Electromagnetic Waves MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Identify the common phase:}\) The electric and magnetic fields of the specified plane wave share the phase \[ \phi=kx-\omega t \] At Event P, \[ \phi_P=\frac{\pi}{3} \] Both corresponding field components are positive at this phase. \(\textbf{Calculate the phase change between events:}\) The spatial displacement is \[ \Delta x=\frac{\lambda}{6} \] The time difference is \[ \Delta t=\frac T4 \] The wave-number and angular-frequency relations are \[ k=\frac{2\pi}{\lambda},\qquad\omega=\frac{2\pi}{T} \] Therefore, \[ \Delta\phi=k\Delta x-\omega\Delta t \] Substituting, \[ \Delta\phi=\frac{2\pi}{\lambda}\frac{\lambda}{6}-\frac{2\pi}{T}\frac T4 \] \[ \Delta\phi=\frac{\pi}{3}-\frac{\pi}{2} \] \[ \Delta\phi=-\frac{\pi}{6} \] The phase at Event Q is \[ \phi_Q=\phi_P+\Delta\phi \] \[ \phi_Q=\frac{\pi}{3}-\frac{\pi}{6} \] \[ \phi_Q=\frac{\pi}{6} \] \(\textbf{Compare magnetic-field components:}\) At Event P, \[ B_P=B_0\sin\frac{\pi}{3} \] \[ B_P=\frac{\sqrt3}{2}B_0 \] At Event Q, \[ B_Q=B_0\sin\frac{\pi}{6} \] \[ B_Q=\frac12B_0 \] Therefore, \[ \frac{B_Q}{B_P}=\frac{B_0/2}{\sqrt3B_0/2} \] \[ \frac{B_Q}{B_P}=\frac1{\sqrt3} \] \(\textbf{Compare electromagnetic energy densities:}\) For a freely propagating plane wave in vacuum, \[ u=\varepsilon_0E^2 \] At Event P, \[ u_P=\varepsilon_0E_0^2\sin^2\frac{\pi}{3} \] \[ u_P=\frac34\varepsilon_0E_0^2 \] At Event Q, \[ u_Q=\varepsilon_0E_0^2\sin^2\frac{\pi}{6} \] \[ u_Q=\frac14\varepsilon_0E_0^2 \] Consequently, \[ \frac{u_Q}{u_P}=\frac{1/4}{3/4} \] \[ \frac{u_Q}{u_P}=\frac13 \] \(\textbf{Physical interpretation:}\) The spatial displacement increases the wave phase, while the later observation time decreases it. Both contributions must be included before evaluating the fields. The magnetic-field component changes in proportion to the sinusoidal factor. The energy density changes according to the square of that factor. Consequently, the magnetic-field ratio and energy-density ratio are different. Both events have positive field components, so no sign reversal occurs between them. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{B_Q}{B_P},\frac{u_Q}{u_P}\right)=\left(\frac1{\sqrt3},\frac13\right)} \]

Explanation: All regions of the electromagnetic spectrum share the same fundamental electromagnetic-wave nature. Radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays can all propagate through vacuum. Their vacuum propagation speed is the same fundamental constant \(c\). However, this does not require their frequencies to be equal. The wave relation is \[ c=f\lambda. \] For two radiations travelling through vacuum, a higher frequency corresponds to a shorter wavelength. Therefore, radiations with substantially different frequencies can have different wavelengths while maintaining identical propagation speeds. Their interactions with matter depend on the properties of the radiation and the material. A material may absorb one frequency band strongly while transmitting or reflecting another band more effectively. Such differences can arise from the physical response of the material to electromagnetic fields at different frequencies. The interactions are not determined solely by the radiation's speed in vacuum. Absorption also does not establish that either radiation is a mechanical wave. Both can remain electromagnetic despite producing different effects in matter. The student incorrectly treats a universal propagation property as though it determined every material interaction. The correct conclusion is that equal vacuum speed and different absorption behaviour are entirely compatible. This is one reason different electromagnetic-spectrum regions can have different practical applications despite obeying the same basic vacuum wave-speed law.

Explanation: \(\textbf{Interpret the graph slopes:}\) For a travelling electromagnetic wave, \[ v=\frac{\omega}{k} \] Therefore, the slope of an \(\omega\)-versus-\(k\) graph passing through the origin represents wave speed. From Graph P, \[ v_P=3.0\times10^8\,\text{m s}^{-1} \] From Graph Q, \[ v_Q=2.0\times10^8\,\text{m s}^{-1} \] Consequently, \[ \frac{v_P}{v_Q}=\frac32 \] \(\textbf{First comparison — Equal wave numbers:}\) The two selected waves have the same wave number \(k_*\). Using \[ \omega=vk \] we obtain \[ \omega_P=v_Pk_* \] \[ \omega_Q=v_Qk_* \] Therefore, \[ \frac{\omega_P}{\omega_Q}=\frac{v_P}{v_Q} \] \[ \frac{\omega_P}{\omega_Q}=\frac32 \] The time period satisfies \[ T=\frac{2\pi}{\omega} \] Thus, \[ \frac{T_P}{T_Q}=\frac{\omega_Q}{\omega_P} \] \[ \frac{T_P}{T_Q}=\frac23 \] \(\textbf{Second comparison — Equal angular frequencies:}\) The selected waves now have the same angular frequency \(\omega_*\). The wavelength is related to frequency and wave speed by \[ \lambda=\frac vf \] Since both angular frequencies are equal, their ordinary frequencies are equal. Therefore, \[ \frac{\lambda_P}{\lambda_Q}=\frac{v_P}{v_Q} \] \[ \frac{\lambda_P}{\lambda_Q}=\frac32 \] \(\textbf{Physical interpretation:}\) Equal wave number means equal wavelength, but it does not imply equal frequency when the propagation speeds differ. Equal angular frequency means equal time period, but it does not imply equal wavelength when the propagation speeds differ. In the first comparison, P has the greater frequency and hence the shorter period. In the second comparison, P has the greater speed and hence the longer wavelength. The same two graph slopes therefore lead to different ratios depending on the quantity held fixed. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{T_P}{T_Q},\frac{\lambda_P}{\lambda_Q}\right)=\left(\frac23,\frac32\right)} \]

