The potential difference between two points in a wire `75.0 cm` apart is `0.938 V`, when the current density is `4.40 xx 10^7 A// m^2`. What is
(a) the magnitude of `E` in the wire?
(b) the resistivity of the material of which the wire is made?

Explanation

To solve the given problem, we will break it down into two parts as specified in the question.

### Given Data:
- Distance (D) = 75.0 cm = 0.75 m (convert cm to m)
- Potential Difference (V) = 0.938 V
- Current Density (J) = 4.40 × 10^7 A/m²

### Part (a): Calculate the Magnitude of Electric Field (E)

1. **Use the formula for Electric Field (E)**:
\[ E = \frac{V}{D} \]
where:
- \( V \) is the potential difference,
- \( D \) is the distance between the two points.

2. **Substituting the values**:
\[ E = \frac{0.938 \, \text{V}}{0.75 \, \text{m}} \]

3. **Calculate E**:
\[ E = 1.24 \, \text{V/m} \]

### Part (b): Calculate the Resistivity (ρ) of the Material

1. **Use the relationship between resistivity (ρ), resistance (R), area (A), and length (L)**:
\[ \rho = R \cdot \frac{A}{L} \]

2. **Use Ohm's Law to express resistance (R)**:
\[ R = \frac{V}{I} \]
where \( I \) is the current.

3. **Relate current density (J) to current (I)**:
\[ J = \frac{I}{A} \quad \Rightarrow \quad I = J \cdot A \]

4. **Substituting for I in the resistance formula**:
\[ R = \frac{V}{J \cdot A} \]

5. **Substituting R into the resistivity formula**:
\[ \rho = \left(\frac{V}{J \cdot A}\right) \cdot \frac{A}{L} \]
Simplifying gives:
\[ \rho = \frac{V}{J} \cdot \frac{1}{L} \]

6. **Substituting the known values**:
\[ \rho = \frac{0.938 \, \text{V}}{4.40 \times 10^7 \, \text{A/m}^2} \cdot 0.75 \, \text{m} \]

7. **Calculate ρ**:
\[ \rho = \frac{0.938}{4.40 \times 10^7} \cdot 0.75 \approx 1.27 \times 10^{-8} \, \Omega \cdot \text{m} \]

### Final Answers:
(a) The magnitude of \( E \) in the wire is \( 1.24 \, \text{V/m} \).
(b) The resistivity of the material of which the wire is made is approximately \( 1.27 \times 10^{-8} \, \Omega \cdot \text{m} \).