201. A wave expression contains the phase \(kx-\omega t\). If the phase is to be dimensionless, the units of \(k\) and \(\omega\) should be
ⓐ. \(k:\text{m}\), \(\omega:\text{s}\)
ⓑ. \(k:\text{s}^{-1}\), \(\omega:\text{m}^{-1}\)
ⓒ. \(k:\text{m s}^{-1}\), \(\omega:\text{Hz m}\)
ⓓ. \(k:\text{m}^{-1}\), \(\omega:\text{s}^{-1}\)
Correct Answer: \(k:\text{m}^{-1}\), \(\omega:\text{s}^{-1}\)
Explanation: The argument of a sine or cosine function must be dimensionless. In the term \(kx\), the coordinate \(x\) has unit \(\text{m}\), so \(k\) must have unit \(\text{m}^{-1}\). In the term \(\omega t\), time \(t\) has unit \(\text{s}\), so \(\omega\) must have unit \(\text{s}^{-1}\), commonly written as \(\text{rad s}^{-1}\). These unit requirements make both \(kx\) and \(\omega t\) pure phase terms. The units are determined by the phase expression, not by the field amplitude.
202. A graph is drawn between frequency \(\nu\) on the vertical axis and \(\frac{1}{\lambda}\) on the horizontal axis for electromagnetic waves in vacuum. The slope of the graph represents
ⓐ. \(\frac{1}{c}\)
ⓑ. \(c\)
ⓒ. \(2\pi c\)
ⓓ. \(\frac{c}{2\pi}\)
Correct Answer: \(c\)
Explanation: The wave relation in vacuum is \(c=\nu\lambda\). Rearranging gives \(\nu=c\left(\frac{1}{\lambda}\right)\). If \(\nu\) is plotted on the vertical axis and \(\frac{1}{\lambda}\) on the horizontal axis, the equation has the form \(y=mx\). The slope is therefore \(c\). This graph shows the inverse relation between \(\nu\) and \(\lambda\) without making the curve non-linear.
203. A wave in vacuum has \(\lambda=3.0\,\text{cm}\). Its frequency lies at
ⓐ. \(1.0\times10^8\,\text{Hz}\)
ⓑ. \(1.0\times10^{10}\,\text{Hz}\)
ⓒ. \(1.0\times10^9\,\text{Hz}\)
ⓓ. \(9.0\times10^{10}\,\text{Hz}\)
Correct Answer: \(1.0\times10^{10}\,\text{Hz}\)
Explanation: \( \textbf{Given:} \) \(\lambda=3.0\,\text{cm}\) and \(c=3.0\times10^8\,\text{m s}^{-1}\).
\( \textbf{Unit conversion:} \)
\[
3.0\,\text{cm}=3.0\times10^{-2}\,\text{m}
\]
\( \textbf{Wave relation:} \)
\[
c=\nu\lambda
\]
\( \textbf{Rearrange:} \)
\[
\nu=\frac{c}{\lambda}
\]
\( \textbf{Substitution:} \)
\[
\nu=\frac{3.0\times10^8}{3.0\times10^{-2}}\,\text{Hz}
\]
\( \textbf{Coefficient simplification:} \)
\[
\frac{3.0}{3.0}=1.0
\]
\( \textbf{Power of ten:} \)
\[
10^8\div10^{-2}=10^{10}
\]
\( \textbf{Final answer:} \) The frequency is \(1.0\times10^{10}\,\text{Hz}\).
204. An electromagnetic wave has \(\omega=6.0\pi\times10^9\,\text{rad s}^{-1}\). Its ordinary frequency is
ⓐ. \(6.0\times10^9\,\text{Hz}\)
ⓑ. \(12.0\times10^9\,\text{Hz}\)
ⓒ. \(3.0\times10^9\,\text{Hz}\)
ⓓ. \(\frac{3.0}{\pi}\times10^9\,\text{Hz}\)
Correct Answer: \(3.0\times10^9\,\text{Hz}\)
Explanation: \( \textbf{Given:} \) \(\omega=6.0\pi\times10^9\,\text{rad s}^{-1}\).
\( \textbf{Required:} \) Ordinary frequency \(\nu\).