Explanation: \(\textbf{Identify the radiation requirement:}\) The required vacuum wavelength is \[ \lambda=500\,\text{nm} \] Converting to SI units, \[ \lambda=500\times10^{-9}\,\text{m} \] \[ \lambda=5.0\times10^{-7}\,\text{m} \] The wave frequency is related to wavelength by \[ c=f\lambda \] Therefore, \[ f=\frac c\lambda \] \(\textbf{Calculate the required source frequency:}\) Substituting, \[ f=\frac{3.0\times10^8}{5.0\times10^{-7}} \] \[ f=0.60\times10^{15}\,\text{Hz} \] \[ f=6.0\times10^{14}\,\text{Hz} \] An oscillating charge emits electromagnetic radiation at the frequency of its harmonic oscillation. Therefore, the proposed source must operate at approximately \(6.0\times10^{14}\,\text{Hz}\). \(\textbf{Compare with the oscillator limit:}\) The maximum available oscillator frequency is \[ f_{\max}=1.0\times10^{11}\,\text{Hz} \] The required frequency factor is \[ \frac{f}{f_{\max}}=\frac{6.0\times10^{14}}{1.0\times10^{11}} \] \[ \frac{f}{f_{\max}}=6.0\times10^3 \] \[ \frac{f}{f_{\max}}=6000 \] \(\textbf{Check the shortest available wavelength:}\) At the oscillator's maximum frequency, \[ \lambda_{\min}=\frac c{f_{\max}} \] \[ \lambda_{\min}=\frac{3.0\times10^8}{1.0\times10^{11}} \] \[ \lambda_{\min}=3.0\times10^{-3}\,\text{m} \] \[ \lambda_{\min}=3.0\,\text{mm} \] This is much longer than the requested \(500\,\text{nm}\). \(\textbf{Physical interpretation:}\) Electromagnetic radiation produced by ordinary oscillatory circuits typically occupies much lower frequency regions than visible light. The source-frequency limitation explains why the experimental demonstration of electromagnetic waves was initially achieved using comparatively long-wavelength radiation. The difficulty is not that visible light and radio waves have different fundamental wave natures. Both are electromagnetic waves and propagate through vacuum at the same speed. The difficulty is producing the much higher oscillation frequency required for visible radiation using the specified electronic oscillator. \(\textbf{Final answer:}\) \[ \boxed{\frac{f_{\text{required}}}{f_{\max}}=6000} \]

Explanation: The student incorrectly identifies a zero instantaneous field value with the stopping of a travelling wave. For a sinusoidal plane electromagnetic wave travelling along positive \(x\), the fields may be written as \[ \vec E=E_0\hat j\sin(kx-\omega t) \] and \[ \vec B=B_0\hat k\sin(kx-\omega t). \] Both field vectors vanish when the common sinusoidal factor becomes zero. At such an event, the instantaneous electric and magnetic energy densities are also zero. The local electromagnetic energy-flow vector, proportional to \(\vec E\times\vec B\), consequently vanishes at that event. However, the travelling-wave phase remains \[ \phi=kx-\omega t. \] A point of fixed phase moves according to \[ kx-\omega t=\text{constant}. \] Therefore, \[ \frac{dx}{dt}=\frac{\omega}{k}=c. \] The phase disturbance continues travelling at the vacuum wave speed. A momentary zero of the field does not alter this propagation speed. At neighbouring positions or later instants, the fields generally have nonzero values. Furthermore, the time-averaged electromagnetic energy transport of the travelling wave remains nonzero. The direction of propagation is determined by the complete travelling-wave solution, not by dividing or normalising the zero vector product at one particular event. When the common sinusoidal factor changes sign, both field vectors reverse together. Their vector-product direction during the subsequent nonzero half-cycle remains the original positive \(x\)-direction. Thus, a momentary zero in instantaneous field-energy flow is compatible with continuous propagation of the electromagnetic wave.

