Question

One of Maxwell's equations for electromagnetic waves is \(\nabla \times \mathbf{B}=C \frac{\partial \mathbf{E}}{\partial t},\) where \(\mathbf{E}\) is the electric field, \(\mathbf{B}\) is the magnetic field, and \(C\) is a constant. a. Show that the fields \(\mathbf{E}(z, t)=A \sin (k z-\omega t) \mathbf{i}\) and \(\mathbf{B}(z, t)=A \sin (k z-\omega t) \mathbf{j}\) satisfy the equation for constants \(A, k,\) and \(\omega,\) provided \(\omega=k / C\). b. Make a rough sketch showing the directions of \(\mathbf{E}\) and \(\mathbf{B}\).

Short answer

Based on the following step-by-step solution, provide a short answer to the question: To check if the provided electric and magnetic fields satisfy the given Maxwell's equation, we first calculated the curl of the magnetic field B and the partial derivative of the electric field E with respect to time. After substituting these values into Maxwell's equation, we found that the given fields satisfy the equation provided that the condition \(\omega=k / C\) is met. For the sketch, both electric and magnetic fields are sinusoidal and perpendicular to each other, with the electric field E pointing along the x-axis and the magnetic field B pointing along the y-axis. They propagate together in the z-direction, forming an electromagnetic wave.

Step-by-step solution

Find curl of B

To find the curl of B, simply calculate the cross product of the del operator with B: \(\nabla \times \mathbf{B}\). Using the given field B(z, t): \(\mathbf{B}(z, t)=A \sin (k z-\omega t) \mathbf{j}\) We can write the curl as follows: \(\nabla \times \mathbf{B} = \begin{bmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & 0 \\ 0 & A \sin (k z-\omega t) & 0 \end{bmatrix}\) Now, compute the cross product: \(\nabla \times \mathbf{B}= \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & A \sin (k z-\omega t) & 0 \end{array}\right|\) \(\nabla \times \mathbf{B}=A k \cos (k z-\omega t) \mathbf{i}\)

Find the partial derivative of E

To find the partial derivative of E with respect to time, simply differentiate E(z, t) with respect to time t: \(\mathbf{E}(z, t)=A \sin (k z-\omega t) \mathbf{i}\) \(\frac{\partial \mathbf{E}}{\partial t}=-A \omega \cos (k z-\omega t) \mathbf{i}\)

Substitute into Maxwell's equation

Now, substitute the curl of B and the partial derivative of E with respect to time t into Maxwell's equation: \(\nabla \times \mathbf{B}=C \frac{\partial \mathbf{E}}{\partial t}\) \(A k \cos (k z-\omega t) \mathbf{i}=C (-A \omega \cos (k z-\omega t) \mathbf{i})\) Comparing both sides, we have: \(A k = -A \omega C\) This equation holds true if \(\omega=k / C\). Therefore, the given fields satisfy Maxwell's equation provided \(\omega=k / C\). ##Part b##

Sketch E and B field directions

Now, let's make a rough sketch of the directions for the electric field E and the magnetic field B. Electric field E is given by: \(\mathbf{E}(z, t)=A \sin (k z-\omega t) \mathbf{i}\) This is pointing in the x-direction (i direction). Similarly, the given magnetic field B is: \(\mathbf{B}(z, t)=A \sin (k z-\omega t) \mathbf{j}\) This is pointing in the y-direction (j direction). In the sketch, we will represent the electric field E with a red arrow pointing in the positive x-direction and the magnetic field B with a blue arrow pointing in the positive y-direction. The electric and magnetic fields are both sinusoidal and propagate in the z-direction, meaning that they are perpendicular to each other, with E along the x-axis and B along the y-axis. Since both propagate in the same direction along the z-axis, they are in phase and form an electromagnetic wave.

Key concepts

Electromagnetic Waves

Electromagnetic waves are waves that consist of oscillating electric and magnetic fields. These fields are perpendicular to each other and propagate through space. Maxwell's equations describe how these fields interact and change over time, leading to the formation of electromagnetic waves. This interaction is the reason why light waves, radio waves, and X-rays can travel through space.
Understanding electromagnetic waves involves recognizing that they do not require a medium to travel. Unlike sound waves, they can move through the vacuum of space. This fascinating ability makes electromagnetic waves crucial for communication technologies, such as broadcasting and wireless internet.

• They consist of both electric and magnetic fields.
• Both fields oscillate at the same frequency and amplitude.
• The waves propagate in a direction that is perpendicular to both fields.

Electric Field

An electric field is a region around a charged particle where a force would be experienced by other charges. It is a vector field and is represented in physics by the letter \( \mathbf{E} \). In the given exercise, the electric field is described by the expression:

\( \mathbf{E}(z, t)=A \sin (k z-\omega t) \mathbf{i} \).
This indicates that the field is oscillating in the direction of the \( \mathbf{i} \), or x-direction.
The electric field's behavior is pivotal in electromagnetism as it determines how electric charges interact at a distance.

• The magnitude of the force experienced by a charge is proportional to the intensity of the electric field.
• Direction of the electric field is the direction of force a positive test charge would experience.
• Sinusoidal functions, like \( \sin(kz - \omega t) \), illustrate how the electric field varies with time and position.

Magnetic Field

A magnetic field is a vector field around a magnetic material or a moving electric charge, within which the force of magnetism acts. In the case of our exercise, the magnetic field \( \mathbf{B} \) is given by:

\( \mathbf{B}(z, t)=A \sin (k z-\omega t) \mathbf{j} \),
indicating that the magnetic field oscillates in the \( \mathbf{j} \), or y-direction.
Magnetic fields, like electric fields, are essential in electromagnetism and affect any moving charges, including the currents in wires.

• The magnetic field influences the direction and velocity of charged particles.
• In this scenario, the \( \sin(kz - \omega t) \) term implies a periodic fluctuation of the field as it propagates.
• Applications of magnetic fields include electric motors, generators, and magnetic resonance imaging (MRI).

Partial Derivatives

Partial derivatives are a tool used in calculus to understand how a function changes as one particular variable changes, keeping all other variables constant. In the context of the given problem, they are used to show the rate of change of the electric field \( \mathbf{E} \) with respect to time \( t \). This is a critical step in verifying whether the fields satisfy Maxwell's equation.

The partial derivative of \( \mathbf{E}(z, t) \) with respect to time \( t \) is:
\( \frac{\partial \mathbf{E}}{\partial t} = -A \omega \cos(kz - \omega t) \mathbf{i} \).
This expression provides insight into the instantaneous rate of change of \( \mathbf{E} \), an important aspect in electromagnetic theory.

• Allows us to analyze dynamic systems where variables change independently.
• The negative sign in \( -A \omega \cos(kz - \omega t) \) represents a phase shift in the oscillation.
• Partial derivatives are crucial for dealing with functions of several variables in physics and engineering models.