\( \textbf{Relation:} \)
\[
\omega=2\pi\nu
\]
\( \textbf{Rearrange:} \)
\[
\nu=\frac{\omega}{2\pi}
\]
\( \textbf{Substitution:} \)
\[
\nu=\frac{6.0\pi\times10^9}{2\pi}\,\text{Hz}
\]
\( \textbf{Cancel \(\pi\):} \)
\[
\nu=\frac{6.0\times10^9}{2}\,\text{Hz}
\]
\( \textbf{Result:} \)
\[
\nu=3.0\times10^9\,\text{Hz}
\]
\( \textbf{Final answer:} \) The ordinary frequency is \(3.0\times10^9\,\text{Hz}\).
205. A wave quantity record is shown below.
| Row | Quantity | Relation |
|---|
| P | Time period | \(T=\frac{1}{\nu}\) |
| Q | Angular frequency | \(\omega=2\pi\nu\) |
| R | Wave number | \(k=\frac{2\pi}{\lambda}\) |
| S | Speed in vacuum | \(c=\frac{\lambda}{\nu}\) |
The row that contains an incorrect relation is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: The correct wave-speed relation in vacuum is \(c=\nu\lambda\). Row S writes \(c=\frac{\lambda}{\nu}\), which has the wrong dependence and wrong unit. Rows P, Q, and R correctly give \(T=\frac{1}{\nu}\), \(\omega=2\pi\nu\), and \(k=\frac{2\pi}{\lambda}\). The error in row S would make speed decrease when frequency increases at fixed wavelength, which is not the wave-speed formula. The relation \(c=\nu\lambda\) is the one that keeps dimensions as \(\text{m s}^{-1}\).
206. Electromagnetic waves transport energy from one region to another mainly through
ⓐ. their electric and magnetic fields
ⓑ. the permanent flow of air particles along the wave path
ⓒ. a stream of conduction current through empty space
ⓓ. isolated magnetic poles carried by the wave
Correct Answer: their electric and magnetic fields
Explanation: An electromagnetic wave carries energy in its electric and magnetic fields. These fields vary with time and space as the wave travels. The energy transport does not require air particles or any other material medium. It also does not require conduction current flowing through the vacuum from source to receiver. The fields themselves form the travelling physical disturbance that carries energy.
207. Sunlight warms a surface on Earth after travelling through space. This is possible because electromagnetic waves
ⓐ. carry energy even through vacuum
ⓑ. require air molecules between the Sun and Earth
ⓒ. carry only magnetic flux but no energy
ⓓ. are identical to sound waves
Correct Answer: carry energy even through vacuum
Explanation: Sunlight is electromagnetic radiation. It travels through the near-vacuum of space and still delivers energy to surfaces on Earth. This shows that electromagnetic waves can transport energy without a material medium. The warming of a surface is evidence of energy transfer by radiation. Treating sunlight as a sound wave would be wrong because sound cannot travel through vacuum.
208. Consider the following statements about energy in electromagnetic waves.
Statement I: The electric field of an electromagnetic wave stores energy.
Statement II: The magnetic field of an electromagnetic wave stores energy.
Statement III: Energy transport by an electromagnetic wave requires matter to vibrate along the path.
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Both electric and magnetic fields can store energy. In an electromagnetic wave, energy is carried through the combined travelling field pattern. Statement III is false because electromagnetic waves can transport energy through vacuum. Mechanical waves need matter to vibrate, but electromagnetic waves are not mechanical vibrations. The energy transport is field-based rather than medium-particle-based.
209. A spacecraft receives a radio signal from Earth. The signal is weak but still detected after crossing space. The physical quantity carried by the radio wave, besides information, is
ⓐ. stationary charge only
ⓑ. a continuous metal wire current
ⓒ. sound pressure in air
ⓓ. energy
Correct Answer: energy
Explanation: Radio waves are electromagnetic waves and carry energy. A receiving antenna can detect the wave because the arriving fields can transfer energy to charges in the antenna. The signal does not require a metal wire connection through space. It also is not a sound pressure wave in air. The weak strength of the signal only means the delivered energy per unit area is small, not that the wave carries no energy.
210. A claim says, "An electromagnetic wave cannot exert any effect on matter because it has no material particles." The claim is
ⓐ. valid because electromagnetic fields cannot transfer energy
ⓑ. valid because radiation pressure exists only for sound waves
ⓒ. invalid because vacuum always blocks electromagnetic waves
ⓓ. invalid because fields can carry energy and interact with matter
Correct Answer: invalid because fields can carry energy and interact with matter
Explanation: Electromagnetic waves do not need material particles as their medium, but they still carry energy. Their electric and magnetic fields can interact with charges and matter. A surface may absorb electromagnetic radiation and become warmer. A receiving antenna can also respond to radio-wave fields. The absence of a material medium does not mean absence of energy or absence of physical effects.
