1. An electromagnetic wave is best described as a wave made of
ⓐ. time-varying electric and magnetic fields
ⓑ. only a vibrating electric field \(\vec{E}\)
ⓒ. only a vibrating magnetic field \(\vec{B}\)
ⓓ. compressions and rarefactions of air particles
Correct Answer: time-varying electric and magnetic fields
Explanation: An electromagnetic wave consists of electric and magnetic fields that vary with time and space. The electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are both essential parts of the wave. It is not a sound-like wave made from pressure changes in a material. It is also not a wave of only one field, because the changing fields support each other during propagation. This coupled-field idea is the starting point for understanding why electromagnetic waves can travel even where ordinary mechanical waves cannot.
2. A wave travelling from the Sun reaches Earth through the nearly empty region of space. This observation supports the idea that electromagnetic waves
ⓐ. need air as a material medium
ⓑ. can travel through vacuum
ⓒ. need water as a material medium
ⓓ. can move only through solid matter
Correct Answer: can travel through vacuum
Explanation: Sunlight is an electromagnetic wave, and it crosses a huge region of space before reaching Earth. Space between the Sun and Earth is not filled with an ordinary material medium like air or water. This shows that electromagnetic waves do not need particles of matter to carry their disturbance. Mechanical waves such as sound need a medium, but electromagnetic waves are carried by changing fields. The sunlight example is a direct way to separate electromagnetic waves from medium-dependent waves.
3. Visible light belongs to the family of
ⓐ. only mechanical transverse waves
ⓑ. pressure waves in air
ⓒ. electromagnetic waves
ⓓ. waves that cannot travel in vacuum
Correct Answer: electromagnetic waves
Explanation: Visible light is one part of the electromagnetic spectrum. It has electric and magnetic fields associated with it, so it is an electromagnetic wave. Its ability to travel through vacuum is consistent with the nature of electromagnetic waves. Calling visible light a pressure wave would confuse it with sound, which requires a material medium. The familiar nature of light makes it a useful first example of electromagnetic radiation.
4. Match the basic field symbols with their usual meanings.
| Column I | Column II |
|---|
| P. \(\vec{E}\) | 1. Magnetic field |
| Q. \(\vec{B}\) | 2. Electric field |
| R. \(\lambda\) | 3. Wavelength |
| S. \(\nu\) | 4. Frequency |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-1, R-3, S-4
ⓒ. P-2, Q-1, R-4, S-3
ⓓ. P-3, Q-4, R-2, S-1
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: The symbol \(\vec{E}\) represents electric field, while \(\vec{B}\) represents magnetic field. The arrows indicate that these fields are vector quantities. The symbol \(\lambda\) is used for wavelength, the distance over which the wave pattern repeats. The symbol \(\nu\) represents frequency, the number of oscillations per unit time. Mixing \(\lambda\) and \(\nu\) later causes errors in the relation \(c=\nu\lambda\), so the symbol meanings should be kept separate from the beginning.
5. A sound wave and an electromagnetic wave are compared in the same region of space. The key difference in their medium requirement is that
ⓐ. both must have air to travel
ⓑ. sound can travel in vacuum, but electromagnetic waves cannot
ⓒ. both are stopped whenever air is removed
ⓓ. electromagnetic waves can travel in vacuum, but sound cannot
Correct Answer: electromagnetic waves can travel in vacuum, but sound cannot
Explanation: Sound is a mechanical wave and needs particles of a medium to transfer energy. In vacuum, there are no material particles available for the compressions and rarefactions of sound. Electromagnetic waves are different because their changing electric and magnetic fields can propagate through empty space. This is why light from distant stars can reach us. The comparison is not about whether a wave is transverse or longitudinal first; it is mainly about whether a material medium is necessary.