Explanation: \(\textbf{Infer frequencies from the observed oscillations:}\) Frequency is the number of complete oscillations per unit time. For a wave completing \(N\) cycles in time \(\Delta t\), \[ f=\frac N{\Delta t} \] The common observation interval is \[ \Delta t=8.0\times10^{-9}\,\text{s} \] For Wave P, \[ f_P=\frac4{8.0\times10^{-9}} \] \[ f_P=5.0\times10^8\,\text{Hz} \] For Wave Q, \[ f_Q=\frac8{8.0\times10^{-9}} \] \[ f_Q=1.0\times10^9\,\text{Hz} \] For Wave R, \[ f_R=\frac2{8.0\times10^{-9}} \] \[ f_R=2.5\times10^8\,\text{Hz} \] Thus, the increasing frequency order is \[ f_R\lt f_P\lt f_Q \] \(\textbf{Determine the wavelength order:}\) All three waves propagate through vacuum at the same speed. Using \[ c=f\lambda \] we obtain \[ \lambda=\frac cf \] Wavelength is therefore inversely proportional to frequency. Consequently, the increasing wavelength order is the reverse of the increasing frequency order: \[ \lambda_Q\lt\lambda_P\lt\lambda_R \] \(\textbf{Check using numerical wavelengths:}\) Taking \(c=3.0\times10^8\,\text{m s}^{-1}\), \[ \lambda_P=\frac{3.0\times10^8}{5.0\times10^8}=0.60\,\text{m} \] \[ \lambda_Q=\frac{3.0\times10^8}{1.0\times10^9}=0.30\,\text{m} \] \[ \lambda_R=\frac{3.0\times10^8}{2.5\times10^8}=1.20\,\text{m} \] The numerical values confirm the required order. \(\textbf{Compare arrival times:}\) For each wave, the propagation time over distance \(L\) is \[ t=\frac Lc \] The distance and speed are identical for all three waves. Therefore, their propagation times are identical. Since they are emitted simultaneously, they also arrive simultaneously. The higher frequency of Q does not make it travel faster in vacuum. Similarly, the longer wavelength of R does not delay its arrival. The correct conclusion combines an inverse frequency–wavelength ordering with equal vacuum propagation times.

Explanation: \(\textbf{Use the measured energy-density equality:}\) For a sinusoidal plane electromagnetic wave in vacuum, the time-averaged total electromagnetic energy density is \[ \langle u\rangle=\frac12\varepsilon_0E_0^2 \] The two waves have equal average energy densities. Therefore, \[ \frac12\varepsilon_0E_{0P}^2=\frac12\varepsilon_0E_{0Q}^2 \] Canceling the common factors, \[ E_{0P}^2=E_{0Q}^2 \] Electric-field amplitudes are nonnegative magnitudes. Consequently, \[ E_{0P}=E_{0Q} \] \(\textbf{Infer the magnetic-field amplitudes:}\) For each freely propagating plane wave in vacuum, \[ E_0=cB_0 \] Therefore, \[ B_{0P}=\frac{E_{0P}}c \] and \[ B_{0Q}=\frac{E_{0Q}}c \] Since the electric-field amplitudes are equal, \[ B_{0P}=B_{0Q} \] Thus, equality of average electromagnetic energy density implies equality of both field amplitudes for the specified vacuum plane waves. \(\textbf{Apply the changed-frequency condition:}\) The frequencies satisfy \[ f_Q=4f_P \] For vacuum electromagnetic waves, \[ \lambda=\frac cf \] Therefore, \[ \frac{\lambda_Q}{\lambda_P}=\frac{f_P}{f_Q} \] \[ \frac{\lambda_Q}{\lambda_P}=\frac14 \] Hence, \[ \lambda_Q=\frac{\lambda_P}{4} \] \(\textbf{Physical interpretation:}\) The energy-density measurement constrains the field amplitudes. It does not require the two waves to have equal frequencies. Frequency controls the temporal and spatial repetition of the electromagnetic oscillations. For equal field amplitudes, changing the frequency does not introduce an independent frequency factor into the classical energy-density expression. Thus, Wave Q can have the same average energy density and field amplitudes as P while possessing one-fourth its vacuum wavelength. The distinction is between electromagnetic-field strength and oscillation frequency. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{E_{0Q}}{E_{0P}},\frac{B_{0Q}}{B_{0P}},\frac{\lambda_Q}{\lambda_P}\right)=\left(1,1,\frac14\right)} \]

