A wire 8.00 m long and 6.00 mm in diameter has a resistance of 30.0 `mOmega`. A potential difference of 23.0 V is applied between the ends. (a) What is the current in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the wire material.

Answer

To solve the problem step by step, we will address each part of the question systematically.

### Given Data:
- Length of the wire, \( L = 8.00 \, \text{m} \)
- Diameter of the wire, \( d = 6.00 \, \text{mm} = 6.00 \times 10^{-3} \, \text{m} \)
- Resistance of the wire, \( R = 30.0 \, \text{m}\Omega = 30.0 \times 10^{-3} \, \Omega \)
- Potential difference, \( V = 23.0 \, \text{V} \)

### Part (a): Calculate the Current in the Wire
Using Ohm's Law, which states that \( V = IR \), we can rearrange it to find the current \( I \):
\[ I = \frac{V}{R} \]
Substituting the given values:
\[ I = \frac{23.0 \, \text{V}}{30.0 \times 10^{-3} \, \Omega} \]
Calculating the current:
\[ I = \frac{23.0}{0.030} = 766.67 \, \text{A} \]
Thus, the current in the wire is approximately:
\[ I \approx 767 \, \text{A} \]

### Part (b): Calculate the Magnitude of the Current Density
Current density \( J \) is defined as the current per unit area. The formula is:
\[ J = \frac{I}{A} \]
where \( A \) is the cross-sectional area of the wire. The area \( A \) of a circular wire is given by:
\[ A = \frac{\pi d^2}{4} \]
Substituting the diameter:
\[ A = \frac{\pi (6.00 \times 10^{-3})^2}{4} \]
Calculating the area:
\[ A = \frac{\pi (36.00 \times 10^{-6})}{4} = \frac{36.00 \pi \times 10^{-6}}{4} \approx 28.27 \times 10^{-6} \, \text{m}^2 \]
Now substituting \( I \) and \( A \) into the current density formula:
\[ J = \frac{767 \, \text{A}}{28.27 \times 10^{-6} \, \text{m}^2} \]
Calculating the current density:
\[ J \approx 2.71 \times 10^7 \, \text{A/m}^2 \]

### Part (c): Calculate the Resistivity of the Wire Material
The resistivity \( \rho \) can be calculated using the formula for resistance:
\[ R = \rho \frac{L}{A} \]
Rearranging to find \( \rho \):
\[ \rho = R \frac{A}{L} \]
Substituting the known values:
\[ \rho = (30.0 \times 10^{-3} \, \Omega) \cdot \frac{28.27 \times 10^{-6} \, \text{m}^2}{8.00 \, \text{m}} \]
Calculating:
\[ \rho = (30.0 \times 10^{-3}) \cdot (3.535875 \times 10^{-6}) \approx 10.6 \times 10^{-8} \, \Omega \cdot \text{m} \]

### Final Answers:
- (a) Current in the wire: \( I \approx 767 \, \text{A} \)
- (b) Current density: \( J \approx 2.71 \times 10^7 \, \text{A/m}^2 \)
- (c) Resistivity of the wire material: \( \rho \approx 10.6 \times 10^{-8} \, \Omega \cdot \text{m} \)