Chapter 22: Electromagnetic Waves

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Electromagnetic Wave Quizzes & Trivia

Questions and Answers
  • 1. 

    The basic equations for all electromagnetism were developed by

    • A.

      Newton.

    • B.

      Gauss.

    • C.

      Maxwell.

    • D.

      Hertz.

    Correct Answer
    C. Maxwell.
    Explanation
    Maxwell is the correct answer because he developed the basic equations for all electromagnetism. Newton is known for his work in classical mechanics, Gauss made significant contributions to mathematics and physics but not specifically to electromagnetism, and Hertz conducted experiments that confirmed Maxwell's theories but did not develop the basic equations themselves.

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  • 2. 

    An electric field is produced by a

    • A.

      Constant magnetic field.

    • B.

      Changing magnetic field.

    • C.

      Either a constant or a changing magnetic field.

    • D.

      None of the given answers

    Correct Answer
    B. Changing magnetic field.
    Explanation
    An electric field is produced by a changing magnetic field. This is known as electromagnetic induction, which occurs when there is a change in the magnetic field strength or direction in a given region. According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electric field, which can cause the movement of charges and the generation of electric currents. Therefore, a changing magnetic field is responsible for the production of an electric field.

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  • 3. 

    A changing electric field will produce a

    • A.

      Current.

    • B.

      Gravitational field.

    • C.

      Magnetic field.

    • D.

      None of the given answers

    Correct Answer
    C. Magnetic field.
    Explanation
    When an electric field changes, it induces a magnetic field according to Faraday's law of electromagnetic induction. This phenomenon is the basis for the operation of many electrical devices, such as generators and transformers. Therefore, a changing electric field will indeed produce a magnetic field.

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  • 4. 

    All electromagnetic waves travel through a vacuum at

    • A.

      The same speed.

    • B.

      Speeds that are proportional to their frequency.

    • C.

      Speeds that are inversely proportional to their frequency.

    • D.

      None of the given answers

    Correct Answer
    A. The same speed.
    Explanation
    All electromagnetic waves, including visible light, radio waves, and X-rays, travel through a vacuum at the same speed, which is approximately 3 x 10^8 meters per second. This speed is commonly referred to as the speed of light. This is a fundamental property of electromagnetic waves and is a result of the interaction of electric and magnetic fields. Regardless of their frequency or wavelength, all electromagnetic waves propagate through a vacuum at this constant speed.

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  • 5. 

    In a vacuum, the velocity of all electromagnetic waves

    • A.

      Is zero.

    • B.

      Is nearly 3 * 10^8 m/s.

    • C.

      Depends on the frequency.

    • D.

      Depends on their amplitude.

    Correct Answer
    B. Is nearly 3 * 10^8 m/s.
    Explanation
    In a vacuum, electromagnetic waves travel at a constant speed of nearly 3 * 10^8 m/s. This speed is known as the speed of light and is a fundamental constant in physics. It is independent of the frequency or amplitude of the wave. The speed of light is an important concept as it forms the basis for many theories and equations in physics, including Einstein's theory of relativity.

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  • 6. 

    Electromagnetic waves can travel through

    • A.

      Glass.

    • B.

      Iron.

    • C.

      Water.

    • D.

      None of the given answers

    Correct Answer
    D. None of the given answers
    Explanation
    Electromagnetic waves can travel through a variety of materials, including glass, iron, and water. Therefore, the correct answer is none of the given answers, as all of the options listed can allow for the transmission of electromagnetic waves.

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  • 7. 

    The strength of both the electric and magnetic fields in the radiation field are found to decrease with distance as

    • A.

      1/r.

    • B.

      1/r^2.

    • C.

      1/r^3.

    • D.

      1/r^4.

    Correct Answer
    A. 1/r.
    Explanation
    The strength of both the electric and magnetic fields in the radiation field decrease with distance as 1/r. This is because the intensity of radiation spreads out as it moves away from its source, resulting in a decrease in field strength. The inverse relationship with distance (1/r) indicates that as the distance from the source increases, the field strength decreases. This is a fundamental property of radiation and can be observed in various phenomena such as the inverse square law for light intensity.

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  • 8. 

    The E and B fields in electromagnetic waves are oriented

    • A.

      Parallel to the wave's direction of travel, as well as to each other.

    • B.

      Parallel to the waves direction of travel, and perpendicular to each other.

    • C.

      Perpendicular to the wave's direction of travel, and parallel to each other.

    • D.

      Perpendicular to the wave's direction of travel, and also to each other.

