A current of 1.8 A flows through a wire of cross-sectional area 0.5 `mm^(2)`? Find the current density in the wire. If the number density of conduction electrons in the wire is `8.8 xx 10^(28) m^(-3)`, find the drift speed of electrons.

Answer

To solve the problem, we need to find the current density in the wire and the drift speed of the electrons. Let's break it down step by step.

### Step 1: Convert the cross-sectional area from mm² to m²
The cross-sectional area of the wire is given as 0.5 mm². We need to convert this to square meters (m²) for consistency in SI units.

\[ \text{Area} = 0.5 \, \text{mm}^2 = 0.5 \times 10^{-6} \, \text{m}^2 \]

### Step 2: Calculate the current density (J)
Current density (J) is defined as the current (I) flowing per unit area (A) of the wire. The formula for current density is:

\[ J = \frac{I}{A} \]

Given:
- Current (I) = 1.8 A
- Area (A) = \(0.5 \times 10^{-6} \, \text{m}^2\)

Substituting the values into the formula:

\[ J = \frac{1.8 \, \text{A}}{0.5 \times 10^{-6} \, \text{m}^2} = \frac{1.8}{0.5 \times 10^{-6}} = 3.6 \times 10^{6} \, \text{A/m}^2 \]

### Step 3: Calculate the drift speed (V_d) of electrons
The drift speed (V_d) can be calculated using the formula:

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

Where:
- \(I\) = current (1.8 A)
- \(n\) = number density of conduction electrons = \(8.8 \times 10^{28} \, \text{m}^{-3}\)
- \(e\) = charge of an electron = \(1.6 \times 10^{-19} \, \text{C}\)
- \(A\) = cross-sectional area = \(0.5 \times 10^{-6} \, \text{m}^2\)

Substituting the values into the formula:

\[ V_d = \frac{1.8}{(8.8 \times 10^{28}) \cdot (0.5 \times 10^{-6}) \cdot (1.6 \times 10^{-19})} \]

Calculating the denominator:

\[ n \cdot A \cdot e = (8.8 \times 10^{28}) \cdot (0.5 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) = 7.04 \times 10^{4} \]

Now, substituting back into the drift speed formula:

\[ V_d = \frac{1.8}{7.04 \times 10^{4}} \approx 2.56 \times 10^{-4} \, \text{m/s} \]

### Final Answers:
- Current Density (J) = \(3.6 \times 10^{6} \, \text{A/m}^2\)
- Drift Speed (V_d) = \(2.56 \times 10^{-4} \, \text{m/s}\)

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