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Let \(\mu=3 \times 10^{-5} \mathrm{H} / \mathrm{m}, \epsilon=1.2 \times 10^{-10} \mathrm{~F} / \mathrm{m}\), and \(\sigma=0\) everywhere. If \(\mathbf{H}=2 \cos \left(10^{10} t-\beta x\right) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\), use Maxwell's equations to obtain expressions for \(\mathbf{B}, \mathbf{D}, \mathbf{E}\), and \(\beta\)

Short Answer

Expert verified
Question: Find the expressions for B, E, D, and β based on the given values of μ, ε, σ, and magnetic field strength (H). Answer: 1. B = \(6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_z\, \mathrm{T}\) 2. E = \(1.897366596101028 \times 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{V/m}\) 3. D = \(7.2 \times 10^{-15}\times 1.897366596101028\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{C/m^2}\) 4. β = 1.897366596101028

Step by step solution

01

Calculate B from H

Considering the relationship between B and H, which is given by \(B = \mu H\), we have: \(B = 3 \times 10^{-5}\times 2\cos\left(10^{10}t - \beta x\right)\) This simplifies to: \(B = 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_z\, \mathrm{T}\)
02

Find E using Faraday's Law

Faraday's Law states: \(\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\) Taking the partial derivative of B with respect to t, we get: \(\frac{\partial B}{\partial t} = 6 \times 10^{-5}\times-10^{10}\sin\left(10^{10}t-\beta x\right)\mathbf{a}_z\) From Faraday's law, we can express E in the following form: \(E = \beta \times 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{V/m}\)
03

Calculate D from E

Considering the relationship between D and E, which is given by \(D = \epsilon E\): \(D = 1.2 \times 10^{-10} \times \beta \times 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right)\) This simplifies to: \(D = 7.2 \times 10^{-15}\beta\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{C/m^2}\)
04

Find β in terms of μ and ε

Using the equation for wave propagation in a non-conductive medium, which is given by \(\beta^2 = \mu\epsilon\omega^2\), where \(\omega\) is the angular frequency: Since \(\omega = 10^{10}\, \mathrm{rad/s}\), we have: \(\beta^2 = 3\times 10^{-5} \times 1.2 \times 10^{-10} \times \left(10^{10}\right)^2\) This simplifies to: \(\beta^2 = 3.6\) Taking the square root, we get: \(\beta = \sqrt{3.6} = 1.897366596101028\) In summary, we have found the required expressions: 1. \(\mathbf{B} = 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_z\, \mathrm{T}\) 2. \(\mathbf{E} = 1.897366596101028 \times 6 \times 10^{-5}\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{V/m}\) 3. \(\mathbf{D} = 7.2 \times 10^{-15}\times 1.897366596101028\cos\left(10^{10}t-\beta x\right) \mathbf{a}_y\, \mathrm{C/m^2}\) 4. \(\beta = 1.897366596101028\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic wave propagation
Understanding the propagation of electromagnetic waves is a fundamental aspect of modern physics and essential for various technological applications such as wireless communication and radar systems. Electromagnetic waves are ripples in the electromagnetic field, consisting of oscillating electric and magnetic fields that propagate through space. These waves can travel through different media, including vacuum, air, and other substances.

Mathematically, the propagation of electromagnetic waves is governed by Maxwell's equations, which describe how electric and magnetic fields are generated and altered by each other and by charges and currents. In the absence of free charges (like in the provided exercise where \(\sigma = 0\)), the waves can be described as sinusoidal variations, as shown in the magnetic field intensity (\textbf{H}). The speed and direction of the propagation are fundamentally determined by the properties of the medium through which the waves are traveling.

Maxwell's equations lead us to a vital quantity, the wave number denoted as \((\beta\)), which is dependent on the medium's permeability and permittivity. In our exercise, \(\beta\) is crucial for expressing all fields involved in the electromagnetic wave, including \(\mathbf{B}\), \(\mathbf{E}\), and \(\mathbf{D}\), showing how interconnected these fields are in the propagation of the wave.
Magnetic field intensity (H)
Magnetic field intensity, often represented by \(\mathbf{H}\), is a vector quantity that describes the concentration and strength of a magnetic field at any point in space. It is related to but not the same as the magnetic flux density (\textbf{B}). \(\mathbf{H}\) helps in determining the magnetizing force provided by a magnetic source, and in our scenario, it is described by a cosine function that varies with time (t) and position (x).

In the exercise, the provided equation for \(\mathbf{H}\) demonstrates an oscillatory pattern typical for electromagnetic waves. It is also directly proportional to the magnetic flux density, as dictated by the simple yet profound relationship \(\mathbf{B} = \mu \mathbf{H}\). By knowing the permeability of the medium (\(\mu\)), we can effortlessly find the corresponding \(\mathbf{B}\) field, which further illustrates the intrinsic link between these vector fields and their role in electromagnetic wave propagation.
Electric flux density (D)
The electric flux density, represented by \(\mathbf{D}\), is an essential parameter that relates to how an electric field \(\mathbf{E}\) interacts with a material medium. It indicates the amount of electric flux passing through a particular area and is influenced by both the electric field present and the medium's permittivity (\(\epsilon\)). In the exercise, we are given an expression for \(\mathbf{E}\) that we can use, along with \(\epsilon\), to calculate \(\mathbf{D}\) utilizing the relationship \(\mathbf{D} = \epsilon \mathbf{E}\).

