Electromagnetic Waves

Question 11:

Q: What is the energy of one photon of visible light with a wavelength of $450 \, \text{nm}$ ? Also, find the frequency of the light.

Solution: First, calculate the frequency $\nu$ of the photon using the relation:

$$c = \lambda \nu$$

Given:

• $\lambda = 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m}$ ,
• $c = 3 \times 10^8 \, \text{m/s}$ .

The frequency is:

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{450 \times 10^{-9}} = 6.67 \times 10^{14} \, \text{Hz}$$

Now, calculate the energy of the photon using $E = h \nu$ , where $h = 6.626 \times 10^{-34} \, \text{J·s}$ :

$$E = 6.626 \times 10^{-34} \times 6.67 \times 10^{14} = 4.42 \times 10^{-19} \, \text{J}$$

Thus, the energy of one photon is $4.42 \times 10^{-19} \, \text{J}$ .

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Question 12:

Q: Derive the expression for the magnetic field inside a long solenoid.

Solution: The magnetic field inside a solenoid can be derived using Ampère’s Law. For a solenoid with $n$ turns per unit length and current $I$ , the magnetic field $B$ inside the solenoid is given by:

$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$$

The path of integration is chosen along a rectangular loop inside the solenoid. The magnetic field inside the solenoid is uniform and parallel to the axis, so:

$$B \cdot l = \mu_0 n I \cdot l$$

Thus:

$$B = \mu_0 n I$$

This is the expression for the magnetic field inside a long solenoid.

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Question 13:

Q: A plane electromagnetic wave has an electric field $E = E_0 \cos(kx - \omega t)$ . Calculate the magnetic field associated with this wave.

Solution: For a plane electromagnetic wave, the magnetic field $B$ is related to the electric field $E$ by:

$$B = \frac{E_0}{c} \cos(kx - \omega t)$$

where $E_0$ is the amplitude of the electric field and $c$ is the speed of light.

Thus, the magnetic field associated with the wave is:

$$B = \frac{E_0}{c} \cos(kx - \omega t)$$

This is the required magnetic field expression.

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Question 14:

Q: Show that the energy density of an electromagnetic wave is equally divided between its electric and magnetic fields.

Solution: The energy density of an electromagnetic wave is given by the sum of the energy densities due to the electric and magnetic fields.

The energy density due to the electric field is:

$$u_E = \frac{\epsilon_0 E^2}{2}$$

The energy density due to the magnetic field is:

$$u_B = \frac{B^2}{2 \mu_0}$$

Since the electric and magnetic fields are related by $B = \frac{E}{c}$ , where $c$ is the speed of light, we can express $B^2$ as:

$$B^2 = \frac{E^2}{c^2}$$

Substitute this into the expression for $u_B$ :

$$u_B = \frac{1}{2 \mu_0} \times \frac{E^2}{c^2}$$

Using the relation $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ , we can show that:

$$u_B = \frac{\epsilon_0 E^2}{2}$$

Thus, the total energy density is:

$$u = u_E + u_B = \frac{\epsilon_0 E^2}{2} + \frac{\epsilon_0 E^2}{2} = \epsilon_0 E^2$$

Therefore, the energy density is equally divided between the electric and magnetic fields.

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Question 15:

Q: An electromagnetic wave has an intensity of $10 \, \text{W/m}^2$ and a frequency of $5 \times 10^{14} \, \text{Hz}$ . Calculate the amplitude of the electric field.

Solution: The intensity $I$ of an electromagnetic wave is related to the electric field amplitude $E_0$ by the formula:

$$I = \frac{1}{2} \epsilon_0 c E_0^2$$

Rearranging for $E_0$ :

$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$

Given:

• $I = 10 \, \text{W/m}^2$ ,
• $c = 3 \times 10^8 \, \text{m/s}$ ,
• $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$ .

Substituting the values:

$$E_0 = \sqrt{\frac{2 \times 10}{8.854 \times 10^{-12} \times 3 \times 10^8}} = 500 \, \text{V/m}$$

Thus, the amplitude of the electric field is $500 \, \text{V/m}$ .

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Question 16:

Q: Derive the expression for the energy carried by an electromagnetic wave per unit volume.

Solution: The energy density $u$ of an electromagnetic wave is the energy per unit volume. It is given by the sum of the energy densities due to the electric and magnetic fields:

$$u = u_E + u_B$$

The energy density due to the electric field is:

$$u_E = \frac{\epsilon_0 E^2}{2}$$

The energy density due to the magnetic field is:

$$u_B = \frac{B^2}{2 \mu_0}$$

Since $B = \frac{E}{c}$ , we substitute this into the expression for $u_B$ :

$$u_B = \frac{1}{2 \mu_0} \times \frac{E^2}{c^2}$$

Using $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ , we can show that:

$$u_B = \frac{\epsilon_0 E^2}{2}$$

Thus, the total energy density is:

$$u = \epsilon_0 E^2$$

This is the energy carried by the electromagnetic wave per unit volume.

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Question 17:

Q: Explain the concept of polarization of electromagnetic waves.

Solution: Polarization of electromagnetic waves refers to the orientation of the oscillations of the electric field vector in the wave. In an unpolarized wave, the electric field oscillates in all directions perpendicular to the direction of propagation. However, in a polarized wave, the electric field oscillates in a specific direction.

Polarization is typically observed in transverse waves. In the case of electromagnetic waves, polarization refers to the direction of the electric field oscillation. Polarizing filters can be used to block certain orientations of the electric field, allowing only waves polarized in a specific direction to pass through.

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Question 18:

Q: If an electromagnetic wave travels in a vacuum, what is the direction of propagation of the wave if the electric field oscillates in the $x$ -direction and the magnetic field oscillates in the $y$ -direction?