211. Radiation pressure is possible because electromagnetic waves carry
ⓐ. only electric charge
ⓑ. momentum as well as energy
ⓒ. only temperature
ⓓ. only wavelength without energy
Correct Answer: momentum as well as energy
Explanation: Electromagnetic waves carry energy, and they also carry momentum. When radiation is absorbed or reflected by a surface, momentum transfer can produce pressure. This pressure is usually small in everyday situations but it is physically real. The effect does not require the wave to carry electric charge as a whole. Radiation pressure connects field energy transport with mechanical action on matter.
212. A perfectly absorbing surface and a perfectly reflecting surface are exposed to the same beam of electromagnetic radiation. The radiation pressure is greater on the reflecting surface because
ⓐ. reflection removes all energy from the wave before contact
ⓑ. absorbing surfaces cannot receive energy
ⓒ. reflection transfers a larger change in momentum
ⓓ. radiation pressure exists only for sound waves
Correct Answer: reflection transfers a larger change in momentum
Explanation: Radiation pressure comes from momentum transfer by electromagnetic waves. When radiation is absorbed, its momentum is transferred to the surface. When radiation is reflected back, the momentum changes direction, so the change in momentum is larger. A larger momentum change per unit time gives a larger pressure. This is a qualitative comparison of momentum transfer, not a claim that absorbing surfaces receive no energy.
213. A light sail in space is pushed very slightly by sunlight. The best explanation is that sunlight
ⓐ. carries no momentum but exerts radiation pressure
ⓑ. carries momentum only inside a material medium
ⓒ. produces pressure only if converted into sound
ⓓ. carries momentum and can exert radiation pressure
Correct Answer: carries momentum and can exert radiation pressure
Explanation: Sunlight is an electromagnetic wave and carries momentum. When it reflects from or is absorbed by a sail, it transfers momentum to the surface. This transfer produces a small radiation pressure. Since the effect occurs in space, it cannot depend on air pushing the sail. The example shows that electromagnetic waves can produce mechanical effects even though they are field waves.
214. Study the table and select the row that correctly describes a common property of electromagnetic waves.
| Row | Property | Correct description |
|---|
| P | Energy transport | Possible through vacuum |
| Q | Momentum | Never associated with electromagnetic waves |
| R | Medium | Always required for propagation |
| S | Radiation pressure | Possible only for sound waves |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row P
Explanation: Electromagnetic waves can transport energy through vacuum, so row P is correct. They can also carry momentum, so row Q is false. They do not require a material medium, making row R false. Radiation pressure is an electromagnetic radiation effect as well as a broader wave-momentum idea, so row S is not suitable. The correct row keeps the field nature of electromagnetic waves connected with energy transport.
215. A surface absorbs electromagnetic radiation continuously. The most suitable qualitative conclusion is that the surface may experience
ⓐ. no effect because the wave has no material medium
ⓑ. only frequency transfer without energy transfer
ⓒ. energy transfer and momentum transfer from the radiation
ⓓ. conduction current through vacuum as the only possible effect
Correct Answer: energy transfer and momentum transfer from the radiation
Explanation: Electromagnetic radiation carries both energy and momentum. When a surface absorbs the radiation, energy is transferred to the surface and may produce heating or other effects. The absorbed momentum can also produce radiation pressure. The effect is not based on conduction current flowing through vacuum. This is why electromagnetic radiation can interact with matter even after travelling through empty space.
216. The energy density stored in an electric field of magnitude \(E\) in vacuum is
ⓐ. \(u_E=\frac{1}{2}\varepsilon_0E^2\)
ⓑ. \(u_E=\frac{1}{2}\mu_0E^2\)
ⓒ. \(u_E=\frac{E^2}{2\varepsilon_0}\)
ⓓ. \(u_E=\frac{1}{2}\varepsilon_0B^2\)
Correct Answer: \(u_E=\frac{1}{2}\varepsilon_0E^2\)
Explanation: The electric field stores energy, and its energy density in vacuum is \(u_E=\frac{1}{2}\varepsilon_0E^2\). The square on \(E\) is important because energy density depends on the magnitude of the field, not on its sign. The constant involved is \(\varepsilon_0\), because this is the electric-field energy term. The expression \(\frac{B^2}{2\mu_0}\) belongs to magnetic-field energy density, not electric-field energy density. Confusing \(\varepsilon_0\) and \(\mu_0\) changes the physical field being described.
