Problem 53 - University Physics with Modern Physics

Question

When a particle of charge \(q>0\) moves with a velocity of \(\overrightarrow{\boldsymbol{v}}_{1}\) at \(45.0^{\circ}\) from the \(+x\) -axis in the \(x y\) -plane, a uniform magnetic field exerts a force \(\overrightarrow{\boldsymbol{F}}_{1}\) along the \(-z\) -axis (Fig. \(27.58 ) .\) When the same particle moves with a velocity \(\overrightarrow{\boldsymbol{v}}_{2}\) with the same magnitude as \(\overrightarrow{\boldsymbol{v}}_{1}\) but along the \(+z\) -axis, a force \(\overrightarrow{\boldsymbol{F}}_{2}\) of magnitude \(\boldsymbol{F}_{2}\) is exerted on it along the \(+x\) -axis. (a) What are the magnitude (in terms of \(q\) , \(v_{1},\) and \(F_{2} )\) and direction of the magnetic field? (b) What is the magnitude of \(\overrightarrow{\boldsymbol{F}}_{1}\) in terms of \(F_{2} ?\)

Step-by-Step Solution

Answer

The magnitude of \( \vec{B} = \sqrt{2} \frac{F_2}{q v_1} \) and direction is in the xy-plane; \( |\vec{F}_1| = F_2 \).

Step 1: Understanding the Force Equation

The force on a charged particle moving in a magnetic field is given by the Lorentz force equation: \( \vec{F} = q \vec{v} \times \vec{B} \). Here, \( \vec{v} \) is the velocity of the particle, \( \vec{B} \) is the magnetic field, and \( q \) is the charge of the particle. The cross-product implies that the force is perpendicular to both the velocity and magnetic field vectors.

Step 2: Analyzing the First Scenario

For \( \vec{v}_1 \), the velocity is at 45 degrees to the +x axis in the xy-plane, so \( \vec{v}_1 = v_1(\hat{i} \cos 45^\circ + \hat{j} \sin 45^\circ) = \frac{v_1}{\sqrt{2}} (\hat{i} + \hat{j}) \). The force \( \vec{F}_1 \) is along the -z axis, which implies \( \vec{B} \) has a component in the plane of \( \hat{i} \) and \( \hat{j} \).

Step 3: Using Force Equation for First Scenario

From \( \vec{F}_1 = q \vec{v}_1 \times \vec{B} \), since \( \vec{F}_1 \) is along -z, \( \vec{B} \) should be in the xy-plane only. \( \vec{B} = B_x \hat{i} + B_y \hat{j} \). By solving \( \vec{v}_1 \times \vec{B} \), we get \( \vec{F}_1 = q \frac{v_1}{\sqrt{2}}(B_x \hat{k} - B_y \hat{k}) = -q \frac{v_1}{\sqrt{2}} (B_y - B_x) \hat{k} \), and identify the magnitude of \( \vec{F}_1 \).

Step 4: Analyzing the Second Scenario

For \( \vec{v}_2 = v_1 \hat{k} \) (moving along +z), the force \( \vec{F}_2 \) is along +x, suggesting \( \vec{B} \) must have a component in \( \hat{j} \) (i.e., \( B_y eq 0 \)). This implies from \( \vec{F}_2 = q \vec{v}_2 \times \vec{B} = q v_1(B_y \hat{i}) \), \( B_y = \frac{F_2}{q v_1} \).

Step 5: Computing Magnitude of Magnetic Field

Given \( B_y \) from the second scenario and that \( \vec{F}_1 \) implies \( B_x = B_y \) since it simplifies force along z (equilibrium), then total magnetic field vector is \( \vec{B} = \frac{F_2}{q v_1}(\hat{i} + \hat{j}) \) with magnitude \( |\vec{B}| = \sqrt{2} \frac{F_2}{q v_1} \).

Step 6: Solving for the Magnitude of \( \vec{F}_1 \)

From initial force equation: \( \vec{F}_1 = -q \frac{v_1}{\sqrt{2}} (B_y - B_x) \hat{k} \) and using \( B_x = B_y = \frac{F_2}{q v_1} \), \( |\vec{F}_1| = q v_1 \frac{F_2}{q v_1} = F_2 \).

Key Concepts

Magnetic Field

A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It is typically represented by the symbol \( \vec{B} \). Magnetic fields exert forces on moving charges, influencing their motion. The direction of the magnetic field is represented by field lines flowing from the north to the south pole of a magnet.

In our problem, the magnetic field is uniform. This means its strength and direction are the same at all points in the region the charged particle moves through. Understanding how the magnetic field interacts with charged particles is crucial. This interaction can be mathematically determined using the Lorentz Force Equation. The angle at which the magnetic field acts on the particle influences the resulting force.

Charged Particle

A charged particle is any particle with an electric charge, denoted by \( q \). In this exercise, we specifically deal with a particle that possesses a positive charge, \( q > 0 \).

Charged particles in motion can be influenced by both electric and magnetic fields. Here, the particle's velocity impacts the force it experiences within the magnetic field. Two important aspects are direction and velocity magnitude. They determine how the Lorentz force will act. For instance, in Scenario 1 of the given problem, the velocity \( \vec{v}_1 \) forms a 45-degree angle with the \( x \)-axis in the \( xy \)-plane, leading to a specific force orientation. This moving charge allows the magnetic field to exert a perpendicular force, demonstrating this concept in action.

Force Equation

The Force Equation in the context of electromagnetism, particularly for a moving charged particle, is the Lorentz force equation: \( \vec{F} = q \vec{v} \times \vec{B} \).

This equation indicates that the force \( \vec{F} \) is the result of the charge \( q \), the velocity \( \vec{v} \) of the particle, and the magnetic field \( \vec{B} \). Here, the cross product \( \times \) implies that the force is perpendicular to both the velocity and the magnetic field vectors.

It's crucial to understand how the force's direction and magnitude are influenced by the charge's velocity and the magnetic field. This ensures correct predictions of the motion path of a charged particle. In the problem, the force \( \vec{F}_1 \) appears along the negative \( z \)-axis as a response to \( \vec{v}_1 \), while \( \vec{F}_2 \) acts along the positive \( x \)-axis when \( \vec{v}_2 \) is along the \( z \)-axis. Such scenarios exhibit the interplay between the given variables in the force equation.

Cross Product

The cross product, a mathematical operation used in vector algebra, is vital in understanding electromagnetic forces. Given two vectors \( \vec{a} \) and \( \vec{b} \), their cross product \( \vec{a} \times \vec{b} \) produces a third vector that is perpendicular to the plane formed by the first two. The magnitude of this vector is \( |\vec{a}||\vec{b}||\sin(\theta)| \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \).

In the context of the Lorentz force, this operation determines the direction of the force vector \( \vec{F} \). The direction of this force is given by the right-hand rule, meaning if you curl the fingers of your right hand from \( \vec{v} \) to \( \vec{B} \), your thumb points in the direction of \( \vec{F} \).

This cross product calculation allows us to figure out how the magnetic field exerts a force on a moving charged particle, illustrated by the examples of \( \vec{F}_1 \) and \( \vec{F}_2 \) in the exercise. It underscores the peculiar property of magnetic forces always acting orthogonal to the velocity of charged particles.