Problem 79 - Fundamentals of Physics

Question

In a hypothetical fusion research lab, high temperature helium gas is completely ionized and each helium atom is separated into two free electrons and the remaining positively charged nucleus, which is called an alpha particle. An applied electric field causes the alpha particles to drift to the east at \(25.0 \mathrm{~m} / \mathrm{s}\) while the electrons drift to the west at \(88.0 \mathrm{~m} / \mathrm{s}\). The alpha particle density is \(2.80 \times 10^{15} \mathrm{~cm}^{-3}\). What are (a) the net current density and (b) the current direction?

Step-by-Step Solution

Answer

(a) 61.824 \( \mathrm{~A/m}^{2} \) east, (b) current flows east.

Step 1: Understand the problem

The problem is asking for the net current density and the direction of the current in a helium gas where alpha particles and electrons drift due to an electric field. The drift velocities of alpha particles and electrons, along with the alpha density, are provided.

Step 2: Write out the necessary formulas

The current density \( J \) is given by the sum of the contributions from positive and negative charges: \( J = n_q \, q \, v_q + n_e \, e \, v_e \), where \( n_q \) is the density of alpha particles, \( q \) is the charge of an alpha particle, \( v_q \) is the drift velocity of alpha particles, \( n_e \) is the density of electrons, \( e \) is the charge of an electron, and \( v_e \) is the drift velocity of electrons.

Step 3: Convert the density to appropriate units

Convert the alpha particle density from \( \mathrm{cm}^{-3} \) to \( \mathrm{m}^{-3} \): \( 2.80 \times 10^{15} \mathrm{~cm}^{-3} = 2.80 \times 10^{21} \mathrm{~m}^{-3} \).

Step 4: Calculate charges for particles

The charge of an alpha particle (which has 2 protons) is twice that of a proton: \( q = 2e = 2 \times 1.6 \times 10^{-19} \mathrm{~C} \). The electron charge is \( e = -1.6 \times 10^{-19} \mathrm{~C} \).

Step 5: Calculate individual current contributions

For the alpha particles: \( J_{\alpha} = n_q \, q \, v_q = 2.80 \times 10^{21} \, \mathrm{m}^{-3} \times 3.2 \times 10^{-19} \mathrm{~C} \times 25 \mathrm{~m/s} = 22.4 \mathrm{~A/m}^{2} \) to the east. For the electrons: \( J_{e} = n_e \, e \, v_e = 2.80 \times 10^{21} \mathrm{~m}^{-3} \times (-1.6 \times 10^{-19} \mathrm{~C}) \times (-88) \mathrm{~m/s} = 39.424 \mathrm{~A/m}^{2} \) to the east (since both signs are negative, direction is opposite to original).

Step 6: Combine contributions for net current density

Add the contributions from alpha particles and electrons: \( J_{net} = J_{\alpha} + J_{e} = 22.4 \mathrm{~A/m}^{2} + 39.424 \mathrm{~A/m}^{2} = 61.824 \mathrm{~A/m}^{2} \). The net current is directed towards the east.

Key Concepts

Drift Velocity

Drift velocity is a key concept when studying the motion of charged particles in response to an electric field. It refers to the average velocity that a charged particle such as an electron or ion attains due to an electric field. In our example, both alpha particles and electrons have specific drift velocities:

• Alpha particles drift eastward at 25.0 m/s.
• Electrons drift westward at 88.0 m/s, but since electrons have a negative charge, their contribution to current is considered in the opposite direction of their physical motion.

The drift velocity plays a crucial role in determining the net current density, as it is directly proportional to the current: larger drift velocities result in higher current contributions from those particles.
In metals or plasmas, where electric fields influence the motion of charged particles, understanding drift velocities helps in analyzing and predicting electrical behavior.

Alpha Particle

An alpha particle, which is commonly found in nuclear reactions and fusion labs, can be understood as a helium nucleus. It consists of two protons and two neutrons, giving it a positive charge due to the protons. In the scenario described, the helium gas is completely ionized:

• This means each helium atom loses its two electrons, resulting in a positively charged alpha particle.
• Alpha particles are significant in determining the current because they add to the positive charge moving through the electric field.

Being positively charged, these alpha particles move in the direction of the electric field applied, contributing directly to the current's direction and magnitude.
Because the charge of an alpha particle is twice that of a single proton, they have a charge of approximately 3.2 × 10^{-19} C, which is crucial in calculating current density.

Electric Field

Electric fields are a fundamental concept in electromagnetism. They exert force on charged particles, causing them to move, and hence are the driving force behind drift velocity.

• In this problem, an electric field causes alpha particles to drift to the east and electrons to drift to the west.
• The direction of the electric field aligns with the direction of positive charge movement, hence to the east in our case.
• This field is pivotal in causing the separation and motion of charged particles, which in turn affects current density.

Studying electric fields allows us to understand not only the motion of individual charge carriers but also the collective effect on macroscopic properties such as current flow.

Ionization

Ionization is the process by which atoms or molecules gain or lose electrons, creating ions. In the context of high-temperature fusion environments, ionization plays a significant role:

• Helium gas being fully ionized means that all its electrons are removed, allowing the remaining helium nucleus to become an alpha particle.
• This transformation into charged ions enables the external electric field to influence the gas significantly.
• Ionization increases the number of charge carriers available in the system, amplifying current flow.

Understanding ionization is essential because it facilitates widespread movements of electrons and positive ions, leading to pronounced electrical effects under an electric field. The complete ionization of helium gas is a critical step that determines the conduction properties in this scenario.