# Quiz: Trivia Questions On Electromagnetic Wave!

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Kbagayas
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Quizzes Created: 4 | Total Attempts: 15,496
Questions: 33 | Attempts: 2,937  Settings  Think you know everything there is to know about electromagnetic waves? This quiz can help. You should be aware of what ground wave communications are most effective, what ionization is, the greatest range that the ionosphere has on signals, what type of radio waves are responsible for long-distance communications, and an electromagnetic wave consist of, and what is a wavefront. This quiz can assist you in your learning about electromagnetic waves. All the best.

• 1.

### Ground-wave communications is most effective in what frequency range?

• A.

300 kHz to 3 MHz

• B.

3 to 30 MHz

• C.

30 to 300 MHz

• D.

above 300 MHz

A. 300 kHz to 3 MHz
Explanation
Ground-wave communications is most effective in the frequency range of 300 kHz to 3 MHz. This is because at these lower frequencies, the ground wave can propagate for longer distances by following the curvature of the Earth's surface. Ground waves are able to travel over obstacles such as hills, buildings, and trees, making them ideal for long-range communication. As the frequency increases, the ground wave becomes less effective and other methods such as skywave or line-of-sight propagation are used.

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• 2.

### The ionization causes the signal to be

• A.

Diffused

• B.

Absorbed

• C.

refracted

• D.

Reflected

C. refracted
Explanation
When an ionization process occurs, it typically involves the addition or removal of an electron from an atom or molecule. This can lead to changes in the behavior of electromagnetic waves passing through the ionized medium. In the case of refraction, the bending of the waves occurs as they pass from one medium to another with different refractive indices. Therefore, it is reasonable to conclude that ionization can cause the signal to be refracted.

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• 3.

### The critical frequency at a particular time is 11.6 MHz. What is the MUF for a transmitting station if the required angle of incidence for propagation to a desired destination is 70°?

• A.

12.3 MHz

• B.

33.9 Hz

• C.

33.9 MHz

• D.

12.3 Hz

C. 33.9 MHz
Explanation
The maximum usable frequency (MUF) for a transmitting station is the highest frequency that can be used for reliable communication at a specific time and location. In this case, the critical frequency is given as 11.6 MHz, which represents the frequency at which the ionosphere can reflect radio waves back to Earth. The MUF is typically higher than the critical frequency and depends on factors such as the angle of incidence. Since the required angle of incidence is 70°, the MUF is calculated to be 33.9 MHz, which is the highest frequency that can be used for reliable communication to the desired destination.

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• 4.

### The ionosphere has the greatest effect on signals in what frequency range?

• A.

300 kHz

• B.

3 to 30 MHz

• C.

30 to 300 MHz

• D.

Above 300 MHz

B. 3 to 30 MHz
Explanation
The ionosphere is a layer of the Earth's atmosphere that contains charged particles. These charged particles can affect the propagation of radio waves. In particular, they can cause the radio waves to be refracted or reflected back to Earth. The effect of the ionosphere is most pronounced in the frequency range of 3 to 30 MHz. In this frequency range, the radio waves can be absorbed, refracted, or reflected by the ionosphere, leading to changes in the strength and direction of the signals. Therefore, the ionosphere has the greatest effect on signals in the 3 to 30 MHz frequency range.

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• 5.

### The type of radio wave responsible for long-distance communications by multiple skips is the

• A.

ground wave

• B.

direct-wave

• C.

surface wave

• D.

Sky wave

D. Sky wave
Explanation
Sky wave is the correct answer because it refers to radio waves that are reflected off the ionosphere and can travel long distances by multiple skips. This phenomenon allows for long-distance communications as the waves bounce between the ionosphere and the Earth's surface. Ground wave, direct-wave, and surface wave do not involve the reflection off the ionosphere and are not capable of long-distance communications by multiple skips.

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• 6.

### Microwave signals propagate by way of the

• A.

direct wave

• B.

