Problem 8 - Fundamentals of Physics

Question

A small but measurable current of \(1.2 \times 10^{-10}\) A exists in a copper wire whose diameter is \(2.5 \mathrm{~mm}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

Step-by-Step Solution

Answer

(a) The current density is approximately \(2.44 \times 10^{-5} \text{ A/m}^2\). (b) The electron drift speed is approximately \(1.79 \times 10^{-5} \text{ m/s}\).

Step 1: Calculate the Cross-sectional Area of the Wire

The wire has a circular cross-section. The diameter of the wire is given as 2.5 mm. First, convert this diameter into meters: 2.5 mm = 0.0025 m. The radius is half of the diameter. Therefore, radius \( r = 0.0025 / 2 = 0.00125 \) m. The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle: \[ A = \pi r^2 \] Plug in the radius: \[ A = \pi (0.00125)^2 \approx 4.91 \times 10^{-6} \text{ m}^2 \].

Step 2: Calculate Current Density

Current density \( J \) is defined as the current \( I \) per unit area \( A \). Use the formula: \[ J = \frac{I}{A} \] where \( I = 1.2 \times 10^{-10} \text{ A} \) and \( A = 4.91 \times 10^{-6} \text{ m}^2 \). Substitute these values in the formula: \[ J = \frac{1.2 \times 10^{-10}}{4.91 \times 10^{-6}} \approx 2.44 \times 10^{-5} \text{ A/m}^2 \].

Step 3: Using the Formula of Electron Drift Speed

The electron drift speed \( v_d \) is related to current density \( J \) and the number of charge carriers \( n \) per unit volume by the equation: \[ J = n e v_d \] where \( e \) is the elementary charge \(\approx 1.6 \times 10^{-19} \) C. We need to rearrange this equation to solve for \( v_d \): \[ v_d = \frac{J}{ne} \].

Step 4: Calculate Electron Drift Speed

Plug in the values into the formula \( v_d = \frac{J}{ne} \): \( J = 2.44 \times 10^{-5} \text{ A/m}^2 \), \( n = 8.49 \times 10^{28} \text{ m}^{-3} \), and \( e = 1.6 \times 10^{-19} \text{ C} \). Calculate: \[ v_d = \frac{2.44 \times 10^{-5}}{8.49 \times 10^{28} \times 1.6 \times 10^{-19}} \approx 1.79 \times 10^{-5} \text{ m/s} \].

Key Concepts

Current Density

In electromagnetism, **current density** is a crucial concept that helps us understand how electric current is distributed over a given area. Specifically, it is defined as the amount of electric current flowing per unit area of cross-section of a conductor. This provides insight into how tightly packed the flow of electrons is within that conductor.

To calculate current density (\( J \)), you use the formula:

• \( J = \frac{I}{A} \)

Where:

• \( I \) is the current in amperes.
• \( A \) is the cross-sectional area in square meters.

In the exercise, the provided example shows how to calculate the current density of a copper wire with a given diameter using the area of a circle formula for its cross-section. By dividing the current by this area, we derived a current density of \( 2.44 \times 10^{-5} \) A/m².

Understanding current density is vital because it affects the conductor's ability to carry electric current efficiently. Higher current density can lead to increased temperature in the material, possibly requiring cooling or a larger cross-section to manage the heat adequately.

Electron Drift Speed

The concept of **electron drift speed** is essential for understanding how quickly electrons move through a conductor when an electric current is present. Even though electrons travel rapidly due to thermal energy, their drift speed under an electric field is much slower.

We calculate electron drift speed (\( v_d \)) using the relationship:

• \( J = n e v_d \)
Here:
• \( J \) is the current density in A/m².
• \( n \) is the number of charge carriers per unit volume.
• \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) C.
• \( v_d \) is what we're trying to find.

Rearranging for \( v_d \), we have:
• \( v_d = \frac{J}{ne} \)
In the exercise, the values for \( J \), \( n \), and \( e \) were substituted into this equation to calculate an electron drift speed of approximately \( 1.79 \times 10^{-5} \) m/s. This velocity is indeed slow because electrons frequently collide with atoms within the conductor, which impedes their movement.

Understanding electron drift speed can help predict how changes in materials or electrical conditions will affect the speed of electron flow, which in turn impacts the functionality of electric circuits.

Charge Carriers

In the world of electromagnetism, **charge carriers** are particles or holes that carry electric charge, and they are crucial for the conduction of current in materials. In metals, the charge carriers are generally electrons. Understanding charge carriers helps explain how materials conduct electricity.

Charge carriers are characterized by their **number density** (\( n \)), which is the number of carriers per unit volume. In the exercise, this value is given as \( 8.49 \times 10^{28} \) m³ for a copper wire. This high number is typical for a good conductor like copper, which has many available electrons for each atom.

The role of charge carriers is vital, as they determine how well a material can transport electric charge. When an electric field is applied to a conductor, these carriers move, creating an electric current. The efficiency of this movement and the resulting current depend on both the quantity and mobility of these carriers.

Conductors with higher charge carrier density tend to conduct electricity more efficiently, as more carriers can contribute to carrying the current. This principle underlies the selection of materials for different electrical applications, ensuring that the material can handle the required electrical load.