Problem 12 - College Physics

Question

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Step-by-Step Solution

Answer

(a) Largest acceleration is non-zero (full formula), smallest is zero. (b) Calculate angle with given conditions.

Step 1: Understand the formula for magnetic force

The magnetic force on a charged particle is given by the formula: \[ F = qvB \sin \theta \]where \( q \) is the charge of the particle, \( v \) is the velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between velocity and the magnetic field.

Step 2: Calculate the largest force (\(\theta = 90^\circ\))

The largest possible force occurs when \( \sin \theta = 1 \) (i.e., \( \theta = 90^\circ \)). So, the maximum force is:\[ F_{max} = qvB \]Using \( q = 1.6 \times 10^{-19} \text{ C} \) and given \( v = 2.50 \times 10^6 \text{ m/s} \) and \( B = 7.40 \times 10^{-2} \text{ T} \), calculate the force:\[ F_{max} = (1.6 \times 10^{-19})(2.50 \times 10^6)(7.40 \times 10^{-2}) \]

Step 3: Calculate acceleration with maximum force

Using Newton's second law, \( F = ma \), the maximum acceleration is:\[ a_{max} = \frac{F_{max}}{m} \]where \( m = 9.11 \times 10^{-31} \text{ kg} \). Use the \( F_{max} \) calculated in Step 2 to find \( a_{max} \).

Step 4: Calculate smallest acceleration (\(\theta = 0^\circ\))

The smallest force occurs when \( \sin \theta = 0 \) (i.e., \( \theta = 0^\circ \)), which means \( F = 0 \). Therefore, the smallest acceleration is zero, \( a_{min} = 0 \).

Step 5: Relate actual acceleration to maximum force

Let the actual acceleration \( a_{actual} = \frac{a_{max}}{4} \). Substitute \( a_{actual} \) into the formula \( a = \frac{F}{m} \) to find \( F_{actual} \).

Step 6: Calculate angle for given acceleration

Substitute \( F_{actual} \) into the magnetic force formula:\[ F_{actual} = qvB \sin \theta \]Solve for \( \sin \theta \) using:\[ \sin \theta = \frac{F_{actual}}{qvB} \]Use the actual force calculated and solve for \( \theta \).

Key Concepts

Electron Velocity

The velocity of an electron is a measure of how fast it moves through space. In the given problem, the electron has a velocity of \(2.50 \times 10^{6} \text{ m/s}\). This high speed is typical for electrons moving in electromagnetic fields.
The velocity of a charged particle like an electron is crucial when determining the magnetic force it experiences. This force depends on the velocity's magnitude and the angle it makes with the magnetic field, affecting the particle's motion significantly. Understanding electron velocity is key to predicting how magnetic fields interact with electrons.

Magnetic Field

A magnetic field is a vector field that influences the motion of charged particles within its reach. In this exercise, the magnetic field has a strength of \(7.40 \times 10^{-2} \text{ T}\) (teslas). However, its direction is unspecified, which is a common scenario in physics exercises to test understanding of principles.
Key characteristics of a magnetic field include:

• Magnitude: The strength of the field, influencing how much force acts on a charged particle.
• Direction: A crucial factor in determining the actual force experienced by a moving charge.

When a charged particle such as an electron moves through a magnetic field, it can be subject to a magnetic force, given by the equation \(F = qvB \sin \theta\). The particle's path is altered depending on its velocity and the direction of the magnetic field.

Acceleration Due to Magnetic Field

The acceleration of a charged particle due to a magnetic field is influenced by the Lorentz force. This force arises when there is a component of the charged particle’s velocity that is perpendicular to the magnetic field.
The formula for magnetic force \(F = qvB \sin \theta\) shows that the force, and thus the acceleration, varies with the angle \(\theta\). In the case where \(\theta = 90^\circ\), the magnetic force is at its maximum, leading to maximum acceleration, computed by \( a_{max} = \frac{F_{max}}{m} \). Conversely, when \(\theta = 0^\circ\), the force and acceleration are zero. This scenario shows how directionality plays a critical role.
In this exercise, we noted that the actual acceleration was one-fourth of the maximum acceleration. Using these principles, the respective angle \(\theta\) between the electron's velocity and the magnetic field can be determined by solving \(\sin \theta = \frac{F_{actual}}{qvB}\). This emphasizes how magnetic fields can manipulate the movement and acceleration of charged particles based on direction and other aspects.