Question

Answer the following questions about force on a moving charge. A proton \(\left(q=1.6 \times 10^{-19} \mathrm{C}\right)\) with velocity \(2 \times 10^{6} \mathrm{j} \mathrm{m} / \mathrm{s}\) experiences a force in newtons of \(\mathbf{F}=5 \times 10^{-12} \mathbf{k}\) as it passes through the origin. Find the magnitude and direction of the magnetic field at that instant.

Short answer

Question: Determine the magnitude and direction of the magnetic field that acts on a proton with a charge of \(1.6 \times 10^{-19} \, \mathrm{C}\) and a velocity of \(2 \times 10^6 \, \mathrm{j} \, \mathrm{m} / \mathrm{s}\), when it experiences a force of \(5 \times 10^{-12} \, \mathrm{k} \, \mathrm{N}\). Answer: The magnitude of the magnetic field that acts on the proton is 15.625 T, and its direction is in the negative x direction.

Step-by-step solution

Write down the given information

We are given the charge of a proton \(q = 1.6 \times 10^{-19} \, \mathrm{C}\), its velocity \(\mathbf{v} = 2 \times 10^6 \, \mathrm{j} \, \mathrm{m} / \mathrm{s}\), and the force it experiences \(\mathbf{F} = 5 \times 10^{-12} \, \mathrm{k} \, \mathrm{N}\).

Use the formula for the force on a moving charge due to a magnetic field

The formula has the form: \(\mathbf{F} = q \mathbf{v} \times \mathbf{B}\). Substitute the given values into this equation: \(5 \times 10^{-12} \, \mathrm{k} \, \mathrm{N} = (1.6 \times 10^{-19} \, \mathrm{C}) \, (2 \times 10^6 \, \mathrm{j} \, \mathrm{m} / \mathrm{s}) \times \mathbf{B}\).

Calculate the cross product to find the magnetic field

To find the magnetic field, \(B\), we need to calculate the cross product of \(\mathbf{v}\) and \(\mathbf{B}\). Thus, we have: $\mathbf{F} = 1.6 \times 10^{-19} \, \mathrm{C} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 \times 10^6 & 0 \\ B_x & B_y & B_z \end{vmatrix}$

Solve for the components of the magnetic field

We can now solve for the components of the magnetic field: $5 \times 10^{-12} \, \mathrm{k} \, \mathrm{N} = 1.6 \times 10^{-19} \, \mathrm{C} \left(\begin{vmatrix} 2 \times 10^6 & 0 \\ B_y & B_z \end{vmatrix} \mathbf{i} - \begin{vmatrix} 0 & 0 \\ B_x & B_z \end{vmatrix} \mathbf{j} + \begin{vmatrix} 0 & 2 \times 10^6 \\ B_x & B_y \end{vmatrix} \mathbf{k} \right)$ Since the only force component given is in the k-direction, we can focus on that component: $5 \times 10^{-12} \, \mathrm{N} = 1.6 \times 10^{-19} \, \mathrm{C} \begin{vmatrix} 0 & 2 \times 10^6 \\ B_x & B_y \end{vmatrix}$

Calculate the magnitude and direction of the magnetic field

Divide both sides of the equation by the charge \(1.6 \times 10^{-19} \, \mathrm{C}\): $3.125 \times 10^7 \, \mathrm{N/C} = \begin{vmatrix} 0 & 2 \times 10^6 \\ B_x & B_y \end{vmatrix}$ Now, we can solve for \(B_x\) and \(B_y\): \(3.125 \times 10^7 \, \mathrm{N/C} = - B_x(2 \times 10^6)\) \(B_x = \dfrac{-3.125 \times 10^7}{2 \times 10^6} = -15.625 \, \mathrm{T}\) The magnetic field has only a component in the x direction, given that the velocity is entirely in the y direction and the force is entirely in the z direction. Therefore, the magnetic field can be written as: \(\mathbf{B} = -15.625 \, \mathrm{T} \, \mathrm{i}\) The magnitude of the magnetic field is: \(|\mathbf{B}| = |-15.625 \, \mathrm{T}| = 15.625 \, \mathrm{T}\) The direction of the magnetic field is in the negative x direction.

Key concepts

Lorentz Force

The Lorentz Force is a fundamental concept in physics that describes the force experienced by a charged particle as it moves through an electromagnetic field. Named after the physicist Hendrik Lorentz, it combines both electric and magnetic force into a single expression that affects charged particles. This force is crucial for understanding how charged particles behave in various fields, particularly in magnetic fields.
The mathematical expression for Lorentz force is\[ \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]Where:

• \( \mathbf{F} \) is the total Lorentz Force on a particle.
• \( q \) is the electric charge of the particle.
• \( \mathbf{E} \) represents the electric field.
• \( \mathbf{v} \) is the velocity of the particle.
• \( \mathbf{B} \) indicates the magnetic field.

In scenarios where only the magnetic field is considered, such as in the exercise given, the Lorentz force simplifies to:\[ \mathbf{F} = q \mathbf{v} \times \mathbf{B} \]This equation tells us that the force depends on the velocity of the particle, the magnetic field, and the charge of the particle. The force is always perpendicular to both the velocity and the magnetic field directions.

Cross Product

The Cross Product is a mathematical operation performed on two vectors in three-dimensional space, yielding another vector that is perpendicular to the plane of the input vectors. It is a crucial part of the Lorentz Force calculation and is represented as \( \mathbf{a} \times \mathbf{b} \).
In general, for vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product is calculated as:\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix} \]This determinant provides the components of the resulting vector in terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), representing the x, y, and z directions respectively. The magnitude of the cross product is given by:\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}| \sin \theta \]Where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). Thus, it shows how the cross product is maximum when the vectors are perpendicular, and zero when they are parallel. Cross product not only helps in determining the magnitude but also the direction of vectors, making it indispensable in calculating the force in magnetic fields.

Magnetic Force on a Proton

The magnetic force on a proton specifically refers to the force exerted by a magnetic field on a moving proton. Protons, being positively charged particles, interact with magnetic fields according to the concept of the Lorentz Force.
In the given exercise, we examine a proton with a charge \( q = 1.6 \times 10^{-19} \mathrm{C} \) moving at a velocity \( \mathbf{v} = 2 \times 10^6 \mathrm{j} \mathrm{m/s} \). The force it experiences \( \mathbf{F} = 5 \times 10^{-12} \mathrm{k} \mathrm{N} \) is entirely due to the magnetic field influence, as there is no electric field involved. Calculating the magnetic field requires rearranging the Lorentz force equation:\[ \mathbf{F} = q \mathbf{v} \times \mathbf{B} \]To isolate \( \mathbf{B} \), the cross product of velocity and the unknown magnetic field vectors must equal the given force. This involves matrix determinants, focusing only on components that result in the observed force direction, which in this exercise is exclusively in the \( \mathbf{k} \) direction.
After solving, the magnetic field \( \mathbf{B} = -15.625 \mathrm{T} \mathrm{i} \) indicates that the field acts solely in the negative x direction. Understanding this interplay of vectors helps to illustrate how magnetic fields can precisely control the path of charged particles like protons, which is an essential principle in technologies such as cyclotrons and MRI machines.