Question

A current of \(0.123 \mathrm{~mA}\) flows in a silver wire whose cross-sectional area is \(0.923 \mathrm{~mm}^{2}\) a) Find the density of electrons in the wire, assuming that there is one conduction electron per silver atom. b) Find the current density in the wire assuming that the current is uniform. c) Find the electron's drift speed.

Short answer

Answer: The density of electrons in the silver wire is approximately 5.86 x 10^28 electrons/m³. The current density is approximately 1.33 x 10² A/m², and the electron's drift speed is approximately 1.44 x 10^-5 m/s.

Step-by-step solution

Finding the density of electrons in the wire

First, we must calculate the number of conduction electrons per unit volume of the wire. We know that silver (Ag) has one conduction electron per atom. The molar mass of silver (Ag) is approximately \(M_{Ag} = 107.87 \ \frac{g}{mol}\). We also have Avogadro's number, \(N_A = 6.022 \times 10^{23} \ \frac{atoms}{mol}\). Now, we need to find the density of silver. This value is \(\rho_{Ag} = 10.49 \ \frac{g}{cm^3}\). Using these values, we can determine the number of conduction electrons per unit volume: \(N = \frac{\rho_{Ag} \cdot N_A}{M_{Ag}}\) \(N = \frac{10.49 \ \frac{g}{cm^3} \cdot 6.022 \times 10^{23} \ \frac{atoms}{mol}}{107.87 \ \frac{g}{mol}}\) \(N \approx 5.86 \times 10^{28} \ \frac{electrons}{m^3}\)

Finding the current density

Next, let's find the current density, \(J\), in the wire. The formula to calculate it is \(J = \frac{I}{A}\), where \(I\) is the current and \(A\) is the cross-sectional area. We're given that \(I = 0.123 \ \text{mA}\) and \(A = 0.923 \ \text{mm}^2\). First, convert the current to amperes (A) and the cross-sectional area to square meters (m^2): \(I = 0.123 \ \text{mA} \cdot \frac{1 \ \text{A}}{1000 \ \text{mA}} = 1.23 \times 10^{-4} \ \text{A}\) \(A = 0.923 \ \text{mm}^2 \cdot \frac{1 \ \text{m}}{1000 \ \text{mm}} \cdot \frac{1 \ \text{m}}{1000 \ \text{mm}} = 9.23 \times 10^{-7} \ \text{m}^2\) Now, we can find the current density: \(J = \frac{1.23 \times 10^{-4} \ \text{A}}{9.23 \times 10^{-7} \ \text{m}^2} \approx 1.33 \times 10^2 \ \mathrm{\frac{A}{m^2}}\)

Finding the electron's drift speed

Lastly, we need to find the drift speed of the electrons, \(v_d\). The formula to calculate the drift speed is given by \(v_d = \frac{J}{nq}\), where \(J\) is the current density, \(n\) is the density of conduction electrons, and \(q\) is the elementary charge (charge of an electron). We know that the charge of an electron is \(q = 1.6 \times 10^{-19} \ \mathrm{C}\). Using the values for \(J\) and \(N\) that we calculated in the previous steps, and the value of \(q\), we can find the drift speed: \(v_d = \frac{1.33 \times 10^2 \ \mathrm{\frac{A}{m^2}}}{5.86 \times 10^{28} \ \frac{electrons}{m^3} \cdot 1.6 \times 10^{-19} \ \mathrm{C}}\) \(v_d \approx 1.44 \times 10^{-5} \ \mathrm{\frac{m}{s}}\) So, the drift speed of the electrons in the silver wire is approximately \(1.44 \times 10^{-5} \ \mathrm{\frac{m}{s}}\).

Key concepts

Density of Electrons

Understanding the density of electrons in a conductor such as a silver wire is pivotal to comprehend how electrical conductivity works. In our exercise, we calculate the number of conduction electrons per unit volume by considering silver's properties. It is significant to remember that every silver atom contributes one conduction electron—the particle responsible for conducting electricity through the metal. The density of electrons is derived using the atomic mass of silver, Avogadro's number, and the density of the material. These quantities forge a path to calculate the overall electron density.

The concept of electron density is akin to measuring the population density of a city - it reflects how many individuals (in this case, electrons) occupy a certain volume. The higher the electron density, the more 'crowded' our city of conduction electrons becomes. This 'crowdedness' has direct implications on the wire's ability to conduct electricity. In a nutshell, a greater density of electrons typically signifies a material with better conductive properties, given that there are more charge carriers available to transport the current.

Current Density

When students encounter the term 'current density', they can visualize it as the flow of traffic moving through a highway. Current density, represented by the symbol J, measures how much current I flows through a unit area A of the conductor. In our exercise, we examined a silver wire and calculated its current density given the current and cross-sectional area. We converted both into consistent units to facilitate our calculation - from milliamperes to amperes for current, and from square millimeters to square meters for area. The resulting value of current density provides us an idea of the intensity of the current flow.

In practical applications, knowing the current density helps engineers design conductors that can handle the required current without overheating or suffering from adverse effects due to high traffic of electrons. It's a delicate balance - too much traffic and there may be proverbial 'accidents' or 'traffic jams', in the form of overheating and energy loss.

Conduction Electrons

Conduction electrons are the free agents in the world of electrical conductivity, capable of moving under the influence of an electric field. In metals like silver, these electrons do not belong to any specific atom; instead, they move freely throughout the metal's lattice structure. This freedom is what allows electrical current to pass through. They are, in essence, the vehicles that drive current along the highway of the conductor.

Each type of conductor has a specific number of conduction electrons. In the case of silver, with each atom contributing one free electron, we can ascertain a high electron density conducive to efficient conduction. However, despite having many free electrons, their individual drift speed, as seen in our problem, is surprisingly slow. When a voltage is applied, the electron traffic starts moving, albeit at a snail's pace, and yet the current flow is collective and rapid enough to power electrical devices and systems.