Problem 7 - University Physics with Modern Physics

Question

A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) (7.60 \(\times\) 10\(^{-3}\) N)\(\hat{\imath}\) - (5.20 \(\times\) 10\(^{-3}\) N)\(\hat{k}\). (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\overrightarrow{B}\) and \(\overrightarrow{F}\)?

Step-by-Step Solution

Answer

\( B_x, B_z \) are calculated; \( B_y \) undetermined; angle = \(90^\circ\).

Step 1: Write down the Lorentz Force Equation

The magnetic force on a charged particle is given by the Lorentz force 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.

Step 2: Express the Known Quantities

Given: Charge \( q = 7.80 \, \mu C = 7.80 \times 10^{-6} \, C \), velocity \( \overrightarrow{v} = -13.80 \times 10^{3} \, m/s \, \hat{\jmath} \), and force \( \overrightarrow{F} = (7.60 \times 10^{-3} \, N) \, \hat{\imath} - (5.20 \times 10^{-3} \, N) \, \hat{k} \).

Step 3: Set Up Cross Product for Force

The force is calculated by the cross product, \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \). Let \( \overrightarrow{B} = B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k} \). Compute \( \overrightarrow{v} \times \overrightarrow{B} \) using the determinant formula for the cross product.

Step 4: Solve for Magnetic Field Components

From the cross product \( \overrightarrow{v} \times \overrightarrow{B} \), \( \overrightarrow{F} = q (B_z \hat{\imath} + B_x \hat{k}) = (7.60 \times 10^{-3} \, N) \hat{\imath} - (5.20 \times 10^{-3} \, N) \hat{k} \). Solve these equations: - \( 7.80 \times 10^{-6} \, C \times -13.80 \times 10^3 \, m/s \times B_z = 7.60 \times 10^{-3} \, N \) - \( 7.80 \times 10^{-6} \, C \times -13.80 \times 10^3 \, m/s \times B_x = -5.20 \times 10^{-3} \, N \).Obtain \( B_z \) and \( B_x \), noting that \( B_y \) cannot be determined from the force.

Step 5: Check Undetermined Components

Since the velocity is in the \( \hat{\jmath} \) direction, the component of the magnetic field in the same direction, \( B_y \), produces no force and is thus undetermined by the force measurement.

Step 6: Calculate Scalar Product and Angle

The scalar product of \( \overrightarrow{B} \cdot \overrightarrow{F} \) results in zero since the cross product and force are orthogonal. Therefore, \( \cos \theta = 0 \) implies the angle \( \theta = 90^\circ \).

Key Concepts

Magnetic Field Components

When a charged particle, like the one in this exercise, moves through a magnetic field, it experiences a force. According to the Lorentz Force Equation, this force is a cross product of the velocity and the magnetic field vectors. Given the force and velocity, we can calculate some components of the magnetic field but not all.

With the force (\(\overrightarrow{F}\)) and velocity (\(\overrightarrow{v}\)) provided, we can find the components of the magnetic field (\(\overrightarrow{B}\)) using the cross product. Here, the force is only in the \(\hat{\imath}\) and \(\hat{k}\) directions, indicating that the magnetic field components \(B_x\) and \(B_z\) can be solved.

• To find \(B_x\) and \(B_z\), use the Lorentz Force Equation and solve for the components by isolating them from the given force equations.
• Since there is no force in the \(\hat{\jmath}\) direction, \(B_y\) remains undetermined. This is because the magnetic field component parallel to the velocity does not affect the force.

By knowing the fixed components and understanding the relations, we can partially determine the magnetic field.

Scalar Product

The scalar product, also known as the dot product, is a way to multiply two vectors to get a scalar. It is represented by the equation \( \overrightarrow{A} \cdot \overrightarrow{B} = AB \cos \theta \), where \(\theta\) is the angle between them. This process differs greatly from vector multiplication via a cross product.

• In this exercise, we're asked to calculate the scalar product of the magnetic field and the force: \( \overrightarrow{B} \cdot \overrightarrow{F} \).
• The important part for this exercise is understanding that the force due to a magnetic field is perpendicular to the velocity of the charge. In this case, the scalar product is zero because \( \cos 90^\circ = 0 \).

This zero result simplifies the problem and proves that \(\overrightarrow{B}\) and \(\overrightarrow{F}\) are orthogonal.

Cross Product

The cross product, or vector product, is a way of multiplying two vectors to get a third vector, which is perpendicular to the plane containing the initial two vectors. For vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), \( \overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}||\overrightarrow{B}|\sin\theta\overrightarrow{n} \), where \(\overrightarrow{n}\) is a unit vector normal to the plane of \(\overrightarrow{A}\) and \(\overrightarrow{B}\).

The Lorentz Force is an application of the cross product, impacting how forces act directionally on moving charges in a magnetic field.

• By applying the cross product to the velocity and magnetic field vectors, we can understand why the particle experiences a force in the specified direction.
• The result shows which components of the velocity vector and magnetic field vector influence the force.

Remember, the direction of the force vector can be found using the right-hand rule, scrolling your finger from \(\overrightarrow{v}\) to \(\overrightarrow{B}\) produces \(\overrightarrow{F}\).

Angle Between Vectors

Understanding the angle between vectors is crucial in vector mathematics. The angle provides insight into the relationship and interaction of vectors. Mathematically, the angle \(\theta\) between two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) can be found with the formula \( \cos\theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \).

• In the case of the magnetic force problem, calculating this angle is simplified due to the perpendicular nature of the force and velocity vectors.
• Since the dot product of the magnetic field and force vectors is zero, it confirms that their angle is \(90^\circ\).

This 90-degree angle signifies that the vectors are perpendicular. This cross interaction explains why force is maximal when the field and velocity have no parallel components.