Question

Answer the following questions about force on a moving charge. A particle with a positive unit charge \((q=1)\) enters a constant magnetic field \(\mathbf{B}=\mathbf{i}+\mathbf{j}\) with a velocity \(\mathbf{v}=20 \mathbf{k} .\) Find the magnitude and direction of the force on the particle. Make a sketch of the magnetic field, the velocity, and the force.

Short answer

Question: Calculate the magnitude and direction of the force on a moving charged particle with velocity \(\mathbf{v}=20\mathbf{k}\) and charge \(q=1\) in a magnetic field \(\mathbf{B}=\mathbf{i}+\mathbf{j}\). Sketch the magnetic field, velocity, and force vectors. Answer: The magnitude of the force on the particle is \(20\sqrt{2}\) units, and the direction is along the vector \(\frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\). The sketch shows the magnetic field vector pointing to the top right, the velocity vector pointing up, and the force vector pointing down and to the right.

Step-by-step solution

Calculate the cross product \(\mathbf{v} \times \mathbf{B}\)

To find the force, we first need to calculate the cross product of the velocity vector and the magnetic field vector: \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B}) = (\mathbf{v} \times \mathbf{B})\) \(\mathbf{v} = \begin{pmatrix} 0 \\ 0 \\ 20 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) \(\mathbf{v} \times \mathbf{B} = \begin{pmatrix} (0\cdot0)-(20\cdot1) \\ (20\cdot1)-(0\cdot0) \\ (0\cdot1)-(0\cdot1) \end{pmatrix} = \begin{pmatrix} -20 \\ 20 \\ 0 \end{pmatrix}\) Thus, the force vector \(\mathbf{F}\) on the particle is \(\mathbf{F} = \begin{pmatrix} -20 \\ 20 \\ 0 \end{pmatrix}\).

Calculate the magnitude of the force

The magnitude of the force vector is given by: \(|{\mathbf{F}}| = \sqrt{(-20)^2 + (20)^2 + (0)^2} = 20\sqrt{2}\) The magnitude of the force on the particle is \(20\sqrt{2}\) units.

Calculate the direction of the force

To find the direction of the force vector, we can calculate its unit vector: \(\hat{\mathbf{F}} = \frac{\mathbf{F}}{|{\mathbf{F}}|} = \frac{\begin{pmatrix} -20 \\ 20 \\ 0 \end{pmatrix}}{20\sqrt{2}} = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\) The direction of the force is along the vector \(\frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\).

Visualize the magnetic field, velocity, and force

To sketch the magnetic field, velocity, and force, you can draw the vectors as arrows starting from the same origin point: - For the magnetic field \(\mathbf{B}=\mathbf{i}+\mathbf{j}\), draw an arrow pointing to the top right (along the line y=x with a slope of 45 degrees). - For the velocity \(\mathbf{v}=20\mathbf{k}\), draw an arrow pointing up (along the z-axis). - For the force \(\mathbf{F}\), draw an arrow pointing down and to the right (along the line y=-x with a slope of -45 degrees). In this sketch, the velocity vector is perpendicular to the magnetic field vector, and the force vector is perpendicular to both the velocity and magnetic field vectors.

Key concepts

Cross Product

When dealing with magnetic forces on moving charges, the concept of the cross product is crucial. It helps us determine the perpendicular force exerted on the charge. In our example, a positive unit charge moves with velocity \( \mathbf{v} = 20 \mathbf{k} \) entering a magnetic field \( \mathbf{B} = \mathbf{i} + \mathbf{j} \). Here, the cross product \( \mathbf{v} \times \mathbf{B} \) gives us the force direction.
This operation ensures that the resulting vector is orthogonal to both the velocity and magnetic field vectors. The calculation follows the determinant method, and the resulting force vector is \( \mathbf{F} = \begin{pmatrix} -20 \ 20 \ 0 \end{pmatrix} \). Remember, the cross product of two vectors gives a third vector that is perpendicular to the plane formed by the first two.

Vector Magnitude

The magnitude of a vector provides its length or size in space. For a force vector \( \mathbf{F} = \begin{pmatrix} -20 \ 20 \ 0 \end{pmatrix} \), calculating its magnitude involves finding the square root of the sum of the squares of its components. This is given by the formula:

• \( |\mathbf{F}| = \sqrt{(-20)^2 + (20)^2 + (0)^2} \)
• The magnitude = \( 20\sqrt{2} \)

Understanding vector magnitude is important for quantifying how strong the force is. A larger magnitude indicates a stronger force.

Unit Vector

A unit vector gives the direction of a vector through a magnitude normalization process, resulting in a vector of length 1. To find the unit vector of the force \( \mathbf{F} \), we divide each component by its magnitude \( 20\sqrt{2} \). This makes the direction independent of the original magnitude. The calculation is:

• \( \hat{\mathbf{F}} = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} \)

This unit vector tells us the precise direction of the force with equal emphasis on its components, ensuring that no single component dominates.

Force Direction

The direction of the force on a charge in a magnetic field is determined by the right-hand rule. In our scenario, the velocity is along the positive z-axis and the magnetic field is in the xy-plane. The force, resulting from the cross product, lies in the xy-plane perpendicular to both. Visualizing:

• The magnetic field \( \mathbf{B} = \mathbf{i} + \mathbf{j} \) points towards the top right in a 2D plane.
• The velocity \( \mathbf{v} = 20 \mathbf{k} \) points upwards.
• The force \( \mathbf{F} = \begin{pmatrix} -20 \ 20 \ 0 \end{pmatrix} \) points down and to the right.

This arrangement illustrates that the force is always orthogonal to both the velocity and magnetic field, thereby resulting in circular motion when in a uniform field.