Question

A vector potential is given as \(\mathbf{A}=A_{0} \cos (\omega t-k z) \mathbf{a}_{y} .(a)\) Assuming as many components as possible are zero, find \(\mathbf{H}, \mathbf{E}\), and \(V .(b)\) Specify \(k\) in terms of \(A_{0}, \omega\), and the constants of the lossless medium, \(\epsilon\) and \(\mu\).

Short answer

#tag_title#Short Answer#tag_content# The magnetic field H can be found as 𝑯_z = A₀ωxsin(ωt - kz). The electric field E has a non-zero x-component given by 𝑬_x = -μA₀2ω²xtzcos(ωt - kz). The vector potential V has a non-zero x-component given by 𝑉_x = (1/2)μA₀2ω²xtz²cos(ωt - kz). Finally, k is determined in terms of A₀, ω, ε, and μ as k = √(ω²εμ).

Step-by-step solution

Determine the magnetic field H

For vector potential 𝐴 = A₀cos(ωt - kz) 𝑎_𝑦, we can determine the magnetic field H by applying Ampere's Law, which states that the curl of H is equal to the current density J plus the time derivative of the electric field E. In a lossless medium, the current density J is zero, so we have: ∇ × 𝑯 = −∂𝑬/∂t Since 𝑯 and 𝑬 both lie in the 𝑥𝑧-plane and 𝐴 lies in the 𝑦-direction, we only need to compute the curl in terms of the x and z components: (∇ × 𝑯)_𝑦 = ∂𝑯_z/∂x - ∂𝑯_𝑥/∂z But since we are assuming as many components as possible are zero, 𝑯_x is zero, and we only have: (∇ × 𝑯)_𝑦 = ∂𝑯_z/∂x Now with the given vector potential 𝐴, we can calculate the time derivative of the electric field: ∂𝑬/∂t = ∂/∂t (A₀cos(ωt - kz)a_y) ∂𝑬/∂t = A₀ωsin(ωt - kz)a_y Now we can equate the curl of 𝑯 and the time derivative of 𝑬: ∂𝐻_𝑧/∂x = A₀ωsin(ωt - kz) Integrate with respect to x: 𝑯_z = A₀ωxsin(ωt - kz) + C But assuming as many components as possible are zero, so we ignore the constant and have: 𝑯_z = A₀ωxsin(ωt - kz)

Determine the electric field E

Now we will determine the electric field by using Faraday's Law, which states that the curl of E is equal to the negative time derivative of the magnetic field, or: ∇ × 𝑬 = -∂𝑩/∂t Since 𝑩 = μ𝑯, we can rewrite the expression as: ∇ × 𝑬 = -μ∂𝑯/∂t Now, taking the curl of 𝑬 and operating on the previously found 𝑯, we have: (∇ × 𝑬)_𝑦 = ∂𝑬_z/∂x - ∂𝑬_x/∂z = -μA₀2ω² xtcos(ωt - kz) However, since 𝑬_x and 𝑬_z are the only non-zero components, we just have: ∂𝑬_x/∂z = -μA₀2ω²xtcos(ωt - kz) Integrate with respect to z: 𝑬_x = -μA₀2ω²xtcos(ωt - kz)z + C Again, we will ignore the constant, so we have: 𝑬_x = -μA₀2ω²xtzcos(ωt - kz)

Determine the vector potential V

Since the vector potential V is related to the electric field E by 𝑉 = -∫𝑬d𝑟, we can find V from 𝑬. For simplicity, we only consider the x-component, as 𝑉_y and 𝑉_z are zero: 𝑉_x = -∫𝑬_xd𝑟 = -∫(-μA₀2ω²xtzcos(ωt - kz))dr Considering the integral along the x-direction, we have: 𝑉_x = μA₀2ω²xtcos(ωt - kz)∫zdx Integrating with respect to x, we have: 𝑉_x = (1/2)μA₀2ω²xtz²cos(ωt - kz)

Determine k in terms of A₀, ω, ε, and μ

Now we need to express k in terms of A₀, ω, ε, and μ. To do this, we can use the wave equations for both the electric field 𝑬 and magnetic field 𝑯: ∇²𝑬 - εμ∂²𝑬/∂t² = 0 ∇²𝑯 - εμ∂²𝑯/∂t² = 0 Plugging the expressions for 𝑬 and 𝑯 into these equations and simplifying, we have: k² = ω²εμ So, k is given by: k = √(ω²εμ)

