Problem 81 - Fundamentals of Physics

Question

A \(5.0 \mu \mathrm{C}\) particle moves through a region containing the uniform magnetic field \(-20 \hat{\mathrm{i}} \mathrm{mT}\) and the uniform electric field \(300 \mathrm{j} \mathrm{V} / \mathrm{m} .\) At a certain instant the velocity of the particle is \((17 \hat{i}-11 \hat{j}+7.0 \hat{\mathrm{k}}) \mathrm{km} / \mathrm{s} .\) At that instant and in unit-vector nota- tion, what is the net electromagnetic force (the sum of the electric and magnetic forces) on the particle?

Step-by-Step Solution

Answer

The net electromagnetic force is \(2.2 \times 10^{-3} \hat{\mathbf{j}} \text{ N}\).

Step 1: Identify the Given Values

We begin by identifying and listing the quantities that are given in the problem:- Charge of the particle: \( q = 5.0 \times 10^{-6} \text{ C} \)- Magnetic field: \( \mathbf{B} = -20 \hat{\mathbf{i}} \text{ mT} = -20 \times 10^{-3} \hat{\mathbf{i}} \text{ T} \)- Electric field: \( \mathbf{E} = 300 \hat{\mathbf{j}} \text{ V/m} \)- Velocity of the particle: \( \mathbf{v} = (17 \hat{\mathbf{i}} - 11 \hat{\mathbf{j}} + 7.0 \hat{\mathbf{k}}) \text{ km/s} = (17000 \hat{\mathbf{i}} - 11000 \hat{\mathbf{j}} + 7000 \hat{\mathbf{k}}) \text{ m/s} \)

Step 2: Calculate the Electric Force

The electric force \( \mathbf{F}_e \) on a charge in an electric field is given by:\[ \mathbf{F}_e = q \mathbf{E} \]Substitute the given values:\[ \mathbf{F}_e = (5.0 \times 10^{-6} \text{ C})(300 \hat{\mathbf{j}} \text{ V/m}) = 1.5 \times 10^{-3} \hat{\mathbf{j}} \text{ N} \]

Step 3: Calculate the Magnetic Force

The magnetic force \( \mathbf{F}_m \) is calculated using the formula:\[ \mathbf{F}_m = q (\mathbf{v} \times \mathbf{B}) \]First, determine the cross product \( \mathbf{v} \times \mathbf{B} \):- \( \mathbf{v} = 17000 \hat{\mathbf{i}} - 11000 \hat{\mathbf{j}} + 7000 \hat{\mathbf{k}} \)- \( \mathbf{B} = -0.020 \hat{\mathbf{i}} \)Compute the cross product:\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 17000 & -11000 & 7000 \ -0.020 & 0 & 0 \end{vmatrix} = -(0 \hat{\mathbf{i}} - 0.020 \times 7000 \hat{\mathbf{j}} - 0 \hat{\mathbf{k}}) = 140 \hat{\mathbf{j}} \text{ m/s} \]Thus, the magnetic force is:\[ \mathbf{F}_m = (5.0 \times 10^{-6} \text{ C})(140 \hat{\mathbf{j}} \text{ m/s}) = 0.7 \times 10^{-3} \hat{\mathbf{j}} \text{ N} \]

Step 4: Calculate the Net Electromagnetic Force

The net force \( \mathbf{F}_{\text{net}} \) is the sum of the electric and magnetic forces:\[ \mathbf{F}_{\text{net}} = \mathbf{F}_e + \mathbf{F}_m = 1.5 \times 10^{-3} \hat{\mathbf{j}} \text{ N} + 0.7 \times 10^{-3} \hat{\mathbf{j}} \text{ N} \]Combine the forces:\[ \mathbf{F}_{\text{net}} = 2.2 \times 10^{-3} \hat{\mathbf{j}} \text{ N} \]

Key Concepts

Electric Force

Electric force is a fundamental concept in physics that describes the interaction between charged particles. It arises when charged particles interact with an electric field.

• For a particle with charge \( q \), placed in an electric field \( \mathbf{E} \), the electric force \( \mathbf{F}_e \) experienced by the particle is given by \( \mathbf{F}_e = q \mathbf{E} \).
• The direction of the electric force is the same as that of the electric field for positive charges, and opposite for negative charges.
• In our given problem, the charge is \( 5.0 \times 10^{-6} \text{ C} \), and the electric field is \( 300 \hat{\mathbf{j}} \text{ V/m} \). Therefore, the electric force becomes \( 1.5 \times 10^{-3} \hat{\mathbf{j}} \text{ N} \).

This calculation shows how the electric field influences the particle, moving it along the field's direction. Electric forces are key across many applications, from powering our homes to driving cellular action in biology.

Magnetic Force

Magnetic force, unlike electric force, depends not only on the charge and field strength but also on the velocity of the charged particle. This force acts perpendicular to both the velocity of the particle and the magnetic field.

• The magnetic force \( \mathbf{F}_m \) is calculated using the equation \( \mathbf{F}_m = q(\mathbf{v} \times \mathbf{B}) \).
• Here, \( \mathbf{v} \) is the velocity vector of the charge, and \( \mathbf{B} \) is the magnetic field vector.
• In the exercise, this results in a magnetic force calculation showing direction and magnitude that contributes to the total electromagnetic force.

Because the magnetic force is perpendicular to the velocity, it can change the direction of the particle but not its speed, making it crucial in devices like cyclotrons and magnetic cranes.

Cross Product

The cross product is a mathematical operation used to calculate a vector that is perpendicular to two other vectors. It is crucial in computing the magnetic force.

• The vectors involved here are the velocity \( \mathbf{v} \) of the particle and the magnetic field \( \mathbf{B} \).
• Mathematically, the cross product is represented by determinants, as seen in the step provided: \( \mathbf{v} \times \mathbf{B} \).
• It results in a vector \( 140 \hat{\mathbf{j}} \text{ m/s} \) in the example problem, which is key for determining the direction of the magnetic force.

The cross product forms the core of determining the magnetic influence on moving charges and is an essential tool in vector calculus, particularly in physics.

Uniform Magnetic Field

A uniform magnetic field is one where the magnetic field lines are parallel and equally spaced, resulting in a consistent strength and direction everywhere. This simplifies the calculations of forces on charged particles or currents.

• In the problem, the magnetic field was \( -20 \hat{\mathbf{i}} \text{ mT} \), indicating a uniform field directed in the negative i direction.
• Such fields provide a predictable environment for analyzing particle movement, as seen in applications like mass spectrometers and magnetic resonance imaging (MRI).
• The uniformity ensures that any changes in the particle path are due to the velocity and field interactions rather than variations in field strength.

Understanding how particles behave in a uniform magnetic field is fundamental to achieving precision in scientific and engineering applications.