Problem 54 - University Physics with Modern Physics

Question

A particle with charge \(9.45 \times 10^{-8} \mathrm{C}\) is moving in a region where there is a uniform magnetic field of 0.450 \(\mathrm{T}\) in the \(+x\) - direction. At a particular instant of time the velocity of the particle has components \(v_{x}=-1.68 \times 10^{4} \mathrm{m} / \mathrm{s}, v_{y}=-3.11 \times 10^{4} \mathrm{m} / \mathrm{s}\) and \(v_{z}=5.85 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the components of the force on the particle at this time?

Step-by-Step Solution

Answer

The force components are \(0\) N, \(-2.487 \times 10^{-3}\) N, and \(-1.32225 \times 10^{-3}\) N in the \(x, y,\) and \(z\) directions, respectively.

Step 1: Understanding the Given Values

The problem gives us the following values: charge \(q = 9.45 \times 10^{-8} \text{ C}\), magnetic field \(\mathbf{B} = 0.450 \text{ T}\) in the \(+x\)-direction, and velocity components \(v_x = -1.68 \times 10^4 \text{ m/s}\), \(v_y = -3.11 \times 10^4 \text{ m/s}\), and \(v_z = 5.85 \times 10^4 \text{ m/s}\). We need to find the force on the particle using these values.

Step 2: Formula for Magnetic Force

The magnetic force \(\mathbf{F}\) on a charged particle is given by the formula \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\), where \(\times\) denotes the cross product of the velocity \(\mathbf{v}\) and the magnetic field \(\mathbf{B}\).

Step 3: Determine the Cross Product

To calculate \(\mathbf{v} \times \mathbf{B}\), set \(\mathbf{v} = (-1.68 \times 10^4, -3.11 \times 10^4, 5.85 \times 10^4)\) and \(\mathbf{B} = (0.450, 0, 0)\). Using the determinant formula for the cross product:\[\mathbf{v} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1.68 \times 10^4 & -3.11 \times 10^4 & 5.85 \times 10^4 \ 0.450 & 0 & 0 \end{vmatrix}\]

Step 4: Calculation of the Cross Product

Calculate each component of \(\mathbf{v} \times \mathbf{B}\):- The \(x\)-component is \(0\) because both vectors' components in the \(y\) and \(z\) directions are multiplied to get zero.- The \(y\)-component is \((0 - 0.450 \times 5.85 \times 10^4) = -2.6325 \times 10^4\).- The \(z\)-component is \(0.450 \times -3.11 \times 10^4 - 0 = -1.3995 \times 10^4\).Thus, \(\mathbf{v} \times \mathbf{B} = (0, -2.6325 \times 10^4, -1.3995 \times 10^4)\).

Step 5: Calculate Force Components

Using \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\), we find the force:- The \(x\)-component is \(9.45 \times 10^{-8} \times 0 = 0\).- The \(y\)-component is \(9.45 \times 10^{-8} \times -2.6325 \times 10^4 = -2.487\times 10^{-3} \text{ N}\).- The \(z\)-component is \(9.45 \times 10^{-8} \times -1.3995 \times 10^4 = -1.32225 \times 10^{-3} \text{ N}\).

Key Concepts

Charged Particle Motion

When a charged particle, like the one described in the exercise, moves within a magnetic field, it experiences a force that influences its motion. This phenomenon is quite common in both natural and technological processes.

Key points to consider include:

• The motion of such a particle is dictated entirely by its charge, speed, direction, and the magnetic field it traverses.
• The resultant path tends to be curved rather than linear, allowing scientists to predict its trajectory based on specific inputs. A prime example is the circular or helical path particles typically follow under uniform magnetic fields.

In this scenario, the magnetic force doesn't speed up or slow down the charged particle but merely changes its direction.

Cross Product Calculation

The cross product is an essential mathematical operation when dealing with vectors in physics. Specifically, for a charge moving in a magnetic field, the cross product helps calculate the magnetic force direction and magnitude.

Consider these details:

• The cross product is denoted by \( \mathbf{v} \times \mathbf{B} \), where \( \mathbf{v} \) represents the velocity vector and \( \mathbf{B} \) the magnetic field vector.
• This operation results in a new vector that is perpendicular to both initial vectors. Its direction is determined by the right-hand rule, a handy tool where you use your fingers to guide the vectors' interaction.
• In this exercise, the calculation uses determinant method, indicating how each component of the resulting vector arises from \ the other components of \( \mathbf{v} \) and \( \mathbf{B} \).

Understanding the cross product's role is fundamental in solving such physics problems accurately.

Magnetic Field Direction

Understanding how the direction of the magnetic field affects a charged particle is crucial. The magnetic field in this exercise points along the \(+x\)-axis, directly influencing the particle’s vector motion.

Here’s why the direction matters:

• Magnetic fields exert forces only on moving charges perpendicular to the field lines. Thus, the particle's velocity components perpendicular to the \(+x\)-axis will interact, influencing motion.
• Only these perpendicular components (\(v_{y}\) and \(v_{z}\)) will result in non-zero cross products, causing force components along the \(y\) and \(z\)-axes.
• This interplay between direction and forces finds widespread use in various applications, from MRI machines to particle accelerators.

Through this directional interplay, one can predict the two-dimensional motion resulting from the three-dimensional vector setup.

Vector Components in Physics

Analyzing vector components is a vital step in simplifying and understanding complex physical phenomena. In this context, it helps break down the velocity and force into understandable parts.

Consider these points:

• Each vector in physics typically has components aligned with the chosen coordinate system; here, it involves \(x\), \(y\), and \(z\) directions.
• By dissecting vector quantities like velocity, we can easily analyze how each direction contributes to the final vector or subsequent physical outcomes like force.
• These components allow the application of vector mathematics, including operations such as the cross product, which can derive new vector quantities essential for describing force and motion.

Consequently, understanding vector components enables clearer visualization and manipulation of complex interactions in physics problems.