A copper wire of cross-sectional area `2.40 xx 10^(-6)m^(2)` and length 4.00 m has a current of 2.00 A uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in 30 min?

Answer

To solve the problem step by step, let's break it down into two parts as given in the question.

### Part (a): Magnitude of the Electric Field along the Wire

1. **Identify Given Values**:
- Cross-sectional area, \( A = 2.40 \times 10^{-6} \, \text{m}^2 \)
- Length of the wire, \( L = 4.00 \, \text{m} \)
- Current, \( I = 2.00 \, \text{A} \)

2. **Find the Resistance of the Wire**:
- Use the formula for resistance:
\[ R = \frac{\rho L}{A} \]
- For copper, the resistivity \( \rho \) is approximately \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \).
- Substitute the values:
\[ R = \frac{(1.68 \times 10^{-8}) \times 4.00}{2.40 \times 10^{-6}} \]
- Calculate \( R \):
\[ R = \frac{6.72 \times 10^{-8}}{2.40 \times 10^{-6}} \approx 0.0280 \, \Omega \]

3. **Calculate the Potential Difference (V)**:
- Use Ohm's Law:
\[ V = I \times R \]
- Substitute the values:
\[ V = 2.00 \times 0.0280 \approx 0.0560 \, \text{V} \]

4. **Calculate the Electric Field (E)**:
- Use the formula:
\[ E = \frac{V}{L} \]
- Substitute the values:
\[ E = \frac{0.0560}{4.00} \approx 0.0140 \, \text{V/m} \]

5. **Final Answer for Part (a)**:
- The magnitude of the electric field along the wire is:
\[ E \approx 1.40 \times 10^{-2} \, \text{V/m} \]

### Part (b): Electrical Energy Transferred to Thermal Energy in 30 Minutes

1. **Identify Given Values**:
- Current, \( I = 2.00 \, \text{A} \)
- Potential difference, \( V \approx 0.0560 \, \text{V} \) (from Part (a))
- Time, \( t = 30 \, \text{minutes} = 30 \times 60 = 1800 \, \text{s} \)

2. **Calculate the Electrical Energy (E)**:
- Use the formula for electrical energy:
\[ E = V \times I \times t \]
- Substitute the values:
\[ E = 0.0560 \times 2.00 \times 1800 \]
- Calculate \( E \):
\[ E = 0.0560 \times 2.00 \times 1800 \approx 201.6 \, \text{J} \]

3. **Final Answer for Part (b)**:
- The electrical energy transferred to thermal energy in 30 minutes is:
\[ E \approx 201.6 \, \text{J} \]

### Summary of Answers:
- (a) Electric Field \( E \approx 1.40 \times 10^{-2} \, \text{V/m} \)
- (b) Electrical Energy \( E \approx 201.6 \, \text{J} \)