Electromagnetic Waves Questions! Trivia Quiz
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Questions and Answers
- 1.
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
- A.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- B.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- C.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- D.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
Correct Answer
A.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.Rate this question:
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- 2.
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
- A.
Visible region
- B.
Infrared region
- C.
Ultraviolet region
- D.
Microwave region
Correct Answer
C. Ultraviolet regionExplanation
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.Rate this question:
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- 3.
The electric and magnetic field of an electromagnetic wave are
- A.
In opposite phase and perpendicular to each other
- B.
In opposite phase and parallel to each other
- C.
In phase and perpendicular to each other
- D.
In phase and parallel to each other
Correct Answer
C. In pHase and perpendicular to each otherExplanation
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.Rate this question:
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- 4.
The velocity of electromagnetic radiation in a medium of permittivity ∈0 and permeability μ0 is given by
- A.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- B.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- C.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- D.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
Correct Answer
C.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.Rate this question:
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- 5.
The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be represented by
- A.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- B.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- C.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- D.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
Correct Answer
A.Explanation
a right-handed circularly polarized wave.Rate this question:
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- 6.
If μ0 be the permeability and k0, the dielectric constant of a medium, its refractive index is given by
- A.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- B.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- C.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- D.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
Correct Answer
C.Explanation
The refractive index of a medium is given by the square root of the product of its permeability and dielectric constant.Rate this question:
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- 7.
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?
- A.
The speed of wave propagation changes only
- B.
The frequency of the wave changes only
- C.
The wavelength of the wave changes only
- D.
None of the above is true
Correct Answer
D. None of the above is trueExplanation
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.Rate this question:
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- 8.
If a 100-watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3 m is
- A.
0.022 Wm-2
- B.
0.22 Wm-2
- C.
4.4 Wm-2
- D.
0.033 Wm-2
Correct Answer
A. 0.022 Wm-2Explanation
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.Rate this question:
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- 9.
A plane electromagnetic wave is an incident on a material surface. The wave delivers momentum p and energy E. Then
- A.
P ≠ 0, E ≠ 0
- B.
P = 0, E = 0
- C.
P = 0, E ≠ 0
- D.
P ≠ 0, E = 0
Correct Answer
A. P ≠ 0, E ≠ 0Explanation
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.Rate this question:
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- 10.
When a plane electromagnetic wave travels in a vacuum, the average electric energy density is given by (here Eo is the amplitude of the electric field of the wave)
- A.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- B.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- C.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
- D.
![Electromagnetic Waves Questions! Trivia Quiz - Quiz]()
Correct Answer
A.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.Rate this question:
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- 11.
The frequencies of X-rays, γ-rays and ultraviolet rays are respectively a, b and c. Then
- A.
A < b, b > c
- B.
A > b, b > c
- C.
A > b, b < c
- D.
A < b, b < c
Correct Answer
A. A < b, b > cExplanation
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 < b, indicating that the frequency of X-rays is less than the frequency of gamma rays. Additionally, b > c, indicating that the frequency of gamma rays is greater than the frequency of ultraviolet rays.Rate this question:
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- 12.
Poynting vector gives
- A.
The magnetic field at a point due to a current in a straight conductor
- B.
The electric field at a point between the plates of a parallel plate capacitor
- C.
The rate of flow of energy perpendicular to direction of motion through unit area in an electromagnetic field
- D.
Total energy of electromagnetic wave
Correct Answer
C. The rate of flow of energy perpendicular to direction of motion through unit area in an electromagnetic fieldExplanation
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.Rate this question:
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- 13.
The frequency of e.m. wave which is best suited to observe a particle of radius 3 × 10-1 cm is of the order of
- A.
1015 Hz
- B.
1014 Hz
- C.
1013 Hz
- D.
1012 Hz
Correct Answer
A. 1015 HzExplanation
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.Rate this question:
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- 14.
The magnetic field in a plane electromagnetic wave is given by By = 2 × 10-7 sin (0.5 × 103 x + 1.5 × 1011 t). This electromagnetic wave is
- A.
A visible light
- B.
An infrared wave
- C.
A microwave
- D.
A radiowave
Correct Answer
C. A microwaveExplanation
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.Rate this question:
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Current Version
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Mar 21, 2023Quiz Edited by
ProProfs Editorial Team -
Dec 19, 2013Quiz Created by
Tanmay Shankar
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