Problem 10 - Fundamentals of Physics

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=-2.50 \hat{\mathrm{i}} \mathrm{mT} .\) At one instant the velocity of the proton is \(\vec{v}=2000 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} .\) At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) \(4.00 \mathrm{k} \mathrm{V} / \mathrm{m}\) (b) \(-4.00 \hat{\mathrm{k}} \mathrm{V} / \mathrm{m},\) and (c) \(4.00 \hat{\mathrm{i}} \mathrm{V} / \mathrm{m} ?\)

Step-by-Step Solution

Answer

(a) \((6.408 \times 10^{-16})\hat{k}\) N, (b) \((-6.392 \times 10^{-16})\hat{k}\) N, (c) \(6.4 \times 10^{-16}\hat{i} + 8 \times 10^{-19}\hat{k}\) N.

Step 1: Understand the Force Calculations

The net force on a charged particle like a proton moving through electric and magnetic fields can be determined using the Lorentz force equation: \[\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\]where \(q\) is the charge of the proton, \(\vec{E}\) is the electric field, \(\vec{v}\) is the velocity, and \(\vec{B}\) is the magnetic field. The charge of the proton is \(q = 1.6 \times 10^{-19} \text{ C}.\)

Step 2: Calculate the Magnetic Force

To find the magnetic force, use the cross product \(\vec{v} \times \vec{B}\). Given \(\vec{v}=2000 \hat{\mathrm{j}}\ \mathrm{m/s}\) and \(\vec{B}=-2.50 \hat{\mathrm{i}}\ \mathrm{mT} = -2.50 \times 10^{-3} \hat{\mathrm{i}}\ \mathrm{T}\), calculate:\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & 2000 & 0 \-2.50 \times 10^{-3} & 0 & 0\end{vmatrix} = 0 \hat{i} + 0 \hat{j} + 5 \hat{k}\]The magnetic force is \(q \cdot 5 \hat{k} = 1.6 \times 10^{-19} \cdot 5 \hat{k} = 8 \times 10^{-19} \hat{k}\ \text{N}.\)

Step 3: Calculate the Total Force for Electric Field (a)

For \(\vec{E} = 4.00\hat{\mathrm{k}}\ \mathrm{V/m} = 4000\hat{\mathrm{k}}\ \mathrm{N/C}\), the electric force is:\[\vec{F}_e = q \cdot \vec{E} = 1.6 \times 10^{-19} \cdot 4000 \hat{k} = 6.4 \times 10^{-16} \hat{k} \ \text{N}\]The net force is:\[\vec{F}_{\text{net}} = \vec{F}_e + \vec{F}_b = 6.4 \times 10^{-16} \hat{k} + 8 \times 10^{-19} \hat{k} = (6.408 \times 10^{-16}) \hat{k} \ \text{N}\]

Step 4: Calculate the Total Force for Electric Field (b)

For \(\vec{E} = -4.00\hat{\mathrm{k}}\ \mathrm{V/m} = -4000\hat{\mathrm{k}}\ \mathrm{N/C}\), the electric force is:\[\vec{F}_e = q \cdot \vec{E} = 1.6 \times 10^{-19} \cdot (-4000 \hat{k}) = -6.4 \times 10^{-16} \hat{k} \ \text{N}\]The net force is:\[\vec{F}_{\text{net}} = \vec{F}_e + \vec{F}_b = -6.4 \times 10^{-16} \hat{k} + 8 \times 10^{-19} \hat{k} = (-6.392 \times 10^{-16}) \hat{k} \ \text{N}\]

Step 5: Calculate the Total Force for Electric Field (c)

For \(\vec{E} = 4.00 \hat{\mathrm{i}}\ \mathrm{V/m} = 4000 \hat{\mathrm{i}}\ \mathrm{N/C}\), the electric force is:\[\vec{F}_e = q \cdot \vec{E} = 1.6 \times 10^{-19} \cdot 4000 \hat{i} = 6.4 \times 10^{-16} \hat{i} \ \text{N}\]Since the magnetic force solely affects the \(\hat{k}\) direction, the net force is:\[\vec{F}_{\text{net}} = 6.4 \times 10^{-16} \hat{i} + 8 \times 10^{-19} \hat{k}\ \text{N}\]

Key Concepts

Uniform Magnetic Field

A uniform magnetic field is a magnetic field with consistent magnitude and direction throughout a particular region. In this context, the magnetic field is denoted as \( \vec{B} = -2.50 \hat{\mathrm{i}} \mathrm{mT} \), which means it is uniform and points in the negative \( \hat{\mathrm{i}} \) direction with a magnitude of 2.50 milliteslas. The uniformity ensures that the magnetic effects on a charged particle, like a proton, remain constant as long as the field doesn't change. Understanding the characteristics of a uniform magnetic field helps in calculating the magnetic force experienced by a charged particle moving through it, which is an essential part of determining the Lorentz force acting on the particle. With a fixed * magnitude * direction students can predict how the field will interact with moving charges.

Electric Field

The electric field, denoted as \( \vec{E} \), is a vector field that represents the force experienced by a positive test charge placed within its vicinity. In the given exercise, three scenarios describe how the electric field varies: - \( 4.00 \mathrm{kV/m} \hat{\mathrm{k}} \) - \( -4.00 \mathrm{kV/m} \hat{\mathrm{k}} \) - \( 4.00 \mathrm{V/m} \hat{\mathrm{i}} \)Fields are expressed in units of volts per meter (V/m), indicating the potential change across a distance.The electric field influences the proton by applying a force directly proportional to the field's strength and direction. This force is calculated using the expression \( \vec{F}_e = q \times \vec{E} \), where \( q \) is the charge of the proton. Hence, the electric field is crucial in determining how the net force acts on the proton as it travels through space.

Proton Velocity

Velocity is a vector quantity describing a particle's speed and direction of motion. For the proton in this exercise, its velocity is given as \( \vec{v} = 2000 \hat{\mathrm{j}}\ \mathrm{m/s} \). This implies that the proton is traveling exclusively along the positive \( \hat{\mathrm{j}} \) direction with a speed of 2000 meters per second. The* direction of velocity is essential when determining the magnetic force because it aligns or crosses with the magnetic field vectors. Understanding the velocity direction helps us compute the cross product \( \vec{v} \times \vec{B} \), an operation important for calculating the magnetic force component of the Lorentz force.

Net Force Calculation

The net force acting on a charged particle like a proton in combined electric and magnetic fields is determined using the Lorentz force law, given by the equation:\[\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\]This equation describes the sum of the electric force \( \vec{F}_e = q \cdot \vec{E} \) and the magnetic force, obtained from the cross-product \( \vec{v} \times \vec{B} \).The step by step calculation involves:

• Determining the magnetic force, given by the formula \( \vec{F}_b = q \cdot (\vec{v} \times \vec{B}) \).
• Evaluating the electric force by multiplying the electric field vector with the charge.
• Add these two forces vectorially to obtain the net force \( \vec{F}_{\text{net}} \).

This ensures a clear understanding of how both electric and magnetic components influence a proton's trajectory and resultant force in different field circumstances.