Question

In a region where \(\mu_{r}=\epsilon_{r}=1\) and \(\sigma=0\), the retarded potentials are given by \(V=x(z-c t) \mathrm{V}\) and \(\mathbf{A}=x\left(\frac{z}{c}-t\right) \mathbf{a}_{z} \mathrm{~Wb} / \mathrm{m}\), where \(c=1 \sqrt{\mu_{0} \epsilon_{0}}\) (a) Show that \(\nabla \cdot \mathbf{A}=-\mu \epsilon \frac{\partial V}{\partial t} .(b)\) Find \(\mathbf{B}, \mathbf{H}, \mathbf{E}\), and \(\mathbf{D} .(c)\) Show that these results satisfy Maxwell's equations if \(\mathbf{J}\) and \(\rho_{v}\) are zero.

Short answer

a) We have shown that ∇ ⋅ A = x = -με ∂V/∂t, the given equation holds true. b) The magnetic field B = B_y * a_y, the magnetic intensity H = B_y * a_y / μ₀, the electric field E = -[(z - ct)*a_x + xa_z + x*a_y], and the electric displacement D = -ε₀[(z - ct)*a_x + xa_z + x*a_y]. c) We have verified that the given potentials and the derived fields satisfy Maxwell's equations when current density J and volume charge density ρ_v are zero. Specifically, we have shown that these fields satisfy Gauss's law for electric fields, Gauss's law for magnetic fields, Faraday's law, and the Ampere-Maxwell law.

Step-by-step solution

Calculate the time derivatives and gradients of the given potentials

We are given that V = x(z - ct) and A = x((z/c) - t)a_z. We will first find the time derivative of V and the gradient of A. ∂V/∂t = -cx And, ∇ ⋅ A = ∂A_x/∂x + ∂A_y/∂y + ∂A_z/∂z = ∂((z/c)-t)x/∂z = x

Verify the given equation for the divergence of A

We have to show that ∇ ⋅ A = -με ∂V/∂t. Based on our calculations in Step 1: ∇ ⋅ A = x = -(-cx) = -με ∂V/∂t Thus, the given equation holds true.

Calculate the magnetic field B

Now we will find the magnetic field B using the formula: B = ∇ × A. Let's compute the curl of A: ∇ × A = |i j k| |∂/∂x ∂/∂y ∂/∂z| |0 0 (z/c - t)| ∇ × A = -∂A_z/∂y * i + (∂A_z/∂x - 0/∂z) * j + 0 * k The only non-zero component of the curl is the y-component: B_y = ∂A_z/∂x = 1 Thus, the magnetic field is: B = B_y * a_y

Calculate the magnetic intensity H

Since μ_r = 1, μ = μ₀, and H = B/μ. We can calculate magnetic intensity as follows: H = B_y * a_y / μ₀

Calculate the electric field E

For E, we will use the formula: E = -∇V - ∂A/∂t. E = -∂V/∂x * a_x - ∂V/∂z * a_z - ∂A/∂t E_x = -∂V/∂x = -(z - ct) E_z = -∂V/∂z = -x E_t = -∂A_x/∂t - ∂A_y/∂t - ∂A_z/∂t = x Hence, E = -[(z - ct)*a_x + xa_z + x*a_y]

Calculate the electric displacement D

Since ε_r = 1 and σ = 0, ε = ε₀. D = ε₀E= -ε₀[(z - ct)*a_x + xa_z + x*a_y]

Verify Maxwell's equations

Maxwell's equations are given by: 1. ∇ ⋅ D = ρ_v (Gauss's law for electric fields) 2. ∇ ⋅ B = 0 (Gauss's law for magnetic fields) 3. ∇ × E = -∂B/∂t (Faraday's law) 4. ∇ × H = J + ∂D/∂t (Ampere-Maxwell law) It is mentioned that ρ_v = 0 and J = 0. For equation 1: ∇ ⋅ D = -ε₀∇ ⋅ [(z - ct)*a_x + xa_z + x*a_y] The divergence is 0, which satisfies equation 1, since ρ_v = 0. For equation 2: ∇ ⋅ B = 0 This is already true, as previously calculated. For equation 3: ∇ × E = -∂B/∂t We have already calculated that E has only y-component, and it is constant, ∂B/∂t = 0, so Faraday's law is satisfied. For equation 4: ∇ × H = J + ∂D/∂t Since J = 0, the left-hand side will be ∂D/∂t. ∂D/∂t = -ε₀[-ca_x - a_y] = ε₀[ca_x + a_y] On the right side, we have ∇ × H = 0, since H has only the y-component. So the Ampere-Maxwell law is satisfied. Thus, we have shown that the given potentials and the derived fields satisfy Maxwell's equations when J = 0 and ρ_v = 0.