Explanation: \(\textbf{Determine the average energy-transfer rate:}\) For a sinusoidal plane electromagnetic wave in vacuum, \[ \langle u\rangle=\frac12\varepsilon_0E_0^2 \] The average energy crossing a unit area per unit time equals the average energy density multiplied by the wave speed. Therefore, \[ \langle S\rangle=c\langle u\rangle \] \[ \langle S\rangle=\frac12c\varepsilon_0E_0^2 \] For a normally illuminated surface of area \(A\), the average rate of energy transfer is \[ P_{\text{av}}=\frac12c\varepsilon_0E_0^2A \] \(\textbf{Compare the two waves:}\) Both waves propagate through vacuum. Their electric-field amplitudes and illuminated areas are identical. Therefore, \[ P_{\text{av},P}=P_{\text{av},Q} \] The different frequencies do not independently alter this average rate when the field amplitudes are fixed. \(\textbf{Determine the emission durations:}\) The time period of Wave P is \[ T_P=\frac1f \] The time period of Wave Q is \[ T_Q=\frac1{2f} \] Thus, \[ T_Q=\frac{T_P}{2} \] Each wave is emitted for exactly 20 complete oscillations. Therefore, \[ \Delta t_P=20T_P \] and \[ \Delta t_Q=20T_Q \] Substituting the period relationship, \[ \Delta t_Q=10T_P \] Hence, \[ \frac{\Delta t_P}{\Delta t_Q}=\frac{20T_P}{10T_P} \] \[ \frac{\Delta t_P}{\Delta t_Q}=2 \] \(\textbf{Compare total absorbed energies:}\) Since both waves are completely absorbed, \[ U=P_{\text{av}}\Delta t \] Therefore, \[ \frac{U_P}{U_Q}=\frac{P_{\text{av},P}\Delta t_P}{P_{\text{av},Q}\Delta t_Q} \] The average powers are equal, giving \[ \frac{U_P}{U_Q}=\frac{\Delta t_P}{\Delta t_Q} \] \[ \frac{U_P}{U_Q}=2 \] \(\textbf{Physical interpretation:}\) The two waves transport energy at the same average rate while they are being emitted. However, 20 cycles of the lower-frequency wave last twice as long as 20 cycles of the higher-frequency wave. The longer emission interval allows P to transfer twice as much total energy. A fixed number of wave cycles does not correspond to a fixed duration when frequencies differ. The result follows from classical electromagnetic energy transport and does not require assigning photon energies to individual cycles. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{\text{av},P}}{P_{\text{av},Q}}=1,\qquad\frac{U_P}{U_Q}=2} \]

Explanation: A freely propagating plane electromagnetic wave in vacuum has several mutually compatible properties. Its electric and magnetic fields are perpendicular to each other and to the propagation direction. Their amplitudes satisfy \[ \frac{E_0}{B_0}=c. \] Their ordered vector product points along the propagation direction. However, these relationships describe a particular travelling electromagnetic-wave field configuration. They are not, individually or collectively, sufficient to prove that any two perpendicular fields constitute an electromagnetic wave. In the given laboratory arrangement, both fields are explicitly static. The electric-field magnitude and direction do not vary with time. Similarly, the magnetic-field magnitude and direction remain constant. The magnitude ratio happens to equal the vacuum wave speed numerically. The cross product also happens to point along positive \(x\). Nevertheless, there is no specified sinusoidal variation, travelling phase or propagating electromagnetic disturbance. A static electric field can be maintained by stationary charge distributions. A static magnetic field can be maintained by steady currents or other magnetostatic sources. Such independently maintained fields can coexist in perpendicular directions. The existence of \(\vec E\times\vec B\) as a mathematical vector does not automatically establish the existence of a freely propagating electromagnetic wave. The student has confused necessary field relationships for a vacuum plane wave with sufficient evidence of wave propagation. To identify a travelling plane electromagnetic wave, the observed fields must also exhibit appropriate spatial and temporal variations consistent with propagation. Therefore, the numerical magnitude ratio and right-hand-rule orientation alone do not justify the student's claim.

Explanation: Electromagnetic waves can be produced by accelerated electric charges. Oscillating currents in suitable transmitting conductors provide a source of accelerated charges and time-varying electromagnetic fields. The generated electromagnetic disturbances can propagate away from the source. Their propagation does not require a continuous conducting connection between the transmitter and receiver. For a harmonically oscillating source, the emitted electromagnetic-wave frequency is the same as the source oscillation frequency. The vacuum relationship between frequency and wavelength is \[ c=f\lambda. \] Therefore, different source frequencies can produce different electromagnetic wavelengths. A metre-wavelength wave has a lower frequency than a millimetre-wavelength wave when both are compared at the same propagation speed. However, the difference in frequency does not imply a difference in the fundamental electromagnetic nature of the waves. Nor does it require the shorter-wavelength wave to propagate faster. The historical experiments of Heinrich Hertz established the production and detection of electromagnetic waves in the comparatively long-wavelength region. Jagdish Chandra Bose subsequently produced and observed electromagnetic waves of much shorter, millimetre-scale wavelengths. These developments supported the understanding that electromagnetic radiation exists across different wavelength regions. They also provided experimental support for Maxwell's theoretical prediction of electromagnetic waves. The absence of a conducting connection between transmitting and receiving equipment is consistent with electromagnetic-wave propagation through space. The important physical inference is that electromagnetic radiation of different frequencies and wavelengths obeys the same basic propagation principles. Thus, both experiments can demonstrate electromagnetic waves even though their source frequencies and observed wavelengths differ.

