Electromagnetic Waves MCQs with Answers – Part 6 (NEET Physics)

Explanation: \(\textbf{Superpose the two coherent electric fields:}\) At \(t=0\), \[ \cos0=1 \] Therefore, \[ \vec E_1=E_0\hat j \] \[ \vec E_2=-E_0\hat k \] The resultant electric field is \[ \vec E=E_0(\hat j-\hat k) \] \(\textbf{Determine the response of the first aerial:}\) The aerial is aligned along \[ \hat a=\frac{\hat j+\hat k}{\sqrt2} \] Its signed electric-field projection is \[ E_a=\vec E\cdot\hat a \] Substituting, \[ E_a=E_0(\hat j-\hat k)\cdot\frac{\hat j+\hat k}{\sqrt2} \] \[ E_a=\frac{E_0}{\sqrt2}(1-1) \] \[ E_a=0 \] \(\textbf{Determine the response of the second aerial:}\) The second aerial is aligned along \[ \hat b=\frac{\hat j-\hat k}{\sqrt2} \] Therefore, \[ E_b=\vec E\cdot\hat b \] \[ E_b=E_0(\hat j-\hat k)\cdot\frac{\hat j-\hat k}{\sqrt2} \] \[ E_b=\frac{E_0}{\sqrt2}(1+1) \] \[ E_b=\sqrt2E_0 \] \(\textbf{Reconstruct the magnetic field:}\) Both waves propagate along positive \(x\). Therefore, \[ \vec B=\frac1c(\hat i\times\vec E) \] Substituting, \[ \vec B=\frac{E_0}{c}[\hat i\times(\hat j-\hat k)] \] \[ \vec B=\frac{E_0}{c}(\hat k+\hat j) \] Its magnitude is \[ B=\frac{\sqrt2E_0}{c} \] \(\textbf{Determine the total electromagnetic energy density:}\) The full electric-field magnitude satisfies \[ E^2=2E_0^2 \] For the resultant co-propagating plane wave, \[ u=\varepsilon_0E^2 \] Therefore, \[ u=2\varepsilon_0E_0^2 \] \(\textbf{Physical interpretation:}\) The resultant electromagnetic field is nonzero. However, the first aerial is perpendicular to its instantaneous electric-field direction. In fact, because the two electric-field components have the same time dependence, that aerial remains insensitive throughout the oscillation. The second aerial is aligned along the resultant electric-field direction and can detect a strong signal. The absence of one receiver's response therefore does not imply zero electromagnetic energy density. This example distinguishes field superposition from the orientation-dependent response of a particular detector. \(\textbf{Final answer:}\) \[ \boxed{E_a=0,\qquad E_b=\sqrt2E_0,\qquad u=2\varepsilon_0E_0^2} \]

Explanation: \(\textbf{Determine the capacitor's displacement current:}\) The signed charge is \[ Q(t)=Q_0\cos(\omega t) \] For the ideal capacitor, \[ I_d=\frac{dQ}{dt} \] Differentiating, \[ I_d(t)=-\omega Q_0\sin(\omega t) \] \(\textbf{Determine the magnetic flux through the pickup coil:}\) The problem specifies \[ \Phi_B(t)=MI_d(t) \] Therefore, \[ \Phi_B(t)=-M\omega Q_0\sin(\omega t) \] \(\textbf{Determine the induced emf:}\) Faraday's law gives \[ \mathcal E=-\frac{d\Phi_B}{dt} \] Differentiating the specified flux expression, \[ \frac{d\Phi_B}{dt}=-M\omega^2Q_0\cos(\omega t) \] Therefore, \[ \mathcal E(t)=+M\omega^2Q_0\cos(\omega t) \] \(\textbf{Evaluate the initial instant:}\) At \(t=0\), \[ \sin0=0,\qquad\cos0=1 \] Thus, \[ I_d(0)=0 \] and \[ \mathcal E(0)=+M\omega^2Q_0 \] The displacement current is zero, but it is changing at its maximum rate. Therefore, the pickup-coil magnetic flux is changing at its maximum rate. \(\textbf{Evaluate one-quarter period later:}\) At \[ t=\frac T4, \] the phase is \[ \omega t=\frac{\pi}{2} \] Therefore, \[ I_d(T/4)=-\omega Q_0 \] The current has its largest negative magnitude. However, \[ \mathcal E(T/4)=M\omega^2Q_0\cos\frac{\pi}{2} \] \[ \mathcal E(T/4)=0 \] \(\textbf{Physical interpretation:}\) The displacement current is the rate of change of capacitor charge. The magnetic flux linked with the pickup coil is proportional to that current under the specified coupling condition. The induced emf is the negative rate of change of that magnetic flux. Consequently, the pickup-coil emf need not be greatest when the displacement-current magnitude is greatest. At a displacement-current extremum, the linked flux is instantaneously stationary and the induced emf vanishes. At a displacement-current zero crossing, the linked flux can change most rapidly. The problem combines the Ampere–Maxwell displacement-current concept with Faraday induction through an explicitly stated coupling relation. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} t=0:\quad&I_d=0,\quad\mathcal E=+M\omega^2Q_0\\ t=T/4:\quad&I_d=-\omega Q_0,\quad\mathcal E=0 \end{aligned}} \]

