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. `"0.1 mm s"^(-1)`
B. `"0.2 mm s"^(-1)`
C. `"0.3 mm s"^(-1)`
D. `"0.2 mm s"^(-1)`

Answer

To find the drift velocity of electrons in a copper wire carrying a current of 1.1 A, we can follow these steps:

### Step 1: Calculate the number density of free electrons (n)
The number density of free electrons can be calculated using the formula:
\[ n = \frac{N_A \cdot \rho}{M} \]
where:
- \(N_A\) = Avogadro's number = \(6.023 \times 10^{23} \, \text{mol}^{-1}\)
- \(\rho\) = density of copper = \(9 \times 10^3 \, \text{kg/m}^3\)
- \(M\) = molar mass of copper = 63 g/mol = \(63 \times 10^{-3} \, \text{kg/mol}\)

Substituting the values:
\[ n = \frac{(6.023 \times 10^{23}) \cdot (9 \times 10^3)}{63 \times 10^{-3}} \]

### Step 2: Calculate the value of n
Calculating the above expression:
\[ n = \frac{(6.023 \times 10^{23}) \cdot (9 \times 10^3)}{63 \times 10^{-3}} = \frac{(6.023 \times 9) \times 10^{26}}{63} = \frac{54.207 \times 10^{26}}{63} \approx 0.86 \times 10^{29} \, \text{m}^{-3} \]

### Step 3: Calculate the cross-sectional area (A) of the wire
The diameter of the wire is given as 1 mm, so the radius \(r\) is:
\[ r = \frac{1 \, \text{mm}}{2} = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \]
The cross-sectional area \(A\) is given by:
\[ A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = \pi (0.25 \times 10^{-6}) \approx 7.85 \times 10^{-7} \, \text{m}^2 \]

### Step 4: Calculate the drift velocity (v_d)
The drift velocity can be calculated using the formula:
\[ v_d = \frac{I}{n \cdot e \cdot A} \]
where:
- \(I\) = current = 1.1 A
- \(e\) = charge of an electron = \(1.6 \times 10^{-19} \, \text{C}\)

Substituting the values:
\[ v_d = \frac{1.1}{(0.86 \times 10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7})} \]

### Step 5: Calculate the drift velocity
Calculating the denominator:
\[ n \cdot e \cdot A = (0.86 \times 10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7}) \approx 1.08 \times 10^{4} \]
Now substituting back into the drift velocity equation:
\[ v_d = \frac{1.1}{1.08 \times 10^{4}} \approx 0.000102 \, \text{m/s} = 0.1 \, \text{mm/s} \]

### Final Answer
The drift velocity \(v_d\) is approximately \(0.1 \, \text{mm/s}\).

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