Problem 8 - College Physics

Question

A wire 6.50 \(\mathrm{m}\) long with diameter of 2.05 \(\mathrm{mm}\) has a resistance of 0.0290\(\Omega .\) What material is the wire most likely made of?

Step-by-Step Solution

Answer

The wire is most likely made of copper.

Step 1: Calculate the cross-sectional area

First, convert the wire's diameter from millimeters to meters by dividing by 1000, which gives 0.00205 meters. Use the formula for the area of a circle, \( A = \pi \times \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter. This yields \( A = \pi \times \left(\frac{0.00205}{2}\right)^2 = 3.30 \times 10^{-6} \text{ m}^2 \).

Step 2: Use resistance formula

Use the resistance formula \( R = \rho \times \frac{L}{A} \), where \( R = 0.0290 \ \Omega \), \( L = 6.50 \ \text{m} \), and \( A = 3.30 \times 10^{-6} \ \text{m}^2 \). Solving for \( \rho \) gives \( \rho = R \times \frac{A}{L} \).

Step 3: Solve for resistivity \( \rho \)

Plug in the values into the formula: \( \rho = 0.0290 \times \frac{3.30 \times 10^{-6}}{6.50} = 1.47 \times 10^{-8} \ \Omega \cdot \text{m} \).

Step 4: Identify the material from resistivity

Compare the calculated resistivity \( 1.47 \times 10^{-8} \ \Omega \cdot \text{m} \) with known resistivities of materials. This value closely matches the resistivity of copper, which is approximately \( 1.68 \times 10^{-8} \ \Omega \cdot \text{m} \).

Key Concepts

Cross-Sectional Area Calculation

To find out the cross-sectional area of a wire, we begin by understanding that it's essentially the area of a circle. A circle's area can be calculated using the formula \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) represents the diameter.
Convert the diameter from millimeters to meters by dividing by 1000. For example, 2.05 millimeters becomes 0.00205 meters. This conversion is crucial because it ensures consistent units when performing further calculations.

When applying these values, the calculation becomes \( A = \pi \left(0.001025\right)^2 = 3.30 \times 10^{-6} \text{ m}^2 \).
This small number reflects the tiny area of a typical wire, highlighting why precise calculations are necessary in these situations. Always remember, precision in converting units and calculating area directly impacts the accuracy of subsequent electrical calculations.

Wire Resistance

Wire resistance is a critical factor in electrical circuits and is determined by the wire's material, length, and cross-sectional area. The formula to calculate resistance \( R \) is \( R = \rho \times \frac{L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire.
Resistance increases with the length of the wire, because electrons encounter more interactions along a longer distance. Conversely, resistance decreases with a larger cross-sectional area, as a larger area provides more space for the electrons to flow, reducing interaction.

In our example, with a length of 6.50 meters and a calculated area of \( 3.30 \times 10^{-6} \text{ m}^2 \), the known resistance is 0.0290 Ω. Using this information, you can compute resistivity if it's not provided. This relationship helps in designing effective circuits by choosing optimal wire dimensions and materials.

Material Identification

Material identification through resistivity is a smart way to ascertain a wire’s material. Each material has a characteristic resistivity value, which makes it unique. By comparing calculated resistivity values to standard material values, you can identify the material.
In this exercise, we first calculate resistivity using \( \rho = R \times \frac{A}{L} \). Plugging in the known values gives \( \rho = 0.0290 \times \frac{3.30 \times 10^{-6}}{6.50} = 1.47 \times 10^{-8} \ \Omega \cdot \text{m} \).

Next, compare this calculated resistivity with known values for common wire materials. In our case, the value \( 1.47 \times 10^{-8} \Omega \cdot \text{m} \) is close to the resistivity of copper, approximately \( 1.68 \times 10^{-8} \Omega \cdot \text{m} \), suggesting the wire is likely copper.
This method is reliable as it leverages standard resistivity data, which is widely available and essential for identifying materials in electrical components.