Sky wave

• C.

surface wave

• D.

Standing wave

A. direct wave
Explanation
Microwave signals propagate by way of the direct wave. This is because microwave signals are high-frequency electromagnetic waves that travel in a straight line from the transmitting antenna to the receiving antenna. The direct wave does not require any reflection or refraction and is the most efficient way for microwave signals to propagate over short distances.

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• 7.

### Line-of-sight communications is not a factor in which frequency range?

• A.

VHF

• B.

UHF

• C.

HF

• D.

Microwave

C. HF
Explanation
High Frequency (HF) is the correct answer because line-of-sight communications is not a factor in this frequency range. HF signals can travel long distances by bouncing off the ionosphere, allowing for long-range communication beyond the line of sight. VHF, UHF, and Microwave frequencies are typically used for line-of-sight communications, where the transmitting and receiving antennas must have a direct line of sight to each other.

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• 8.

### A microwave-transmitting antenna is 550 ft high. The receiving antenna is 200 ft high. The maximum transmission distance is

• A.

53.2 km

• B.

53.2 ft

• C.

53.2 mi

• D.

53.2 m

C. 53.2 mi
Explanation
The maximum transmission distance is 53.2 mi. This can be determined by subtracting the height of the receiving antenna (200 ft) from the height of the transmitting antenna (550 ft), which gives a height difference of 350 ft. The maximum transmission distance can then be calculated using the formula D = 1.23 * √H, where D is the distance in miles and H is the height difference in feet. Plugging in the values, we get D = 1.23 * √350 = 53.2 mi.

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• 9.

### To increase the transmission distance of a UHF signal, which of the following should be done?

• A.

Increase antenna gain

• B.

Increase antenna height

• C.

Increase transmitter power

• D.

B. Increase antenna height
Explanation
Increasing the antenna height can help increase the transmission distance of a UHF signal. This is because a higher antenna height allows for a clearer line of sight between the transmitter and receiver, reducing obstructions and improving signal strength. By raising the antenna, the signal can travel further without significant loss or interference, resulting in an increased transmission distance.

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• 10.

### _________ is the amount of voltage induced in a wave by an electromagnetic wave.

• A.

• B.

Field strength

• C.

Magnetic induction

• D.

Power density

B. Field strength
Explanation
Field strength is the amount of voltage induced in a wave by an electromagnetic wave. This term refers to the intensity or magnitude of the electric or magnetic field produced by the wave. It measures the strength of the wave and is directly related to the amount of voltage induced. Therefore, field strength is the correct answer to the given question.

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• 11.

### An electromagnetic wave consists of ____________.

• A.

Both electric field and magnetic field

• B.

An electric field only

• C.

A magnetic field only

• D.

Non-magnetic field only

A. Both electric field and magnetic field
Explanation
An electromagnetic wave consists of both an electric field and a magnetic field. These fields are perpendicular to each other and oscillate in a direction perpendicular to the direction of wave propagation. The electric field is created by the movement of charged particles, while the magnetic field is created by the changing electric field. Together, these fields propagate through space, carrying energy and information.

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• 12.

### If a radio receiver needs 1 nW/m2 of power density to function, how far away from a 1-W point source will it continue to work?

• A.

8.9 km

• B.

8.9 m

• C.

9.8 km

• D.

9.8 m

A. 8.9 km
Explanation
A radio receiver needs a minimum power density of 1 nW/m2 to function. A point source with a power of 1 W will emit radiation that spreads out as it propagates. As the distance from the source increases, the power density decreases. In order for the radio receiver to continue working, the power density at its location needs to be at least 1 nW/m2. Since the power density decreases with distance, the receiver will continue to work at a greater distance from the source. Among the given options, the only distance that would provide a power density of at least 1 nW/m2 is 8.9 km.

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• 13.

### What causes tropospheric ducting of radio waves?

• A.

A very low pressure area

• B.

An aurora to the north

• C.

