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(a) If the electric field and magnetic field in a sinusoidal plane wave were interchanged, in which direction relative to before would the energy propagate? (b) What if the electric and the magnetic fields were both changed to their negatives?

Short Answer

Expert verified
(a) If the electric field and magnetic field are interchanged, the energy will propagate in the opposite direction to before. (b) If both the electric and magnetic fields are negated, the energy's direction of propagation remains the same as before.

Step by step solution

01

Understanding Interchanging Electric and Magnetic Fields

According to Poynting's theorem, the direction of energy propagation in an electromagnetic wave is given by the vector product of the Electric field \(E\) and the Magnetic field \(B\). This is referred to as the Poynting vector \(S\). Mathematically, \(S = E \times B\). For a right-handed coordinate system, if electric field \(E\) is in the x-direction and magnetic field \(B\) is in the y-direction, then the energy propagates in the positive z-direction. If the electric field and magnetic field are interchanged, \(E\) will be in the y-direction and \(B\) in the x-direction.
02

Determine the Energy Propagation After Interchange

Now that the fields are interchanged, we calculate the cross product of \(B\) and \(E\). However, this gives the Poynting vector \(S'= B \times E\). Calculating the cross product with \(B\) in the x-direction and \(E\) in the y-direction results in a Poynting vector in the negative z-direction. This means that when the electric and magnetic fields are swapped, the direction of energy propagation is opposed to the original direction.
03

Understanding the Negation of Electric and Magnetic Fields

Now we examine the effect of changing both the electric and magnetic fields to their negatives. Therefore, the new fields are \(-E\) and \(-B\). Remember that the direction of the cross product changes if either or both vectors are negated because \(E \times -B = -(E \times B)\).
04

Determine the Energy Propagation After Negation

Now that both fields are negated, let's calculate the resultant Poynting vector \(-S= -E \times -B\). Simplifying, this gives us \(-S=E \times B\). Hence, when both fields are negated, the direction of energy propagation remains the same as the original direction.

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Key Concepts

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

Poynting's theorem
Poynting's theorem is a cornerstone in understanding electromagnetic wave propagation. It provides us with a way to calculate the rate of energy transfer per unit area in an electromagnetic field. This theorem bridges the gap between Maxwell's equations and energy transfer concepts, allowing us to determine how much power is being transported and in what direction.

The theorem states that the rate of energy transfer, or the power through a surface, is represented by the Poynting vector \( \mathbf{S} \), also known as the power flux density. Formally, the Poynting vector is calculated as the cross product of the electric field (\(\mathbf{E}\)) and the magnetic field (\(\mathbf{B}\)), mathematically expressed as \( \mathbf{S} = \mathbf{E} \times \mathbf{B} \). In a vacuum, Poynting's theorem can be described by the equation \( abla \cdot \mathbf{S} + \frac{\partial u}{\partial t} = -\mathbf{J} \cdot \mathbf{E} \), where \( u \) represents the energy density of the electromagnetic field, and \( \mathbf{J} \) is the current density. This expression shows that the divergence of the Poynting vector and the rate of change of energy density is equal to the negative of the work done on charges, emphasizing the conservation of energy in electromagnetic systems.
Poynting vector
The Poynting vector, denoted as \( \mathbf{S} \), acts as a directional energy flux that tells us not only the magnitude but also the direction in which electromagnetic energy propagates. It is a key concept for visualizing the flow of energy in electromagnetic waves and is especially important in various applications, such as wireless communications, power transmission, and photonics.

As defined by Poynting's theorem, the vector is found using the cross product of the electric and magnetic fields: \( \mathbf{S} = \mathbf{E} \times \mathbf{B} \). It's crucial to note that \( \mathbf{S} \) is always perpendicular to both \( \mathbf{E} \) and \( \mathbf{B} \), and its direction is in accordance with the right-hand rule—a fundamental principle in vector analysis. For example, if \( \mathbf{E} \) points east and \( \mathbf{B} \) points north, then \( \mathbf{S} \) will point upwards. This orientation is intuitive if you consider the thumb, index, and middle fingers on your right hand to represent \( \mathbf{E} \), \( \mathbf{B} \), and \( \mathbf{S} \) respectively.
Cross product of electric and magnetic fields
The cross product of electric and magnetic fields is fundamental to the formation of the Poynting vector and thus the propagation of electromagnetic waves. In vector mathematics, the cross product is a binary operation on two vectors in three-dimensional space, and it results in a third vector that is perpendicular to the plane of the initial vectors.

When we take the cross product of the electric field (\(\mathbf{E}\)) and the magnetic field (\(\mathbf{B}\)), according to the right-hand rule, we place the fingers of our right hand in the direction of the electric field, fold them towards the magnetic field, and the extended thumb points in the direction of the Poynting vector (\

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Most popular questions from this chapter

What is the physical significance of the Poynting vector?

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of \(6.0 \mathrm{V} / \mathrm{m}\) and points eastward. What are the magnitude and direction of the magnetic field at this instant?

A potential difference \(V(t)=V_{0} \sin \omega t \quad\) is maintained across a parallel-plate capacitor with capacitance \(C\) consisting of two circular parallel plates. A thin wire with resistance \(R\) connects the centers of the two plates, allowing charge to leak between plates while they are charging. (a) Obtain expressions for the leakage current \(I_{\text {res }}(t)\) in the thin wire. Use these results to obtain an expression for the current \(I_{\text {real }}(t)\) in the wires connected to the capacitor. (b) Find the displacement current in the space between the plates from the changing electric field between the plates. (c) Compare \(I_{\text {real }}(t)\) with the sum of the displacement current \(I_{\mathrm{d}}(t)\) and resistor current \(I_{\mathrm{res}}(t)\) between the plates, and explain why the relationship you observe would be expected.

Distances in space are often quoted in units of light years, the distance light travels in 1 year. (a) How many meters is a light-year? (b) How many meters is it to Andromeda, the nearest large galaxy, given that it is \(2.54 \times 10^{6}\) ly away? (c) The most distant galaxy yet discovered is \(13.4 \times 10^{9}\) ly away. How far is this in meters?

Suppose a spherical particle of mass \(m\) and radius \(R\) in space absorbs light of intensity \(I\) for time \(t\). (a) How much work does the radiation pressure do to accelerate the particle from rest in the given time it absorbs the light? (b) How much energy carried by the electromagnetic waves is absorbed by the particle over this time based on the radiant energy incident on the particle?

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