Problem 7 - Physics

Question

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{~T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\). At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

Step-by-Step Solution

Answer

The angle \( \theta \) is approximately \(19.76^\circ\).

Step 1: Understand the Relationship

The magnetic force acting on a charged particle moving through a magnetic field is given by the formula: \[ F = qvB \sin(\theta) \] where \( F \) is the magnetic force, \( q \) is the charge of the electron, \( v \) is the velocity of the electron, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field. We can solve for \( \sin(\theta) \) once we determine the magnetic force.

Step 2: Calculate the Magnetic Force

Firstly, we need to find the force \( F \) using Newton's second law of motion, which states \( F = ma \), where \( m \) is the mass of the electron and \( a \) is its acceleration. The electron mass \( m \) is approximately \(9.11 \times 10^{-31} \) kg. Thus,\[ F = (9.11 \times 10^{-31} \text{ kg})(3.50 \times 10^{14} \text{ m/s}^2) \]Calculating this gives:\[ F = 3.1885 \times 10^{-16} \text{ N} \]

Step 3: Substitute Values into the Magnetic Force Formula

Substitute the known values into the magnetic force formula:\[ F = qvB \sin(\theta) \] Where the charge of the electron \( q \) is \(-1.6 \times 10^{-19} \text{ C}\), then:\[ 3.1885 \times 10^{-16} = (1.6 \times 10^{-19})(6.80 \times 10^6)(8.70 \times 10^{-4}) \sin(\theta) \]Simplifying this equation:\[ 3.1885 \times 10^{-16} = 9.4464 \times 10^{-16} \sin(\theta) \]

Step 4: Solve for \( \sin(\theta) \)

Rearrange the equation to solve for \( \sin(\theta) \):\[ \sin(\theta) = \frac{3.1885 \times 10^{-16}}{9.4464 \times 10^{-16}} \]Calculating the right side gives:\[ \sin(\theta) = 0.337 \]

Step 5: Determine the Angle \( \theta \)

Now, find \( \theta \) using the arcsine function:\[ \theta = \arcsin(0.337) \]Using a calculator, compute:\[ \theta \approx 19.76^\circ \]

Key Concepts

Electron Acceleration

Acceleration is a key concept in understanding how particles, like electrons, respond to forces. When a force acts on an electron, it causes the electron to accelerate. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the electron. In this exercise, the electron is given an acceleration of magnitude \(3.50 \times 10^{14} \text{ m/s}^2\).
This high acceleration indicates that the electron is undergoing a substantial change in velocity in response to a magnetic force. To calculate this force, we apply Newton's second law of motion:

• Force \( F = ma \), where \( m \) is the electron's mass (\(9.11 \times 10^{-31} \text{ kg}\)).
• We calculate the force as \( F = (9.11 \times 10^{-31}) (3.50 \times 10^{14}) \), resulting in a force of \(3.1885 \times 10^{-16} \text{ N}\).

Understanding electron acceleration helps us comprehend how charged particles behave under electromagnetic influences, which is crucial for fields such as electronics and electromagnetism.

Magnetic Field

A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. In this problem, the magnetic field has a magnitude of \(8.70 \times 10^{-4} \, T\) (Tesla).
This magnetic field produces a force on the electron as it moves through it, resulting in the electron's acceleration discussed earlier. The magnetic force experienced by a charged particle like an electron in a magnetic field is described mathematically by:

• \( F = qvB \sin(\theta) \)
• Here, \( F \) is the force, \( q \) is the electron charge, \( v \) is velocity, \( B \) is the magnetic field magnitude, and \( \theta \) is the angle between velocity and magnetic field.

The strength of the magnetic field determines how much the electron will be deflected, and solving for this force helps us determine other critical parameters like the angle at which the electron travels relative to the magnetic field.

Velocity of Electron

The velocity of an electron is essential in determining how it interacts with a magnetic field. In this exercise, our electron travels at a speed of \(6.80 \times 10^{6} \text{ m/s}\).
The velocity is a vector quantity with both magnitude and direction. When calculating the magnetic force on a moving charge, knowing both these aspects is crucial, as the direction affects how the magnetic field influences the electron.

• The magnetic force becomes maximum if the electron's velocity is perpendicular to the magnetic field \((\theta = 90^\circ)\).
• It is zero if the velocity is parallel to the field \((\theta = 0^\circ)\).

In this exercise, the combination of the electron's high speed and the given magnetic field determines the magnetic force, which is used to find the angle \( \theta \) through which it moves.

Angle Calculation

Calculating the angle \( \theta \) between the electron's velocity and the magnetic field is critical for understanding the specifics of its motion through the field. This angle determines how much of the magnetic force affects the electron's trajectory.
To solve for the angle:

• We rearrange the magnetic force formula \( F = qvB \sin(\theta) \) to express \( \sin(\theta) \).
• Using the calculated magnetic force \(3.1885 \times 10^{-16} \text{ N}\), charge \( q = -1.6 \times 10^{-19} \text{ C}\), velocity \( v = 6.80 \times 10^6 \text{ m/s}\), and magnetic field \( B = 8.70 \times 10^{-4} \text{ T}\), we find:
• \[\sin(\theta) = \frac{3.1885 \times 10^{-16}}{(1.6 \times 10^{-19})(6.80 \times 10^6)(8.70 \times 10^{-4})} = 0.337\]
• Using the arcsin function, we find \( \theta = \arcsin(0.337) \approx 19.76^\circ \).

This calculation illustrates the interplay of various forces and vectors, which is fundamental in many technological applications such as electric motors and particle accelerators.