Problem 4 - University Physics with Modern Physics

Question

A particle with mass \(1.81 \times 10^{-3} \mathrm{kg}\) and a charge of \(1.22 \times\) \(10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\vec{v}=\left(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{j}\) What are the magnitude and direction of the particle's accoleration produced by a uniform magnetic field \(\vec{B}=(1.63 \mathrm{T}) \hat{\imath}+\) \((0.980 \mathrm{T}) \hat{\jmath} ?\)

Step-by-Step Solution

Answer

The acceleration is 0.330 m/s² in the negative z-axis direction.

Step 1: Identify the Known Values

Start by listing all the known values we have from the problem:- Mass of the particle, \( m = 1.81 \times 10^{-3} \text{ kg} \)- Charge of the particle, \( q = 1.22 \times 10^{-8} \text{ C} \)- Velocity of the particle, \( \vec{v}=(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}) \hat{j} \)- Magnetic field, \( \vec{B}=(1.63 \mathrm{T}) \hat{\imath}+(0.980 \mathrm{T}) \hat{\jmath} \)

Step 2: Formula for Magnetic Force

We use the Lorentz force formula to find the magnetic force on a charged particle in a magnetic field: \[ \vec{F} = q(\vec{v} \times \vec{B}) \]This force will then be used to find the acceleration of the particle.

Step 3: Cross Product Calculation

Calculate the cross product \( \vec{v} \times \vec{B} \) using the components:1. \( \vec{v} = (0)\hat{i} + (3.00 \times 10^{4})\hat{j} + (0)\hat{k} \)2. \( \vec{B} = (1.63)\hat{i} + (0.980)\hat{j} + (0)\hat{k} \)The cross product is:\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & 3.00 \times 10^{4} & 0 \1.63 & 0.980 & 0\end{vmatrix}= (0)\hat{i} + (0)\hat{j} + (-4.89 \times 10^{4})\hat{k} \]

Step 4: Calculate the Magnetic Force

Substitute the cross product into the Lorentz force formula:\[ \vec{F} = (1.22 \times 10^{-8} \text{ C}) \times (0 \hat{i} + 0 \hat{j} - 4.89 \times 10^{4} \hat{k}) \]This gives \[ \vec{F} = (0) \hat{i} + (0) \hat{j} + (-5.96 \times 10^{-4} \text{ N}) \hat{k} \]

Step 5: Calculate the Acceleration

To find the acceleration, use Newton's second law, \( \vec{F} = m\vec{a} \).\[ \vec{a} = \frac{\vec{F}}{m} = \frac{-5.96 \times 10^{-4} \hat{k}}{1.81 \times 10^{-3}} \]This results in:\[ \vec{a} = 0 \hat{i} + 0 \hat{j} - 0.330 \hat{k} \text{ m/s}^2\]

Step 6: Determine the Magnitude and Direction

The magnitude of the acceleration vector is calculated as:\[ |\vec{a}| = \sqrt{(0)^2 + (0)^2 + (-0.330)^2} = 0.330 \, \text{m/s}^2 \]Since only the \(\hat{k}\) component is present, the direction of the acceleration is along the negative \(z\)-axis.

Key Concepts

Lorentz Force

The concept of Lorentz force is central to understanding how charged particles behave in electromagnetic fields. It is the force exerted on a charged particle moving within these fields. The formula for Lorentz force is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), where \(q\) is the charge of the particle, \(\vec{v}\) is its velocity, and \(\vec{B}\) represents the magnetic field. This formula shows that the force depends on the velocity and magnetic field's orientation.
The cross product \(\vec{v} \times \vec{B}\) implies that the force is perpendicular to both the velocity of the particle and the magnetic field. This perpendicularity is a key point, meaning the particle's path will curve rather than align directly along any one of those vectors. The Lorentz force does not work in isolation; rather, it is a fundamental component when calculating acceleration in magnetic contexts. Understanding this concept is crucial for solving many physics problems involving charged particles.

Cross Product

The cross product in vector mathematics allows us to calculate a vector that is orthogonal to two given vectors. In the context of magnetic forces, this operation is used to find the direction of the force acting on a charged particle within a magnetic field.
Given vectors \(\vec{v} = (0)\hat{i} + (3.00 \times 10^{4})\hat{j} + (0)\hat{k}\) and \(\vec{B} = (1.63)\hat{i} + (0.980)\hat{j} + (0)\hat{k}\), the cross product \(\vec{v} \times \vec{B}\) is evaluated using a determinant:

• Place vectors in a matrix along with unit vectors \(\hat{i}, \hat{j}, \hat{k}\)
• Determinant calculation gives \((0)\hat{i} + (0)\hat{j} + (-4.89 \times 10^{4})\hat{k}\)

Here, the resulting vector from the cross product indicates the direction of the magnetic force, showing it acts along the negative \(z\)-axis. This vector operation is essential for interpreting the physical direction of forces in fields.

Acceleration Calculation

Once the force on the particle is determined using the Lorentz force law, the next step is to calculate the particle's acceleration. According to Newton's second law, acceleration \(\vec{a}\) is determined by dividing the force \(\vec{F}\) by the mass \(m\) of the particle: \(\vec{a} = \frac{\vec{F}}{m}\).
In this problem, \(\vec{F} = (0)\hat{i} + (0)\hat{j} + (-5.96 \times 10^{-4})\hat{k}\) newtons, and with \(m = 1.81 \times 10^{-3}\) kg, the acceleration is calculated as:

• \(\vec{a} = \frac{-5.96 \times 10^{-4} \hat{k}}{1.81 \times 10^{-3}}\)
• \(\vec{a} = 0 \hat{i} + 0 \hat{j} - 0.330 \hat{k}\, \text{m/s}^2\)

This provides the acceleration's magnitude and confirms its direction is aligned along the negative \(z\)-axis. Understanding this step shows how forces relate to changes in motion, a pivotal concept in mechanics.

Physics Problem Solving

Physics problem solving involves breaking down each part of a complex scenario into manageable steps. For this magnetic force problem, the process is as follows:

• Identify and list known quantities like mass, charge, and velocity.
• Apply key formulas such as the Lorentz force equation to find forces.
• Carry out vector operations like cross products for force direction.
• Use Newton’s laws to relate force and acceleration.

Each of these steps builds on the last, consolidating understanding and leading to a solution. This systematic approach aids in tackling similar physics problems, fostering deeper insights into the behavior of systems under different forces.