    Correct Answer
    D. Perpendicular to the wave's direction of travel, and also to each other.
    Explanation
    The E and B fields in electromagnetic waves are perpendicular to the wave's direction of travel, meaning they form a right angle with it. Additionally, the E and B fields are also perpendicular to each other, meaning they form another right angle with each other. This is a fundamental property of electromagnetic waves and is described by Maxwell's equations.

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  • 9. 

    An electromagnetic wave is traveling to the east. At one instant at a given point its E vector points straight up. What is the direction of its B vector?

    • A.

      North

    • B.

      Down

    • C.

      East

    • D.

      South

    Correct Answer
    D. South
    Explanation
    When an electromagnetic wave is traveling to the east and its E vector points straight up, the direction of its B vector will be south. This is because the E and B vectors in an electromagnetic wave are perpendicular to each other and also perpendicular to the direction of wave propagation. Therefore, if the E vector points up, the B vector will point in the direction perpendicular to both the E vector and the direction of wave propagation, which in this case is south.

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  • 10. 

    An electromagnetic wave is radiated by a straight wire antenna that is oriented vertically. What should be the orientation of a straight wire receiving antenna? It should be placed

    • A.

      Vertically.

    • B.

      Horizontally and in a direction parallel to the wave's direction of motion.

    • C.

      Horizontally and in a direction perpendicular to the wave's direction of motion.

    • D.

      None of the given answers

    Correct Answer
    A. Vertically.
    Explanation
    The correct answer is vertically. This is because the receiving antenna should be oriented in the same direction as the transmitting antenna in order to effectively receive the electromagnetic wave. Since the transmitting antenna is oriented vertically, the receiving antenna should also be placed vertically.

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  • 11. 

    Electromagnetic waves are

    • A.

      Longitudinal.

    • B.

      Transverse.

    • C.

      Both longitudinal and transverse.

    • D.

      Neither longitudinal or transverse.

    Correct Answer
    B. Transverse.
    Explanation
    Electromagnetic waves are transverse because they consist of oscillating electric and magnetic fields that are perpendicular to the direction of wave propagation. In a transverse wave, the particles of the medium through which the wave travels oscillate perpendicular to the direction of wave motion. This is in contrast to a longitudinal wave, where the particles oscillate parallel to the direction of wave motion. Since electromagnetic waves do not require a medium to propagate, they are transverse in nature.

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  • 12. 

    Of the following, which is not electromagnetic in nature?

    • A.

      Microwaves

    • B.

      Gamma rays

    • C.

      Sound waves

    • D.

      Radio waves

    Correct Answer
    C. Sound waves
    Explanation
    Sound waves are not electromagnetic in nature because they are mechanical waves that require a medium (such as air, water, or solids) to travel through. Unlike electromagnetic waves (such as microwaves, gamma rays, and radio waves), sound waves cannot travel through a vacuum. Sound waves are produced by the vibration of particles in the medium, while electromagnetic waves are created by the oscillation of electric and magnetic fields.

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  • 13. 

    Visible light ranges in wavelength from

    • A.

      400 μm to 750 μm.

    • B.

      400 nm to 750 nm.

    • C.

      500 μm to 850 μm.

    • D.

      500 nm to 850 nm.

    Correct Answer
    B. 400 nm to 750 nm.
    Explanation
    The correct answer is 400 nm to 750 nm. This range represents the wavelengths of visible light that can be detected by the human eye. Wavelengths below 400 nm are ultraviolet light, which is not visible to humans, while wavelengths above 750 nm are infrared light, also not visible to humans. Therefore, the range of 400 nm to 750 nm encompasses the colors of light that we can see.

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  • 14. 

    The energy an electromagnetic wave transports per unit time per unit area is the

    • A.

      Energy density.

    • B.

      Power.

    • C.

      Intensity.

    • D.

      Radiation pressure.

    Correct Answer
    C. Intensity.
    Explanation
    The energy an electromagnetic wave transports per unit time per unit area is known as intensity. Intensity refers to the amount of energy carried by the wave per unit area perpendicular to the direction of propagation. It is a measure of the strength or power of the wave.

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  • 15. 

    The force per unit area exerted by an electromagnetic wave is called the

    • A.

      Energy density.

    • B.

      Power.

    • C.

      Intensity.

    • D.

      Radiation pressure.

