An electric current of 16 A exists in a metal wire of cross section `10^(-6) m^(2)` and length 1 m. Assuming one free electron per atom. The drift speed of the free electrons in the wire will be (Density of metal = `5 xx 10^(3) kg//m^(3)` atomic weight = 60)

Options

A. `5 xx 10^(-3) m//s`
B. `2 xx 10^(-3) m//s`
C. `4 xx10^(-3) m//s`
D. `7.5 xx 10^(-3) m//s`

Answer

To find the drift speed of the free electrons in the wire, we can use the formula that relates current (I), electron density (N), charge of an electron (E), cross-sectional area (A), and drift speed (Vd):

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

### Step 1: Calculate the Electron Density (N)

Given:
- Density of the metal, \( \rho = 5 \times 10^3 \, \text{kg/m}^3 \)
- Atomic weight (molar mass), \( M = 60 \, \text{g/mol} = 60 \times 10^{-3} \, \text{kg/mol} \)
- Avogadro's number, \( N_A = 6.02 \times 10^{23} \, \text{atoms/mol} \)

To find the number of atoms per unit volume (N), we can use the formula:

\[ N = \frac{\text{Density}}{\text{Atomic Weight}} \times N_A \]

Substituting the values:

\[ N = \frac{5 \times 10^3 \, \text{kg/m}^3}{60 \times 10^{-3} \, \text{kg/mol}} \times 6.02 \times 10^{23} \, \text{atoms/mol} \]

Calculating:

\[ N = \frac{5 \times 10^3}{60 \times 10^{-3}} \times 6.02 \times 10^{23} \]

\[ N = \frac{5 \times 10^3}{0.06} \times 6.02 \times 10^{23} \]

\[ N = 83333.33 \times 6.02 \times 10^{23} \]

\[ N \approx 5.02 \times 10^{28} \, \text{electrons/m}^3 \]

### Step 2: Use the Current to Find Drift Speed (Vd)

Given:
- Current, \( I = 16 \, \text{A} \)
- Charge of an electron, \( E = 1.6 \times 10^{-19} \, \text{C} \)
- Cross-sectional area, \( A = 10^{-6} \, \text{m}^2 \)

Rearranging the formula for drift speed:

\[ V_d = \frac{I}{N \cdot E \cdot A} \]

Substituting the values:

\[ V_d = \frac{16}{5.02 \times 10^{28} \cdot 1.6 \times 10^{-19} \cdot 10^{-6}} \]

Calculating:

\[ V_d = \frac{16}{5.02 \times 1.6 \times 10^{3}} \]

\[ V_d = \frac{16}{8.032 \times 10^{3}} \]

\[ V_d \approx 1.99 \times 10^{-3} \, \text{m/s} \]

### Final Answer

Thus, the drift speed of the free electrons in the wire is approximately:

\[ V_d \approx 2 \times 10^{-3} \, \text{m/s} \]