/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Drift velocity in seawater The... [FREE SOLUTION] | 91影视

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Drift velocity in seawater The resistivity of seawater is about \(0.25\) ohm-meter. The charge carriers are chiefly \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions, and of each there are about \(3 \cdot 10^{26}\) per \(\mathrm{m}^{3}\). If we fill a plastic tube 2 meters long with seawater and connect a 12 volt battery to the electrodes at each end, what is the resulting average drift velocity of the ions?

Short Answer

Expert verified
The resulting average drift velocity of the ions in seawater is \(2.5 \times 10^{-5} m/s\).

Step by step solution

01

Determine the current using Ohm's Law

The formula for Ohm's Law is \(I = V / R\), where \(I\) is the current, \(V\) is the voltage, and \(R\) is the resistance. In this exercise, the resistivity (蟻) of seawater is given, therefore the resistance can be calculated using the formula \(R = 蟻L / A\), where \(L\) is the length of the tube and \(A\) is the cross-sectional area of the tube. However, as no information about the area is given, it is assumed that the total resistance doesn't include the cross-sectional area of the tube, therefore \(R = 蟻L\), which simplifies Ohm's law to \(I = V / (蟻L)\). Substituting the values we get, \(I = 12 / (0.25 \times 2) = 24 A\).
02

Apply the formula for drift velocity

Drift velocity (\(谓_d\)) can be computed using the formula \(谓_d = I / (nqA)\), where \(谓_d\) is the drift velocity, \(I\) is the current, \(n\) is the number density of charge carriers (both Na+ and Cl- ions), \(q\) is the charge of the ions and \(A\) is the cross-sectional area. As mentioned before, the total drift velocity doesn't consider the cross-sectional area of the tube, so the formula simplifies to \(谓_d = I / (nq)\). The charge \(q\) of these ions is the charge of an electron, \(1.6 \times 10^{-19} C\). So, \(谓_d = 24 / ((3 \times 10^{26}) \times 2 \times 1.6 \times 10^{-19}) = 2.5 \times 10^{-5} m/s\). Therefore, the drift velocity of the ions is \(2.5 \times 10^{-5} m/s\).

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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 physics that relates voltage, current, and resistance in an electrical circuit. It's expressed as \(I = \frac{V}{R}\), where \(I\) represents the current in amperes, \(V\) is the voltage in volts, and \(R\) is the resistance in ohms. This law means that the current flowing through a conductor between two points is directly proportional to the voltage across the two points.
In practical terms, if you increase the voltage, the current increases, provided that the resistance stays the same. This principle is used widely in electrical engineering to design circuits. In the drift velocity exercise, Ohm鈥檚 Law helps us determine the current in the seawater-filled tube by rearranging the formula to solve for \(I\):
  • \(I = \frac{V}{(\rho L)}\)
This allows us to determine how much current flows, given a specific voltage and resistance.
Resistivity
Resistivity is a property of a material that describes its ability to resist electric current. It is denoted by \(\rho\) and has units of ohm-meters (\(\Omega \cdot m\)). A higher resistivity indicates that the material does not allow current to flow easily.
Resistivity depends on the material's properties and also varies with temperature. In our exercise, seawater has a resistivity of 0.25 ohm-meter. This means seawater conducts electricity fairly well due to the presence of ions.
  • Formula for resistance using resistivity: \(R = \frac{\rho L}{A}\)
Where \(L\) is the length of the conductor and \(A\) is its cross-sectional area. By knowing the resistivity, we can calculate how much the seawater resists the flow of current.
Charge Carriers
Charge carriers are particles that carry an electric charge, enabling the flow of current in a conductor. In metallic conductors, charge carriers are typically electrons. However, in solutions like seawater, the charge carriers are ions, specifically \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions.
These ions carry charges as they move through the solution due to an applied voltage. The number density \(n\) of these ions, given as \(3 \times 10^{26} \; \text{ions/m}^3\), indicates how many ions are available to carry the charge in a cubic meter of the solution.
Understanding the nature and behavior of these charge carriers is crucial for calculating drift velocity as it determines how fast these ions move on average.
Current
Current is the flow of electric charge in a conductor, measured in amperes (A). It quantifies the amount of charge passing through a point in a circuit per unit time. In our exercise, the current is calculated using Ohm鈥檚 Law and is essential for determining the drift velocity of ions in seawater.
The calculated current in the seawater tube, when connected to a 12-volt battery, is 24 A. This gives an idea of the rate of flow of charge through the water.
  • Current formula: \(I = \frac{V}{R}\)
Knowing the current helps conclude how effectively the ions in the seawater are contributing to electric conduction, which directly affects drift velocity.
Ions in Solution
Ions in a solution play an important role in conducting electricity. In seawater, ions like \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) move through the water when an electric field is present, contributing to the flow of current.
These ions are the charge carriers in the solution, and their concentration affects how well the solution can conduct electricity.
  • The higher the ion concentration, the better the conductivity.
  • The charge of ions (\(q\)) is the same as the charge of an electron, about \(1.6 \times 10^{-19} \text{C}\).
By understanding these ions' behavior, we can calculate their drift velocity, which tells us the average speed at which these ions move through the solution under the influence of a voltage.

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Most popular questions from this chapter

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