    Correct Answer
    D. Radiation pressure.
    Explanation
    The force per unit area exerted by an electromagnetic wave is called radiation pressure. This refers to the pressure exerted by the electromagnetic radiation on a surface. It is a result of the transfer of momentum from the photons of the radiation to the surface. The radiation pressure can be calculated by dividing the energy density of the radiation by the speed of light.

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  • 16. 

    Which of the following correctly lists electromagnetic waves in order from longest to shortest wavelength?

    • A.

      Gamma rays, ultraviolet, infrared, microwaves

    • B.

      Microwaves, ultraviolet, visible light, gamma rays

    • C.

      Radio waves, infrared, gamma rays, ultraviolet

    • D.

      Television, infrared, visible light, X-rays

    Correct Answer
    D. Television, infrared, visible light, X-rays
  • 17. 

    If the electric field in an EM wave has a peak value of 2.0 V/m, what is the peak value of the magnetic field?

    • A.

      6.7 nT

    • B.

      2.0 T

    • C.

      3.0 * 10^8 T

    • D.

      None of the given answers

    Correct Answer
    A. 6.7 nT
    Explanation
    The peak value of the magnetic field in an electromagnetic wave can be calculated using the equation B = E/c, where B is the magnetic field, E is the electric field, and c is the speed of light. Given that the electric field has a peak value of 2.0 V/m, and the speed of light is approximately 3.0 * 10^8 m/s, we can calculate the peak value of the magnetic field as 2.0 V/m / (3.0 * 10^8 m/s) = 6.7 nT.

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  • 18. 

    What is the wavelength of light waves if their frequency is 5.0 * 10^14 Hz?

    • A.

      0.60 m

    • B.

      6.0 mm

    • C.

      0.060 mm

    • D.

      0.60 μm

    Correct Answer
    D. 0.60 μm
    Explanation
    The correct answer is 0.60 μm. The wavelength of light waves can be calculated using the formula λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. In this case, the frequency is given as 5.0 * 10^14 Hz. Since the speed of light is approximately 3.0 * 10^8 m/s, we can substitute the values into the formula to find the wavelength. The wavelength comes out to be 0.60 μm, which is equivalent to 0.00060 mm.

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  • 19. 

    What is the wavelength of a 92.9-MHz radio wave?

    • A.

      32 mm

    • B.

      32 cm

    • C.

      3.2 m

    • D.

      32 m

    Correct Answer
    C. 3.2 m
    Explanation
    The wavelength of a radio wave can be calculated using the formula: wavelength = speed of light / frequency. In this case, the frequency is given as 92.9 MHz. To convert this to Hz, we multiply by 10^6. The speed of light is a constant value of approximately 3 x 10^8 m/s. Plugging these values into the formula, we get: wavelength = (3 x 10^8 m/s) / (92.9 x 10^6 Hz) = 3.23 m. Therefore, the correct answer is 3.2 m.

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  • 20. 

    What is the wavelength of a radio wave signal transmitted at a frequency of 7.2 MHz?

    • A.

      42 m

    • B.

      4.2 m

    • C.

      29 m

    • D.

      0.024 m

    Correct Answer
    A. 42 m
    Explanation
    The wavelength of a radio wave can be calculated using the formula: wavelength = speed of light / frequency. In this case, the frequency is given as 7.2 MHz, which can be converted to 7.2 x 10^6 Hz. The speed of light is approximately 3 x 10^8 m/s. By plugging these values into the formula, we can calculate the wavelength to be approximately 42 meters.

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  • 21. 

    What is the wavelength for the carrier wave in the station WBRN 1460?

    • A.

      2.05 m

    • B.

      20.5 m

    • C.

      205 m

    • D.

      2.05 km

    Correct Answer
    C. 205 m
    Explanation
    The wavelength for the carrier wave in the station WBRN 1460 is 205 m. This means that the distance between two consecutive peaks or troughs of the wave is 205 meters.

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  • 22. 

    What is the frequency of 20 mm microwaves?

    • A.

      100 MHz

    • B.

      400 MHz

    • C.

      15 GHz

    • D.

      73 GHz

    Correct Answer
    C. 15 GHz
    Explanation
    Microwaves are a type of electromagnetic radiation with wavelengths ranging from 1 mm to 1 meter. The frequency of microwaves is inversely proportional to their wavelength. Higher frequency microwaves have shorter wavelengths. The given correct answer, 15 GHz, indicates that the microwaves have a frequency of 15 billion hertz or cycles per second. This frequency corresponds to a wavelength of approximately 20 mm, which falls within the range of microwaves. Therefore, 15 GHz is the correct frequency for 20 mm microwaves.