Explanation: \(\textbf{Determine the duration of emission:}\) The transmitter emits a finite train containing \[ N=10 \] complete oscillations. The frequency is \[ f=5.0\times10^8\,\text{Hz} \] The time period of one oscillation is \[ T=\frac1f \] \[ T=\frac1{5.0\times10^8} \] \[ T=2.0\times10^{-9}\,\text{s} \] Therefore, the duration of the complete wave train is \[ \Delta t_{\text{pulse}}=NT \] \[ \Delta t_{\text{pulse}}=10(2.0\times10^{-9}) \] \[ \Delta t_{\text{pulse}}=2.0\times10^{-8}\,\text{s} \] \[ \Delta t_{\text{pulse}}=20\,\text{ns} \] \(\textbf{Determine the pulse length in the medium:}\) The transmitted wave retains its frequency and temporal pulse duration. However, its propagation speed in the medium is \[ v_m=2.0\times10^8\,\text{m s}^{-1} \] The spatial extent of the pulse is \[ L_m=v_m\Delta t_{\text{pulse}} \] Substituting, \[ L_m=(2.0\times10^8)(2.0\times10^{-8}) \] \[ L_m=4.0\,\text{m} \] The same result follows from the medium wavelength: \[ \lambda_m=\frac{v_m}{f} \] \[ \lambda_m=\frac{2.0\times10^8}{5.0\times10^8}=0.40\,\text{m} \] Thus, \[ L_m=10\lambda_m=4.0\,\text{m} \] \(\textbf{Compare propagation times:}\) The travel distance is \[ d=120\,\text{m} \] In vacuum, \[ t_0=\frac dc \] \[ t_0=\frac{120}{3.0\times10^8} \] \[ t_0=4.0\times10^{-7}\,\text{s} \] \[ t_0=400\,\text{ns} \] In the medium, \[ t_m=\frac d{v_m} \] \[ t_m=\frac{120}{2.0\times10^8} \] \[ t_m=6.0\times10^{-7}\,\text{s} \] \[ t_m=600\,\text{ns} \] The additional travel time is \[ \Delta t_{\text{delay}}=t_m-t_0 \] \[ \Delta t_{\text{delay}}=600-400 \] \[ \Delta t_{\text{delay}}=200\,\text{ns} \] \(\textbf{Physical interpretation:}\) The pulse contains the same number of cycles in both environments. Its duration is unchanged because the source frequency is unchanged. However, the smaller propagation speed produces a shorter wavelength and therefore a shorter spatial wave train. The propagation delay increases because the signal travels the specified distance more slowly. Pulse duration, spatial pulse length and travel time are three distinct quantities. \(\textbf{Final answer:}\) \[ \boxed{(L_m,\Delta t_{\text{delay}})=(4.0\,\text{m},200\,\text{ns})} \]

Explanation: \(\textbf{Interpret the energy-density oscillation:}\) For a sinusoidal plane electromagnetic wave in vacuum, \[ E=E_0\sin(kx-\omega t+\phi) \] The instantaneous total electromagnetic energy density is \[ u=\varepsilon_0E^2 \] Therefore, \[ u=\varepsilon_0E_0^2\sin^2(kx-\omega t+\phi) \] The squared sine function repeats twice during one complete oscillation of the electric field. Consequently, the interval between consecutive energy-density maxima is half the electromagnetic-wave period. \(\textbf{Recover the wave period:}\) The graph shows consecutive maxima separated by \[ \Delta t_u=5-1=4\,\text{ns} \] Therefore, \[ \frac T2=4\,\text{ns} \] \[ T=8\,\text{ns} \] \[ T=8.0\times10^{-9}\,\text{s} \] \(\textbf{Determine frequency and wavelength:}\) \[ f=\frac1T \] \[ f=\frac1{8.0\times10^{-9}} \] \[ f=1.25\times10^8\,\text{Hz} \] In vacuum, \[ c=f\lambda \] Therefore, \[ \lambda=\frac cf \] \[ \lambda=\frac{3.0\times10^8}{1.25\times10^8} \] \[ \lambda=2.4\,\text{m} \] \(\textbf{Recover the electric-field amplitude:}\) The graph provides the maximum total energy density, not its time average. Therefore, \[ u_{\max}=\varepsilon_0E_0^2 \] Rearranging, \[ E_0=\sqrt{\frac{u_{\max}}{\varepsilon_0}} \] Substituting, \[ E_0=\sqrt{\frac{8.85\times10^{-8}}{8.85\times10^{-12}}} \] \[ E_0=\sqrt{10^4} \] \[ E_0=100\,\text{V m}^{-1} \] \(\textbf{Determine magnetic-field amplitude:}\) For the freely propagating vacuum wave, \[ B_0=\frac{E_0}{c} \] \[ B_0=\frac{100}{3.0\times10^8} \] \[ B_0=3.33\times10^{-7}\,\text{T} \] \(\textbf{Physical check:}\) The electromagnetic energy density is never negative because it depends on squared field magnitudes. It reaches a maximum during both positive and negative field extrema. Therefore, the energy-density graph has twice the frequency of the signed electric-field oscillation. Treating the \(4\,\text{ns}\) separation as a complete field period would incorrectly halve the wavelength. \(\textbf{Final answer:}\) \[ \boxed{(\lambda,B_0)=(2.4\,\text{m},3.33\times10^{-7}\,\text{T})} \]

Explanation: Electromagnetic radiation can be produced by accelerating electric charges. When the switch is initially open, there is no steady conduction current through the circuit. Closing the switch changes the electrical conditions. During the transient, charges respond to the applied electric field and the current develops towards its steady value. This time-dependent process can produce changing electromagnetic fields. Suitable changing fields can generate electromagnetic disturbances that propagate away from the circuit. Therefore, radiation may be produced during the switching transient. After the transient has ended, the current is assumed to be steady. A steady current produces a magnetic field, but its mere existence does not imply continuous electromagnetic radiation. The current may remain nonzero while the associated macroscopic magnetic-field configuration remains time-independent. The original Ampere circuital law describes the magnetic field produced by steady conduction current. Maxwell's displacement-current correction accounts for additional magnetic-field contributions associated with changing electric flux. Neither law implies that every nonzero current must continuously radiate electromagnetic waves. A periodically oscillating current can act as a continuing radiation source because charges undergo repeated acceleration. A switching transient differs because the significant time variation occurs primarily while the circuit changes state. The radiation produced during the transient can continue propagating after the current has settled. Thus, the correct conclusion separates radiation associated with changing source conditions from the static magnetic field maintained by steady DC current.

