Problem 85 - Fundamentals of Physics

Question

At one instant, \(\vec{v}=(-2.00 \hat{\mathrm{i}}+4.00 \mathrm{j}-6.00 \mathrm{k}) \mathrm{m} / \mathrm{s}\) is the velocity of a proton in a uniform magnetic field \(\vec{B}=(2.00 \hat{\mathrm{i}}-\) \(4.00 \hat{\mathrm{j}}+8.00 \hat{\mathrm{k}}) T .\) At that instant, what are (a) the magnetic force \(\vec{F}\) acting on the proton, in unit-vector notation, (b) the angle between \(\vec{v}\) and \(\vec{F},\) and \((\mathrm{c})\) the angle between \(\vec{v}\) and \(\vec{B} ?\)

Step-by-Step Solution

Answer

(a) \((1.28 \times 10^{-18}\hat{\imath} + 0.64 \times 10^{-18}\hat{\jmath}) \mathrm{N}\); (b) \(90^\circ\); (c) calculated using \(\cos\phi = \frac{-68}{\sqrt{56}\sqrt{84}}\)."

Step 1: Calculate the Magnetic Force

The magnetic force \( \vec{F} \) on a charged particle moving through a magnetic field is given by \( \vec{F} = q \cdot (\vec{v} \times \vec{B}) \), where \( q \) is the charge of the proton \( q = +1.6 \times 10^{-19} \, \text{C} \). First, find the cross product \( \vec{v} \times \vec{B} \).\[\vec{v} = (-2.00\hat{\imath} + 4.00\hat{\jmath} - 6.00\hat{k})\, \text{m/s}\] \[\vec{B} = (2.00\hat{\imath} - 4.00\hat{\jmath} + 8.00\hat{k})\, \text{T}\] \[\vec{v} \times \vec{B} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ -2.00 & 4.00 & -6.00 \ 2.00 & -4.00 & 8.00 \end{vmatrix} = \hat{\imath}(4.00 \times 8.00 - (-6.00) \times (-4.00)) - \hat{\jmath}((-2.00) \times 8.00 - (-6.00) \times 2.00) + \hat{k}((-2.00) \times (-4.00) - 4.00 \times 2.00)\] \[ = \hat{\imath}\:(32.00 - 24.00) - \hat{\jmath}\:(-16.00 - (-12.00)) + \hat{k}\:(8.00 - 8.00)\] \[ = 8.00\hat{\imath} - (-4.00)\hat{\jmath} + 0\hat{k} = 8.00\hat{\imath} + 4.00\hat{\jmath} + 0\hat{k} \] Then the magnetic force is \[\vec{F} = 1.6 \times 10^{-19} \cdot (8.00\hat{\imath} + 4.00\hat{\jmath} + 0\hat{k}) = (1.28 \times 10^{-18}\hat{\imath} + 0.64 \times 10^{-18}\hat{\jmath}) \, \text{N}.\]

Step 2: Determine the Angle between \(\vec{v}\) and \(\vec{F}\)

The angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) can be found using the formula \( \cos\theta = \frac{\vec{A} \cdot \vec{B}}{||\vec{A}|| ||\vec{B}||} \). Since \( \vec{F} = \vec{v} \times \vec{B} \), \( \vec{F} \) is perpendicular to both \( \vec{v} \) and \( \vec{B} \). This means the angle is \( 90^\circ \). Thus, \( \theta = 90^\circ \) between \( \vec{v} \) and \( \vec{F} \).

Step 3: Calculate the Angle between \(\vec{v}\) and \(\vec{B}\)

Use the dot product to find the angle \( \phi \) between \( \vec{v} \) and \( \vec{B} \). \( \vec{v} \cdot \vec{B} = (-2)(2) + (4)(-4) + (-6)(8) = -4 - 16 - 48 = -68 \). The magnitudes are given by \( ||\vec{v}|| = \sqrt{(-2)^2 + 4^2 + (-6)^2} = \sqrt{4+16+36} = \sqrt{56}\) and \( ||\vec{B}|| = \sqrt{2^2 + (-4)^2 + 8^2} = \sqrt{4+16+64} = \sqrt{84} \). Thus, \[\cos\phi = \frac{-68}{\sqrt{56}\sqrt{84}}\]. This gives \( \phi = \cos^{-1}\left(\frac{-68}{\sqrt{56}\sqrt{84}}\right) \).

