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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 the result from \( \rho = R \frac{A}{L} \) using the calculated values for \( L \) and \( A \).

Step by step solution

01

Calculate the Length of the Wire

First, find the circumference of one coil, which is the perimeter of a circle. The circumference \( C \) can be calculated using the formula:\[ C = \pi \times \text{diameter} \]The diameter of each coil is 3.50 cm, so the circumference of one coil is \( C = \pi \times 3.50 \, \text{cm} \). Multiply this by the total number of coils to get the total length of the wire \( L \):\[ L = 75 \times \pi \times 3.50 \, \text{cm} \]Convert this length to meters, as standard SI units are used for resistivity calculations. 1 cm = 0.01 m, thus:\[ L = 75 \times \pi \times 0.035 \, \text{m} \]Compute \( L \) to use in further calculations.
02

Calculate the Cross-Sectional Area of the Wire

The cross-sectional area \( A \) of the wire is circular, because the wire itself is cylindrical. The radius of the wire is half the diameter, which is \( 3.25 \, \text{mm} \) or \( 0.00325 \, \text{m} \). Calculate \( A \) using the formula for the area of a circle:\[ A = \pi \times (0.00325 / 2)^2 \]Compute this area in square meters.
03

Use Ohm's Law to Calculate Resistivity

Ohm's law for resistivity states \( R = \rho \frac{L}{A} \), where \( \rho \) is resistivity, \( R \) is resistance, \( L \) is length, and \( A \) is the cross-sectional area. Rearrange this to find resistivity \( \rho \):\[ \rho = R \frac{A}{L} \]Substitute \( R = 1.74 \, \Omega \), and the previously calculated \( A \) and \( L \) to find \( \rho \).

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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 in the field of electrical engineering and physics. It expresses the relationship between voltage, current, and resistance in a circuit. This law is named after Georg Simon Ohm, who discovered that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.

The formula associated with Ohm's Law is \( V = IR \), where \( V \) is the voltage (in volts), \( I \) is the current (in amperes), and \( R \) is the resistance (in ohms). This relationship is a cornerstone for analyzing electrical circuits and forms the basis for more complex theories.

In relation to resistivity, Ohm's Law can be expanded to include properties of materials, particularly how they impede electric flow.

For resistivity \( \rho \), this relationship becomes \( R = \rho \frac{L}{A} \), where \( L \) is the length of the conductor and \( A \) is its cross-sectional area. Rearranging it allows us to derive resistivity:
  • \(\rho = R \frac{A}{L}\)
This formula shows how the material's inherent property affects its resistance, providing valuable insights during conductivity and material testing.
Cross-Sectional Area
The cross-sectional area is a concept frequently encountered in physics and engineering when dealing with three-dimensional shapes, like wires or pipes. When you cut through a cylindrical wire at any point along its length, the exposed face is a circle, which is the cross-section.

The area of this circular cross-section is vital for calculating properties like resistivity and resistance. This is because the amount of material available through which electric currents flow affects the overall resistance experienced in the conductor.

For a wire with a circular cross-section, the formula for calculating the cross-sectional area is:\[ A = \pi \left( \frac{d}{2} \right)^2 \]
where \( d \) is the diameter of the wire. This is useful as it allows you to determine how much space is available for electrons to travel through the wire. The larger the cross-sectional area, the more easily current can flow, usually resulting in lower resistance.
Cylindrical Wire
A cylindrical wire is a common form of electrical conductor due to its efficient geometry in conducting electrical currents. The cylindrical shape allows for uniform distribution of material, contributing to consistent electrical properties throughout its length.

The essential characteristics of a cylindrical wire include its length, diameter, and the material it's composed of. These attributes affect calculations related to electrical properties like resistance and resistivity.

Cylindrical wires are often chosen for their ease of use in calculations and manufacturing. Ohm's Law, the concept of resistivity, and the calculation of cross-sectional areas assume a uniform distribution of the conducting material, and wires are ideally suited to these assumptions.

Understanding the geometry of a cylindrical wire helps in calculating the resistance and resistivity, crucial for designing circuits and ensuring they function properly under desired conditions.
Circumference and Length Calculations
Calculating the circumference and length of a spiral or coil-shaped conductor is an essential step in determining its overall resistance and the resistivity of the material. The circumference of a coil helps determine the total length of the wire, which is a critical factor in resistivity calculations.

The circumference \( C \) of a coil is given by the formula:
  • \( C = \pi \times d \)
where \( d \) is the diameter of the coil. Knowing the circumference aids in calculating the complete length of the wire in a coil, particularly when multiplied by the number of turns or coils.

For example, in a coil with 75 turns each of 3.50 cm diameter, the total length \( L \) can be found using:
  • \( L = 75 \times \pi \times 3.50 \text{ cm} = 75 \times \pi \times 0.035 \text{ m} \)
Accurate length calculations are foundational in electrical engineering tasks as they directly influence the resistance a material exhibits, and thus its function within a circuit.

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Most popular questions from this chapter

A copper wire has a square cross section 2.3 \(\mathrm{mm}\) on a side. The wire is 4.0 \(\mathrm{m}\) long and carries a current of 3.6 \(\mathrm{A}\) . The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

An idealized voltmeter is connected across the terminals of a \(15.0-\mathrm{V}\) battery, and a \(75.0-\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table below shows the maximum current \(I_{\text { max }}\) for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$\begin{array}{ccc}{\text { Wire gauge }} & {\text { Diameter (cm) }} & {I_{\max }(\mathbf{A})} \\ {14} & {0.163} & {18} \\\ {12} & {0.205} & {25} \\ {10} & {0.259} & {30} \\ {8} & {0.326} & {40} \\\ {6} & {0.412} & {60} \\ {5} & {0.462} & {65} \\ {4} & {0.519} & {85}\end{array}$$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of 4200 \(\mathrm{W}\) of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is 120 \(\mathrm{V}\) , determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length 42.0 \(\mathrm{m}\) . At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electric energy is \(\$ 0.11\) per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

A \(12.0-\mathrm{V}\) battery has an internal resistance of 0.24 \(\mathrm{s}\) and a capacity of 50.0 \(\mathrm{A} \cdot \mathrm{h}\) (see Exercise 25.47\() .\) The battery is charged by passing a 10 -A current through it for 5.0 \(\mathrm{h}\) . (a) What is the terminal voltage during charging? (b) What total electricalenergy is supplied to the battery during charging? (c) What electrical energy is dissipated in the internal resistance during charging? (d) The battery is now completely discharged through a resistor, again with a constant current of 10 \(\mathrm{A}\) . What is the external circuit resistance? (e) What total electrical energy is supplied to the external resistor? (f) What total electrical energy is dissipated in the internal resistance? (g) Why are the answers to parts (b) and (e) not the same?

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) accidentally grasps the terminals of a \(14-\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

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