Solution: In an electromagnetic wave, the electric field $E$ , the magnetic field $B$ , and the direction of propagation are all mutually perpendicular. Using the right-hand rule, if the electric field oscillates in the $x$ -direction and the magnetic field oscillates in the $y$ -direction, the wave must propagate in the $z$ -direction.

Thus, the direction of propagation of the wave is along the $z$ -axis.

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Question 19:

Q: How do electromagnetic waves carry energy and momentum?

Solution: Electromagnetic waves carry both energy and momentum. The energy of the wave is stored in the electric and magnetic fields, and it propagates through space as the wave moves. The energy density of the wave is the sum of the energy densities due to the electric and magnetic fields.

The momentum $p$ of an electromagnetic wave is related to its energy $E$ by:

$$p = \frac{E}{c}$$

where $c$ is the speed of light. When electromagnetic waves interact with matter, they can transfer momentum to the matter, which can lead to effects like radiation pressure. This is the basis for phenomena like the solar sail, where radiation pressure from sunlight can be used to propel a spacecraft.

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Question 20:

Q: What is the speed of an electromagnetic wave in a medium with relative permittivity $\epsilon_r$ and relative permeability $\mu_r$ ?

Solution: The speed of light in a medium is given by the formula:

$$v = \frac{c}{\sqrt{\epsilon_r \mu_r}}$$

where:

• $c$ is the speed of light in vacuum,
• $\epsilon_r$ is the relative permittivity of the medium,
• $\mu_r$ is the relative permeability of the medium.

Thus, the speed of the electromagnetic wave in the medium is $v = \frac{c}{\sqrt{\epsilon_r \mu_r}}$ .

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Question 21:

Q: How do electromagnetic waves behave when they encounter a conductor?

Solution: When an electromagnetic wave encounters a conductor, the electric field induces currents in the conductor, which results in the absorption of energy from the wave. This is known as the skin effect, where the wave penetrates only a small distance into the conductor, and most of the energy is dissipated near the surface.

At high frequencies, the wave's penetration depth becomes smaller, and energy is absorbed at a faster rate.

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Question 22:

Q: Explain the concept of the electric displacement vector $\mathbf{D}$ in the context of electromagnetic waves.

Solution: The electric displacement vector $\mathbf{D}$ is related to the electric field $\mathbf{E}$ by:

$$\mathbf{D} = \epsilon \mathbf{E}$$

where $\epsilon$ is the permittivity of the medium. The displacement vector accounts for the polarization of the medium, and in the context of electromagnetic waves, it is used to describe how the electric field interacts with the medium, especially in dielectric materials.

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Question 23:

Q: Derive the relation for the energy density in an electromagnetic wave using Maxwell’s equations.

Solution: We start from Maxwell’s equations, and particularly from the Poynting vector, which describes the power per unit area carried by the wave. The energy density of the wave is the time-averaged energy flux, which is given by:

$$u = \frac{1}{2} \epsilon_0 E^2$$

This expression represents the energy density in an electromagnetic wave. The derivation is based on integrating the Poynting vector over time and using the relations between the electric and magnetic fields.

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Question 24:

Q: What are the differences between the electric field and the magnetic field in terms of their behavior in an electromagnetic wave?

Solution: In an electromagnetic wave, the electric field and magnetic field are perpendicular to each other and to the direction of propagation. However, they differ in their properties:

• The electric field is responsible for the force on charged particles and can exist independently.
• The magnetic field is produced by moving charges (currents) and is perpendicular to the electric field.

Both fields oscillate in phase, and their magnitudes are related by the speed of light $c$ .

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Question 25:

Q: How do electromagnetic waves interact with different materials such as metals, dielectrics, and semiconductors?

Solution: Electromagnetic waves interact with different materials in different ways:

• Metals: In metals, the free electrons respond to the electric field of the wave, causing the wave to be reflected or absorbed. High-frequency waves may exhibit the skin effect.
• Dielectrics: In dielectric materials, the electric field induces polarization, which can lead to the wave being transmitted with some absorption.
• Semiconductors: In semiconductors, the wave interacts with the conduction electrons, and the material's conductivity affects the wave’s transmission.

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Question 26:

Q: What is the significance of the plane of polarization in an electromagnetic wave?

Solution: The plane of polarization refers to the plane in which the electric field oscillates. In unpolarized light, the electric field oscillates in multiple planes. In polarized light, the electric field oscillates in only one plane, which can be controlled using polarizing filters.

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Question 27:

Q: How do electromagnetic waves carry energy through space?

Solution: Electromagnetic waves carry energy through space in the form of oscillating electric and magnetic fields. These oscillating fields propagate outward from the source as the wave travels. The energy is transported by the wave’s intensity, which is related to the square of the electric field and magnetic field amplitudes.

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Question 28:

Q: What is the role of electromagnetic waves in communication systems?

Solution: Electromagnetic waves are essential for communication systems because they can transmit information over long distances without the need for physical wires. Radio waves, microwaves, and other frequencies of electromagnetic radiation are used to carry signals in radio, television, mobile phones, and satellite communication systems.

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Question 29:

Q: How does the refractive index of a medium affect the speed of an electromagnetic wave?

Solution: The refractive index $n$ of a medium is given by:

$$n = \frac{c}{v}$$

where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the medium. The refractive index indicates how much the wave will slow down when it enters the medium. A higher refractive index results in a slower wave speed.

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Question 30:

Q: How do electromagnetic waves exhibit both wave-like and particle-like properties?

Solution: Electromagnetic waves exhibit wave-like properties such as interference and diffraction, which are typical of waves. At the same time, they also exhibit particle-like properties, such as the ability to transfer energy in discrete packets called photons. This dual nature is described by the theory of quantum electrodynamics.