) is the free electron density (in \( m^{-3} \)),
- \( A \) is the cross-sectional area (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: Rearranging the formula to solve for drift velocity \( v_d \)

We can rearrange the formula to isolate \( v_d \):

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

### Step 2: Plugging in the known values

Given:
- Current \( I = 16 \, A \)
- Cross-sectional area \( A = 10^{-5} \, m^2 \)
- Free electron density \( n = 4 \times 10^{28} \, m^{-3} \)
- Charge of an electron \( e = 1.6 \times 10^{-19} \, C \)

Now substitute these values into the equation:

\[ v_d = \frac{16}{(4 \times 10^{28}) \cdot (10^{-5}) \cdot (1.6 \times 10^{-19})} \]

### Step 3: Calculating the denominator

First, calculate the product in the denominator:

\[ n \cdot A \cdot e = (4 \times 10^{28}) \cdot (10^{-5}) \cdot (1.6 \times 10^{-19}) \]

Calculating it step by step:

1. \( 4 \times 10^{28} \cdot 10^{-5} = 4 \times 10^{23} \)
2. \( 4 \times 10^{23} \cdot 1.6 \times 10^{-19} = 6.4 \times 10^{4} \)

So, the denominator is \( 6.4 \times 10^{4} \).

### Step 4: Completing the calculation

Now substitute back into the equation for \( v_d \):

\[ v_d = \frac{16}{6.4 \times 10^{4}} = \frac{16}{64000} = 0.00025 \, m/s \]

### Step 5: Converting to the required format

To express \( v_d \) in the form of \( x \times 10^{-4} \):

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

### Final Answer

Thus, the drift velocity \( v_d \) is:

\[ \boxed{2.5} \]