Electromagnetic Waves Questions! Trivia Quiz

1. The velocity of electromagnetic radiation in a medium of permittivity ∈₀ and permeability μ₀ is given by

The velocity of electromagnetic radiation in a medium of permittivity ε0 and permeability μ0 is given by the equation v = 1/√(ε0μ0). This equation is derived from Maxwell's equations, which describe the behavior of electromagnetic waves. It shows that the velocity of electromagnetic radiation depends on the properties of the medium it is traveling through. In a vacuum, where ε0 and μ0 have specific values, the velocity of electromagnetic radiation is equal to the speed of light. However, in other media, the velocity can be different due to the different values of ε0 and μ0.

Explanation

The velocity of electromagnetic radiation in a medium of permittivity ε0 and permeability μ0 is given by the equation v = 1/√(ε0μ0). This equation is derived from Maxwell's equations, which describe the behavior of electromagnetic waves. It shows that the velocity of electromagnetic radiation depends on the properties of the medium it is traveling through. In a vacuum, where ε0 and μ0 have specific values, the velocity of electromagnetic radiation is equal to the speed of light. However, in other media, the velocity can be different due to the different values of ε0 and μ0.

2. The electric and magnetic field of an electromagnetic wave are

In opposite phase and perpendicular to each other

In opposite phase and parallel to each other

In phase and perpendicular to each other

In phase and parallel to each other

The electric and magnetic fields of an electromagnetic wave are in phase and perpendicular to each other. This means that when the electric field is at its maximum, the magnetic field is also at its maximum, and vice versa. Additionally, the electric and magnetic fields are oriented at right angles to each other, meaning they are perpendicular. This relationship between the electric and magnetic fields is a fundamental characteristic of electromagnetic waves and is described by Maxwell's equations.

Explanation

The electric and magnetic fields of an electromagnetic wave are in phase and perpendicular to each other. This means that when the electric field is at its maximum, the magnetic field is also at its maximum, and vice versa. Additionally, the electric and magnetic fields are oriented at right angles to each other, meaning they are perpendicular. This relationship between the electric and magnetic fields is a fundamental characteristic of electromagnetic waves and is described by Maxwell's equations.

3. The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be represented by

a right-handed circularly polarized wave.

Explanation

a right-handed circularly polarized wave.

4. The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. the electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance will be greater than the electric field intensity produced by the radiations coming from a 50 W bulb at the same distance. This is because the electric field intensity is directly proportional to the power of the source and inversely proportional to the square of the distance. Therefore, a higher power source will produce a stronger electric field intensity at the same distance.

Explanation

The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance will be greater than the electric field intensity produced by the radiations coming from a 50 W bulb at the same distance. This is because the electric field intensity is directly proportional to the power of the source and inversely proportional to the square of the distance. Therefore, a higher power source will produce a stronger electric field intensity at the same distance.

5. Poynting vector gives

The magnetic field at a point due to a current in a straight conductor

The electric field at a point between the plates of a parallel plate capacitor

The rate of flow of energy perpendicular to direction of motion through unit area in an electromagnetic field

Total energy of electromagnetic wave

The Poynting vector gives the rate of flow of energy perpendicular to the direction of motion through unit area in an electromagnetic field. It represents the power per unit area carried by an electromagnetic wave. It is calculated by taking the cross product of the electric field vector and the magnetic field vector, and its direction gives the direction of energy flow.

Explanation

The Poynting vector gives the rate of flow of energy perpendicular to the direction of motion through unit area in an electromagnetic field. It represents the power per unit area carried by an electromagnetic wave. It is calculated by taking the cross product of the electric field vector and the magnetic field vector, and its direction gives the direction of energy flow.

6. The frequencies of X-rays, γ-rays and ultraviolet rays are respectively a, b and c. Then

A < b, b > c

A > b, b > c

A > b, b < c

A < b, b < c

The frequencies of electromagnetic waves are inversely proportional to their wavelengths. X-rays have higher frequencies than gamma rays, and ultraviolet rays have lower frequencies than gamma rays. Therefore, a c, indicating that the frequency of gamma rays is greater than the frequency of ultraviolet rays.

Explanation

The frequencies of electromagnetic waves are inversely proportional to their wavelengths. X-rays have higher frequencies than gamma rays, and ultraviolet rays have lower frequencies than gamma rays. Therefore, a c, indicating that the frequency of gamma rays is greater than the frequency of ultraviolet rays.

7. A plane electromagnetic wave is an incident on a material surface. The wave delivers momentum p and energy E. Then

P ≠ 0, E ≠ 0

P = 0, E = 0

P = 0, E ≠ 0

P ≠ 0, E = 0

When a plane electromagnetic wave is incident on a material surface, it transfers both momentum and energy to the surface. This means that both the momentum (p) and energy (E) of the wave are not equal to zero. Therefore, the correct answer is p ≠ 0, E ≠ 0.

Explanation

When a plane electromagnetic wave is incident on a material surface, it transfers both momentum and energy to the surface. This means that both the momentum (p) and energy (E) of the wave are not equal to zero. Therefore, the correct answer is p ≠ 0, E ≠ 0.

8. The amplitude of an electromagnetic wave in the vacuum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is correct?