Key Concepts

Proton in Magnetic Field

When we talk about a proton moving through a magnetic field, we are dealing with an interesting physical phenomenon. Protons are charged particles, each carrying a positive charge of approximately \( +1.6 \times 10^{-19} \, \text{C} \). When a charged particle like a proton moves through a magnetic field, it experiences a force that acts perpendicular to both the direction of the magnetic field and its velocity. This is known as the magnetic force, denoted as \( \vec{F} \).

This scenario requires us to understand how the magnetic force is calculated. The formula used is \( \vec{F} = q \cdot (\vec{v} \times \vec{B}) \), where \( q \) represents the charge of the proton, \( \vec{v} \) is the velocity vector, and \( \vec{B} \) is the magnetic field vector. Here, the cross product \( \vec{v} \times \vec{B} \) gives a vector that is perpendicular to the plane formed by the velocity and the magnetic field vectors. This is a crucial point because it leads to the Lorentz force law.

In essence, the proton's trajectory is affected by this force, resulting in a curved path known as a circular or helical trajectory. Understanding these interactions is key to grasping fundamental concepts in electromagnetism.

Cross Product of Vectors

The cross product \( \vec{v} \times \vec{B} \) is an essential operation in vector algebra, especially in physics when dealing with magnetic fields. Defined for three-dimensional vectors, the cross product results in a new vector that is perpendicular to the plane of the original vectors. This concept can be visualized using the right-hand rule: if you point your index finger in the direction of \( \vec{v} \) and your middle finger in the direction of \( \vec{B} \), your thumb points in the direction of the cross product \( \vec{v} \times \vec{B} \).

The calculation of the cross product involves a determinant, particularly one that includes the unit vectors \( \hat{\imath}, \hat{\jmath}, \hat{k} \). The determinant gives us:

- An \( \hat{\imath} \) component equal to \(( v_2B_3 - v_3B_2 )\)
- A \( \hat{\jmath} \) component equal to \(( v_3B_1 - v_1B_3 )\)
- And a \( \hat{k} \) component equal to \(( v_1B_2 - v_2B_1 )\)

In the given algebraic expression, the cross product results in a vector with components \( 8.00\hat{\imath} + 4.00\hat{\jmath} + 0\hat{k} \), which is then multiplied by the proton's charge to find the magnetic force.

Angle Between Vectors

Understanding the angle between two vectors is crucial, as it helps explain the directionality and relationships between forces and motions. There are two situations to consider in this exercise: the angle between the velocity vector \( \vec{v} \) and the magnetic force vector \( \vec{F} \), and the angle between \( \vec{v} \) and the magnetic field vector \( \vec{B} \).

For \( \vec{v} \) and \( \vec{F} \), the nature of the cross product tells us that these two vectors are orthogonal, or at a right angle to each other (\( 90^\circ \). This is an inherent trait of cross products because they produce vectors that define a plane perpendicular to the original vectors involved.

To find the angle between \( \vec{v} \) and \( \vec{B} \), we apply the dot product. The dot product formula \( \vec{v} \cdot \vec{B} = ||\vec{v}|| \cdot ||\vec{B}|| \cdot \cos\phi \) helps us find \( \phi \), the angle between these two vectors. Solving the equation \( \cos\phi = \frac{\vec{v} \cdot \vec{B}}{||\vec{v}|| ||\vec{B}||} \) gives us the specific angle, showcasing the unique relationship between the trajectory of the proton and the magnetic field.

Unit-Vector Notation

Unit-vector notation is a method used in physics to express vectors in a clear, concise way. By breaking down vectors into components along x, y, and z axes, described with unit vectors \( \hat{\imath}, \hat{\jmath}, \hat{k} \), we gain a complete and easy-to-interpret picture of directions and magnitudes.

This notation is particularly effective when dealing with problems involving vector sums and differences, the cross product, and other vector operations. Each component of the vector is expressed as a multiple of a unit vector, which represents one unit of measurement in its respective direction. For instance, in the vector \( \vec{v} = (-2\hat{\imath} + 4\hat{\jmath} - 6\hat{k}) \), \(-2\hat{\imath}\) indicates 2 units in the negative x-direction.

Such clarity makes it easier to understand the problem of a proton moving through a magnetic field, translating the problem into manageable algebraic operations for solving the magnetic force acting upon it. Thus, unit-vector notation provides a foundation for more intricate calculations involving vector quantities in physics.