For students, it's important to understand that while \(\mathbf{E}\) can tell us the force that would be experienced by a charge in the field, \(\mathbf{D}\) captures how that field is distributed in space, considering the medium's influence. In our exercise case, the sinusoidal form of \(\mathbf{E}\) and \(\mathbf{D}\) underscores the variability of these quantities with time and space, characteristic of propagating waves.
Electric field intensity (E)
Electric field intensity, denoted \(\mathbf{E}\), is a vector field representing the electric force per unit charge exerted on a positive test charge at any point in space. It's a fundamental concept in electromagnetism that facilitates the understanding of how charges interact within an electric field. In the context of our exercise, the \(\mathbf{E}\) field can be derived using Faraday's Law of electromagnetic induction, which links the change in the magnetic field (\textbf{B}) over time to the induced electric field.

The relationship \(abla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\) signifies that a time-varying magnetic field will induce a circulating electric field. Here, Faraday's Law is used to deduce the \(\mathbf{E}\) component due to a changing \(\mathbf{B}\) field. Consequently, the sinusoidal nature of the magnetic field propagates to the electric field, which is an illustration of how the alteration of one field induces the other, reinforcing the concept of electromagnetic wave propagation.

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Most popular questions from this chapter

Consider the region defined by \(|x|,|y|\), and \(|z|<1\). Let \(\epsilon_{r}=5, \mu_{r}=4\), and \(\sigma=0 .\) If \(J_{d}=20 \cos \left(1.5 \times 10^{8} t-b x\right) \mathbf{a}_{y} \mu \mathrm{A} / \mathrm{m}^{2}(a)\) find \(\mathbf{D}\) and \(\mathbf{E} ;(b)\) use the point form of Faraday's law and an integration with respect to time to find \(\mathbf{B}\) and \(\mathbf{H} ;(c)\) use \(\nabla \times \mathbf{H}=\mathbf{J}_{d}+\mathbf{J}\) to find \(\mathbf{J}_{d} \cdot(d)\) What is the numerical value of \(b\) ?

In region \(1, z<0, \epsilon_{1}=2 \times 10^{-11} \mathrm{~F} / \mathrm{m}, \mu_{1}=2 \times 10^{-6} \mathrm{H} / \mathrm{m}\), and \(\sigma_{1}=\) \(4 \times 10^{-3} \mathrm{~S} / \mathrm{m} ;\) in region \(2, z>0, \epsilon_{2}=\epsilon_{1} / 2, \mu_{2}=2 \mu_{1}\), and \(\sigma_{2}=\sigma_{1} / 4\). It is known that \(\mathbf{E}_{1}=\left(30 \mathbf{a}_{x}+20 \mathbf{a}_{y}+10 \mathbf{a}_{z}\right) \cos 10^{9} t \mathrm{~V} / \mathrm{m}\) at \(P\left(0,0,0^{-}\right) \cdot(a)\) Find \(\mathbf{E}_{N 1}, \mathbf{E}_{t 1}, \mathbf{D}_{N 1}\), and \(\mathbf{D}_{t 1}\) at \(P_{1} \cdot(b)\) Find \(\mathbf{J}_{N 1}\) and \(\mathbf{J}_{t 1}\) at \(P_{1} \cdot(c)\) Find \(\mathbf{E}_{t 2}\), \(\mathbf{D}_{t 2}\), and \(\mathbf{J}_{t 2}\) at \(P_{2}\left(0,0,0^{+}\right) .(d)\) (Harder) Use the continuity equation to help show that \(J_{N 1}-J_{N 2}=\partial D_{N 2} / \partial t-\partial D_{N 1} / \partial t\), and then determine \(\mathbf{D}_{N 2}\) \(\mathbf{J}_{N 2}\), and \(\mathbf{E}_{N 2}\).

Let the internal dimensions of a coaxial capacitor be \(a=1.2 \mathrm{~cm}, b=4 \mathrm{~cm}\), and \(l=40 \mathrm{~cm}\). The homogeneous material inside the capacitor has the parameters \(\epsilon=10^{-11} \mathrm{~F} / \mathrm{m}, \mu=10^{-5} \mathrm{H} / \mathrm{m}\), and \(\sigma=10^{-5} \mathrm{~S} / \mathrm{m}\). If the electric field intensity is \(\mathbf{E}=\left(10^{6} / \rho\right) \cos 10^{5} t \mathbf{a}_{\rho} \mathrm{V} / \mathrm{m}\), find \((a) \mathbf{J} ;(b)\) the total conduction current \(I_{c}\) through the capacitor; \((c)\) the total displacement current \(I_{d}\) through the capacitor; \((d)\) the ratio of the amplitude of \(I_{d}\) to that of \(I_{c}\), the quality factor of the capacitor.

Write Maxwell's equations in point form in terms of \(\mathbf{E}\) and \(\mathbf{H}\) as they apply to a sourceless medium, where \(\mathbf{J}\) and \(\rho_{v}\) are both zero. Replace \(\epsilon\) by \(\mu, \mu\) by \(\epsilon, \mathbf{E}\) by \(\mathbf{H}\), and \(\mathbf{H}\) by \(-\mathbf{E}\), and show that the equations are unchanged. This is a more general expression of the duality principle in circuit theory.

(a) Show that the ratio of the amplitudes of the conduction current density and the displacement current density is \(\sigma / \omega \epsilon\) for the applied field \(E=\) \(E_{m} \cos \omega t\). Assume \(\mu=\mu_{0} .(b)\) What is the amplitude ratio if the applied field is \(E=E_{m} e^{-t / \tau}\), where \(\tau\) is real?

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