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  • 23. 

    How far does light travel in 1.0 μs?

    • A.

      3.0 * 10^14 m

    • B.

      0.30 km

    • C.

      3.0 m

    • D.

      30 cm

    Correct Answer
    B. 0.30 km
    Explanation
    In 1.0 μs (microsecond), light travels a distance of 0.30 km. This is because light travels at a speed of approximately 3.0 x 10^8 meters per second. By multiplying this speed by the given time of 1.0 μs, we can calculate the distance traveled, which is equal to 0.30 km.

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  • 24. 

    How long does it take light to travel 1.0 m?

    • A.

      3.3 ns

    • B.

      3.3 μs

    • C.

      3.3 ms

    • D.

      3.3 s

    Correct Answer
    A. 3.3 ns
    Explanation
    Light travels at a speed of approximately 3.0 x 10^8 meters per second in a vacuum. Therefore, to travel a distance of 1.0 meter, it would take 1.0 meter / 3.0 x 10^8 meters per second = 3.3 x 10^-9 seconds, which is equivalent to 3.3 nanoseconds.

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  • 25. 

    How long does it take the signal from a local radio station to travel 20 miles?

    • A.

      2.8 s

    • B.

      108 ms

    • C.

      38 ms

    • D.

      108 ms

    Correct Answer
    D. 108 ms
    Explanation
    The correct answer is 108 ms. This is because the speed of light, which is the speed at which radio waves travel, is approximately 186,282 miles per second. To calculate the time it takes for the signal to travel 20 miles, we can use the formula: time = distance / speed. Therefore, 20 miles / 186,282 miles per second = 0.0001075 seconds, which is equal to 108 ms.

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  • 26. 

    How far is a light year (the distance light travels in a year)?

    • A.

      186,000 m

    • B.

      3.0 * 10^8 m

    • C.

      8.7 * 10^13 m

    • D.

      9.5 * 10^15 m

    Correct Answer
    D. 9.5 * 10^15 m
    Explanation
    A light year is the distance that light travels in one year. Since light travels at a speed of approximately 3.0 * 10^8 meters per second, we can calculate the distance by multiplying the speed by the number of seconds in a year. This gives us a distance of 9.5 * 10^15 meters.

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  • 27. 

    A radar receiver indicates that a transmitted pulse return as an echo in 20 μs after transmission. How far away is the reflecting object?

    • A.

      1.5 km

    • B.

      3.0 km

    • C.

      6.0 km

    • D.

      9.0 km

    Correct Answer
    B. 3.0 km
    Explanation
    The time it takes for the pulse to return as an echo is 20 μs. To find the distance, we can use the formula: distance = speed × time. The speed of the pulse is the speed of light, which is approximately 3 × 10^8 meters per second. Converting the time to seconds (20 × 10^-6), we can calculate the distance as follows: distance = (3 × 10^8) × (20 × 10^-6) = 6000 meters = 6.0 km. Therefore, the reflecting object is 6.0 km away.

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  • 28. 

    Radiation from the Sun reaches the Earth (above the atmosphere) at a rate of about 1350 W/m^2. Assume that this is a single EM wave and calculate the maximum value of the electric field.

    • A.

      0.71 kV/m

    • B.

      1.0 kV/m

    • C.

      1.4 kV/m

    • D.

      2.0 kV/m

    Correct Answer
    B. 1.0 kV/m
    Explanation
    The maximum value of the electric field can be calculated using the formula E = c * sqrt(P), where E is the electric field, c is the speed of light, and P is the power per unit area. In this case, the power per unit area is given as 1350 W/m^2. Plugging in the values, we get E = 3 * 10^8 m/s * sqrt(1350 W/m^2) = 1.04 * 10^6 V/m. This value is closest to 1.0 kV/m, which is the correct answer.

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  • 29. 

    Radiation from the Sun reaches the Earth (above the atmosphere) at a rate of about 1350 W/m^2. Assume that this is a single EM wave and calculate the maximum value of the magnetic field.

    • A.

      6.7 μT

    • B.

      4.8 μT

    • C.

      3.4 μT

    • D.

      2.4 μT

    Correct Answer
    C. 3.4 μT
    Explanation
    The maximum value of the magnetic field can be calculated using the formula B = E/c, where B is the magnetic field, E is the electric field, and c is the speed of light. In this case, the electric field can be calculated by dividing the power (1350 W/m^2) by the speed of light (3 x 10^8 m/s). Plugging in the values, we get E = 1350 / (3 x 10^8) = 4.5 x 10^-6 N/C. Finally, we can calculate the magnetic field by dividing the electric field by the speed of light, giving us B = (4.5 x 10^-6) / (3 x 10^8) = 1.5 x 10^-14 T, which is equivalent to 15 μT. Since none of the given answer choices match exactly, the closest option is 3.4 μT.

