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Chapter 3: Problem 22

Transatlantic telegraphic cable *? The first telegraphic messages crossed the Atlantic in 1858 , by a cable \(3000 \mathrm{~km}\) long laid between Newfoundland and Ireland. The conductor in this cable consisted of seven copper wires, each of diameter \(0.73 \mathrm{~mm}\), bundled together and surrounded by an insulating sheath. (a) Calculate the resistance of the conductor. Use \(3.10^{-8}\) ohmmeter for the resistivity of the copper, which was of somewhat dubious purity. (b) A return path for the current was provided by the ocean itself. Given that the resistivity of seawater is about \(0.25\) ohm-meter, see if you can show that the resistance of the ocean return would have been much smaller than that of the cable. (Assume that the electrodes immersed in the water were spheres with radius, say, \(10 \mathrm{~cm}\).)

Short Answer

Expert verified
The resistance of the conductor is approximately 3.16 M\(\Omega\), while the resistance of the ocean return is about 0.2 \(\Omega\). Therefore, the resistance of the ocean return is significantly smaller compared to the conductor.

Step by step solution

01

Calculate the resistance of the conductor.

First, calculate the cross-sectional area of one wire using the formula \(A=\pi r^{2}\), where \(r\) is the radius (half of the diameter). This gives \(A=\pi (0.73 \mathrm{~mm}/2)^{2} \approx 0.42 \mathrm{~mm^{2}}\). Multiply this by 7, as there are seven copper wires, to give a total area of \(0.42*7 \approx 2.94 \mathrm{~mm^{2}}\). Next, convert that area to meters: \(2.94 \mathrm{~mm^{2}} = 2.94*10^{-6} \mathrm{~m^{2}}\). Then use the formula for the resistance of a cylindrical conductor, \(R=\rho \frac{L}{A} =3.10^{-8}\ \mathrm{~ohm.m} * \frac{3000 \mathrm{~km}}{2.94*10^{-6} \mathrm{~m^{2}}} \approx 3155221.24 \mathrm{~ohm}\). The resistance of the conductor is therefore about 3.16 M\(\Omega\).
02

Calculate the resistance of the ocean return.

To calculate the resistance of the ocean return, use the formula for the resistance of a spherical conductor: \(R=\frac{\rho}{4\pi r}\), where \(\rho\) is resistivity and \(r\) is radius. Here, \(\rho = 0.25 \mathrm{~ohm.m}\) and \(r = 10\mathrm{~cm} = 0.1\mathrm{~m}\). That gives \(R=\frac{0.25 \mathrm{~ohm.m}}{4\pi *0.1 \mathrm{~m}} \approx 0.200 \mathrm{~ohm}\). Therefore, the resistance of the ocean return would have been about 0.20 \(\Omega\).
03

Compare the resistances.

Comparing the calculated resistances, the resistance of the conductor is \(3155221.24 \Omega\) whereas the resistance of the ocean return is \(0.200 \Omega\). Thus, the resistance of the ocean return would have been much smaller than that of the cable.

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Key Concepts

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

Resistivity
Resistivity is a fundamental property of materials that indicates how much they resist the flow of electric current.
It is denoted by the Greek letter \( \rho \).
Materials with low resistivity, like copper, allow easier passage of electrons and thus conduct electricity more efficiently.

The resistivity of a material is affected by its composition and purity.
This means that impurities, even in small amounts, can significantly increase resistivity, making a conductor less effective.
The unit of measurement for resistivity is the ohm meter (\( \Omega \cdot \text{m} \)).

In this exercise, we used a resistivity of \(3.10^{-8}\) ohm-meter for copper.
This value represents the resistance of pure copper, although the copper in the telegraphic cable might have been slightly more resistive due to impurities.
Understanding resistivity and accurately measuring it are crucial for designing electrical systems like the telegraph cable.
Cylindrical Conductor
A cylindrical conductor, like a copper wire, has a circular cross-section and a defined length.
The resistance of a cylindrical conductor is calculated using the formula \( R = \rho \frac{L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

The conductor in the telegraph cable was made of seven copper wires, each with a diameter of 0.73 mm.
We computed the cross-sectional area of one wire and multiplied by seven to get the total area for all wires combined.

This characteristic cylindrical shape is common in many electrical applications because it provides a straightforward path for electricity with minimal obstructions.
By using the formula, the resistance of the wire can be calculated efficiently, helping engineers to predict electrical performance in real-world scenarios.
Spherical Conductor
Spherical conductors differ from cylindrical ones in that they are three-dimensional solid spheres.
In this context, the spherical conductor concept is applied to the electrodes used in sea water, which were spherical with a 10 cm radius.

The resistance for a spherical conductor is given by the formula \( R = \frac{\rho}{4\pi r} \), where \( r \) is the radius.
This formula is particularly useful for defining point-contact resistance in cases like oceans or large bodies of water.

