Problem 80 - Fundamentals of Physics

Question

An electron is moving at \(7.20 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a magnetic field of strength \(83.0 \mathrm{mT}\). What is the (a) maximum and (b) minimum magnitude of the force acting on the electron due to the field? (c) At one point the electron has an acceleration of magnitude \(4.90 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) What is the angle between the electron's velocity and the magnetic field?

Step-by-Step Solution

Answer

(a) Maximum force is \(9.55 \times 10^{-15} \, \text{N}\). (b) Minimum force is \(0\). (c) Angle is \(2.68^\circ\).

Step 1: Understand the Magnetic Force Formula

The magnetic force exerted on a moving charge like an electron can be calculated using the formula \( F = qvB \sin\theta \), where \( q \) is the charge of the electron \((-1.6 \times 10^{-19} \, C)\), \( v \) is the velocity of the electron, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Step 2: Calculate Maximum Force (\( \theta = 90^\circ \))

For the maximum force, \( \sin\theta = 1 \) (since \( \theta = 90^\circ \)), so the formula simplifies to \( F_{\text{max}} = qvB \). Substituting the given values, the calculation becomes:\[F_{\text{max}} = (1.6 \times 10^{-19} \, C)(7.20 \times 10^{6} \, \text{m/s})(83.0 \times 10^{-3} \, \text{T}) = 9.55 \times 10^{-15} \, \text{N}\]

Step 3: Calculate Minimum Force (\( \theta = 0^\circ \))

For the minimum force, \( \sin\theta = 0 \) (since \( \theta = 0^\circ \)), so the formula shows that \( F_{\text{min}} = 0 \). This occurs because the electron's velocity is parallel to the magnetic field lines, resulting in no force.

Step 4: Relate Magnetic Force to Acceleration

Given the magnitude of acceleration \( a = 4.90 \times 10^{14} \, \text{m/s}^2 \), we can use Newton's second law \( F = ma \) to find the force. Using the mass of an electron \( m = 9.11 \times 10^{-31} \, \text{kg} \), the force is calculated as:\[F = (9.11 \times 10^{-31} \, \text{kg})(4.90 \times 10^{14} \, \text{m/s}^2) = 4.46 \times 10^{-16} \, \text{N}\]

Step 5: Solve for Angle \( \theta \)

Using \( F = qvB \sin\theta \) and the force calculated in Step 4, solve for the sine of the angle:\[4.46 \times 10^{-16} \, = (1.6 \times 10^{-19} \, C)(7.20 \times 10^{6} \, \text{m/s})(83.0 \times 10^{-3} \, \text{T}) \sin\theta\]\[\sin\theta = \frac{4.46 \times 10^{-16}}{9.55 \times 10^{-15}} = 0.0467\]The angle \( \theta \) is \( \sin^{-1}(0.0467) \approx 2.68^\circ \).

Key Concepts

Magnetic Field Strength

Magnetic field strength, denoted by the symbol \( B \), is a measure of the magnetizing force. It determines how strongly the magnetic field can influence moving charges, like electrons. In our exercise, the magnetic field strength is given as \( 83.0 \, \text{mT} \), or milliTesla, which is equal to \( 83.0 \times 10^{-3} \text{T} \). It is essential to understand that the magnetic field exerts a force on a moving charge perpendicular to both the magnetic field and the velocity of the charge. The

• This relationship is governed by the formula \( F = qvB \sin\theta \).
• Here, \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity, and \( \theta \) is the angle between velocity and the magnetic field.

The field's strength directly influences the force. Thus, a stronger field means more force will act on the charge for the same velocity and magnetic orientation. This concept is vital for applications where manipulating charged particles is necessary, such as in cyclotrons and many types of magnetic sensors.

Electron Velocity

The velocity of an electron plays a crucial role in determining the magnetic force it experiences. In the given exercise, the electron moves with a velocity of \( 7.20 \times 10^6 \, \text{m/s} \). Velocity is a vector quantity, meaning it has both magnitude (value) and direction. This direction is significant when calculating magnetic force since the force depends on the angle \( \theta \) between the electron's velocity vector and the magnetic field vector.

• When the electrons move perpendicular to the magnetic field, \( \theta = 90^\circ \), the force is maximized because \( \sin \theta = 1 \).
• If the electron moves parallel to the field, \( \theta = 0^\circ \), and \( \sin \theta = 0 \), resulting in no magnetic force acting on the electron.

Thus, controlling velocity and its direction can be crucial in applications where it's important to manage or control the forces on electrons, such as in electron beams used in cathode ray tubes or other scientific instrumentation.

Newton's Second Law

Newton's second law of motion connects the force acting on an object and the object's acceleration. Formally, it is expressed as \( F = ma \), where \( F \) is the net force acting on the object, \( m \) is its mass, and \( a \) is the acceleration.In the context of an electron moving in a magnetic field, knowing the electron's mass (\( 9.11 \times 10^{-31} \, \text{kg} \)) and its acceleration \( 4.90 \times 10^{14} \, \text{m/s}^2 \) due to the magnetic force allows us to calculate the force using this principle.The calculated force helps further analyze the scenario:- It enables us to determine the angle \( \theta \) between the velocity and the magnetic field using the magnetic force formula \( F = qvB \sin\theta \).- Solving for \( \theta \) using the provided values and computed force reveals the exact orientation needed to reach the given acceleration condition.Understanding and applying Newton's second law in the context of magnetic forces helps bridge classical mechanics with electromagnetism, facilitating deeper insights into how charged particles behave under various forces.