The speed of wave propagation changes only

The frequency of the wave changes only

The wavelength of the wave changes only

None of the above is true

The amplitude of an electromagnetic wave is not related to the speed of wave propagation, frequency, or wavelength. Doubling the amplitude of the wave will not cause any changes in these properties. Therefore, none of the statements are true.

Explanation

The amplitude of an electromagnetic wave is not related to the speed of wave propagation, frequency, or wavelength. Doubling the amplitude of the wave will not cause any changes in these properties. Therefore, none of the statements are true.

9. One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

Visible region

Infrared region

Ultraviolet region

Microwave region

The given question states that 11eV of energy is required to dissociate a carbon monoxide molecule into carbon and oxygen atoms. According to the electromagnetic spectrum, different regions of radiation have different frequencies and energies. The ultraviolet region of the spectrum has higher frequencies and energies compared to the visible and infrared regions. Since the dissociation requires a relatively high amount of energy, it is reasonable to assume that the minimum frequency of the appropriate electromagnetic radiation lies in the ultraviolet region.

Explanation

The given question states that 11eV of energy is required to dissociate a carbon monoxide molecule into carbon and oxygen atoms. According to the electromagnetic spectrum, different regions of radiation have different frequencies and energies. The ultraviolet region of the spectrum has higher frequencies and energies compared to the visible and infrared regions. Since the dissociation requires a relatively high amount of energy, it is reasonable to assume that the minimum frequency of the appropriate electromagnetic radiation lies in the ultraviolet region.

10. The magnetic field in a plane electromagnetic wave is given by B_y = 2 m 10^-7 sin (0.5 m 10³ x + 1.5 m 10¹¹ t). This electromagnetic wave is

A visible light

An infrared wave

A microwave

A radiowave

The given equation represents a magnetic field in a plane electromagnetic wave. The frequency of the wave is determined by the coefficient in front of 't', which is 1.5 × 10^11. This frequency falls within the range of microwave frequencies, which typically range from 300 MHz to 300 GHz. Therefore, the electromagnetic wave described by the equation is a microwave.

Explanation

The given equation represents a magnetic field in a plane electromagnetic wave. The frequency of the wave is determined by the coefficient in front of 't', which is 1.5 × 10^11. This frequency falls within the range of microwave frequencies, which typically range from 300 MHz to 300 GHz. Therefore, the electromagnetic wave described by the equation is a microwave.

11. If μ₀ be the permeability and k₀, the dielectric constant of a medium, its refractive index is given by

The refractive index of a medium is given by the square root of the product of its permeability and dielectric constant.

Explanation

The refractive index of a medium is given by the square root of the product of its permeability and dielectric constant.

12. The frequency of e.m. wave which is best suited to observe a particle of radius 3 m 10^-1 cm is of the order of

1015 Hz

1014 Hz

1013 Hz

1012 Hz

The frequency of an electromagnetic wave that is best suited to observe a particle of radius 3 × 10-1 cm is of the order of 1015 Hz. This is because the wavelength of the wave should be on the same order of magnitude as the size of the particle in order to effectively observe it. By using a high frequency wave, we can achieve a smaller wavelength, allowing us to observe smaller particles with greater precision. Therefore, a frequency of 1015 Hz is the most suitable for observing a particle of this size.

Explanation

The frequency of an electromagnetic wave that is best suited to observe a particle of radius 3 × 10-1 cm is of the order of 1015 Hz. This is because the wavelength of the wave should be on the same order of magnitude as the size of the particle in order to effectively observe it. By using a high frequency wave, we can achieve a smaller wavelength, allowing us to observe smaller particles with greater precision. Therefore, a frequency of 1015 Hz is the most suitable for observing a particle of this size.

13. If a 100-watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3 m is

0.022 Wm-2

0.22 Wm-2

4.4 Wm-2

0.033 Wm-2

The intensity of the bulb at 3 m can be calculated by multiplying the power of the bulb (100 watts) by its efficiency (2.5% or 0.025) and dividing it by the surface area of a sphere with a radius of 3 m (4πr^2). Therefore, the intensity is (100 watts * 0.025) / (4π * 3^2) = 0.022 Wm-2.

Explanation

The intensity of the bulb at 3 m can be calculated by multiplying the power of the bulb (100 watts) by its efficiency (2.5% or 0.025) and dividing it by the surface area of a sphere with a radius of 3 m (4πr^2). Therefore, the intensity is (100 watts * 0.025) / (4π * 3^2) = 0.022 Wm-2.

14. When a plane electromagnetic wave travels in a vacuum, the average electric energy density is given by (here E_o is the amplitude of the electric field of the wave)

The average electric energy density of a plane electromagnetic wave in a vacuum is directly proportional to the square of the amplitude of the electric field of the wave (Eo). This means that as the amplitude of the electric field increases, the average electric energy density also increases. The electric energy density represents the amount of energy carried by the electric field per unit volume. Therefore, a higher amplitude of the electric field indicates a higher concentration of energy in the wave.

Explanation

The average electric energy density of a plane electromagnetic wave in a vacuum is directly proportional to the square of the amplitude of the electric field of the wave (Eo). This means that as the amplitude of the electric field increases, the average electric energy density also increases. The electric energy density represents the amount of energy carried by the electric field per unit volume. Therefore, a higher amplitude of the electric field indicates a higher concentration of energy in the wave.