Problem 7 - University Physics with Modern Physics

Question

A particle with charge 7.80\(\mu \mathrm{C}\) is moving with velocity \(\vec{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{\imath}}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{k}}\) (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{B} \cdot \vec{F} .\) What is the angle between \(\vec{B}\) and \(\vec{\boldsymbol{F}} ?\)

Step-by-Step Solution

Answer

Magnetic field components: \( B_x = 0.176 \ \mathrm{T}, B_z = 0.257 \ \mathrm{T} \). Unknown: \( B_y \). Angle: \( 90^\circ \).

Step 1: State the Given Values

Given, the charge of the particle is \( q = 7.80 \times 10^{-6} \mathrm{C} \), and its velocity is \( \vec{v} = -3.80 \times 10^{3} \hat{J} \ \mathrm{m/s} \). The magnetic force \( \vec{F} \) experienced by the charge is \( \vec{F} = +7.60 \times 10^{-3} \hat{I} - 5.20 \times 10^{-3} \hat{K} \ \mathrm{N} \).

Step 2: Use Lorentz Force Equation

The Lorentz force equation is \( \vec{F} = q \vec{v} \times \vec{B} \), where \( \times \) denotes the cross product. Since force \( \vec{F} \) is given, and \( \vec{v} = -3.80 \times 10^3 \hat{J} \), we express the cross product in terms of unknown \( \vec{B} = B_x \hat{I} + B_y \hat{J} + B_z \hat{K} \).

Step 3: Compute the Cross Product

Given that \( \vec{v} \times \vec{B} = (-3.80 \times 10^3 \hat{J}) \times (B_x \hat{I} + B_y \hat{J} + B_z \hat{K}) \), we apply the cross product rules: \( \hat{J} \times \hat{I} = -\hat{K}, \hat{J} \times \hat{J} = 0, \hat{J} \times \hat{K} = \hat{I} \). Thus: \(-3.80 \times 10^3 \hat{J} \times B_x \hat{I} = -3.80 \times 10^3 B_x \hat{K} \) and \(-3.80 \times 10^3 \hat{J} \times B_z \hat{K} = 3.80 \times 10^3 B_z \hat{I} \).

Step 4: Apply the Force Components

Based on the Lorentz force \( \vec{F} = q ( 3.80 \times 10^3 B_z \hat{I} - 3.80 \times 10^3 B_x \hat{K} ) \), set this equal to the given force \( \vec{F} = +7.60 \times 10^{-3} \hat{I} - 5.20 \times 10^{-3} \hat{K} \). Thus:\[7.80 \times 10^{-6} \times 3.80 \times 10^3 B_z = 7.60 \times 10^{-3} \]\[7.80 \times 10^{-6} \times 3.80 \times 10^3 B_x = 5.20 \times 10^{-3} \].

Step 5: Solve for Magnetic Field Components

Solve the equations for \( B_x \) and \( B_z \):\[B_z = \frac{7.60 \times 10^{-3}}{7.80 \times 10^{-6} \times 3.80 \times 10^3} = 0.257 \ \mathrm{T} \]\[B_x = \frac{5.20 \times 10^{-3}}{7.80 \times 10^{-6} \times 3.80 \times 10^3} = 0.176 \ \mathrm{T} \]

Step 6: Determine Unknown Components

The component \( B_y \) does not affect the force on the particle, as the cross product between velocity along \( \hat{J} \) and a magnetic field along \( \hat{J} \) is zero. Hence, \( B_y \) remains undetermined from the force measurement.

Step 7: Calculate Scalar Product and Angle

Since \( \vec{v} \) and \( \vec{F} \) are perpendicular, \( \vec{B} \cdot \vec{F} = B_x F_x + B_y F_y + B_z F_z = 0 \) as none of the components corresponds. The angle \( \theta \) between \( \vec{B} \) and \( \vec{F} \) is calculated using the fact that the scalar product is zero, hence \( \theta = 90^\circ \).

Step 8: Final Observation

Therefore, the components of the magnetic field we determined are \( B_x = 0.176 \ \mathrm{T} \) and \( B_z = 0.257 \ \mathrm{T} \), while \( B_y \) cannot be calculated from the force information provided. The angle between \( \vec{B} \) and \( \vec{F} \) is \( 90^\circ \), meaning they are perpendicular.

Key Concepts

Lorentz Force Equation

The Lorentz force equation is an essential tool for understanding how charged particles move in electric and magnetic fields. Given by the formula \( \vec{F} = q(\vec{v} \times \vec{B}) \), where:

• \( \vec{F} \) is the force experienced by the charge.
• \( q \) is the charge of the particle.
• \( \vec{v} \) represents the particle's velocity.
• \( \vec{B} \) stands for the magnetic field.
• \( \times \) denotes the cross product.

This equation reveals that the force is perpendicular to both the velocity and the magnetic field, making it a perfect fit for the problem at hand. We know the charge, velocity, and force; our goal is to uncover the magnetic field components using this relationship.

Cross Product

Understanding the cross product is crucial for solving this type of problem. The cross product of two vectors, \( \vec{a} \times \vec{b} \), results in a third vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \). Its magnitude is calculated as:\[|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin(\theta)\]where \( \theta \) is the angle between the two vectors. In our problem, the velocity vector \( \vec{v} \) is along the \( \hat{J} \) direction. When crossed with the magnetic field \( \vec{B} \) (with an unknown direction), the cross product yields the components of force \( \vec{F} \) parallel to the \( \hat{I} \) and \( \hat{K} \) axes. Thus, we can connect these components back to the field strengths along those directions.

Magnetic Force Components

The magnetic force experienced by a charged particle is defined by how it splits along the coordinate axes when broken into components. When calculating these components using the Lorentz force, we find that:

• The force component \( F_x \) is due to the velocity and the magnetic field component \( B_z \).
• The force component \( F_z \) is influenced by the velocity and the magnetic field component \( B_x \).
• The component \( F_y \), associated with \( B_y \), does not appear as it results in no force (because the cross product with a parallel component is zero).

This step is a fantastic application of breaking complex interactions into understandable parts. By computing the field components, \( B_x = 0.176 \ \mathrm{T} \) and \( B_z = 0.257 \ \mathrm{T} \), we understand how each field component influences the magnetic force.

Angle between Vectors

The angle between vectors is calculated using their scalar or dot product, which helps understand their geometric orientation. When two vectors are perpendicular, the scalar product is zero. Mathematically, the scalar product of two vectors \( \vec{A} \cdot \vec{B} \) is given by:\[\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta)\]This formula aligns perfectly with our exercise, where the angle \( \theta \) between the magnetic field \( \vec{B} \) and force \( \vec{F} \) vectors leads to a zero scalar product. As a result, the angle between \( \vec{B} \) and \( \vec{F} \) is found to be \( 90^\circ \), indicating they are perpendicular. This perpendicular nature simplifies the process of calculation and aids in visualizing the alignment of vectors.