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Chapter 17: Problem 30

Electromagnetic waves do not transport (A) Energy (B) Charge (C) Momentum (D) Information

Short Answer

Expert verified
Electromagnetic waves do not transport charge (Option B). They can travel through space and matter without carrying any charged particles, allowing them to move through a vacuum.

Step by step solution

01

Understanding Electromagnetic Waves

Electromagnetic waves are waves that can travel through space and matter without any medium. They are produced when an electric field interacts with a magnetic field, creating oscillations in both fields. As a result, they transport energy and information from one place to another. Examples of electromagnetic waves include visible light, radio waves, microwaves, and X-rays.
02

Analyzing Option A - Energy

Electromagnetic waves do transport energy. The energy of an electromagnetic wave is related to its frequency, and higher frequency waves have more energy. For example, ultraviolet light has higher energy than visible light, which is why it can cause sunburns and other harm to our skin. Hence, option A is false.
03

Analyzing Option B - Charge

Electromagnetic waves do not transport charge. They are a result of oscillating electric and magnetic fields, and they do not carry any charged particles with them. This is the reason electromagnetic waves can travel through a vacuum, where there are no charged particles. Therefore, option B is true.
04

Analyzing Option C - Momentum

Though it may seem counter-intuitive, electromagnetic waves do carry momentum. However, the momentum of an electromagnetic wave is much smaller than that of a particle, like an electron or a proton. When electromagnetic waves interact with matter, they can transfer their momentum to the particles in the matter, causing a force known as radiation pressure. Therefore, option C is false.
05

Analyzing Option D - Information

Electromagnetic waves do transport information. Many modern communication technologies, like television, radio, and cell phones, rely on electromagnetic waves to send signals and hence transfer information from one place to another. Visible light allows us to see and interpret the world around us by carrying information on the colors and shapes of objects. Consequently, option D is false.
06

Conclusion

After going through each option, we can conclude that electromagnetic waves do not transport charge (Option B), making it the correct answer for this problem.

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Most popular questions from this chapter

The rms value of the electric field of the light coming from the sun is \(720 \mathrm{~N} / \mathrm{C}\). The average total energy density of the electromagnetic wave is [2006] (A) \(3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3}\) (B) \(4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}\) (C) \(6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^{3}\) (D) \(81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}\)

The frequency from 3 to \(\mathrm{MHz}\) is known as (A) Audio band (B) Medium frequency band (C) Very high frequency band (D) High frequency band

The charge on a parallel plate capacitor is varying as \(q=q_{0} \sin 2 \pi \cdot n t\) The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is (A) \(\frac{q}{\varepsilon_{0} A}\) (B) \(\frac{q_{0}}{\varepsilon_{0}} \sin 2 \pi n t\) (C) \(2 \pi n q_{0} \cos \pi n t\) (D) \(\frac{2 \pi n q_{0}}{\varepsilon_{0}} \cos 2 \pi n t\)

The Maxwell's four equations are written as (i) \(\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q_{0}}{\varepsilon_{0}}\) (ii) \(\oint \vec{B} \cdot \overrightarrow{d s}=0\) (iii) \(\oint \vec{E} \cdot \overrightarrow{d l}=\frac{d}{d t} \oint \vec{B} \cdot \overrightarrow{d s}\) (iv) \(\oint \vec{B} \cdot \overrightarrow{d s}=\mu_{0} \varepsilon_{0} \frac{d}{d t} \oint \vec{E} \cdot \overrightarrow{d s}\) The equations which have sources of \(\vec{E}\) and \(\vec{B}\) (A) (i), (ii) and (iii) (B) (i) and (ii) (C) (i) and (iii) (D) (i) and (iv)

An electromagnetic wave in vacuum has the electric and magnetic fields \(\vec{E}\) and \(\vec{B}\), which are always perpendicular to each other. The direction of polarizations is given by \(\vec{X}\) and that of wave propagation by \(\vec{k}\) Then \([2012]\) (A) \(\vec{X} \| \vec{E}\) and \(\vec{k} \| \vec{E} \times \vec{B}\) (B) \(\vec{X} \| \vec{B}\) and \(\vec{K} \| \vec{E} \times \vec{B}\) (C) \(\vec{X} \| \vec{E}\) and \(\vec{k} \| \vec{B} \times \vec{E}\) (D) \(\vec{X} \| \vec{B}\) and \(\vec{k} \| \vec{B} \times \vec{E}\)
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