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A tightly coiled spring having 75 coils, each 3.50 \(\mathrm{cm}\) in diameter, is made of insulated metal wire 3.25 \(\mathrm{mm}\) in diameter. An ohmmeter connected across its opposite ends reads 1.74\(\Omega\) . What is the resistivity of the metal?

Short Answer

Expert verified
The resistivity of the metal is approximately 1.75 \( \times 10^{-6} \) \( \Omega \cdot \mathrm{m} \).

Step by step solution

01

Calculate the Total Length of the Wire

To find the total length of the wire, use the formula for the circumference of each coil and multiply by the number of coils. The circumference of each coil is given by the formula: \( C = \pi \times d \). Here, \( d = 3.50 \, \mathrm{cm} = 0.035 \, \mathrm{m} \). So, \( C = \pi \times 0.035 = 0.110 \, \mathrm{m} \). Therefore, the total length of the wire is \( L = 75 \times 0.110 = 8.25 \, \mathrm{m} \).
02

Determine the Cross-sectional Area of the Wire

The wire is a cylinder with a diameter of 3.25 mm, or 0.00325 m. The cross-sectional area, \( A \), of a wire can be calculated using \( A = \pi \times \left(\frac{d}{2}\right)^2 \). Substituting the given diameter: \( A = \pi \times \left( \frac{0.00325}{2} \right)^2 \approx 8.30 \times 10^{-6} \, \mathrm{m^2} \).
03

Use Ohm's Law to Find Resistivity

Ohm's law in terms of resistivity is \( R = \rho \frac{L}{A} \), where \( R \) is resistance, \( L \) is length, \( A \) is cross-sectional area, and \( \rho \) is resistivity. Solving for \( \rho \) gives us: \( \rho = R \frac{A}{L} \). Substituting the known values \( R = 1.74 \, \Omega \), \( L = 8.25 \, \mathrm{m} \), and \( A = 8.30 \times 10^{-6} \, \mathrm{m^2} \), we find \( \rho = 1.74 \frac{8.30 \times 10^{-6}}{8.25} \approx 1.75 \times 10^{-6} \, \Omega \cdot \mathrm{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle used to relate voltage, current, and resistance in electrical circuits. It is usually expressed in the formula \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. This law is key in determining how electrical components will behave under certain conditions. However, in situations dealing with materials specifically, Ohm's Law is adjusted to relate resistance with material properties: the relation becomes \( R = \rho \frac{L}{A} \), where \( R \) is still the resistance, but \( \rho \) is the resistivity of the material, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area. Understanding Ohm's Law can help you calculate the resistivity of a material if you know the dimensions of the conductor and its resistance. This is crucial for applications like designing electrical circuits, where knowing the material's resistivity can determine how conductive a wire is. Using Ohm's Law, we can easily rearrange the formula to solve for resistivity: \( \rho = R \frac{A}{L} \). This rearrangement shows how resistivity depends directly on resistance and inversely on both length and cross-sectional area.
Cross-sectional Area
The cross-sectional area of a conductor is a measure of how large the wire is in terms of its width. For a cylindrical wire, such as in this problem, the cross-sectional area \( A \) is calculated using the formula \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter of the wire. The cross-sectional area is important because it impacts the wire's resistance and thus affects how much current can flow through it. A larger area means there is more space for the current to pass through, which typically reduces resistance. Conversely, a smaller cross-sectional area increases resistance because there is less space for current flow.In practice, when designing circuits and selecting materials, engineers and physicists choose wires with appropriate cross-sectional areas to ensure that electrical currents can flow efficiently without overheating or causing excessive energy loss. This concept is crucial when we rearrange Ohm's Law to solve for resistivity, as seen above, because the area \( A \) plays a pivotal role in determining how resistive the material will be when a specific current is applied.
Resistance Measurement
Resistance measurement is a core activity in analyzing electrical circuits and understanding material properties. Resistance indicates how much a material opposes the flow of electric current, measured in ohms (\( \Omega \)). In practical terms, resistance measurement helps determine if a circuit component will function correctly and safely. To measure resistance, devices called ohmmeters are used. An ohmmeter is connected across the component, and it directly measures how much resistance the component presents against the flow of current. In the problem, the wire's resistance is measured as 1.74 \( \Omega \). Knowing the resistance allows us to use Ohm's Law to compute other relevant properties like resistivity, given certain dimensions of a wire, as we've done. Accurate resistance measurements are vital in quality control processes in manufacturing electronic components and engineering excellent designs in electrical devices. When considering the resistance of materials, it's also critical to comprehend that resistance depends not just on the material, but also on environmental factors like temperature, which can cause changes in the resistance observed.

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Most popular questions from this chapter

A hollow aluminum cylinder is 2.50 \(\mathrm{m}\) long and has an inner radius of 3.20 \(\mathrm{cm}\) and an outer radius of 4.60 \(\mathrm{cm}\) . Treat each surface (inner, outer, and the two end faces) as an equipotential surface. At room temperature, what will an ohmmeter read if it is connected between (a) the opposite faces and (b) the inner and outer surfaces?

A strand of wire has resistance 5.60\(\mu \Omega\) . Find the net resistance of 120 such strands if they are (a) placed side by side to form a cable of the same length as a single strand, and (b) connected end to end to form a wire 120 times as long as a single strand.

A Nonideal Ammeter. Unlike the idealized ammeter described in Section 25.4, any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{A}\) is connected in series with a resistor \(R\) and a battery of emf \(\mathcal{E}\) and internal resistance \(r .\) The current measured by the ammeter is \(I_{A}\) . Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_{A}, r, R_{A},\) and \(R\) . The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{A}\) . (b) If \(R=3.80 \Omega\) , \(\mathcal{E}=7.50 \mathrm{V},\) and \(r=0.45 \Omega,\) find the maximum value of the ammeter resistance \(R_{A}\) so that \(I_{A}\) is within 1.0\(\%\) of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

A typical cost for electric power is 12.0\(\phi\) per kilowatt-hour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75-\mathrm{W}\) bulb burning day and night? (b) Suppose your refrigerator uses 400 \(\mathrm{W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

An 18 -gauge wire (diameter 1.02 \(\mathrm{mm} )\) carries a current with a current density of \(1.50 \times 10^{6} \mathrm{A} / \mathrm{m}^{2}\) . Calculate (a) the current in the wire and \((\mathrm{b})\) the drift velocity of electrons in the wire.

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