Problem 52 - University Physics with Modern Physics

Question

A particle with charge 7.26 \(\times\) 10\(^{-8}\) C is moving in a region where there is a uniform 0.650-T magnetic field in the +\(x\)-direction. At a particular instant, the velocity of the particle has components \(v_x =\) -1.68 \(\times\) 10\(^4\) m/s, \(v_y =\) -3.11 \(\times\) 104 m/s, and \(v_z =\) 5.85 \(\times\) 10\(^4\) m/s. What are the components of the force on the particle at this time?

Step-by-Step Solution

Answer

The force components are \(F_y = 2.83 \times 10^{-3} \text{ N}\) and \(F_z = 1.47 \times 10^{-3} \text{ N}\).

Step 1: Understand the Formula for Magnetic Force

The magnetic force on a charged particle moving in a magnetic field is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), where \(q\) is the charge, \(\vec{v}\) is the velocity vector, and \(\vec{B}\) is the magnetic field vector.

Step 2: Identify and Write Down Given Values

We are given the charge \(q = 7.26 \times 10^{-8} \text{ C}\), the magnetic field \(\vec{B} = 0.650 \text{ T in the } +x\text{-direction}\), and the velocity components: \(v_x = -1.68 \times 10^4 \text{ m/s}\), \(v_y = -3.11 \times 10^4 \text{ m/s}\), \(v_z = 5.85 \times 10^4 \text{ m/s}\).

Step 3: Set Up the Cross Product

Since \(\vec{B} = 0.650 \hat{i}\), the cross product \(\vec{v} \times \vec{B}\) simplifies to:\[\left( -1.68 \times 10^4 \hat{i} + (-3.11 \times 10^4 \hat{j}) + 5.85 \times 10^4 \hat{k} \right) \times 0.650 \hat{i}.\]Since \(\hat{i} \times \hat{i} = 0\), only the remaining components are needed.

Step 4: Calculate the Cross Product Components

Using "right-hand rule" for cross product basis vectors, we find:\[\\hat{j} \times \hat{i} = -\hat{k}, \ \hat{k} \times \hat{i} = \hat{j}.\]Therefore, \[\vec{v} \times \vec{B} = 0.650 \left( (-3.11 \times 10^4 \hat{j}) \times \hat{i} + (5.85 \times 10^4 \hat{k}) \times \hat{i} \right) = \ 0.650 \left(-3.11 \times 10^4 (-\hat{k}) + 5.85 \times 10^4 \hat{j} \right), \]which results in: \[\vec{v} \times \vec{B} = (3.90 \times 10^4) \hat{j} + (2.02 \times 10^4) \hat{k} \text{ T} \cdot \text{m/s}.\]

Step 5: Determine the Force Using the Charge

Now that we have the cross product, apply the force formula \(\vec{F} = q(\vec{v} \times \vec{B})\):\[\vec{F} = 7.26 \times 10^{-8} \text{ C} \times ((3.90 \times 10^4) \hat{j} + (2.02 \times 10^4) \hat{k}).\]Calculating this results in:\[\vec{F} = 2.83 \times 10^{-3} \hat{j} + 1.47 \times 10^{-3} \hat{k} \text{ N}.\]

Step 6: Present the Components of the Force

The components of the force on the particle at this time are \[F_x = 0, \, F_y = 2.83 \times 10^{-3} \text{ N}, \, F_z = 1.47 \times 10^{-3} \text{ N}.\]

Key Concepts

Lorentz force

The Lorentz force is a fundamental concept in electromagnetism, governing the force on a charged particle moving through a magnetic field. This force can be expressed mathematically by the formula \( \vec{F} = q(\vec{v} \times \vec{B}) \). Here, \( q \) represents the charge of the particle, expressed in coulombs. \( \vec{v} \) is the velocity vector of the particle, and \( \vec{B} \) indicates the magnetic field vector, both represented in meters per second for velocity and teslas for magnetic field strength, respectively.

Notably, this formula highlights a unique quality: the magnetic force is always perpendicular to the plane formed by the particle's velocity and the magnetic field. This is why magnetic fields do not do work on the particles; they only change the direction of the velocity, not its magnitude. Consequently, the Lorentz force plays a pivotal role in various applications like cyclotrons or mass spectrometers, where it helps steer charged particles along circular paths.

Cross product

A cross product is a vital mathematical operation used in physics to determine the perpendicular direction to two given vectors. When it comes to the magnetic force, the cross product between the velocity vector \( \vec{v} \) and the magnetic field vector \( \vec{B} \) is essential. The resulting vector, \( \vec{v} \times \vec{B} \), is perpendicular to both \( \vec{v} \) and \( \vec{B} \).

One can calculate the cross product using the right-hand rule, a physical demonstration where you point your fingers in the direction of \( \vec{v} \), curl them towards \( \vec{B} \), and your thumb will show the direction of \( \vec{v} \times \vec{B} \). The calculation of the cross product follows a specific pattern with the basis vectors (\( \hat{i}, \hat{j}, \hat{k} \)):

• \( \hat{i} \times \hat{j} = \hat{k} \)
• \( \hat{j} \times \hat{k} = \hat{i} \)
• \( \hat{k} \times \hat{i} = \hat{j} \)

In our example exercise, the cross product results in a new vector with non-zero components only in the \( \hat{j} \) and \( \hat{k} \) directions.

Vector components

In understanding vector components, consider vectors as quantities made up of both magnitude and direction. In three-dimensional space, vectors are often broken down into components along the \(x\), \(y\), and \(z\)-axes for clarity and ease of calculation.

For instance, if a particle's velocity \( \vec{v} \) is given, each component \( v_x, v_y, \) and \( v_z \) corresponds to the particle's velocity in the directions of the \(x\), \(y\), and \(z\) axes, respectively. Knowing these components allows you to understand how the particle moves through space. Similarly, for a magnetic field \( \vec{B} \), the focus may lie on a particular direction, like the \(+x\)-direction in this case.

By calculating the vector components of the cross product, you can find how each axis was affected by the force. The sum of these components reflects the total force experienced by the particle, providing a complete view of its movement dynamics.