Problem 5 - Fundamentals of Physics

Question

A beam contains \(2.0 \times 10^{8}\) doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). What are the (a) magnitude and (b) direction of the current density \(\vec{J} ?\) (c) What additional quantity do you need to calculate the total current \(i\) in this ion beam?

Step-by-Step Solution

Answer

(a) 6.4 A/m², (b) north, (c) cross-sectional area.

Step 1: Understand the Concept of Current Density

The current density \( \vec{J} \) is defined as the amount of electric current flowing per unit area of a cross-section. It can be calculated using the formula: \[ \vec{J} = nq\vec{v} \] where \( n \) is the number of charge carriers per unit volume, \( q \) is the charge of each carrier, and \( \vec{v} \) is the drift velocity of the charge carriers.

Step 2: Identify Known Quantities

Given in the problem are the following: - Number density of ions, \( n = 2.0 \times 10^8 \) ions/cm³ (or \( 2.0 \times 10^{14} \) ions/m³ after converting units). - Velocity of ions, \( v = 1.0 \times 10^5 \) m/s. - Charge of each doubly charged ion, which is \( q = 2e \), where \( e = 1.6 \times 10^{-19} \) C.

Step 3: Calculate the Magnitude of Current Density

Using the formula \( \vec{J} = nq\vec{v} \), substitute in the known quantities to find the magnitude of current density:\[ \begin{align*}\vec{J} &= n \cdot q \cdot v \&= 2.0 \times 10^{14} \text{ ions/m}^3 \times 2 \times 1.6 \times 10^{-19} \text{ C/ion} \times 1.0 \times 10^5 \text{ m/s} \&= 6.4 \text{ A/m}^2.\end{align*} \]

Step 4: Determine the Direction of Current Density

The direction of the current density is the same as the direction of the flow of positive charges, which in this case is north. Thus, \( \vec{J} \) points north.

Step 5: Identify Additional Quantity for Total Current

To calculate the total current \( i \) in the ion beam, you need to know the cross-sectional area \( A \) through which the ions are flowing. Given the current density \( \vec{J} \), the total current can be calculated using the relation \( i = J \cdot A \).

Key Concepts

Charge Carriers

Charge carriers are the particles that carry electric charge through a conductive medium, enabling the flow of electrical current. In many materials, such as metals, the charge carriers are electrons, which are negatively charged particles. In the case of the ion beam mentioned in the exercise, the charge carriers are ions. These ions can have either positive or negative charges. Here, the ions are doubly charged positive ions, which means each ion carries twice the elementary charge (i.e., the charge of a proton). The elementary charge, denoted as \(e\), has an approximate value of \(1.6 \times 10^{-19}\) Coulombs. Doubling this value gives us the charge of each ion in the beam. Charge carriers are essential for understanding how current is generated and what affects its magnitude and direction.

Drift Velocity

Drift velocity is the average velocity that a particle, such as an electron or ion, acquires due to an electric field. In a conductor, when an electric field is applied, the charge carriers begin to move, and this movement is characterized by a drift velocity. It is different from the random thermal motion of these particles, as it is directional. For the ion beam scenario described, all ions move northward with a drift velocity of \(1.0 \times 10^5\) m/s. This consistent northward motion contributes to the current density in that direction. Drift velocity is crucial in determining the timing and means by which charge moves through a medium, directly affecting the current flow and its characteristics.

Current Formula

The current formula is a fundamental relationship in electromagnetism, relating the flow of charge to time and space. In its most basic form, the current \( i \) can be calculated as the product of the charge density per volume, the charge of each carrier, and the drift velocity. Mathematically, current density \( \vec{J} \) is expressed as: \[ \vec{J} = nq\vec{v} \] where:

• \( n \) is the charge carrier density,
• \( q \) is the charge per carrier,
• \( \vec{v} \) is the drift velocity.

By multiplying these three quantities, you find the current density, which is the amount of current per unit area. This formula is pivotal when working with engineering and scientific applications involving currents in conductive materials.

Cross-Sectional Area

The concept of cross-sectional area is important when calculating the total current flowing through any medium. It refers to the area of the slice or cross-section of the medium that the current flows through. For a circular wire, this would be the area of the circle that the wire forms. Since current density \( \vec{J} \) is defined as the current per unit area, knowing the cross-sectional area \( A \) is key in determining the total current \( i \) using the formula: \[ i = J \cdot A \] Here, \( J \) represents the current density, and \( A \) represents the cross-sectional area. The total current is essentially an accumulation of all the micro-currents over the entire area. When solving problems involving the flow of electricity, such as the ion beam exercise, calculating the total current requires understanding and applying the concept of cross-sectional area effectively.