Problem 8 - Physics

Question

One component of a magnetic field has a magnitude of \(0.048 \mathrm{~T}\) and points along the \(+x\) axis, while the other component has a magnitude of \(0.065 \mathrm{~T}\) and points along the \(-y\) axis. A particle carrying a charge of \(+2.0 \times 10^{-5} \mathrm{C}\) is moving along the \(+z\) axis at a speed of \(4.2 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the \(+x\) axis.

Step-by-Step Solution

Answer

The net magnetic force is \(6.79 \times 10^{-3}\, \text{N}\), at an angle of \(36.4^\circ\) with the +x axis.

Step 1: Understand the Problem

The problem involves a charged particle moving through a magnetic field with components in the x and y directions. We need to find the magnetic force on the particle and its angle with the x-axis.

Step 2: Break Down the Components

The magnetic field has two components: \(B_x = 0.048 \, \text{T}\) (along +x) and \(B_y = 0.065 \, \text{T}\) (along -y). The charged particle moves along the z-axis with a speed \(v = 4.2 \times 10^3 \, \text{m/s}\) and carries a charge \(q = 2.0 \times 10^{-5} \, \text{C}\).

Step 3: Calculate the Magnetic Force Using Lorentz Force Law

The magnetic force on a charge moving in a magnetic field is given by the Lorentz Force Law: \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\). Since the velocity is along the z-axis (\(\mathbf{v} = v \hat{z}\)):- The magnetic force from \(B_x\) is \(F = q v B_x = 2.0 \times 10^{-5} \, \text{C} \times 4.2 \times 10^3 \, \text{m/s} \times 0.048 \, \text{T} = 4.032 \times 10^{-3} \, \text{N}\) in the y-direction.- The magnetic force from \(B_y\) is \(F = q v B_y = 2.0 \times 10^{-5} \, \text{C} \times 4.2 \times 10^3 \, \text{m/s} \times 0.065 \, \text{T} = 5.46 \times 10^{-3} \, \text{N}\) in the x-direction.

Step 4: Determine the Net Magnetic Force

The net magnetic force \(\mathbf{F}_{net}\) is the vector sum of the individual forces:\[ F_{net} = \sqrt{(5.46 \times 10^{-3})^2 + (4.032 \times 10^{-3})^2} \]\[ F_{net} = \sqrt{(29.82 \times 10^{-6}) + (16.256 \times 10^{-6})} \]\[ F_{net} = \sqrt{46.076 \times 10^{-6}} = 6.79 \times 10^{-3} \text{ N} \]

Step 5: Calculate the Angle with the +x Axis

The angle \(\theta\) between the net force and the x-axis is given by:\[ \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{4.032 \times 10^{-3}}{5.46 \times 10^{-3}} \right) \]\[ \theta = \tan^{-1}(0.738) = 36.4^\circ \]

Key Concepts

Lorentz Force

The Lorentz force is a fundamental concept in electromagnetism. It describes the force experienced by a charged particle moving through an electric and magnetic field. This force can be determined using the equation \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \), where \( \mathbf{F} \) is the force exerted on the particle, \( q \) is the charge of the particle, \( \mathbf{v} \) is the velocity, and \( \mathbf{B} \) is the magnetic field.
Within this equation, the cross product \( \mathbf{v} \times \mathbf{B} \) indicates that the magnetic force is perpendicular both to the direction of velocity \( \mathbf{v} \) and to the magnetic field \( \mathbf{B} \).

• If \( \mathbf{v} \) and \( \mathbf{B} \) are parallel or antiparallel, the force is zero as there is no perpendicular component.
• This force can result in circular or helical motion of particles, depending on additional forces or fields acting.

For the exercise, by plugging in the appropriate values on each magnetic field component separately, we obtain the total force experienced by the charged particle.

Vector Addition

In physics, vector addition is crucial to finding the resultant vector from two or more vectors. When dealing with perpendicular vectors such as the magnetic force components in our problem, the net force is calculated using the Pythagorean theorem.
This method is used because magnetic forces in the "+y" and "+x" directions are orthogonal. To find the net magnetic force \( \mathbf{F}_{net} \), we compute:
\[ \mathbf{F}_{net} = \sqrt{F_x^2 + F_y^2} \] This formula holds because we are essentially finding the hypotenuse of a right triangle formed by the component vectors.

• Start by determining each component: the force due to the x-component of the magnetic field and the y-component.
• Square each component's magnitude.
• Add the squares together.
• Finally, take the square root of that sum to find the magnitude of the net force.

This gives a clear understanding of how forces combine to produce a resultant force vector.

Magnetic Field Components

Magnetic fields are vector fields that have both a magnitude and a directional component. In this exercise, the magnetic field is described as having distinct components along the x and y axes. Specifically:
- The x-component \( B_x \) has a magnitude of \( 0.048 \, \text{T} \) and points along the positive x-axis.- The y-component \( B_y \) has a magnitude of \( 0.065 \, \text{T} \) and points along the negative y-axis.
These components imply that while the overall magnetic field originates from these combined vectors, each part affects the particle differently as it travels along the z-axis.

• The x-component affects the particle motion in the y-direction, contributing to a resulting magnetic force in that direction.
• Conversely, the y-component influences motion along the x-axis, applying force in that direction.
• These interactions are critical in calculating the net force that acts on the charged particle.

Understanding these components allows one to deconstruct complex interactions in magnetic fields.

Angle Calculation

Calculating the angle of a resultant vector, like the net force in this exercise, with respect to one of its components or another axis can be achieved using trigonometric functions. In this specific problem, we use the tangent function:
The angle \( \theta \) is determined by:\[ \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \]Here, \( \theta \) represents the angle between the net force and the positive x-axis.

• To find \( \theta \), divide the y-component force, \( F_y \), by the x-component force, \( F_x \).
• Then, apply the arctangent function to this ratio, yielding the angle in degrees or radians.

Through this approach, we gain both a visual and numerical appreciation of how forces interact and how their resultant orientation relates to individual component directions.