Explanation: \(\textbf{Use electromagnetic-energy accounting:}\) The energy initially directed at the target is \[ U_i=0.30\,\text{J} \] Of this, the absorbed and transmitted energies are \[ U_a=0.12\,\text{J},\qquad U_t=0.06\,\text{J} \] With the stipulated absence of other losses, energy conservation gives \[ U_i=U_a+U_t+U_r \] Therefore, the reflected energy is \[ U_r=0.30-0.12-0.06 \] \[ U_r=0.12\,\text{J} \] \(\textbf{Determine the reflection fraction:}\) The fraction of the incident electromagnetic energy returned towards the transmitter is \[ R=\frac{U_r}{U_i} \] Substituting, \[ R=\frac{0.12}{0.30}=0.40=40\% \] This energy fraction does not determine the echo time by itself. \(\textbf{Determine the radar propagation distance:}\) The target is \(450\,\text{m}\) from the transmitter. The pulse first travels to the target and the reflected pulse then returns over the same distance. The total travelled distance is therefore \[ s_{\text{round trip}}=2(450)=900\,\text{m} \] \(\textbf{Calculate the earliest echo arrival:}\) Electromagnetic waves propagate through the specified air path at \[ v=3.0\times10^8\,\text{m s}^{-1} \] The return time is \[ t=\frac{s_{\text{round trip}}}{v} \] \[ t=\frac{900}{3.0\times10^8} \] \[ t=3.0\times10^{-6}\,\text{s}=3.0\,\mu\text{s} \] \(\textbf{Physical check:}\) Using only the outward distance would underestimate the echo time by a factor of two. Using only the absorbed energy would underestimate the reflected energy, since the transmitted portion must also be subtracted. Reflection is the feature exploited by radar; its signal amplitude and return delay describe different aspects of the same event. \(\textbf{Final answer:}\) \[ \boxed{R=40\%,\qquad t_{\text{echo}}=3.0\,\mu\text{s}} \]

