A current 0.5 amperes flows in a conductor of cross-sectional area of `10^(-2) m^2`. If the electron density is `0.3 * 10^(28) m^(-3)` , then the drift velocity of free electrons is

Options

A. `2.1 * 10^(-5) ms^(-1)`
B. `2.5 * 10^(-3) ms^(-1)`
C. `1.5 * 10^(-6) ms^(-1)`
D. `1.04 * 10^(-7) ms^(-1)`

Answer

To find the drift velocity of free electrons in the conductor, we can use the formula:

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

Where:
- \( I \) is the current (in amperes),
- \( n \) is the electron density (in electrons per cubic meter),
- \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs),
- \( A \) is the cross-sectional area of the conductor (in square meters),
- \( v_d \) is the drift velocity (in meters per second).

Given:
- \( I = 0.5 \, \text{A} \)
- \( A = 10^{-2} \, \text{m}^2 \)
- \( n = 0.3 \times 10^{28} \, \text{m}^{-3} \)

### Step 1: Rearranging the Formula
We need to solve for the drift velocity \( v_d \). Rearranging the formula gives:

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

### Step 2: Substituting the Values
Now, we can substitute the known values into the equation:

1. Current \( I = 0.5 \, \text{A} \)
2. Electron density \( n = 0.3 \times 10^{28} \, \text{m}^{-3} \)
3. Charge of an electron \( e = 1.6 \times 10^{-19} \, \text{C} \)
4. Cross-sectional area \( A = 10^{-2} \, \text{m}^2 \)

Substituting these values into the equation for \( v_d \):

\[ v_d = \frac{0.5}{(0.3 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-2})} \]

### Step 3: Calculating the Denominator
Now, we calculate the denominator:

1. Calculate \( n \cdot e \cdot A \):

\[ n \cdot e \cdot A = (0.3 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-2}) \]

Calculating step by step:

- \( 0.3 \times 1.6 = 0.48 \)
- \( 0.48 \times 10^{28} \times 10^{-2} = 0.48 \times 10^{26} \)
- \( 0.48 \times 10^{26} = 4.8 \times 10^{25} \)

### Step 4: Final Calculation for Drift Velocity
Now substitute this back into the equation for \( v_d \):

\[ v_d = \frac{0.5}{4.8 \times 10^{25}} \]

Calculating this gives:

\[ v_d = \frac{0.5}{4.8} \times 10^{-25} = 0.1041667 \times 10^{-25} \approx 1.04 \times 10^{-26} \, \text{m/s} \]

### Final Answer
Thus, the drift velocity \( v_d \) is approximately:

\[ v_d \approx 1.04 \times 10^{-7} \, \text{m/s} \]