Explanation: The total electric charge of the system is \[ Q_{\text{total}}=q+(-q)=0. \] This quantity remains zero throughout the motion. However, the total charge is not the only property relevant to electromagnetic radiation. The spatial arrangement and motion of the positive and negative charges also matter. An electric dipole consists of equal and opposite charges separated by a distance. Its electric dipole moment is directed from the negative charge towards the positive charge. For the stated fixed axis, its magnitude is \[ p(t)=qd(t). \] Substituting the separation, \[ p(t)=q[d_0+a\cos(\omega t)]. \] Therefore, \[ p(t)=qd_0+qa\cos(\omega t). \] The first term is constant, whereas the second term varies harmonically. The charges consequently form a time-varying electric dipole. Their oscillatory motion involves acceleration, allowing the system to act as an electromagnetic-radiation source. The harmonic component has angular frequency \(\omega\), so the corresponding emitted radiation can have frequency \(\omega/(2\pi)\). The static component of the dipole moment does not itself generate the oscillatory radiation. The fact that total charge is zero does not require the electromagnetic disturbances produced by the accelerating positive and negative charges to cancel everywhere. Radiation also does not require charged particles to escape from the system and travel to the detector. The energy can propagate through space in electromagnetic fields. Thus, an electrically neutral oscillating dipole can radiate even though the algebraic sum of its charges remains zero.

Explanation: \(\textbf{Identify the controlled conditions:}\) Both waves have the same frequency and electric-field amplitude. The propagation speeds differ according to \[ v_Q=\frac{v_P}{2} \] The appropriate electric-to-magnetic amplitude relation is \[ \frac{E_0}{B_0}=v \] Therefore, \[ B_0=\frac{E_0}{v} \] \(\textbf{Compare magnetic-field amplitudes:}\) For Wave P, \[ B_{0P}=\frac{E_0}{v_P} \] For Wave Q, \[ B_{0Q}=\frac{E_0}{v_Q} \] Substituting \(v_Q=v_P/2\), \[ B_{0Q}=\frac{E_0}{v_P/2} \] \[ B_{0Q}=\frac{2E_0}{v_P} \] Therefore, \[ \frac{B_{0Q}}{B_{0P}}=2 \] \(\textbf{Compare wavelengths:}\) The wavelength relation is \[ \lambda=\frac vf \] Since frequency is the same, \[ \frac{\lambda_Q}{\lambda_P}=\frac{v_Q}{v_P} \] \[ \frac{\lambda_Q}{\lambda_P}=\frac12 \] \(\textbf{Determine the product relationship:}\) Combining the two wave relations, \[ B_0\lambda=\frac{E_0}{v}\frac vf \] The propagation speed cancels: \[ B_0\lambda=\frac{E_0}{f} \] Since both \(E_0\) and \(f\) are unchanged, this product is identical for the two waves. Therefore, \[ \frac{B_{0Q}\lambda_Q}{B_{0P}\lambda_P}=1 \] \(\textbf{Physical interpretation:}\) At fixed electric-field amplitude, a lower propagation speed corresponds to a larger magnetic-field amplitude under the stated plane-wave relation. At fixed frequency, the same lower speed corresponds to a shorter wavelength. The two changes compensate in the product \(B_0\lambda\). The equality of this product follows from the specified controlled quantities, not from a universal rule requiring all electromagnetic waves to have the same magnetic-field amplitude or wavelength. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{B_{0Q}}{B_{0P}},\frac{\lambda_Q}{\lambda_P},\frac{B_{0Q}\lambda_Q}{B_{0P}\lambda_P}\right)=\left(2,\frac12,1\right)} \]