Key concepts

Understanding Ampere's Law in Electromagnetics

Ampere's Law is a fundamental principle in electromagnetics that relates the magnetic field in space to the electric current that produces it. In essence, it states that the integral of the magnetic field (H) around a closed loop is equal to the total electric current passing through the loop, plus the rate change of electric displacement over time. This is captured mathematically by the integral form of Ampere's Law:
\[\begin{equation}\oint \mathbf{H} \cdot d\mathbf{l} = I + \frac{d\Phi_E}{dt},\end{equation}\]where \( \mathbf{H} \) is the magnetic field, \( d\mathbf{l} \) is an infinitesimal vector along the loop, and \( \Phi_E \) is the electric flux. If we apply Ampere's Law in the context of the given problem, where the magnetic field \( \mathbf{H} \) is derived from a vector potential \( \mathbf{A} \) and there's no current present (a lossless medium), the integral reduces to the curl of \( \mathbf{H} \) being equal to the rate change of electric field \( \mathbf{E} \) over time. This is used to determine the magnetic field generated by the given vector potential.

In our exercise, we have a vector potential \( \mathbf{A} = A_{0} \cos (\omega t-k z)\mathbf{a}_{y} \) which means that the resulting magnetic field \( \mathbf{H} \) and electric field \( \mathbf{E} \) will result from the variations in this potential. By applying the right-hand rule and considering the geometry of the problem, along with the assumption that as many components of \( \mathbf{H} \) and \( \mathbf{E} \) are zero, the step-by-step solution handles the simplification expertly and calculates the magnetic field correctly.

Deciphering Faraday's Law of Electromagnetic Induction

Faraday's Law of Electromagnetic Induction illustrates how a change in magnetic flux can induce an electric field. This concept is pivotal in understanding how electric generators and transformers operate. Mathematically, Faraday's Law can be expressed using the curl of the electric field \( \mathbf{E} \) as proportional to the negative change over time of the magnetic flux density \( \mathbf{B} \) as follows:
\[\begin{equation}abla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t},\end{equation}\]where \( \mathbf{B} \) is related to the magnetic field \( \mathbf{H} \) through the magnetic permeability \( \mu \) of the medium: \( \mathbf{B} = \mu\mathbf{H} \).

In the step by step explanation of the exercise, the use of Faraday's Law comes into play when determining the electric field \( \mathbf{E} \) that is associated with the time-varying magnetic field we found from Ampere's Law. The approach used in the solution correctly applies Faraday's Law by taking the negative time derivative of the magnetic field to find the curl of the electric field, then integrating this result to determine the electric field components.By using Faraday's Law, we account for the induced electric field that is caused by a changing magnetic field, which is a key concept for students to grasp in order to fully understand the behavior of electromagnetic waves and their propagation.

Wave Equations and Their Role in Electromagnetics

Wave equations are differential equations that describe the propagation of waves through a medium. In electromagnetics, they play a crucial role in detailing how electric and magnetic fields disperse and interact. The wave equations for electric and magnetic fields in free space or any non-conductive medium are given by Maxwell's equations:
\[\begin{equation}abla^2 \mathbf{E} - \mu\epsilon\frac{\partial^2 \mathbf{E}}{\partial t^2} = 0,\end{equation}\]\[\begin{equation}abla^2 \mathbf{H} - \mu\epsilon\frac{\partial^2 \mathbf{H}}{\partial t^2} = 0,\end{equation}\]where \( \mu \) represents the magnetic permeability, and \( \epsilon \) denotes the electric permittivity of the medium. These parameters dictate how electric and magnetic fields propagate through space.

For the given exercise, by solving the wave equations for \( \mathbf{E} \) and \( \mathbf{H} \) given the vector potential \( \mathbf{A} \) and its corresponding derivatives, we can examine the propagation characteristics, such as the wave number \( k \) which defines the spatial frequency of the wave. The solution ties \( k \) back to the other given quantities (\( A_{0} \), \( \omega \), \( \epsilon \) and \( \mu \)) which directly influence the wave’s behavior in the medium. Understanding wave equations facilitates deeper insights into topics such as wave reflection, refraction, diffraction, and interference, which are pivotal for grasping the physics behind many modern-day applications like wireless communications, radar systems, and even medical imaging techniques.