Problem 8 - University Physics with Modern Physics

Question

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}} .\) The magnetic force on the particle is measured to be \(\overrightarrow{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{i}}+\left(7.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{j}}\) (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{F}}\) . What is the angle between \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{F}} ?\)

Step-by-Step Solution

Answer

The velocity components are \( v_x = -106 \, \text{m/s} \), \( v_y = 48.6 \, \text{m/s} \), and \( v_z \) is undetermined. The angle between \( \overrightarrow{\boldsymbol{v}} \) and \( \overrightarrow{\boldsymbol{F}} \) is 90 degrees.

Step 1: Understand the Force Equation

The magnetic force on a charged particle is given by \( \overrightarrow{\boldsymbol{F}} = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \) where \( q \) is the charge, \( \overrightarrow{\boldsymbol{v}} \) is the velocity, and \( \overrightarrow{\boldsymbol{B}} \) is the magnetic field. Given \( q = -5.60 \times 10^{-9} \) C and \( \overrightarrow{\boldsymbol{B}} = -1.25 \hat{\boldsymbol{k}} \), the force equation can be expanded to find components of \( \overrightarrow{\boldsymbol{v}} \).

Step 2: Solve for Velocity Components

The force equation in components is:\[\begin{align*}F_x &= q(v_y B_z - v_z B_y) = -3.40 \times 10^{-7} \F_y &= q(v_z B_x - v_x B_z) = 7.40 \times 10^{-7} \F_z &= q(v_x B_y - v_y B_x) = 0\end{align*}\]Since \( \overrightarrow{\boldsymbol{B}} = -1.25 \hat{\boldsymbol{k}} \), \( B_x = B_y = 0 \) and \( B_z = -1.25 \), then:\[\begin{align*}F_x &= q(-v_y B_z) = -3.40 \times 10^{-7} \F_y &= q(v_x (-B_z)) = 7.40 \times 10^{-7}\end{align*}\]Solving these, we find:\( v_x = \frac{7.40 \times 10^{-7}}{q \cdot 1.25} \) and \( v_y = \frac{-3.40 \times 10^{-7}}{q \cdot -1.25} \)

Step 3: Calculate Known Components of Velocity

Given \( q = -5.60 \times 10^{-9} \) C, substitute this into the above equations\[\begin{align*}v_x &= \frac{7.40 \times 10^{-7}}{(-5.60 \times 10^{-9}) \times 1.25} = -1.06 \times 10^{2} \, \text{m/s} \v_y &= \frac{-3.40 \times 10^{-7}}{-5.60 \times 10^{-9} \times (-1.25)} = 4.86 \times 10^{1} \, \text{m/s} \v_z &\text{is not determined from the force equation}\end{align*}\]

Step 4: Determine Unknown Velocity Component

Since the force in the \( z \)-direction is zero, this does not affect the force and must be perpendicular to the magnetic field. Thus, \( v_z \) is undetermined since it does not contribute to the magnetic force in \( x \) and \( y \) components.

Step 5: Calculate Scalar Product F•F

To calculate \( \overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{F}} \), use the formula \( F_x^2 + F_y^2 + F_z^2 \):\[(3.40 \times 10^{-7})^2 + (7.40 \times 10^{-7})^2 + 0^2 = 6.49 \times 10^{-13} \, \text{N}^2\]

Step 6: Angle Between Velocity and Force

The velocity and force vector are perpendicular because the force result from the cross product of velocity and magnetic field, which always produces a vector perpendicular to both velocity and magnetic field. Thus, the angle between \( \overrightarrow{\boldsymbol{v}} \) and \( \overrightarrow{\boldsymbol{F}} \) is 90 degrees.

Key Concepts

Velocity Components

When a charged particle moves through a magnetic field, it experiences a force that affects its velocity components. The velocity vector of a particle is made up of three components: \( v_x \), \( v_y \), and \( v_z \). Each of these represents the speed in the respective directions of the coordinate axes.
In the context of a uniform magnetic field, as given in our exercise, the magnetic force equation is expressed as \( \overrightarrow{\boldsymbol{F}} = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \). This formula shows that the force depends on the cross product of velocity and the magnetic field vectors.

By solving the force equations step by step, we can find the values for specific components of the velocity. For instance:

• Using \( F_x = q(-v_y B_z) \), we determined \( v_x = -1.06 \times 10^2 \text{ m/s} \).
• From \( F_y = q(v_x (-B_z)) \), we found \( v_y = 4.86 \times 10^1 \text{ m/s} \).

The \( v_z \) component remains undetermined from force equations, indicating that movements along the \( z \)-axis do not influence the force experienced in this setup.

Vector Cross Product

The vector cross product is a fundamental concept in calculating the magnetic force on a charged particle. It is expressed in the equation \( \overrightarrow{\boldsymbol{F}} = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \). This cross product between velocity \( \overrightarrow{\boldsymbol{v}} \) and magnetic field \( \overrightarrow{\boldsymbol{B}} \) is key to understanding why force is always perpendicular to the plane formed by these two vectors.
A cross product results in a third vector that is orthogonal to the original pair:

• The direction of this vector follows the right-hand rule, which helps determine the direction of the force vector \( \overrightarrow{\boldsymbol{F}} \).
• The magnitude is computed using the formula \( |\overrightarrow{\boldsymbol{v}}| |\overrightarrow{\boldsymbol{B}}| \sin(\theta) \), where \( \theta \) is the angle between the velocity and magnetic field vectors.

In our exercise, the force is calculated in such a way that proves the angle between the velocity and force is \( 90 \) degrees. This attribute of the cross product fundamentally establishes how the velocity vector components interact with the magnetic field.

Uniform Magnetic Field

A uniform magnetic field is one where the magnetic field lines are parallel and equally spaced. This implies that the strength and direction of the magnetic field remain constant across the space it's acting within.
In practical terms, a uniform magnetic field is represented mathematically as a constant vector, such as the one in our example, \( \overrightarrow{\boldsymbol{B}} = -1.25 \hat{\boldsymbol{k}} \). In this scenario, the field is directed along the negative \( z \)-axis and has a uniform magnitude of \( 1.25 \text{ T} \).

This uniformity greatly simplifies calculations as it means:

• No direction change in the force as the particle moves, making it easier to compute velocity components affected by the magnetic force.
• The effects of the magnetic field can be precisely calculated irrespective of the particle’s position in the field.

As a result, in a uniform magnetic field, the task of predicting the motion of charged particles becomes more manageable, allowing for a clear calculation of forces like in our exercise.