Explanation: \(\textbf{Identify the electric-force contribution:}\) The charge is positive. The wave electric field is \[ \vec E=E_0\hat j \] Therefore, the electric force is \[ \vec F_E=qE_0\hat j \] It points along positive \(y\). \(\textbf{Determine the magnetic-force direction:}\) The charge velocity is \[ \vec v=v\hat i \] The wave magnetic field is \[ \vec B=\frac{E_0}{c}\hat k \] Therefore, \[ \vec v\times\vec B =v\hat i\times\frac{E_0}{c}\hat k \] Using \[ \hat i\times\hat k=-\hat j, \] we obtain \[ \vec v\times\vec B=-\frac{vE_0}{c}\hat j \] The magnetic force therefore opposes the electric force. \(\textbf{Construct the net force:}\) \[ \vec F=q(\vec E+\vec v\times\vec B) \] Substituting, \[ \vec F=q\left(E_0\hat j-\frac{vE_0}{c}\hat j\right) \] Factoring, \[ \vec F=qE_0\left(1-\frac vc\right)\hat j \] \(\textbf{Examine the proposed cancellation:}\) Exact cancellation would require \[ 1-\frac vc=0 \] Therefore, \[ \frac vc=1 \] \[ v=c \] However, the problem specifies \[ 0\lt v\lt c \] Thus, \[ 1-\frac vc\gt 0 \] The net force cannot vanish within the permitted speed range. \(\textbf{Physical interpretation:}\) For motion along the propagation direction, the magnetic force opposes the wave electric force. Its magnitude increases with the charge's speed. However, for every permitted speed below \(c\), the magnetic contribution remains smaller than the electric contribution. Therefore, the net transverse force remains along positive \(y\). The equality \(E=cB\) is essential to this conclusion. The student has correctly identified the opposing force directions but incorrectly assumed that a subluminal speed can produce exact cancellation. This demonstrates how the linked electric and magnetic amplitudes of an electromagnetic wave constrain the force experienced by a moving charged particle. \(\textbf{Final answer:}\) \[ \boxed{\vec F=qE_0\left(1-\frac vc\right)\hat j,\qquad |\vec F|\gt 0\ \text{for}\ 0\lt v\lt c} \]

Explanation: \(\textbf{Distinguish radar echo time from transmission time:}\) A radar echo travels from the transmitter to the target and then returns. For target distance \(R\), the propagation time is \[ \Delta t=\frac{2R}{c} \] The relevant interval is the difference between the echo reception time and the emission time of its original pulse. \(\textbf{Test the first possible emission time:}\) If the echo at \(11.0\,\mu\text{s}\) originated from the pulse emitted at \(t=0\), \[ \Delta t_1=11.0-0 \] \[ \Delta t_1=11.0\,\mu\text{s} \] Therefore, \[ R_1=\frac{c\Delta t_1}{2} \] Substituting, \[ R_1=\frac{(3.0\times10^8)(11.0\times10^{-6})}{2} \] \[ R_1=1650\,\text{m} \] \(\textbf{Test the later possible emission time:}\) If the echo originated from the pulse emitted at \(8.0\,\mu\text{s}\), \[ \Delta t_2=11.0-8.0 \] \[ \Delta t_2=3.0\,\mu\text{s} \] The corresponding range would be \[ R_2=\frac{(3.0\times10^8)(3.0\times10^{-6})}{2} \] \[ R_2=450\,\text{m} \] \(\textbf{Apply the independent distance constraint:}\) The target is known to lie between \(1000\) and \(2000\,\text{m}\). The \(450\,\text{m}\) possibility is therefore excluded. The \(1650\,\text{m}\) result lies within the permitted range. \(\textbf{Check the second echo:}\) A stationary target at \(1650\,\text{m}\) gives a round-trip delay of \[ \Delta t=\frac{2(1650)}{3.0\times10^8} \] \[ \Delta t=11.0\,\mu\text{s} \] The pulse emitted at \(t=8.0\,\mu\text{s}\) should therefore return at \[ t_{\mathrm{return}}=8.0+11.0 \] \[ t_{\mathrm{return}}=19.0\,\mu\text{s} \] This agrees with the second recorded echo. \(\textbf{Physical interpretation:}\) A periodic radar transmission can make one echo reception time compatible with more than one possible pulse of origin. The independent distance constraint removes this ambiguity. The second echo then supplies a consistency check using the pulse repetition interval. The radar uses reflected microwave radiation to obtain distance information. Neither the target distance nor the pulse repetition interval changes the microwave propagation speed in vacuum. \(\textbf{Final answer:}\) \[ \boxed{R=1650\,\text{m},\qquad \Delta t_{\mathrm{round\ trip}}=11.0\,\mu\text{s}} \]