6. The speed of electromagnetic waves in vacuum is usually represented by \(c\). Its approximate value is
ⓐ. \(3.0\times10^8\,\text{m s}^{-1}\)
ⓑ. \(3.0\times10^6\,\text{m s}^{-1}\)
ⓒ. \(3.0\times10^8\,\text{Hz}\)
ⓓ. \(3.0\times10^{-8}\,\text{m s}^{-1}\)
Correct Answer: \(3.0\times10^8\,\text{m s}^{-1}\)
Explanation: In vacuum, electromagnetic waves travel with speed \(c\approx3.0\times10^8\,\text{m s}^{-1}\). The unit \(\text{m s}^{-1}\) is required because \(c\) is a speed, not a frequency. The option with \(\text{Hz}\) gives a unit for frequency, so it cannot represent wave speed. The value \(3.0\times10^8\,\text{m s}^{-1}\) is the same for all electromagnetic waves in vacuum, whether they are radio waves, visible light, or gamma rays. Their differences lie in frequency and wavelength, not in their vacuum speed.
7. The relation connecting speed, frequency, and wavelength of an electromagnetic wave in vacuum is
ⓐ. \(c=\nu\lambda\)
ⓑ. \(c=\frac{\lambda}{\nu}\)
ⓒ. \(c=\nu+\lambda\)
ⓓ. \(c=\frac{\nu}{\lambda}\)
Correct Answer: \(c=\nu\lambda\)
Explanation: The wave speed equals frequency multiplied by wavelength, so the relation is \(c=\nu\lambda\). Frequency \(\nu\) tells how many oscillations occur per second, and wavelength \(\lambda\) tells the spatial length of one oscillation. Multiplying \(\nu\) in \(\text{Hz}\) by \(\lambda\) in \(\text{m}\) gives \(\text{m s}^{-1}\), the unit of speed. The reciprocal forms would give the wrong dimensional meaning. This relation later helps compare different parts of the electromagnetic spectrum.
8. A beam of visible light travels from a distant star to an observer through space. The most suitable reason for its successful travel is that light
ⓐ. is carried by dust particles across space
ⓑ. becomes a mechanical sound wave in empty space
ⓒ. is electromagnetic and can cross vacuum
ⓓ. stops being a wave while crossing vacuum
Correct Answer: is electromagnetic and can cross vacuum
Explanation: Visible light is an electromagnetic wave, so it can propagate through vacuum. It does not need dust, air, or any other material particles to carry it. The wave nature of light remains intact while it travels through empty space. It is not converted into a sound wave, because sound needs a medium and is a mechanical wave. The star-light example is a simple physical proof that electromagnetic waves are field waves rather than material-particle vibrations.
9. The electric and magnetic fields in an electromagnetic wave should not be imagined as
ⓐ. time-varying field quantities
ⓑ. two parts of the same wave
ⓒ. field quantities that can exist in vacuum
ⓓ. particles travelling with the wave
Correct Answer: particles travelling with the wave
Explanation: The fields \(\vec{E}\) and \(\vec{B}\) are not tiny pieces of matter travelling through space. They are physical field quantities that vary with time and position. An electromagnetic wave can carry energy through vacuum because the changing fields themselves propagate. This is different from a mechanical wave, where particles of a medium oscillate about their mean positions. Treating the fields as material particles hides the main reason electromagnetic waves do not require a material medium.
10. A basic description says that an electromagnetic wave carries changing electric and magnetic fields. The word "changing" is essential because
ⓐ. time variation of fields is central to electromagnetic-wave propagation
ⓑ. a constant field alone is the same as a propagating electromagnetic wave
ⓒ. only the electric field changes while the magnetic field must remain constant
ⓓ. the fields must change from vector quantities into scalar quantities
Correct Answer: time variation of fields is central to electromagnetic-wave propagation
Explanation: Electromagnetic waves involve fields that vary with time and space. A constant electric field or a constant magnetic field by itself is not the same thing as a travelling electromagnetic wave. The time-varying nature allows the electric and magnetic aspects of the wave to be connected. Both \(\vec{E}\) and \(\vec{B}\) take part in the wave, so it is not a one-field disturbance. The word "changing" prevents the idea of a static field from being confused with a propagating wave.
