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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.
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:
  • \( R \) 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.
To solve for resistivity, we rearrange the formula: \( \rho = R \frac{A}{L} \). By substituting the computed area and the given resistance and length of the wire, you can find the unknown resistivity, which leads us to the final goal of identifying the material.
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.
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.
Each of these elements plays a crucial role in determining how electricity flows through a wire, and understanding them is essential for designing effective electrical systems.

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Most popular questions from this chapter

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.6 V. You cut off a 20.0-m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a 40.0-m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

A lightning bolt strikes one end of a steel lightning rod, producing a 15,000-A current burst that lasts for 65 \(\mu\)s. The rod is 2.0 m long and 1.8 cm in diameter, and its other end is connected to the ground by 35 m of 8.0-mm-diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

A 3.00-m length of copper wire at 20\(^\circ\)C has a 1.20-mlong section with diameter 1.60 mm and a 1.80-m-long section with diameter 0.80 mm. There is a current of 2.5 mA in the 1.60- mm-diameter section. (a) What is the current in the 0.80-mmdiameter section? (b) What is the magnitude of \(\vec E\) in the 1.60-mm-diameter section? (c) What is the magnitude of \(\vec E\) in the 0.80-mm-diameter section? (d) What is the potential difference between the ends of the 3.00-m length of wire?

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance \(x\) from the left end and obeys the formula \(\rho (x) = a + bx^2\), where a and b are constants. At the left end, the resistivity is 2.25 \(\times\) 10\(^{-8} \Omega\) \(\cdot\) m, while at the right end it is 8.50 \(\times\) 10\(^{-8}\Omega\) \(\cdot \) m. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a 1.75-A current? (c) If we cut the rod into two 75.0-cm halves, what is the resistance of each half?

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