Explanation: \(\textbf{Determine visible transmission:}\) The incident visible power is \[ P_V=120\,\text{W} \] The wall transmits \(90\%\). Therefore, \[ P_{V,\mathrm{transmitted}}=0.90(120) \] \[ P_{V,\mathrm{transmitted}}=108\,\text{W} \] \(\textbf{Determine infrared transmission:}\) The infrared power incident on the wall from inside is \[ P_{\mathrm{IR}}=80\,\text{W} \] The wall transmits \(20\%\). Therefore, \[ P_{\mathrm{IR},\mathrm{transmitted}}=0.20(80) \] \[ P_{\mathrm{IR},\mathrm{transmitted}}=16\,\text{W} \] Statement I is correct. \(\textbf{Interpret the different spectral responses:}\) The wall transmits a much greater fraction of incident visible radiation than infrared radiation. Under the stated no-reflection condition, the remaining fractions are absorbed. Therefore, the wall absorbs \(10\%\) of the incident visible power and \(80\%\) of the incident infrared power. Statement II is correct. The experiment demonstrates wavelength-dependent interaction between electromagnetic radiation and matter. \(\textbf{Examine the temperature claim:}\) Visible light and infrared radiation are both electromagnetic waves. They propagate at the same speed when freely travelling through vacuum. However, equal vacuum speed does not imply equal absorption by a particular material. It also does not require the wall temperature to remain unchanged. Statement III is incorrect. \(\textbf{Assess the equilibrium-temperature inference:}\) The power-transmission measurements do not specify all subsequent energy exchanges. The absorbing wall may transfer energy to the enclosure or surroundings. Its own temperature and thermal-emission properties can affect those exchanges. Therefore, the measurements alone cannot determine the final equilibrium temperature. Statement IV is correct. \(\textbf{Physical interpretation:}\) The greenhouse-effect illustration involves different interactions of incoming radiation and outgoing thermal radiation. It does not require visible and infrared waves to travel at different speeds in vacuum. Nor does the selective absorption measurement, by itself, establish a complete thermal equilibrium. The correct combination is Statements I, II and IV.

Explanation: Gamma rays are high-frequency electromagnetic radiation. Their production is characteristically associated with nuclear transitions. They are not themselves streams of radioactive nuclei. Gamma radiation can interact with matter and produce ionisation. Such interactions can damage biological material, making controlled gamma irradiation useful for sterilisation. The effectiveness of treatment depends on sufficient radiation reaching the material requiring sterilisation. Therefore, the penetration and absorption properties of the packaging and contents matter. A sealed package need not prevent all gamma radiation from reaching its contents. However, receiving gamma radiation is not the same process as becoming a radioactive source. The incident electromagnetic radiation can transfer energy to matter. That energy transfer does not, by itself, establish the creation of unstable radioactive nuclei in the treated material. The external source can therefore be removed after the treatment. Suitable radiation measurements and facility procedures remain important for safety. The given detector check is an additional observation rather than proof that all irradiated objects must emit persistent radiation. The reasoning distinguishes the external source, the electromagnetic radiation emitted by that source, and the material that receives the radiation. It also distinguishes a controlled radiation application from an unjustified conclusion about the treated material's radioactive properties.