217. The magnetic energy density in a magnetic field of magnitude \(B\) is written as
ⓐ. \(u_B=\frac{1}{2}\varepsilon_0B^2\)
ⓑ. \(u_B=\frac{1}{2}\mu_0B^2\)
ⓒ. \(u_B=\frac{B^2}{2\mu_0}\)
ⓓ. \(u_B=\varepsilon_0cB\)
Correct Answer: \(u_B=\frac{B^2}{2\mu_0}\)
Explanation: Magnetic energy density is given by \(u_B=\frac{B^2}{2\mu_0}\). The field quantity in this expression is the magnetic field magnitude \(B\), and the constant in the denominator is \(\mu_0\). The form is not \(\frac{1}{2}\mu_0B^2\), because that has the constant placed incorrectly for vacuum formula. The electric-field energy density uses \(\varepsilon_0E^2\), while the magnetic-field energy density uses \(\frac{B^2}{\mu_0}\). The two formulas look similar in being quadratic, but the constants appear differently.
218. In a plane electromagnetic wave travelling through vacuum, the average electric and magnetic energy densities are
ⓐ. unequal, with the electric part always zero
ⓑ. equal in their average contributions
ⓒ. unequal, with the magnetic part always zero
ⓓ. unrelated because the fields have different phases
Correct Answer: equal in their average contributions
Explanation: A plane electromagnetic wave in vacuum carries energy in both its electric and magnetic fields. The average electric energy density and average magnetic energy density are equal. This equality is consistent with the relation \(E=cB\) and the speed relation \(c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}\). The fields have different units and perpendicular directions, but their average energy contributions match. The equality is about energy sharing, not about making \(E\) and \(B\) numerically equal.
219. A vacuum electromagnetic wave has instantaneous electric field \(E=120\,\text{V m}^{-1}\). Taking \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\), the instantaneous electric energy density is closest to
ⓐ. \(6.37\times10^{-8}\,\text{J m}^{-3}\)
ⓑ. \(1.27\times10^{-7}\,\text{J m}^{-3}\)
ⓒ. \(5.31\times10^{-14}\,\text{J m}^{-3}\)
ⓓ. \(1.44\times10^{4}\,\text{J m}^{-3}\)
Correct Answer: \(6.37\times10^{-8}\,\text{J m}^{-3}\)
Explanation: \( \textbf{Known values:} \) \(E=120\,\text{V m}^{-1}\) and \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\).
\( \textbf{Required quantity:} \) Electric energy density \(u_E\).
\( \textbf{Energy-density relation:} \)
\[
u_E=\frac{1}{2}\varepsilon_0E^2
\]
\( \textbf{Substitution:} \)
\[
u_E=\frac{1}{2}(8.85\times10^{-12})(120)^2
\]
\( \textbf{Square of field:} \)
\[
(120)^2=14400=1.44\times10^4
\]
\( \textbf{Multiplication:} \)
\[
(8.85\times10^{-12})(1.44\times10^4)=12.744\times10^{-8}
\]
\( \textbf{Apply factor \(\frac{1}{2}\):} \)
\[
u_E=6.372\times10^{-8}\,\text{J m}^{-3}
\]
\( \textbf{Rounded value:} \)
\[
u_E\approx6.37\times10^{-8}\,\text{J m}^{-3}
\]
\( \textbf{Final answer:} \) The electric energy density is \(6.37\times10^{-8}\,\text{J m}^{-3}\).
220. If the electric field amplitude of a plane electromagnetic wave is doubled, the average energy density associated with the electric field becomes
ⓐ. four times
ⓑ. two times
ⓒ. three times
ⓓ. one half
Correct Answer: four times
Explanation: Electric-field energy density depends on the square of the electric field. The relation is \(u_E=\frac{1}{2}\varepsilon_0E^2\). If \(E\) becomes \(2E\), then \(E^2\) becomes \((2E)^2=4E^2\). Therefore, the electric energy density becomes four times its earlier value. This square dependence is why field amplitude changes have a stronger effect on energy density than a simple linear change.