Lightning between the transmitting and receiving antenna

• D.

A temperature inversion

D. A temperature inversion
Explanation
Tropospheric ducting of radio waves is caused by a temperature inversion. In a temperature inversion, warm air is trapped above cooler air near the surface, creating a layer of warm air that acts as a duct for radio waves to travel through. This phenomenon occurs when there is a change in the temperature profile of the troposphere, allowing radio waves to be refracted and travel long distances, sometimes even beyond the horizon.

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• 14.

### What is a wavefront?

• A.

A voltage pulse in a conductor

• B.

A current pulse in a conductor

• C.

A voltage pulse across a resistor

• D.

A fixed point in an electromagnetic wave

D. A fixed point in an electromagnetic wave
Explanation
A wavefront refers to a fixed point in an electromagnetic wave. It represents the continuous locus of points in space that are in phase with each other. In other words, it is the imaginary surface formed by connecting all the points of the wave that are at the same phase. This concept is commonly used in the study of optics and wave propagation to understand how waves propagate through space and interact with different mediums.

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• 15.

### A transmitter has a power output of 115 W at a carrier frequency of 525 MHz. It is connected to an antenna with a power gain of 100 . The receiving antenna is 20,000 m away and has a power gain of 10. Calculate the power delivered to the receiver, assuming free-space propagation. Assume also that there are no losses or mismatches in the system.

• A.

595 nW

• B.

595 pW

• C.

56.9 nW

• D.

56.9 W

A. 595 nW
Explanation
The power delivered to the receiver can be calculated using the Friis transmission equation, which states that the power received is equal to the power transmitted multiplied by the product of the antenna gains and the free-space path loss. In this case, the power transmitted is 115 W, the power gain of the transmitting antenna is 100, and the power gain of the receiving antenna is 10. The free-space path loss can be calculated using the formula: 20 * log10(distance) + 20 * log10(frequency) - 147.55, where the distance is 20,000 m and the frequency is 525 MHz. Plugging in these values and performing the calculation, we find that the power delivered to the receiver is approximately 595 nW.

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• 16.

### What causes the ionosphere to absorb radio waves?

• A.

The weather below the ionosphere

• B.

The ionization of the D region

• C.

The presence of the ionized cloud in the E region

• D.

The splitting of the E region

B. The ionization of the D region
Explanation
The ionosphere absorbs radio waves due to the ionization of the D region. The D region is a layer in the ionosphere that is primarily composed of ionized nitrogen molecules. When radio waves pass through this region, the ionized molecules interact with the waves, causing them to be absorbed. This absorption is due to the collisions between the ions and electrons in the D region and the radio waves. As a result, radio waves cannot penetrate the ionosphere and are either reflected or absorbed by it.

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• 17.

### Which ionospheric region is closest to the Earth?

• A.

A region

• B.

E Region

• C.

D region

• D.

F region

C. D region
Explanation
The D region is the ionospheric region closest to the Earth. It is located at an altitude of about 60-90 kilometers above the Earth's surface. This region is characterized by a high electron density and is responsible for absorbing high-frequency radio waves. The D region plays a crucial role in the propagation of radio signals and is important for long-distance communication.

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• 18.

### What causes the VHF radio waves to be propagated several hundred miles over oceans?

• A.

An overcast of cirriform clouds

• B.

A high pressure zone

• C.

A polar air mass

• D.

Explanation
A widespread temperature inversion can cause VHF radio waves to be propagated several hundred miles over oceans. In a temperature inversion, the normal decrease in temperature with increasing altitude is reversed, causing the warmer air to be located above cooler air. This inversion acts as a barrier, trapping and reflecting the radio waves back towards the Earth's surface, allowing them to travel longer distances.

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• 19.

### Microwave and UHF systems have which of the following in common?

• A.

• B.

Both are cheaper than HF radio

• C.

They operate on the same frequency band.

• D.

They are both line-of-sight (LOS) systems.