Explanation: \(\textbf{Identify what remains unchanged:}\) The battery voltage and plate separation remain constant. For an ideal parallel-plate capacitor, \[ E=\frac Vd \] Therefore, \[ E=\frac{100}{1.0\times10^{-3}} \] \[ E=1.0\times10^5\,\text{V m}^{-1} \] The field magnitude inside the changing plate projection remains constant. \(\textbf{Evaluate the fixed inner surface:}\) Path P has radius \(2.0\,\text{cm}\). It remains inside the plate projection throughout the expansion. Its enclosed electric flux is \[ \Phi_{EP}=E\pi r_P^2 \] Both \(E\) and \(r_P\) are constant. Consequently, \[ \Delta\Phi_{EP}=0 \] The average signed displacement current is \[ I_{dP,\mathrm{avg}}=\varepsilon_0\frac{\Delta\Phi_{EP}}{\Delta t} \] \[ I_{dP,\mathrm{avg}}=0 \] \(\textbf{Evaluate the complete flux enclosed by Q:}\) Path Q has radius \(7.0\,\text{cm}\), larger than the plate radius throughout the motion. It encloses the complete field-filled plate projection. Thus, \[ \Phi_{EQ}=E\pi R^2 \] The flux change is \[ \Delta\Phi_{EQ}=E\pi(R_f^2-R_i^2) \] Substituting, \[ \Delta\Phi_{EQ}=10^5\pi[(0.060)^2-(0.040)^2] \] \[ \Delta\Phi_{EQ}=10^5\pi(0.0020) \] \[ \Delta\Phi_{EQ}=200\pi\,\text{V m} \] \(\textbf{Determine Q's average displacement current:}\) The time interval is \[ \Delta t=2.0\times10^{-3}\,\text{s} \] Therefore, \[ I_{dQ,\mathrm{avg}}=\varepsilon_0\frac{\Delta\Phi_{EQ}}{\Delta t} \] \[ I_{dQ,\mathrm{avg}} =\frac{(8.85\times10^{-12})(200\pi)}{2.0\times10^{-3}} \] \[ I_{dQ,\mathrm{avg}}\approx+2.78\times10^{-6}\,\text{A} \] \[ I_{dQ,\mathrm{avg}}\approx+2.78\,\mu\text{A} \] \(\textbf{Compare stored electrical energies:}\) At fixed plate separation, \[ C=\frac{\varepsilon_0\pi R^2}{d} \] At constant voltage, \[ U=\frac12CV^2 \] Therefore, \[ U\propto R^2 \] The energy ratio is \[ \frac{U_f}{U_i}=\left(\frac{R_f}{R_i}\right)^2 \] \[ \frac{U_f}{U_i}=\left(\frac64\right)^2 \] \[ \frac{U_f}{U_i}=\frac94 \] \(\textbf{Physical interpretation:}\) The fixed inner measuring surface encloses an unchanged electric flux. The larger surface encloses a changing total flux because the field-filled plate area increases. The capacitor draws additional charge from the battery despite its constant voltage. Thus, zero displacement current through a particular fixed inner surface does not require zero displacement current through the full changing-flux region. The result also depends on the voltage-controlled condition; an isolated expanding capacitor would behave differently. \(\textbf{Final answer:}\) \[ \boxed{I_{dP,\mathrm{avg}}=0,\qquad I_{dQ,\mathrm{avg}}=+2.78\,\mu\text{A},\qquad \frac{U_f}{U_i}=\frac94} \]

Explanation: \(\textbf{Identify what each filter transmits:}\) Filter P transmits red and infrared radiation. It blocks the green component. Filter Q transmits green and infrared radiation. It blocks the red component. For a spectral component to emerge from both passive filters, it must be transmitted by each. \(\textbf{Track the visible red component:}\) The incident red power is \[ P_R=40\,\text{mW} \] Filter P transmits the red radiation. However, Filter Q blocks it. Therefore, \[ P_{R,\mathrm{out}}=0 \] \(\textbf{Track the visible green component:}\) The incident green power is \[ P_G=60\,\text{mW} \] Filter P blocks this component before it reaches Q. Therefore, \[ P_{G,\mathrm{out}}=0 \] \(\textbf{Track the infrared component:}\) The incident infrared power is \[ P_{\mathrm{IR}}=100\,\text{mW} \] Both filters transmit infrared radiation completely. Consequently, \[ P_{\mathrm{IR,out}}=100\,\text{mW} \] \(\textbf{Determine the final spectral composition:}\) The total emerging power is \[ P_{\mathrm{out}}=P_{R,\mathrm{out}}+P_{G,\mathrm{out}}+P_{\mathrm{IR,out}} \] Substituting, \[ P_{\mathrm{out}}=0+0+100 \] \[ P_{\mathrm{out}}=100\,\text{mW} \] \(\textbf{Interpret the observations:}\) The emerging radiation contains no visible component from the specified source. It therefore does not provide visible red or green illumination to the unaided eye. However, it still contains infrared radiation. A suitable infrared detector can register this emerging beam. The absence of visible illumination must not be interpreted as the absence of all electromagnetic radiation. The filters have not converted visible radiation into infrared radiation. They have simply removed the specified visible components while transmitting the original infrared component. \(\textbf{Final answer:}\) \[ \boxed{P_{\mathrm{visible,out}}=0,\qquad P_{\mathrm{IR,out}}=100\,\text{mW}} \]

