Problem 5 - College Physics

Question

At a given instant, a particle with a mass of \(5.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(3.50 \times 10^{-8} \mathrm{C}\) has a velocity with a magnitude of \(2.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) in the \(+y\) direction. It is moving in a uniform magnetic field that has magnitude 0.8 \(\mathrm{T}\) and is in the \(-x\) direction. What are (a) the magnitude and direction of the magnetic force on the particle and (b) its resulting acceleration?

Step-by-Step Solution

Answer

The magnetic force is \(5.60 \times 10^{-3} \ \mathrm{N}\) in the \(-z\) direction; acceleration is \(1.12 \ \mathrm{m/s^2}\).

Step 1: Identify the Given Values

First, list all the given values for the problem.- Mass of the particle: \( m = 5.00 \times 10^{-3} \ \mathrm{kg} \)- Charge of the particle: \( q = 3.50 \times 10^{-8} \ \mathrm{C} \)- Velocity of the particle: \( v = 2.00 \times 10^{5} \ \mathrm{m/s} \) in the \(+y\) direction- Magnetic field: \( B = 0.8 \ \mathrm{T} \) in the \(-x\) direction.

Step 2: Calculate the Magnetic Force Magnitude

Use the formula for the magnetic force on a charged particle: \[ F = qvB \sin(\theta) \]Since the velocity is perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \).Substitute the values:\[ F = (3.50 \times 10^{-8} \ \mathrm{C})(2.00 \times 10^{5} \ \mathrm{m/s})(0.8 \ \mathrm{T}) \]\[ F = 5.60 \times 10^{-3} \ \mathrm{N} \]

Step 3: Determine the Direction of the Magnetic Force

Use the right-hand rule to find the direction of the magnetic force.- Point your fingers in the direction of the velocity (\(+y\)).- Curl them toward the direction of the magnetic field (\(-x\)).- Your thumb points in the direction of the force, which is in the \(-z\) direction.

Step 4: Calculate the Particle's Acceleration

Use Newton's second law to find the acceleration: \[ F = ma \]Solve for acceleration (\( a \)):\[ a = \frac{F}{m} = \frac{5.60 \times 10^{-3} \ \mathrm{N}}{5.00 \times 10^{-3} \ \mathrm{kg}} \]\[ a = 1.12 \ \mathrm{m/s^2} \]

Step 5: Summarize the Results

The magnetic force has a magnitude of \(5.60 \times 10^{-3} \ \mathrm{N}\) and is directed in the \(-z\) direction. The resulting acceleration of the particle is \(1.12 \ \mathrm{m/s^2}\).

Key Concepts

Charged Particle

When we talk about a charged particle, we are referring to any object that possesses an electric charge. In physics, charge can be positive or negative, and it is a fundamental characteristic that leads to electromagnetic interactions. In the context of the exercise, the charged particle has a very small mass of \(5.00 \times 10^{-3} \ \mathrm{kg}\) and carries a charge of \(3.50 \times 10^{-8} \ \mathrm{C}\).
These particles interact with electric and magnetic fields. Understanding charged particles is essential because every particle behaves differently based on its charge when subject to these fields. The charge of the particle dictates the magnitude of the force it will experience in a magnetic field, which is crucial to solving problems involving magnetic forces.
For instance, the equation \( F = qvB \sin(\theta)\) shows how both charge \(q\) and velocity \(v\) of the particle play a role in determining the magnitude of the magnetic force.

Magnetic Field

Magnetic fields are invisible forces that influence the behavior of charged particles within their vicinity. They are represented by lines that show the direction and strength of the magnetic force at any given point in space. In this problem, the magnetic field has a magnitude of 0.8 \(\mathrm{T}\) and points in the \(-x\) direction.
The strength of a magnetic field is measured in teslas (T). A uniform magnetic field means that the field lines are parallel to each other, creating a consistent force throughout. This consistency makes calculations like the Lorentz force more straightforward, as conditions do not vary with position.
Magnetic fields exert forces on moving charges. The force is perpendicular to both the direction of the field and the velocity of the particle, as captured by the Lorentz force equation. Understanding these interactions is key to analyzing particle motion in contexts like this exercise.

Right-Hand Rule

The right-hand rule is a simple mnemonic device used in physics to determine the direction of vectors in a three-dimensional space. When dealing with magnetic forces, it helps to find the force direction on a charged moving particle.

• First, point your fingers in the direction of the velocity of the particle. In the exercise, this is the \(+y\) direction.
• Next, curl your fingers towards the direction of the magnetic field, which is the \(-x\) direction in the problem.
• Your thumb will point in the direction of the magnetic force. Here, that direction is the \(-z\) axis.

Using the right-hand rule ensures accuracy and consistency when solving problems about forces in magnetic fields. It's a vital concept that makes it easier to understand complex interactions intuitively.

Particle Acceleration

When a magnetic force acts on a charged particle, it causes a change in the particle's motion, known as acceleration. This acceleration is derived from Newton's second law of motion, which states that force equals mass times acceleration \(F = ma\).
The exercise illustrates this by calculating the particle's acceleration. After obtaining the magnetic force (\(5.60 \times 10^{-3} \ \mathrm{N}\)), we use the mass of the particle to find its acceleration:
\[ a = \frac{F}{m} = \frac{5.60 \times 10^{-3} \ \mathrm{N}}{5.00 \times 10^{-3} \ \mathrm{kg}} = 1.12 \ \mathrm{m/s^2} \]
This equation shows that acceleration is directly proportional to the force when mass is constant. Understanding this relationship is essential for predicting how charged particles move in fields, thus solving problems in electromagnetism efficiently.