Problem 56 - Physics

Question

A cylindrical aluminum pipe of length \(1.50 \mathrm{~m}\) has an inner radius of \(2.00 \times 10^{-3} \mathrm{~m}\) and an outer radius of \(3.00 \times 10^{-3} \mathrm{~m}\). The interior of the pipe is completely filled with copper. What is the resistance of this unit?

Step-by-Step Solution

Answer

The resistance is approximately 0.89 mΩ.

Step 1: Understand the Structure

The cylindrical pipe has an inner copper core and an outer aluminum shell. Since the copper fills the interior, the current can flow through both the copper and the aluminum simultaneously. This setup indicates that the copper core and aluminum shell are in parallel.

Step 2: Calculate Cross-sectional Areas

The cross-sectional area for the copper is determined by the inner radius:\( A_{copper} = \pi (2.00 \times 10^{-3} \, \text{m})^2 \).The cross-sectional area for aluminum is the difference between the outer and inner circles:\( A_{aluminum} = \pi ((3.00 \times 10^{-3} \, \text{m})^2 - (2.00 \times 10^{-3} \, \text{m})^2) \).

Step 3: Use Resistance Formula

The resistance formula is given by:\( R = \frac{\rho \cdot L}{A} \), where \( \rho \) is resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

Step 4: Calculate Copper Resistance

Given the resistivity of copper \( \rho_{copper} = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \):\( R_{copper} = \frac{1.68 \times 10^{-8} \, \Omega \cdot \text{m} \times 1.50 \, \text{m}}{\pi (2.00 \times 10^{-3} \, \text{m})^2} \approx 2.00 \, \text{m}\Omega \).

Step 5: Calculate Aluminum Resistance

Given the resistivity of aluminum \( \rho_{aluminum} = 2.82 \times 10^{-8} \, \Omega \cdot \text{m} \):\( R_{aluminum} = \frac{2.82 \times 10^{-8} \, \Omega \cdot \text{m} \times 1.50 \, \text{m}}{\pi \left((3.00 \times 10^{-3} \, \text{m})^2 - (2.00 \times 10^{-3} \, \text{m})^2\right)} \approx 1.59 \, \text{m}\Omega \).

Step 6: Compute Total Resistance in Parallel

The total resistance for resistors in parallel is given by:\( \frac{1}{R_{total}} = \frac{1}{R_{copper}} + \frac{1}{R_{aluminum}} \).Calculate:\( \frac{1}{R_{total}} = \frac{1}{2.00 \text{ m}\Omega} + \frac{1}{1.59 \text{ m}\Omega} \),\( R_{total} \approx 0.89 \, \text{m}\Omega \).

Key Concepts

Cylindrical Pipe

A cylindrical pipe is a three-dimensional geometric shape, resembling a hollow cylinder. Imagine a regular straw; a cylindrical pipe functions in much the same way, but for transporting electricity rather than liquid. In this exercise, the pipe contains two metal components: copper and aluminum.

- **Inner radii and outer radii:** Since the copper is located in the inner section, it has the smaller radius, known as the inner radius. Meanwhile, the outer aluminum shell extends from the inner surface (copper) to the outer radius.
- **Electrical path:** The current can flow through both the metals simultaneously. Thus, determining how these materials conduct electricity is pivotal for calculating the overall resistance of the setup.

Copper and Aluminum in Parallel

When two conductors serve as pathways for electricity, they can either be in series or parallel. In this case of the cylindrical pipe, the copper core entirely surrounded by the aluminum shell signifies that these materials are in parallel.

- **Behavior in parallel:** Here, electricity has multiple paths to follow. It doesn't just depend on one single conductor, like it would in a series setup. This means the lower resistance path will carry more current.
- **Resistance calculation implication:** For materials in parallel, the total resistance can be calculated using the formula: \[ \frac{1}{R_{total}} = \frac{1}{R_{copper}} + \frac{1}{R_{aluminum}} \] Thus, the total resistance will be less than that of the smallest individual resistance.

Resistivity

Resistivity is a fundamental material property that indicates how strongly a material opposes the flow of electric current. The unit of resistivity is ohm-meter ( \(\Omega \cdot \text{m}\)).- **Dependence on material:** Each material has its specific resistivity value. For example, copper, being a highly efficient conductor, has a lower resistivity ( \(1.68 \times 10^{-8} \Omega \cdot \text{m}\)) compared to aluminum ( \(2.82 \times 10^{-8} \Omega \cdot \text{m}\)).
- **Effect on resistance:** The resistance ( \(R\)) of any conductor can be calculated using its resistivity ( \(\rho\)) along with its length ( \(L\)) and cross-sectional area ( \(A\)) using the formula: \[ R = \frac{\rho \cdot L}{A} \] Lower resistivity means that, for the same dimensions, the material will have lower resistance.

Cross-sectional Area

The cross-sectional area is the area through which current flows in a conductor. It's crucial for the resistance calculation since a larger area allows more current to pass through, reducing resistance.

- **Calculating for copper:** As the copper fills the inner region, its cross-sectional area ( \(A_{copper}\)) is the area of the inner circle: \[ A_{copper} = \pi (2.00 \times 10^{-3} \, \text{m})^2 \]- **Calculating for aluminum:** The cross-sectional area ( \(A_{aluminum}\)) of the aluminum is the difference between the outer and inner circular areas, providing the annular region through which electricity flows: \[ A_{aluminum} = \pi \left((3.00 \times 10^{-3} \, \text{m})^2 - (2.00 \times 10^{-3} \, \text{m})^2\right) \]Understanding these calculations helps determine how effectively the pipe can conduct electricity based on the materials present.