Question

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (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 \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

Short answer

(a) \(v_x = -105.71\, \text{m/s}\), \(v_y = 48.57\, \text{m/s}\). (b) \(v_z\) is undetermined. (c) \(\theta \approx 90^\circ\), perpendicular.

Step-by-step solution

Understand the Force Equation

The magnetic force on a charged particle is given by the equation \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \), where \( q \) is the charge, \( \overrightarrow{v} \) is the velocity vector, and \( \overrightarrow{B} \) is the magnetic field vector. Here, \( q = -5.60 \times 10^{-9} \text{ C} \) and \( \overrightarrow{B} = -1.25 \hat{k} \text{ T} \). We know the force \( \overrightarrow{F} = -(3.40 \times 10^{-7}) \hat{i} + (7.40 \times 10^{-7}) \hat{j} \text{ N} \).

Set Up the Cross Product Expression

Considering the force equation \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \), expand this using the known force components and expressions for the velocity components \( \overrightarrow{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \). The cross product results in \( \overrightarrow{v} \times \overrightarrow{B} = -1.25(v_y \hat{i} - v_x \hat{j}) \) because the \( \hat{k} \) component of \( \overrightarrow{B} \) multiplies the \( \hat{i} \) and \( \hat{j} \) components of \( \overrightarrow{v} \).

Solve for Velocity Components

According to \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \), solve for each velocity component:- \( -3.40 \times 10^{-7} = q(-1.25v_y) \Rightarrow v_y = \frac{3.40 \times 10^{-7}}{1.25 \times 5.60 \times 10^{-9}} \approx 48.57 \text{ m/s}\)- \( 7.40 \times 10^{-7} = q(-1.25v_x) \Rightarrow v_x = -\frac{7.40 \times 10^{-7}}{1.25 \times 5.60 \times 10^{-9}} \approx -105.71 \text{ m/s}\).

Identify Undetermined Components

Velocity component \( v_z \) is not involved in the cross product calculation as it cancels out when crossed with \( \hat{k} \). Therefore, \( v_z \) cannot be determined from the given force and magnetic field.

Calculate the Scalar Product and Angle

The scalar product \( \vec{v} \cdot \overrightarrow{F} \) for known components is \( (-105.71)(-3.40 \times 10^{-7}) + (48.57)(7.40 \times 10^{-7}) = 5.67 \times 10^{-5} \text{ N} \cdot \text{m/s}\). The angle \( \theta \) between \( \vec{v} \) and \( \overrightarrow{F} \) is given by \( \cos \theta = \frac{\vec{v} \cdot \overrightarrow{F}}{\|\vec{v}\| \|\overrightarrow{F}\|} \). Compute magnitudes, \( \|\vec{v}\| \approx 116.54 \text{ m/s} \) and \( \|\overrightarrow{F}\| \approx 8.01 \times 10^{-7} \text{ N} \), then solve for \( \theta \).

Key concepts

Velocity Components

The velocity of a charged particle in a magnetic field can be tricky to understand, especially since the movement is influenced by the magnetic force. Velocity is a vector with components in the x, y, and z directions.
By breaking the velocity into components, the problem becomes easier to solve. Knowing two components, such as \( v_x \) and \( v_y \), allows us to proceed with calculations like the cross product with the magnetic field.
In this specific problem, the magnetic force and field direction helped us determine the x and y components of the velocity:

• \( v_x \) was calculated approximately as -105.71 m/s
• \( v_y \) was calculated approximately as 48.57 m/s

However, the z-component (\( v_z \)) remained undetermined for reasons we’ll explore next.

Cross Product

The concept of the cross product is essential in calculating the force acting in a magnetic field. The cross product (\( \overrightarrow{v} \times \overrightarrow{B} \)) combines the velocity and magnetic field vectors.
The magnetic field interacts perpendicularly with the velocity through the cross product, indicating the direction and size of the magnetic force. In mathematical terms, it's calculated as follows:

• Only \( v_x \) and \( v_y \) affect the calculation because \( \hat{k} \) in \( \overrightarrow{B} \) multiplies with the respective components.
• This results in the force components being in the \( \hat{i} \) and \( \hat{j} \), excluding \( v_z \)

In vivid detail:
- The cross product results for this exercise were:
\(\overrightarrow{F} = -1.25(v_y \hat{i} - v_x \hat{j})\).
This is why only two of the three components (\( v_x \) and \( v_y \)) are solved.

Scalar Product

Unlike the cross product, the scalar product (or dot product) compares two vectors for similarity in direction. It’s a single number, reflecting component alignment.
With vectors \( \overrightarrow{v} \) and \( \overrightarrow{F} \), the scalar product helps in understanding their directional relationship.

• The formula is \( \vec{v} \cdot \overrightarrow{F} = v_x F_x + v_y F_y + v_z F_z \)
• Only known components \( v_x \) and \( v_y \) were used in our calculation.

In this problem, it showed the interaction between velocity and magnetic force, summed up by \( 5.67 \times 10^{-5} \text{ N} \cdot \text{m/s} \).
This single value simplifies our understanding of their alignment, with further steps enabling us to find angles of direction.

Charge of Particle

The charge \( q \) of the particle plays a key role in determining how it interacts with the magnetic field. In this exercise, it was given as \(-5.60 \text{ nC} \) (or \( -5.60 \times 10^{-9} \text{ C} \)).
The charge affects the magnitude and direction of the force experienced by the particle.
Here's how charge enters the process:

• The force:\( \overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B}) \)
• Being negative, the force on the particle is in the opposite direction compared to if it were positive.

Therefore, charge not only bears on strength but also on the direction of resulting forces.

Magnetic Field

Magnetic fields influence charged particles and are vector quantities, denoted often by \( \overrightarrow{B} \). In this problem, the magnetic field \( \overrightarrow{B} = -1.25 \hat{k} \text{ T} \) remained constant.
Understanding its direction and magnitude helps in applying the force equation we use:

• It establishes the plane of rotation or force exertion on charged particles.
• Is always perpendicular to the force due to the cross product relationship.

As evidenced, fields are integral in plotting a particle's trajectory across the three-dimensional space it travels through.