A current of 10 A is mainatained in a conductor of cross - section `1 cm^(2)` . If the free electron density in the conductor is ` 9 xx 10^(28) m^(-3)`, then drift velocity of free electrons is .

Options

A. ` 6.94 xx 10^(-6) m//s`
B. `5.94 xx 10 ^(-2) m//s`
C. `1.94 xx 10^(-3) m//s`
D. `2.94 xx 10^(-4) m//s`

Answer

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

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

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

### Step-by-step Solution:

1. **Identify the given values**:
- Current, \( I = 10 \, \text{A} \)
- Electron density, \( N = 9 \times 10^{28} \, \text{m}^{-3} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Cross-sectional area, \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \))

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

3. **Substitute the values into the equation**:
\[ V_D = \frac{10}{(9 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (1 \times 10^{-4})} \]

4. **Calculate the denominator**:
- First, calculate \( N \cdot e \cdot A \):
\[ N \cdot e \cdot A = (9 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (1 \times 10^{-4}) \]
- Calculate \( 9 \times 1.6 = 14.4 \)
- Therefore:
\[ N \cdot e \cdot A = 14.4 \times 10^{28 - 19 - 4} = 14.4 \times 10^{5} = 1.44 \times 10^{6} \]

5. **Now substitute back to find \( V_D \)**:
\[ V_D = \frac{10}{1.44 \times 10^{6}} = 6.94 \times 10^{-6} \, \text{m/s} \]

6. **Final Answer**:
The drift velocity of the free electrons is:
\[ V_D \approx 6.94 \times 10^{-6} \, \text{m/s} \]