D. They are both line-of-sight (LOS) systems.
Explanation
Microwave and UHF systems both have line-of-sight (LOS) characteristics, meaning that they require a direct line of sight between the transmitting and receiving antennas. This is because their signals travel in a straight line and do not bend or follow the curvature of the Earth. This limitation of LOS systems is important to consider when planning the placement and coverage of microwave and UHF systems.

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• 20.

• A.

Uses molecules in the Earth’s troposphere to scatter some radio energy back to Earth

• B.

Has a limited capability for data transmission

• C.

Is not widely used commercially

• D.

All of the above

D. All of the above
Explanation
The correct answer is "all of the above." Tropospheric scatter radio uses molecules in the Earth's troposphere to scatter radio energy back to Earth. This technique has a limited capability for data transmission and is not widely used commercially. Therefore, all the given statements are true.

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• 21.

### Which of the following are electromagnetic?

• A.

• B.

Light

• C.

Gamma rays

• D.

All of the above

D. All of the above
Explanation
Radio waves, light, and gamma rays are all forms of electromagnetic radiation. Electromagnetic radiation is a type of energy that travels in waves and does not require a medium to propagate. It consists of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. Radio waves have the longest wavelength and lowest frequency, while gamma rays have the shortest wavelength and highest frequency. Light falls in between these two extremes. Therefore, all three options mentioned in the answer (radio waves, light, and gamma rays) are examples of electromagnetic radiation.

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• 22.

### Sky wave cannot be “heard”

• A.

Close to the transmitter

• B.

Far from the transmitter

• C.

In the silent zone

• D.

In the skip zone

D. In the skip zone
Explanation
In the skip zone, sky waves are unable to reach the receiver due to a phenomenon called skip distance. Skip distance refers to the area where the sky wave is refracted and bent back towards the Earth's surface, but it does not reach the receiver. This can occur due to various factors such as frequency, time of day, and atmospheric conditions. Therefore, in the skip zone, the sky wave cannot be heard by the receiver.

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• 23.

### What power must a point-source of radio waves transmit so that the power density at 3000 meters from the source is 1 µW/m2?

• A.

113 kW

• B.

113 W

• C.

131 W

• D.

131 kW

A. 113 kW
Explanation
To calculate the power required, we can use the formula: Power = Power density * Area. Since the power density is given as 1 µW/m2 and the distance from the source is 3000 meters, we can assume that the area is a sphere with a radius of 3000 meters. Using the formula for the surface area of a sphere, we can calculate the required power to be 113 kW.

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• 24.

### Helical antennas are often used for satellite tracking at VHF because of

• A.

Troposcatter

• B.

Ionospheric refraction

• C.

Subrefraction

• D.

Explanation
The Faraday effect is the correct answer because helical antennas are often used for satellite tracking at VHF frequencies due to their ability to mitigate the Faraday effect. The Faraday effect refers to the rotation of the polarization plane of an electromagnetic wave as it passes through a medium in the presence of a magnetic field. Helical antennas are designed to minimize the effects of polarization rotation caused by the Faraday effect, making them suitable for satellite tracking at VHF frequencies.

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• 25.

### Diffraction of electromagnetic waves

• A.

is caused by reflections from the ground wave

• B.

arrives only with spherical wavefronts

• C.

Will occur around edge of a sharp obstacle

C. Will occur around edge of a sharp obstacle
Explanation
Diffraction of electromagnetic waves occurs when they encounter an edge or obstacle. This phenomenon causes the waves to bend and spread out around the edge, creating a pattern of interference. The sharper the edge, the more pronounced the diffraction will be. Therefore, the correct answer is that diffraction will occur around the edge of a sharp obstacle.

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• 26.

### The absorption of radio waves by the atmosphere depends on

• A.

Their frequency

• B.

Their distance from the transmitter

• C.

The polarization of the wave

• D.