Explanation: \(\textbf{Relate instantaneous energy density to field magnitude:}\) For a freely propagating plane electromagnetic wave in vacuum, the electric and magnetic energy densities are equal. Therefore, the instantaneous total energy density is \[ u=\varepsilon_0E^2 \] The electric-field component at the specified phase is \[ E=E_0\cos\phi \] Given \[ \cos\phi=-\frac12 \] we have \[ E=-\frac{E_0}{2} \] Thus, \[ u=\varepsilon_0\frac{E_0^2}{4} \] \(\textbf{Recover electric-field amplitude:}\) Rearranging, \[ E_0^2=\frac{4u}{\varepsilon_0} \] Substituting, \[ E_0^2=\frac{4(2.2125\times10^{-8})}{8.85\times10^{-12}} \] \[ E_0^2=\frac{8.85\times10^{-8}}{8.85\times10^{-12}} \] \[ E_0^2=10^4 \] Therefore, \[ E_0=100\,\text{V m}^{-1} \] \(\textbf{Determine the signed instantaneous electric field:}\) \[ E=-\frac{100}{2} \] \[ E=-50\,\text{V m}^{-1} \] \(\textbf{Determine the magnetic-field component:}\) For the corresponding in-phase plane-wave components, \[ B=\frac Ec \] Therefore, \[ B=\frac{-50}{3.0\times10^8} \] \[ B=-1.67\times10^{-7}\,\text{T} \] The negative sign is essential because the common phase factor is negative at the selected instant. \(\textbf{Calculate time-averaged energy density:}\) For a sinusoidal plane wave, \[ \langle u\rangle=\frac12\varepsilon_0E_0^2 \] Substituting, \[ \langle u\rangle=\frac12(8.85\times10^{-12})(100)^2 \] \[ \langle u\rangle=\frac12(8.85\times10^{-8}) \] \[ \langle u\rangle=4.425\times10^{-8}\,\text{J m}^{-3} \] \(\textbf{Physical check:}\) The instantaneous field magnitude is half its amplitude, so the instantaneous energy density is one-fourth its maximum. The time-averaged energy density is one-half the maximum. Therefore, the time average is twice the measured instantaneous energy density at the selected phase. The instantaneous magnetic component is negative, but its energy-density contribution remains positive because it depends on \(B^2\). \(\textbf{Final answer:}\) \[ \boxed{\left(B,\langle u\rangle\right)=\left(-1.67\times10^{-7}\,\text{T},4.425\times10^{-8}\,\text{J m}^{-3}\right)} \]

Explanation: Electromagnetic radiation can be produced by accelerating electric charges. Sustained harmonic motion is one particular kind of accelerated motion. For System P, the charge oscillates with a well-defined frequency \(f\). The associated harmonic electromagnetic radiation can have the same frequency as this source oscillation. Consequently, its vacuum wavelength is related to that frequency through \(c=f\lambda\). System Q also undergoes acceleration, even though the acceleration occurs only during a brief interval. Periodicity is not a necessary condition for the generation of electromagnetic radiation. A non-periodic change in charge motion can produce an electromagnetic disturbance that propagates away from the source. However, the duration \(\tau\) of a single acceleration episode is not automatically the period of a sinusoidal oscillation. Therefore, assigning an exact single radiation frequency \(1/\tau\) is not justified. Unlike sustained harmonic motion, a non-periodic event does not generally define one unique repeating cycle. The emitted disturbance need not be a monochromatic sinusoidal wave. When the charge subsequently moves with constant velocity, its acceleration becomes zero. That later uniform motion does not require continued generation of radiation. Nevertheless, any electromagnetic disturbance already emitted during the acceleration interval can continue propagating through vacuum. The distinction is between the condition for radiation generation and the additional periodicity required to associate the radiation with one well-defined harmonic frequency. Thus, both systems can radiate, but only the stated harmonic source supplies a definite single oscillation frequency from the information given.

Explanation: \(\textbf{Extract the propagation vector:}\) The wave phase is \[ \phi=6\pi x+8\pi y-3\pi\times10^9t \] The spatial part of the phase is \[ \vec k\cdot\vec r=6\pi x+8\pi y \] Therefore, the wave vector is \[ \vec k=6\pi\hat i+8\pi\hat j \] Its magnitude is \[ k=\sqrt{(6\pi)^2+(8\pi)^2} \] \[ k=\sqrt{36\pi^2+64\pi^2} \] \[ k=10\pi\,\text{rad m}^{-1} \] \(\textbf{Determine the propagation direction:}\) The wave propagates along its positive wave-vector direction because the phase has the form \(\vec k\cdot\vec r-\omega t\). The propagation unit vector is \[ \hat n=\frac{\vec k}{k} \] \[ \hat n=\frac{6\pi\hat i+8\pi\hat j}{10\pi} \] \[ \hat n=\frac{3\hat i+4\hat j}{5} \] \(\textbf{Calculate wavelength:}\) \[ \lambda=\frac{2\pi}{k} \] \[ \lambda=\frac{2\pi}{10\pi} \] \[ \lambda=0.20\,\text{m} \] \(\textbf{Recover electric-field direction from the magnetic field:}\) For a freely propagating plane wave in vacuum, \[ \vec B=\frac1c(\hat n\times\vec E) \] Equivalently, \[ \vec E=c(\vec B\times\hat n) \] The magnetic field is \[ \vec B=5.0\times10^{-7}\hat k\cos\phi \] Substituting, \[ \vec E=(3.0\times10^8)(5.0\times10^{-7})\left[\hat k\times\frac{3\hat i+4\hat j}{5}\right]\cos\phi \] \[ \vec E=150\left[\frac{3(\hat k\times\hat i)+4(\hat k\times\hat j)}5\right]\cos\phi \] Using \[ \hat k\times\hat i=\hat j,\qquad\hat k\times\hat j=-\hat i, \] we obtain \[ \vec E=150\left(\frac{3\hat j-4\hat i}{5}\right)\cos\phi \] \[ \vec E=(-120\hat i+90\hat j)\cos\phi\,\text{V m}^{-1} \] \(\textbf{Verify the wave speed:}\) The angular frequency is \[ \omega=3\pi\times10^9\,\text{rad s}^{-1} \] Therefore, \[ v=\frac{\omega}{k} \] \[ v=\frac{3\pi\times10^9}{10\pi} \] \[ v=3.0\times10^8\,\text{m s}^{-1} \] The result agrees with the vacuum propagation speed. \(\textbf{Check field geometry:}\) The electric-field amplitude is \[ E_0=\sqrt{(-120)^2+90^2}=150\,\text{V m}^{-1} \] This satisfies \[E_0=cB_0\] The electric-field vector is perpendicular to the wave vector because \[(-120)(3)+90(4)=0\] The fields share the same phase, and their ordered cross product points along \((3\hat i+4\hat j)/5\). Thus, the reconstructed electric field and wavelength satisfy all the required plane-wave conditions. \(\textbf{Final answer:}\) \[ \boxed{\vec E=(-120\hat i+90\hat j)\cos\phi\,\text{V m}^{-1},\qquad\lambda=0.20\,\text{m}} \]