Explanation: \(\textbf{Identify the electric force:}\) The particle has positive charge \(q\). The wave electric field is \[ \vec E=E_0\hat j \] Therefore, \[ \vec F_E=qE_0\hat j \] The electric force points along positive \(y\). \(\textbf{Determine the magnetic force:}\) The particle velocity is \[ \vec v=\frac{3c}{5}\hat j \] The magnetic field is \[ \vec B=\frac{E_0}{c}\hat k \] The magnetic contribution is \[ \vec F_B=q(\vec v\times\vec B) \] Substituting, \[ \vec F_B=q\left(\frac{3c}{5}\hat j\times\frac{E_0}{c}\hat k\right) \] Since \[ \hat j\times\hat k=\hat i, \] we obtain \[ \vec F_B=\frac35qE_0\hat i \] \(\textbf{Construct the resultant electromagnetic force:}\) \[ \vec F=\vec F_E+\vec F_B \] Therefore, \[ \vec F=qE_0\hat j+\frac35qE_0\hat i \] \[ \vec F=qE_0\left(\frac35\hat i+\hat j\right) \] \(\textbf{Determine the instantaneous power delivered to the charge:}\) The mechanical power delivered by the electromagnetic force is \[ P=\vec F\cdot\vec v \] Substituting, \[ P=qE_0\left(\frac35\hat i+\hat j\right)\cdot\frac{3c}{5}\hat j \] The \(x\)-directed magnetic contribution is perpendicular to the particle velocity. Therefore, \[ P=qE_0\frac{3c}{5} \] \[ P=\frac35qE_0c \] \(\textbf{Check the energy-transfer mechanism:}\) A magnetic force is always perpendicular to the instantaneous particle velocity. Therefore, it contributes no instantaneous mechanical power. Here, the magnetic force changes the direction of the net force relative to the electric-field direction. However, only the component parallel to velocity contributes to instantaneous work. The electric field is parallel to the positive-\(y\) velocity, so the electromagnetic force performs positive work. The result integrates the transverse field arrangement of the electromagnetic wave with the distinct mechanical effects of its electric and magnetic components. \(\textbf{Final answer:}\) \[ \boxed{\vec F=qE_0\left(\frac35\hat i+\hat j\right),\qquad P=\frac35qE_0c} \]

Explanation: \(\textbf{Separate selectivity from absolute acceptance limits:}\) The selectivity ratio compares radiation reaching the target with radiation reaching healthy tissue. A higher selectivity ratio is favourable under the given definition. However, the plan must also satisfy the two independent acceptance limits. \(\textbf{Calculate selectivity for Plan P:}\) The target receives \(60\%\) of the incident beam power. The specified healthy-tissue region receives \(15\%\). Therefore, \[ S_P=\frac{0.60}{0.15} \] \[ S_P=4 \] \(\textbf{Calculate selectivity for Plan Q:}\) The target receives \(40\%\) of the incident power. Healthy tissue receives \(8\%\). Thus, \[ S_Q=\frac{0.40}{0.08} \] \[ S_Q=5 \] Plan Q has the greater selectivity ratio. \(\textbf{Check the target-power requirement:}\) The target must receive at least \(35\%\). For P, \[ 60\%\ge35\% \] For Q, \[ 40\%\ge35\% \] Both satisfy the target-power requirement. \(\textbf{Check the healthy-tissue limit:}\) Healthy tissue must receive no more than \(10\%\). For P, \[ 15\%\gt 10\% \] Therefore, P fails. For Q, \[ 8\%\le10\% \] Therefore, Q passes. \(\textbf{Interpret the result:}\) Plan P delivers more radiation to the target, but it also delivers more to the specified healthy-tissue region. A larger target fraction alone does not make it the acceptable choice. Plan Q satisfies the target minimum, satisfies the healthy-tissue maximum and has the greater specified selectivity ratio. Gamma radiation can interact with biological matter, which explains both its potential medical usefulness and the importance of controlling unwanted exposure. The comparison is an idealised assessment of the stated radiation-power conditions, not a complete clinical treatment calculation. The numerical criteria, rather than an assumption that more radiation is always preferable, determine the accepted plan. \(\textbf{Final answer:}\) \[ \boxed{S_P=4,\qquad S_Q=5} \]