The polarization of the atmosphere

A. Their frequency
Explanation
The absorption of radio waves by the atmosphere depends on their frequency. Different frequencies of radio waves interact differently with the atmosphere. Higher frequencies are more likely to be absorbed by the atmosphere, while lower frequencies can travel longer distances. This absorption is due to various atmospheric factors such as water vapor, oxygen, and other molecules that can absorb and scatter radio waves. Therefore, the frequency of the radio waves plays a significant role in determining how much of the signal is absorbed by the atmosphere.

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• 27.

### The characteristic impedance of free space is

• A.

Not known

• B.

50Ω

• C.

infinite

• D.

377Ω

D. 377Ω
Explanation
The characteristic impedance of free space is 377Ω. This value represents the ratio of the electric field to the magnetic field in an electromagnetic wave propagating through free space. It is a fundamental constant in electromagnetics and is used in various calculations and designs of antennas, transmission lines, and other RF systems.

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• 28.

### The largest frequency that will be returned to the Earth when transmitted vertically under ionospheric conditions is called the

• A.

Critical frequency

• B.

Optimum Working Frequency (OWF)

• C.

Maximum Usable Frequency (MUF)

• D.

Skip zone

A. Critical frequency
Explanation
The critical frequency is the largest frequency that can be returned to Earth when a signal is transmitted vertically under ionospheric conditions. This frequency is determined by the ionospheric conditions and the density of the ionized layer in the atmosphere. It represents the upper limit for reliable communication using skywave propagation.

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• 29.

### As electromagnetic waves travel in free space, only one of the following can happen to them:

• A.

Absorption

• B.

Reflection

• C.

Attenuation

• D.

Refraction

C. Attenuation
Explanation
Electromagnetic waves can undergo several processes as they travel through free space. Absorption refers to the energy of the wave being absorbed by the medium it encounters. Reflection occurs when the wave bounces off a surface without being absorbed or transmitted. Refraction happens when the wave changes direction as it passes from one medium to another. Attenuation, on the other hand, refers to the gradual decrease in the intensity or amplitude of the wave as it travels through a medium. Therefore, attenuation is the only option that accurately describes what can happen to electromagnetic waves in free space.

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• 30.

### In a vacuum, the speed of light

• A.

Depends on its constant

• B.

Depends on its wavelength

• C.

Depends on its electric and magnetic field

• D.

Is a universal constant

D. Is a universal constant
Explanation
The speed of light in a vacuum is a universal constant because it does not depend on any external factors such as its wavelength or electric and magnetic field. Regardless of these variables, the speed of light remains constant at approximately 299,792,458 meters per second. This constant speed of light is a fundamental principle in physics and plays a crucial role in various scientific theories and equations.

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• 31.

### When the magnetic field is perpendicular to the surface of the Earth, what is the polarization of the TEM wave?

• A.

Circular

• B.

Elliptical

• C.

Vertical

• D.

Horizontal

D. Horizontal
Explanation
When the magnetic field is perpendicular to the surface of the Earth, the polarization of the TEM wave is horizontal. This means that the electric field oscillates in a horizontal plane.

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• 32.

### What is the period of the solar cycle in which great electrical disturbance will occur?

• A.

Every 11 years

• B.

Every 4 years

• C.

Every 5 years

• D.

Every 10 years

A. Every 11 years
Explanation
The correct answer is Every 11 years. The solar cycle refers to the approximately 11-year cycle of the Sun's activity, which includes changes in solar radiation, sunspots, and other solar phenomena. During this cycle, the Sun goes through periods of high and low activity, with the peak of activity known as the solar maximum. It is during this solar maximum that great electrical disturbances, such as solar flares and coronal mass ejections, are more likely to occur. Therefore, the period of the solar cycle in which great electrical disturbance will occur is every 11 years.

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• 33.

### Ducting occurs in which region of the atmosphere?

• A.

F2

• B.

Stratosphere

• C.

Troposphere

• D.

Ionosphere Back to top