11. For electric field \(\vec{E}\), two commonly used units are
ⓐ. \(\text{Wb m}^{-2}\) and \(\text{N A}^{-1}\text{m}^{-1}\)
ⓑ. \(\text{N C}^{-1}\) and \(\text{V m}^{-1}\)
ⓒ. \(\text{Hz}\) and \(\text{m s}^{-1}\)
ⓓ. \(\text{A}\) and \(\text{C s}^{-1}\)
Correct Answer: \(\text{N C}^{-1}\) and \(\text{V m}^{-1}\)
Explanation: Electric field can be expressed as force per unit charge, giving the unit \(\text{N C}^{-1}\). It can also be expressed through potential difference per unit distance, giving the unit \(\text{V m}^{-1}\). These two units are equivalent ways of measuring the same physical quantity. The unit \(\text{T}\) belongs to magnetic field, and \(\text{Wb}\) belongs to magnetic flux. Since electromagnetic waves involve both electric and magnetic fields, keeping their units separate avoids confusing \(\vec{E}\) with \(\vec{B}\).
12. The SI unit of magnetic flux \(\Phi_B\) is
ⓐ. \(\text{Wb}\)
ⓑ. \(\text{T}\)
ⓒ. \(\text{Hz}\)
ⓓ. \(\text{N C}^{-1}\)
Correct Answer: \(\text{Wb}\)
Explanation: Magnetic flux \(\Phi_B\) is measured in weber, written as \(\text{Wb}\). Magnetic field \(\vec{B}\) itself is measured in tesla, written as \(\text{T}\). The unit \(\text{Hz}\) belongs to frequency, and \(\text{N C}^{-1}\) is a unit of electric field. Flux and field are related, but they are not the same quantity. This distinction becomes useful when electric flux \(\Phi_E\) is introduced in the displacement-current idea.
13. Consider the following basic quantities used in electromagnetic waves.
Statement I: \(\vec{E}\) and \(\vec{B}\) are vector field quantities.
Statement II: \(\lambda\) represents wavelength and \(\nu\) represents frequency.
Statement III: The unit \(\text{A}\) is used for electric current.
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I, II and III
ⓓ. I and III only
Correct Answer: I, II and III
Explanation: Electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are vector quantities because they have directions as well as magnitudes. Wavelength is denoted by \(\lambda\), while frequency is denoted by \(\nu\). Electric current is measured in ampere, written as \(\text{A}\). These symbols appear repeatedly in electromagnetic-wave relations and displacement-current relations. A symbol error at this stage can later lead to using the right formula with the wrong physical meaning.
14. A quantity list gives \(I_c\), \(I_d\), \(\varepsilon_0\), and \(\mu_0\). The pair that represents currents is
ⓐ. \(\varepsilon_0\) and \(\mu_0\)
ⓑ. \(\mu_0\) and \(I_d\)
ⓒ. \(I_c\) and \(\varepsilon_0\)
ⓓ. \(I_c\) and \(I_d\)
Correct Answer: \(I_c\) and \(I_d\)
Explanation: The symbol \(I_c\) represents conduction current, and \(I_d\) represents displacement current. Both are current-type quantities and are measured in ampere \(\text{A}\). The symbols \(\varepsilon_0\) and \(\mu_0\) are constants associated with electric and magnetic effects in vacuum. They are not current symbols, even though they appear in important electromagnetic formulas. Separating current symbols from constant symbols helps make Maxwell's correction easier to read later.
15. An electromagnetic wave in vacuum has frequency \(1.0\times10^8\,\text{Hz}\). Taking \(c=3.0\times10^8\,\text{m s}^{-1}\), its wavelength is
ⓐ. \(0.30\,\text{m}\)
ⓑ. \(30\,\text{m}\)
ⓒ. \(3.0\,\text{m}\)
ⓓ. \(3.0\times10^8\,\text{m}\)
Correct Answer: \(3.0\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(\nu=1.0\times10^8\,\text{Hz}\) and \(c=3.0\times10^8\,\text{m s}^{-1}\).
\( \textbf{Required:} \) Wavelength \(\lambda\).
\( \textbf{Wave relation:} \)
\[
c=\nu\lambda
\]
\( \textbf{Reason for using it:} \) In vacuum, electromagnetic-wave speed, frequency, and wavelength are connected by \(c=\nu\lambda\).
\( \textbf{Rearrangement:} \)
\[
\lambda=\frac{c}{\nu}
\]
\( \textbf{Substitution:} \)
\[
\lambda=\frac{3.0\times10^8}{1.0\times10^8}\,\text{m}
\]
\( \textbf{Simplification:} \)
\[
\lambda=3.0\,\text{m}
\]
\( \textbf{Unit check:} \) Since \(\text{Hz}=\text{s}^{-1}\), \(\frac{\text{m s}^{-1}}{\text{s}^{-1}}\) gives \(\text{m}\).
