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 free electrons in a metal wire, we can use the formula for current in terms of drift speed:

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

Where:
- \( I \) is the current (in Amperes)
- \( n \) is the number of charge carriers per unit volume (in \( m^{-3} \))
- \( A \) is the cross-sectional area of the wire (in \( m^{2} \))
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, C \))
- \( v_d \) is the drift speed (in \( m/s \))

### Step 1: Calculate the number of free electrons per unit volume, \( n \)

The number of free electrons per unit volume can be calculated using the formula:

\[ n = \frac{\rho \cdot N_A}{M} \]

Where:
- \( \rho \) is the density of the metal (in \( kg/m^{3} \))
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \, mol^{-1} \))
- \( M \) is the atomic weight (in \( g/mol \))

Given:
- Density, \( \rho = 5 \times 10^{3} \, kg/m^{3} \)
- Atomic weight, \( M = 60 \, g/mol = 60 \times 10^{-3} \, kg/mol \)

Now substituting the values into the formula:

\[ n = \frac{5 \times 10^{3} \, kg/m^{3} \cdot 6.022 \times 10^{23} \, mol^{-1}}{60 \times 10^{-3} \, kg/mol} \]

Calculating \( n \):

\[ n = \frac{5 \times 10^{3} \cdot 6.022 \times 10^{23}}{60 \times 10^{-3}} = \frac{5 \times 10^{3} \cdot 6.022 \times 10^{23}}{0.06} \]

\[ n \approx 5.02 \times 10^{28} \, m^{-3} \]

### Step 2: Rearrange the formula to find drift speed \( v_d \)

The drift speed can be rearranged from the current formula:

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

### Step 3: Substitute the known values into the drift speed formula

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

Now substituting the values:

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

Calculating \( v_d \):

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

\[ v_d = \frac{16}{8.032 \times 10^{3}} \approx 1.99 \times 10^{-3} \, m/s \]

### Final Result

The drift speed of the free electrons in the wire is approximately:

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