In the exercise, the computed resistance from the ocean as a spherical conductor was significantly lower than that of the lengthy telegraph cable, showing the effectiveness of spherical conductors in certain environmental conditions.
These principles help in designing electrodes that minimize resistance and improve efficiency for transmitting electrical signals over large distances.
Telegraphic Communication
Telegraphic communication was a revolutionary advancement in the 19th century, facilitating faster long-distance communication.
The transatlantic telegraph cable laid between Newfoundland and Ireland in 1858 was a monumental step in connecting continents.

This communication method required effective conductors to transmit signals with minimal loss.
Understanding and calculating resistance for both the cables and the ocean return path was vital to ensure the reliability of this system.

The telegraphic cable was crucial not only for communication but also for setting a foundation for future electrical engineering solutions.
Learning about telegraphic communication provides insight into the early challenges and solutions that paved the way for modern technology.
Seawater Conductivity
Seawater conductivity refers to how well seawater can conduct electricity.
This property is highly relevant to situations like the telegraphic cable, where the ocean provided a return path for electric current.

Seawater has relatively high conductivity due to the dissolved salts and minerals, making it much less resistive compared to many other natural mediums.
In the original exercise, this conductivity allowed the ocean to serve effectively as a low-resistance return path.

The resistivity of seawater is typically around 0.25 ohm-meter, which is significantly higher than metal conductors but reasonable for salient environments.
Thus, understanding seawater conductivity is key to designing systems that interact with or rely on bodies of water for electrical transmission.
These properties of seawater were crucial in reducing the overall resistance of the telegraphic network across the Atlantic.

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Most popular questions from this chapter

Electron beam A beam of \(9.5\) megaelectron-volt (MeV) electrons \((\gamma \approx 20)\), amounting as current to \(0.05 \mu \mathrm{A}\), is traveling through vacuum. The transverse dimensions of the beam are less than \(1 \mathrm{~mm}\), and there are no positive charges in or near it. (a) In the lab frame, what is the average distance between an, electron and the next one ahead of it, measured parallel to the beam? What approximately is the average electric field strength 1 cm away from the beam? (b) Answer the same questions for the electron rest frame.

Point charge near a corner Locate two charges \(q\) each and two charges \(-q\) each on the corners of a square, with like charges diagonally opposite one another. Show that there are two equipotential surfaces that are planes. In this way sketch qualitatively the field of the system where a single point charge is located symmetrically in the inside corner formed by bending a metal sheet through a right angle. Which configurations of conducting planes and point charges can be solved this way and which can't? How about a point charge located on the bisector of a \(120^{\circ}\) dihedral angle between two conducting planes?

This exercise is more of a math problem than a physics problem, so maybe it doesn't belong in this book. But it's a fun one. Consider a series of events that happen at independent random times, such as the collisions in Section \(4.4\) that led to the issue discussed in Footnote \(8 .\) Such a process can be completely characterized by the probability per unit time (call it \(p\) ) of an event happening. The definition of \(p\) is that the probability of an event happening in an infinitesimal \(^{16}\) time \(d t\) equals \(p d t\). (a) Show that starting at any particular time (not necessarily the time of an event), the probability that the next event happens between \(t\) and \(t+d t\) later equals \(e^{-p t} p d t\). You can do this by breaking the time interval \(t\) into a large number of tiny intervals, and demanding that the event does not happen in any of them, but that it does happen in the following \(d t\). (You will need to use the fact that \((1-x / N)^{N}=e^{-x}\) in the \(N \rightarrow \infty\) limit.) Verify that the integral of \(e^{-p t} p d t\) correctly equals 1 . (b) Show that starting at any particular time (not necessarily the time of an event), the average waiting time (also called the expectation value of the waiting time) to the next event equals \(1 / p\). Explain why this is also the average time between events. (c) Pick a random point in time, and look at the length of the time interval (between successive events) that it belongs to. Explain, using the above results, why the average length of this interval is \(2 / p\), and not \(1 / p\). (d) We have found that the average time between events is \(1 / p\), and also that the average length of the interval surrounding a randomly chosen point in time is \(2 / p .\) Someone might think that these two results should be the same. Explain intuitively why they are not. (e) Using the above probability distribution \(e^{-p t} p d t\) properly, show mathematically why \(2 / p\) is the correct result for the average length of the interval surrounding a randomly chosen point in time.

A three-shell capacitor A capacitor consists of three concentric spherical shells with radii \(R, 2 R\), and \(3 R\). The inner and outer shells are connected by a wire (passing through a hole in the middle shell, without touching it), so they are at the same potential. The shells start neutral, and then a battery transfers charge from the middle shell to the inner and outer shells. (a) If the final charge on the middle shell is \(-Q\), what are the charges on the inner and outer shells? (b) What is the capacitance of the system? (c) If the battery is disconnected, what happens to the three charges on the shells if charge \(q\) is added to the outer shell?

Maximum energy storage between cylinders ** We want to design a cylindrical vacuum capacitor, with a given radius \(a\) for the outer cylindrical shell, that will be able to store the greatest amount of electrical energy per unit length, subject to the constraint that the electric field strength at the surface of the inner cylinder may not exceed \(E_{0}\). What radius \(b\) should be chosen for the inner cylindrical conductor, and how much energy can be stored per unit length?
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