\( \textbf{Final answer:} \) The wavelength is \(3.0\,\text{m}\).
16. Study the table and identify the row with a mismatched unit.
| Row | Quantity | Unit |
|---|
| P | Electric field \(\vec{E}\) | \(\text{N C}^{-1}\) |
| Q | Magnetic field \(\vec{B}\) | \(\text{T}\) |
| R | Frequency \(\nu\) | \(\text{Hz}\) |
| S | Wavelength \(\lambda\) | \(\text{A}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Wavelength \(\lambda\) is a length, so its SI unit is metre, written as \(\text{m}\). The ampere \(\text{A}\) is the SI unit of electric current, not wavelength. Electric field may be measured in \(\text{N C}^{-1}\), magnetic field in \(\text{T}\), and frequency in \(\text{Hz}\). Rows P, Q, and R therefore match the quantities with suitable units. The mismatch in row S comes from confusing a wave-length quantity with a current quantity.
17. Assertion: Electromagnetic waves can travel through vacuum.
Reason: Their propagation is associated with time-varying electric and magnetic fields, not with vibrations of material particles.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Electromagnetic waves can travel through vacuum because they are not mechanical waves. Their propagation is described using changing electric and magnetic fields. A material medium is not required for these fields to carry energy through space. The Reason directly supports the Assertion because it explains why the absence of material particles does not stop the wave. This is the same property that allows light from the Sun and stars to reach Earth.
18. A basic wave record for an electromagnetic wave lists \(c\), \(\nu\), and \(\lambda\). The blank in the relation \(c=\_\_\_\_\_\_\) is best filled by
ⓐ. \(\nu+\lambda\)
ⓑ. \(\nu-\lambda\)
ⓒ. \(\frac{\nu}{\lambda}\)
ⓓ. \(\nu\lambda\)
Correct Answer: \(\nu\lambda\)
Explanation: The speed of a wave is the product of its frequency and wavelength. For electromagnetic waves in vacuum, this is written as \(c=\nu\lambda\). The product has the right unit because \(\nu\) has unit \(\text{s}^{-1}\) and \(\lambda\) has unit \(\text{m}\), giving \(\text{m s}^{-1}\). Addition or subtraction cannot combine quantities with different units into a speed. The relation also shows why higher frequency corresponds to shorter wavelength when \(c\) remains fixed in vacuum.
19. In the quantity \(\Phi_E\), the subscript \(E\) indicates that it is related to
ⓐ. magnetic flux
ⓑ. electric flux
ⓒ. frequency
ⓓ. conduction current
Correct Answer: electric flux
Explanation: The symbol \(\Phi_E\) represents electric flux. The subscript \(E\) connects the flux with the electric field \(\vec{E}\). Magnetic flux is written as \(\Phi_B\), where the subscript \(B\) refers to the magnetic field \(\vec{B}\). This distinction becomes important because displacement current is linked with changing electric flux, not changing magnetic flux. Confusing \(\Phi_E\) with \(\Phi_B\) would lead to the wrong physical source for Maxwell's correction.
20. A displacement current \(I_d\) and a conduction current \(I_c\) are compared only by their SI units. The suitable unit for both is
ⓐ. \(\text{Hz}\)
ⓑ. \(\text{A}\)
ⓒ. \(\text{Wb}\)
ⓓ. \(\text{T}\)
Correct Answer: \(\text{A}\)
Explanation: Both \(I_c\) and \(I_d\) are current-type quantities, so both have the SI unit ampere \(\text{A}\). Conduction current is associated with actual motion of charge carriers in a conductor. Displacement current is not ordinary charge flow through a dielectric or vacuum gap, but it appears in equations with the same unit as current. The unit \(\text{Hz}\) belongs to frequency, \(\text{Wb}\) to magnetic flux, and \(\text{T}\) to magnetic field. The shared unit is one reason \(I_d\) can be added to \(I_c\) in the corrected Ampere-Maxwell law.
