Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper `=9 xx 10^(3) kg m^(-3)` and atomic weight = 63)

Options

A. `0.3 mm//sec`
B. `0.1 mm//sec`
C. `0.2mm//sec`
D. `0.2cm//sec`

Answer

To find the drift velocity of electrons in a copper wire with a given current, we can follow these steps:

### Step 1: Determine the cross-sectional area of the wire
The diameter of the wire is given as 1 mm. First, we convert this to meters:
\[ d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \]
The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \text{ m} \]
The cross-sectional area \( A \) of the wire (circular cross-section) is given by:
\[ A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6} \text{ m}^2 \approx 7.85 \times 10^{-7} \text{ m}^2 \]

### Step 2: Calculate the number of free electrons per unit volume
Given:
- Density of copper \( \rho = 9 \times 10^{3} \text{ kg/m}^3 \)
- Atomic weight of copper \( M = 63 \text{ g/mol} = 0.063 \text{ kg/mol} \)
- Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ atoms/mol} \)

The number of moles of copper per unit volume is:
\[ n = \frac{\rho}{M} = \frac{9 \times 10^{3}}{0.063} \approx 1.42 \times 10^{5} \text{ mol/m}^3 \]
The number of atoms (and hence free electrons, since each atom contributes one free electron) per unit volume is:
\[ n_e = n \times N_A = 1.42 \times 10^{5} \times 6.022 \times 10^{23} \approx 8.56 \times 10^{28} \text{ electrons/m}^3 \]

### Step 3: Use the formula for drift velocity
The formula for drift velocity \( v_d \) is given by:
\[ v_d = \frac{I}{n_e \cdot e \cdot A} \]
Where:
- \( I = 1.1 \text{ A} \) (current)
- \( e = 1.6 \times 10^{-19} \text{ C} \) (charge of an electron)

Substituting the values:
\[ v_d = \frac{1.1}{(8.56 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7})} \]

Calculating the denominator:
\[ n_e \cdot e \cdot A = (8.56 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7}) \approx 1.07 \times 10^{3} \]

Now substituting back into the drift velocity equation:
\[ v_d = \frac{1.1}{1.07 \times 10^{3}} \approx 1.03 \times 10^{-3} \text{ m/s} \]

### Step 4: Convert to mm/s
To convert the drift velocity to mm/s:
\[ v_d \approx 1.03 \text{ mm/s} \]

### Final Answer
The drift velocity (approx.) is \( 1.03 \text{ mm/s} \).

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