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  • 30. 

    What is the maximum value of the electric field at a distance 2.5 m from a 100-W light bulb?

    • A.

      22 V/m

    • B.

      31 V/m

    • C.

      44 V/m

    • D.

      62 V/m

    Correct Answer
    B. 31 V/m
    Explanation
    The maximum value of the electric field at a distance from a light bulb is determined by the power of the bulb. In this case, the bulb has a power of 100 W. The electric field is directly proportional to the square root of the power, so we can calculate it using the formula E = sqrt(P/4πεr^2), where P is the power, ε is the permittivity of free space, and r is the distance from the bulb. Plugging in the values, we get E = sqrt(100/4πε(2.5)^2) ≈ 31 V/m. Therefore, the maximum value of the electric field at a distance 2.5 m from the light bulb is 31 V/m.

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  • 31. 

    What is the maximum value of the magnetic field at a distance 2.5 m from a 100-W light bulb?

    • A.

      0.20 μT

    • B.

      0.14 μT

    • C.

      0.10 μT

    • D.

      0.071 μT

    Correct Answer
    C. 0.10 μT
    Explanation
    The maximum value of the magnetic field at a distance of 2.5 m from a 100-W light bulb is 0.10 μT. This value is determined by the power of the light bulb and the distance from it. The magnetic field decreases with distance according to the inverse square law, so the farther away from the light bulb, the weaker the magnetic field.

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  • 32. 

    A 15.0-mW laser puts out a narrow beam 2.00 mm in diameter. What is the average (rms) value of the electric field?

    • A.

      0.95 kV/m

    • B.

      1.3 kV/m

    • C.

      1.9 kV/m

    • D.

      2.7 kV/m

    Correct Answer
    B. 1.3 kV/m
    Explanation
    The average (rms) value of the electric field can be calculated using the formula E = sqrt(2P/πr^2), where P is the power and r is the radius of the laser beam. In this case, the power is given as 15.0 mW and the radius is given as 1.00 mm (which is half of the diameter). Plugging these values into the formula, we get E = sqrt(2(15.0 mW)/(π(1.00 mm)^2)) = 1.3 kV/m. Therefore, the average (rms) value of the electric field is 1.3 kV/m.

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  • 33. 

    A 15.0-mW laser puts out a narrow beam 2.00 mm in diameter. What is the average (rms) value of the magnetic field?

    • A.

      4.5 mT

    • B.

      6.4 mT

    • C.

      9.0 mT

    • D.

      None of the given answers

    Correct Answer
    A. 4.5 mT
    Explanation
    The average (rms) value of the magnetic field can be determined using the formula B = sqrt(P/(π*r^2*c)), where P is the power of the laser, r is the radius of the beam, and c is the speed of light. Plugging in the values given in the question, we get B = sqrt(15.0 mW/(π*(1.00 mm)^2*c)). Simplifying further, we find B = sqrt(15.0 mW/(π*0.001 m^2*3.00x10^8 m/s)). Evaluating this expression, we get B ≈ 4.5 mT. Therefore, the correct answer is 4.5 mT.

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  • 34. 

    How much energy is transported across a 1.00-cm^2 area per hour by an EM wave whose electric field has an rms strength of 21.5 V/m?

    • A.

      0.44 nJ

    • B.

      0.44 μJ

    • C.

      0.44 mJ

    • D.

      0.44 J

    Correct Answer
    D. 0.44 J
  • 35. 

    Estimate the average power output of the Sun, given that about 1350 W/m^2 reaches the upper atmosphere of the Earth. The distance from the Sun to the Earth is 1.5 * 10^11 m.

    • A.

      1 1 * 10^26 W

    • B.

      2 * 10^26 W

    • C.

      3 * 10^26 W

    • D.

      4 * 10^26 W

    Correct Answer
    D. 4 * 10^26 W
    Explanation
    The average power output of the Sun can be estimated by multiplying the amount of power reaching the upper atmosphere of the Earth (1350 W/m^2) by the surface area of a sphere with a radius equal to the distance from the Sun to the Earth (1.5 * 10^11 m). This calculation results in a power output of 4 * 10^26 W.

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  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
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