Chapter 25: Problem 14
A wire 6.50 m long with diameter of 2.05 mm has a resistance of 0.0290 \(\Omega\). What material is the wire most likely made of?
Short Answer
Expert verified
The wire is most likely made of copper.
Step by step solution
01
Convert Diameter to Radius
To find the cross-sectional area, we first need to convert the diameter of the wire to a radius. The conversion is straightforward: radius is half the diameter. Thus, the radius \( r \) is given by \( r = \frac{2.05}{2} \) mm, which is \( 1.025 \) mm or \( 1.025 \times 10^{-3} \) meters.
02
Calculate Cross-Sectional Area
Using the radius, we calculate the cross-sectional area \( A \) of the wire. The formula for the area of a circle is \( A = \pi r^2 \). Substituting the value of \( r \), \( A = \pi (1.025 \times 10^{-3})^2 \). Calculating this gives \( A \approx 3.30 \times 10^{-6} \) square meters.
03
Use Resistance Formula
We use the resistance formula \( R = \rho \frac{L}{A} \) to solve for the resistivity \( \rho \). We know that \( R = 0.0290 \Omega \), \( L = 6.50 \) meters, and \( A = 3.30 \times 10^{-6} \) square meters. Rearranging for \( \rho \), we get \( \rho = R \frac{A}{L} \).
04
Calculate Resistivity
Substitute the known values into the rearranged formula to calculate the resistivity: \( \rho = 0.0290 \times \frac{3.30 \times 10^{-6}}{6.50} \). Evaluating this expression gives \( \rho \approx 1.47 \times 10^{-8} \Omega \cdot m \).
05
Identify the Material
The calculated resistivity \( 1.47 \times 10^{-8} \Omega \cdot m \) is close to the known resistivity of copper, which is approximately \( 1.68 \times 10^{-8} \Omega \cdot m \). Therefore, the material of the wire is most likely copper.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross-Sectional Area
To comprehend how to determine the cross-sectional area of a wire, let’s start by understanding its basic geometry. The cross-section of a wire, much like a circle, requires us to find its radius to calculate the area.
Given the diameter of the wire is 2.05 mm, the radius is simply half of that. So, you divide 2.05 by 2 to get 1.025 mm, or 1.025 x 10⁻³ meters to keep the units consistent with standard calculations.
To find the area, use the formula for the area of a circle, which is \( A = \pi r^2 \). Substitute the radius into the formula: \( A = \pi (1.025 \times 10^{-3})^2 \). This calculation is crucial as it will impact the value of resistance found in subsequent steps.
Given the diameter of the wire is 2.05 mm, the radius is simply half of that. So, you divide 2.05 by 2 to get 1.025 mm, or 1.025 x 10⁻³ meters to keep the units consistent with standard calculations.
To find the area, use the formula for the area of a circle, which is \( A = \pi r^2 \). Substitute the radius into the formula: \( A = \pi (1.025 \times 10^{-3})^2 \). This calculation is crucial as it will impact the value of resistance found in subsequent steps.
Resistance Formula
The resistance of a wire is determined by several factors, including the material it is made from, its length, and its cross-sectional area. In physics, these relationships are expressed in the resistance formula:
\( R = \rho \frac{L}{A} \), where:
\( R = \rho \frac{L}{A} \), where:
- \( R \) is the resistance in ohms (\(\Omega\)).
- \( \rho \) is the resistivity of the material in ohm meters.
- \( L \) is the length of the wire in meters.
- \( A \) is the cross-sectional area in square meters.
Material Identification
After calculating the resistivity of the wire using the measures given and applying the resistance formula, we end up with a value of approximately \( 1.47 \times 10^{-8} \Omega \cdot m \).
With this resistivity value in hand, we can compare it to known resistivities of various materials. This process is akin to comparing a fingerprint to a database to identify a person.
Copper, commonly used in electrical applications due to its excellent conductive properties, has a known resistivity of roughly \( 1.68 \times 10^{-8} \Omega \cdot m \). Given the close match, we deduce that the wire is most likely composed of copper, confirming its use in many practical situations.
With this resistivity value in hand, we can compare it to known resistivities of various materials. This process is akin to comparing a fingerprint to a database to identify a person.
Copper, commonly used in electrical applications due to its excellent conductive properties, has a known resistivity of roughly \( 1.68 \times 10^{-8} \Omega \cdot m \). Given the close match, we deduce that the wire is most likely composed of copper, confirming its use in many practical situations.
Wire Properties
Understanding the properties of a wire involves examining several key aspects that affect its function and application. The primary properties include:
- Length: Longer wires have higher resistance because electrons encounter more obstacles over longer distances. In our case, the wire is 6.50 meters long.
- Diameter: The wire’s diameter directly influences its cross-sectional area, which in turn affects resistance; a larger area reduces resistance.
- Material: Different materials conduct electricity differently; metals like copper have low resistivity, making them excellent conductors.