A copper wire of radius a forms the inner conductor of a coaxial cable, and an imperfect insulator of conductivity \(\sigma\) and external radius \(b\) fills the space between the inner and outer conductors. Find the resistance per unit length of the insulator for currents flowing in it parallel to the wire. Find also the conductance per unit length for currents flowing radially outwards from the wire. Why is the conductance rather than the resistance asked for in the second case?

Short Answer

Expert verified
Answer: The resistance per unit length for currents flowing parallel to the wire is \(\frac{1}{\sigma}\), and the conductance per unit length for currents flowing radially outwards from the wire is \(\sigma\).

Step by step solution

01

Calculate electric field in the insulator

Let's consider a cylindrical volume of length L and radius 'r' in the insulator region. The electric field E(r) in the insulator between 'a' and 'b' is given by the equation: \(E(r) = \frac{J}{\sigma}\) where J is the current density at radius 'r'.
02

Resistance per unit length for currents flowing parallel to the wire

We can calculate the resistance per unit length when current flows parallel to the wire. The total current I flowing through the given cylindrical volume is: \(I = \int_{a}^{b} J 2 \pi r dr\) From Ohm's law, we have: \(V = IR\) Now, the resistance per unit length is: \(R = \frac{V}{I}\) By substituting the values for V and I in the above equation, we get: \(R = \frac{\int_{a}^{b} E(r) 2 \pi r dr}{\int_{a}^{b} J 2 \pi r dr}\) Since \(J = \sigma E(r)\), we can write: \(R = \frac{\int_{a}^{b} \frac{1}{\sigma} J 2 \pi r dr}{\int_{a}^{b} J 2 \pi r dr} = \frac{1}{\sigma} \frac{\int_{a}^{b} 2 \pi r dr}{\int_{a}^{b} 2 \pi r dr} = \frac{1}{\sigma}\)
03

Conductance per unit length for currents flowing radially outward from the wire

To calculate the conductance per unit length for currents flowing radially outward from the wire, we need to compute the inverse of the resistance described by Ohm's law: \(G = \frac{1}{R} = \sigma\) For the radial case, we are asked to find conductance instead of resistance because conductance is a more convenient parameter to express the property of the insulator when the current flows radially instead of parallel to the wire. So, the resistance per unit length for currents flowing parallel to the wire is \(\frac{1}{\sigma}\), and the conductance per unit length for currents flowing radially outwards from the wire is \(\sigma\).

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Key Concepts

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

Electric Field in Insulator
Understanding the behavior of an electric field in an insulator, such as the one found in a coaxial cable, is crucial for students studying electromagnetism and circuit analysis. An electric field arises due to the presence of electric charges and the electrical interaction between them. In an insulator, or dielectric material, the electric field does not cause a significant flow of electric charge; instead, it polarizes the material.

During polarization, the bound charges within the insulating material get slightly displaced from their equilibrium position, which creates a dipole moment. However, the insulator does not allow for free flowing current. For the case of the coaxial cable in the exercise, the electric field in the insulator region between the inner and outer conductors can be mathematically represented by the equation:\[\begin{equation} E(r) = \frac{J}{\sigma}\end{equation}\]Where 'E' is the electric field at a radius 'r' from the wire, 'J' is the current density, and '\(\sigma\)' signifies the material's conductivity. It's essential to note that the strength and direction of the electric field are influenced by the properties of the insulator and the electric charges on the conductors.
Current Density and Ohm’s Law
Current density is a vital concept in understanding how current is distributed over a cross-sectional area through which it flows. It is defined as the amount of electric current flowing per unit cross-sectional area. When dealing with the cylindrical geometry of a coaxial cable, current density can differ at various points depending on the radial distance from the center.

Ohm's law is a foundational principle in electrical engineering and physics, establishing a direct proportionality between voltage (V), current (I), and resistance (R). It can be represented as:\[\begin{equation} V = IR\end{equation}\]In the context of the provided exercise, when current flows parallel to the wire, the resistance per unit length can be derived using the relationship between the electric field and current density as well as Ohm's law. The Ohm’s law relationship helps us understand that resistance increases with length and decreases with cross-sectional area, a concept that's instrumental in calculating resistance in various electrical components.
Conductivity of Materials
Conductivity, often denoted as '\(\sigma\)', is a measure that tells us how well a material can conduct an electric current. It is the inverse of resistivity, which is a measure of how much a material resists the flow of current. Conductivity depends on the physical properties of the material, such as the presence of free carriers (electrons or holes in semiconductors), and is a significant factor in determining the performance of electrical components.

In the exercise, the conductivity of the insulator is a given value and influences how we calculate both resistance and conductance. Materials with high conductivity have lower resistance and higher conductance, facilitating the flow of electric current. Conversely, materials with low conductivity have higher resistance and lower conductance, inhibiting the current flow. Understanding the conductivity of materials is crucial for designing and analyzing electrical systems, as it impacts the efficiency and effectiveness of the electronic devices.
Resistance and Conductance per Unit Length
Resistance and conductance are two sides of the same coin. Resistance is the opposition to the flow of electric current, whereas conductance is a measure of how easily current can flow through a material. When evaluating the properties of coaxial cables, it is important to consider both resistance and conductance per unit length, as it influences signal integrity and power loss.

The resistance per unit length for currents flowing parallel to the wire in a coaxial cable can be calculated using the integral forms derived from Ohm’s law. On the other hand, for radial currents, as encountered in the provided exercise, conductance offers a more intuitive understanding of the material’s property. It simplifies the analysis, especially when dealing with the radial geometry of coaxial cables. It is often more convenient to express the insulating properties in terms of conductance when current flows perpendicularly to the cylindrical surface of the inner conductor.These concepts provide a deep understanding of how electrical signals and power are managed within different materials and geometries, which is paramount in the fields of electrical engineering and electronics.

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Most popular questions from this chapter

Show that the rate of production of heat per unit volume at a point in a conductor is \(\mathbf{J} \cdot \mathbf{E}\) or \(\sigma \mathbb{E}^{2}\), where \(\mathrm{J}\) is the current density, \(\mathrm{E}\) the electric field and \(\sigma\) the conductivity.

A d.c. source has an electromotance of \(120 \mathrm{~V}\) and a negligible internal resistance. If \(n\) cells, each of electromotance \(2.1 \mathrm{~V}\) and internal resistance \(0.1 \Omega\), are to be charged from this source with a charging current of \(3 \mathrm{~A}\), find the series resistance necessary. If \(n=20\), what proportion of energy delivered by the source is wasted as heat? Find also the potential difference across the 20 cells.

Three of the four arms of a Wheatstone bridge network are of \(40 \Omega\), the fourth is \(41 \Omega\) and the resistance of the detector is \(20 \Omega\). If the driving source has an electromotance of \(2 \mathrm{~V}\) and negligible internal resistance, find the current through the detector using Kirchhoff's laws.

A d.c generator in series with a resistance \(R\) is connected in parallel with both a battery of electromotance \(12 \mathrm{~V}\) and a resistive load. The electromotance of the generator may fluctuate, and \(R\) is of such a value that all the current to the load is supplied by the generator when its electromotance is \(60 \mathrm{~V}\). What fraction of the current does it supply when its electromotance drops to \(50 \mathrm{~V}\) ? Neglect the internal resistances of the generator and battery.

Two \(10 \Omega\) resistors are connected in series and a third resistor \(R\) is connected in parallel with one of them. What value of \(R\) makes the equivalent resistance of the whole combination equal to \(18 \Omega\) ?

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