Explanation: \(\textbf{Identify the invariant frequency:}\) The electromagnetic wave travels through two stationary transparent layers. Under the stated conditions, its frequency remains unchanged at the interfaces. Therefore, \[ f=1.0\times10^8\,\text{Hz} \] The time period is \[ T=\frac1f \] \[ T=\frac1{1.0\times10^8} \] \[ T=1.0\times10^{-8}\,\text{s} \] \[ T=10\,\text{ns} \] \(\textbf{Travel time through Layer P:}\) The thickness is \[ L_P=1.0\,\text{m} \] The propagation speed is \[ v_P=2.0\times10^8\,\text{m s}^{-1} \] Therefore, \[ t_P=\frac{L_P}{v_P} \] \[ t_P=\frac{1.0}{2.0\times10^8} \] \[ t_P=5.0\times10^{-9}\,\text{s} \] \[ t_P=5.0\,\text{ns} \] \(\textbf{Travel time through Layer Q:}\) Similarly, \[ t_Q=\frac{L_Q}{v_Q} \] \[ t_Q=\frac{1.0}{1.0\times10^8} \] \[ t_Q=1.0\times10^{-8}\,\text{s} \] \[ t_Q=10\,\text{ns} \] \(\textbf{Total propagation delay:}\) The total travel time is \[ t_{\text{total}}=t_P+t_Q \] \[ t_{\text{total}}=5.0+10 \] \[ t_{\text{total}}=15\,\text{ns} \] \(\textbf{Determine the phase difference:}\) The angular frequency is \[ \omega=2\pi f \] The phase accumulated during propagation is equivalent to the angular frequency multiplied by travel time. Its magnitude is \[ \Delta\phi=\omega t_{\text{total}} \] Substituting, \[ \Delta\phi=2\pi(1.0\times10^8)(15\times10^{-9}) \] \[ \Delta\phi=3\pi \] A phase difference of \(3\pi\) is equivalent to \(\pi\) modulo \(2\pi\). Therefore, the exit oscillation is opposite in sign to the entrance oscillation at the same clock time, apart from any amplitude change. \(\textbf{Cross-check using wavelengths:}\) In Layer P, \[ \lambda_P=\frac{v_P}{f}=2.0\,\text{m} \] The \(1.0\,\text{m}\) thickness corresponds to half a wavelength. In Layer Q, \[ \lambda_Q=\frac{v_Q}{f}=1.0\,\text{m} \] Its thickness corresponds to one complete wavelength. The combined phase advance is \[ \Delta\phi=2\pi\left(\frac12+1\right)=3\pi \] which agrees with the travel-time method. \(\textbf{Final answer:}\) \[ \boxed{(t_{\text{total}},\Delta\phi)=(15\,\text{ns},\pi\ \text{mod}\ 2\pi)} \]

Explanation: The electric and magnetic fields of a freely propagating sinusoidal plane electromagnetic wave oscillate in phase at the same position. However, the two probes in this experiment are not located at the same position. The electric-field probe is at \(x=0\), while the magnetic-field probe is at \(x=\lambda/4\). An appropriate electric-field expression for the wave is \[ E_y(x,t)=E_0\cos(kx-\omega t). \] The corresponding magnetic field is \[ B_z(x,t)=B_0\cos(kx-\omega t). \] Both field components therefore share exactly the same travelling-wave phase. At the electric-field probe, \[ E_y(0,t)=E_0\cos(-\omega t) \] \[ E_y(0,t)=E_0\cos(\omega t) \] At the magnetic-field probe, \[ B_z(\lambda/4,t)=B_0\cos\left(k\frac{\lambda}{4}-\omega t\right) \] Using \(k=2\pi/\lambda\), \[ B_z(\lambda/4,t)=B_0\cos\left(\frac{\pi}{2}-\omega t\right) \] Therefore, \[ B_z(\lambda/4,t)=B_0\sin(\omega t). \] This reproduces the two measured time traces. Their apparent quarter-cycle difference arises entirely from the spatial separation of the probes. If both probes were placed at \(x=0\), their corresponding signed field components would be proportional to the same cosine function. Thus, a phase comparison between measurements at different positions must account for the travelling-wave spatial phase. A phase difference between two separated probe signals is not automatically an intrinsic phase difference between the local electric and magnetic fields. The student's conclusion is therefore incorrect.