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Chapter 27: Problem 8

Figure \(\mathrm{P} 27.8\) represents a section of a circular conductor of nonuniform diameter carrying a current of 5.00 \(\mathrm{A}\) . The radius of cross section \(A_{1}\) is \(0.400 \mathrm{cm} .\) (a) What is the magnitude of the current density across \(A_{1} ?\) (b) If the current density across \(A_{2}\) is one-fourth the value across \(A_{1},\) what is the radius of the conductor at \(A_{2} ?\)

Short Answer

Expert verified
The current density at \(A_1\) is \(\frac{5.00}{\pi (0.004)^2} \mathrm{A/m^2}\). The radius at \(A_2\) is \(2\) times the radius at \(A_1\), which is \(0.800 \mathrm{cm}\).

Step by step solution

01

Calculate the cross-sectional area at A1

The cross-sectional area of a circle is calculated using the formula \(A = \pi r^2\), where \(r\) is the radius of the circle. For section \(A_1\), plug in the given radius \(r_1 = 0.400 \mathrm{cm} = 0.004 \mathrm{m}\) to find the area in square meters.
02

Compute the current density at A1

Current density \(J\) is defined as the current \(I\) divided by the area \(A\) through which it flows: \(J = \frac{I}{A}\). Use the current \(I = 5.00 \mathrm{A}\) and the area calculated in Step 1 to find the current density at \(A_1\).
03

Find the current density at A2

The problem states that the current density at \(A_2\) is one-fourth of that at \(A_1\). Therefore, divide the current density at \(A_1\) by 4 to determine the current density at \(A_2\).
04

Calculate the cross-sectional area at A2

Since the current at \(A_2\) is the same as at \(A_1\), we can use the current density at \(A_2\) to find the area at \(A_2\) using the formula for current density: \(A_2 = \frac{I}{J_2}\), where \(J_2\) is the current density at \(A_2\).
05

Determine the radius of the conductor at A2

With the area at \(A_2\) found in Step 4, apply the area formula for a circle \(A = \pi r^2\) to solve for the radius at \(A_2\). Take the square root of the result after dividing the area by \(\pi\) to find \(r_2\).

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Key Concepts

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

Cross-Sectional Area of a Conductor
Understanding the cross-sectional area of a conductor is crucial when studying electrical circuits. Essentially, the cross-sectional area refers to the size of the surface where the electric current cuts through the conductor. It can be visualized like the face of a cut pipe, where the electricity flows through.

In technical terms, the cross-sectional area, represented by the letter 'A' in mathematical equations, is the area of the section perpendicular to the flow of current. For conductors with uniform shapes, such as wires, this area is consistently the same along their length. However, if the conductor has a non-uniform diameter, like in our exercise, this area can change along the length of the conductor, affecting the distribution of electric current.

When it comes to mathematics, the cross-sectional area of a circular conductor (like a wire) is calculated using the formula:
\[ A = \.pi r^2 \],
where \( r \) is the radius of the conductor at the particular section. This formula is derived from the geometric properties of a circle and is fundamental in understanding the relationship between current and conductor dimensions.
Electric Current and Flow
Electric current is a fundamental concept in electricity which refers to the flow of electric charge. Typically measured in amperes (A), current represents how much charge is flowing through a conductor per unit of time.

The flow of electric current through a conductor is not always uniform; it can vary based on several factors including the conductor's material, temperature, and particularly its cross-sectional area. The larger the area through which current can flow, the lower the resistance, and vice versa.

Current Density

Linked to electric current is an essential concept known as current density. It is denoted by \( J \) and represents the current flow per unit of cross-sectional area, mathematically expressed as:
\[ J = \frac{I}{A} \], where \( I \) is the electric current and \( A \) is the cross-sectional area.

Understanding current density is vital because it helps in determining how evenly the electric current is distributed across different sections of the conductor. A higher current density indicates a more significant amount of current flowing through a smaller area, which can lead to increased heat and potential damage to the conductor if not managed properly.
Circular Conductor
A circular conductor is a common type of conductor with a cross-section that is circular in shape. This kind of conductor, often seen in wires and cables, has a consistent geometry that simplifies calculations related to electrical resistance, current density, and conductance.

In a circular conductor, the cross-sectional area plays a vital role in determining its electrical properties. As mentioned previously, you can determine this area by using the formula for the area of a circle: \( A = \.pi r^2 \), where \( r \) is the radius. However, in the case of non-uniform conductors, where the diameter varies, this has a direct impact on the calculation of the current density.

Non-Uniform Circular Conductors

In non-uniform circular conductors, like in our textbook exercise, the radius鈥攁nd therefore the cross-sectional area鈥攃an change along the length of the conductor. This change means that the current density also varies at different points, which is why understanding how to calculate for both uniform and non-uniform cross-sections is essential in electrical engineering and physics. The concept also has real-world applications in designing electrical systems that need to handle varying current loads efficiently and safely.

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Most popular questions from this chapter

A current density of \(6.00 \times 10^{-13} \mathrm{A} / \mathrm{m}^{2}\) exists in the atmosphere at a location where the electric field is 100 \(\mathrm{V} / \mathrm{m}\) . Calculate the electrical conductivity of the Earth's atmosphere in this region.

A close analogy exists between the flow of energy by heat because of a temperature difference (see Section 20.7\()\) and the flow of electric charge because of a potential difference. The energy \(d Q\) and the electric charge \(d q\) can both be transported by free electons in the conducting material. Consequently, a good electrical conductor is usually a good thermal conductor as well. Consider a thin conducting slab of thickness \(d x,\) area \(A,\) and electrical conductivity \(\sigma,\) with a potential difference \(d V\) between opposite faces. Show that the current \(I=d q / d t\) is given by the equation on the left below: \(\begin{array}{ll}{} & {\text { Thermal Conduction }} \\ {\text { Charge Conduction }} & {(\text { Eq. } 20.14)} \\ {\frac{d q}{d t}=\sigma A\left|\frac{d V}{d x}\right|} & {\frac{d Q}{d t}=k A\left|\frac{d T}{d x}\right|}\end{array}\) In the analogous thermal conduction equation on the right, the rate of energy flow \(d Q / d t\) (in SI units of joules per second) is due to a temperature gradient \(d T / d x,\) in a material of thermal conductivity \(k\) . State analogous rules relating the direction of the electric current to the change in potential, and relating the direction of energy flow to the change in temperature.

A high-voltage transmission line with a diameter of 2.00 \(\mathrm{cm}\) and a length of 200 \(\mathrm{km}\) carries a steady current of 1000 \(\mathrm{A}\) . If the conductor is copper wire with a free charge density of \(8.49 \times 10^{28}\) electrons/m \(^{3}\) , how long does it take one electron to travel the full length of the line?

A Van de Graaff generator (see Figure 25.29\()\) is operating so that the potential difference between the high-voltage electrode \(B\) and the charging needles at \(A\) is 15.0 \(\mathrm{kV}\) . Calculate the power required to drive the belt against electrical forces at an instant when the effective current delivered to the high-voltage electrode is 500\(\mu A\) .

An \(11.0-\mathrm{W}\) energy-efficient fluorescent lamp is designed to produce the same illumination as a conventional \(40.0-\mathrm{W}\) incandescent lightbulb. How much money does the user of the energy-efficient lamp save during 100 hours of use? Assume a cost of \(\$ 0.0800 / \mathrm{kWh}\) for energy from the power company.
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