Key concepts

Electromagnetic Potentials

Electromagnetic potentials, including the scalar potential 'V' and the vector potential '\textbf{A}', are powerful concepts used to describe the electric and magnetic fields, denoted as '\textbf{E}'-field and '\textbf{B}'-field, respectively. In the context of our exercise, these potentials help express electromagnetic fields in a simpler form. The scalar potential 'V' represents the potential energy per unit charge due to the electric field, while the vector potential '\textbf{A}' correlates with magnetic field lines.

To understand the role these potentials play, consider the electric field '\textbf{E}' which can be found by taking the negative gradient of the scalar potential 'V' and subtracting the time derivative of the vector potential '\textbf{A}'. Mathematically, this relationship is given by the equation \( \textbf{E} = -abla V - \frac{\textbf{A}}{dt} \). Similarly, the magnetic field '\textbf{B}' can be found by taking the curl of the vector potential '\textbf{A}', as expressed by \( \textbf{B} = abla \times \textbf{A} \).

Through the exercise we also encounter an important relation \( abla \textbf{A} = -\textbf{\textup{Continuous Equations}}\), which ties both electric and magnetic fields to their potentials. The exercise solution demonstrates the practical application of this relationship, providing insight into the interconnected nature of electric and magnetic phenomena.

Magnetic Field (B-Field)

The magnetic field, or '\textbf{B}'-field, is a vector field that represents the presence and strength of magnetic forces. A key characteristic of the magnetic field is that it is always solenoidal, meaning that it has no divergence; mathematically \( abla \textbf{B} = 0 \). This is aligned with Gauss's Law for magnetism, one of Maxwell's equations, which confirms that magnetic monopoles do not exist in nature.

In the solution provided, the magnetic field is derived using the curl of the vector potential '\textbf{A}', yielding a non-zero '\textbf{B}'-field aligned with the y-axis. This result is directly linked to the spatial changes in the vector potential. Such a computed '\textbf{B}'-field reflects the magnetic influence of currents and magnetic materials when present, although our scenario assumes a vacuum with no free currents or charges.

Understanding how this '\textbf{B}'-field emerges from Maxwell's equations and potentials is crucial for students, as it pinpoints how magnetic effects manifest from changes in the electromagnetic potentials.

Electric Field (E-Field)

The electric field, denoted as '\textbf{E}'-field, illustrates the force that would be exerted on a positive charge at any point in space. It is a vector field that can be described completely by two components: the conservative field derived from the scalar potential 'V' and the solenoidal field associated with the vector potential '\textbf{A}'. As per Faraday's law, a changing magnetic field induces an electric field, which is reflected by the second term in the equation \( \textbf{E} = -abla V - \frac{\textbf{A}}{dt} \).

In our exercise, the electric field is computed from both the gradient of 'V' and the time rate of change of '\textbf{A}', resulting in components along the x and z axes. This treatment allows students to engage with the dynamic nature of electromagnetic fields. It shows that time-varying magnetic fields are inherently linked to electric fields, a concept emphasizing the inseparable nature of electricity and magnetism in electromagnetic wave propagation, which Maxwell's equations elegantly unify.

Electromagnetic Wave Propagation

Electromagnetic wave propagation pertains to how waves carrying electric and magnetic fields spread through space. This phenomenon is a cornerstone of modern physics and underpins technologies like radio, television, and cellular communications. Maxwell's equations describe this propagation by revealing how time-varying electric fields generate magnetic fields and vice versa. These changing fields transport energy and momentum through space, often at the speed of light in a vacuum.

In the exercise, one of Maxwell's equations, specifically Faraday's Law \( abla \times \textbf{E} = -\frac{\textbf{B}}{dt} \), demonstrates how a time-varying magnetic field is linked to the curl of the electric field. Similarly, the Ampere-Maxwell Law involving the curl of the magnetic intensity '\textbf{H}' field and the change in electric displacement '\textbf{D}' showcases electromagnetic wave interactions. When solving the problem, we observe no free current or charge, simplifying the situation and focusing on the intrinsic properties of the electromagnetic wave's propagation through space.