Explanation: \(\textbf{Determine the spatial wavelength:}\) The given frequency is \[ f=50\,\text{MHz}=5.0\times10^7\,\text{Hz} \] In vacuum, \[ c=f\lambda \] Therefore, \[ \lambda=\frac cf=\frac{3.0\times10^8}{5.0\times10^7} \] \[ \lambda=6.0\,\text{m} \] The wave period is \[ T=\frac1f=20\,\text{ns} \] \(\textbf{Convert the measurement position and time into phase:}\) For the stated wave, \[ \phi=kx-\omega t=2\pi\left(\frac x\lambda-\frac tT\right) \] At the selected time, \[ \frac tT=\frac{15}{20}=\frac34 \] For detector P, \[ \frac{x_P}{\lambda}=\frac{1.5}{6.0}=\frac14 \] Therefore, \[ \phi_P=2\pi\left(\frac14-\frac34\right)=-\pi \] For detector Q, \[ \frac{x_Q}{\lambda}=\frac{3.0}{6.0}=\frac12 \] Therefore, \[ \phi_Q=2\pi\left(\frac12-\frac34\right)=-\frac{\pi}{2} \] \(\textbf{Evaluate the signed electric fields:}\) At P, \[ \vec E_P=120\hat j\cos(-\pi)=-120\hat j\,\text{V m}^{-1} \] At Q, \[ \vec E_Q=120\hat j\cos(-\pi/2)=\vec0 \] \(\textbf{Use the magnetic-field amplitude and direction:}\) In a forward-propagating vacuum plane wave, \[ \vec B=\frac1c(\hat i\times\vec E) \] At P, \[ \vec B_P=\frac1{3.0\times10^8}\left[\hat i\times(-120\hat j)\right] \] \[ \vec B_P=-4.0\times10^{-7}\hat k\,\text{T} \] At Q, the electric and magnetic fields oscillate in phase and both vanish: \[ \vec B_Q=\vec0 \] \(\textbf{Physical checks:}\) The field signs at P reverse together, so \(\vec E_P\times\vec B_P\) still points along positive \(x\). At Q, the common sinusoidal phase gives a simultaneous zero of both transverse fields. A time shift and a position shift must both be included; ignoring either gives the wrong phase. \(\textbf{Final answer:}\) \[ \boxed{\vec E_P=-120\hat j\,\text{V m}^{-1},\quad\vec B_P=-4.0\times10^{-7}\hat k\,\text{T},\quad \vec E_Q=\vec B_Q=\vec0} \]

Explanation: \(\textbf{Recover Wave Q's magnetic field:}\) Both waves travel along positive \(z\): \[ \hat n=\hat k \] For Wave Q, \[ \vec E_Q=80\hat j \] The corresponding magnetic field is \[ \vec B_Q=\frac1c(\hat k\times80\hat j) \] Using \[ \hat k\times\hat j=-\hat i, \] we obtain \[ \vec B_Q=-\frac{80}{c}\hat i \] \(\textbf{Superpose the electric fields:}\) \[ \vec E_{\mathrm{total}}=\vec E_P+\vec E_Q \] \[ \vec E_{\mathrm{total}}=60\hat i+80\hat j \] Its magnitude is \[ E=\sqrt{60^2+80^2} \] \[ E=\sqrt{10000} \] \[ E=100\,\text{V m}^{-1} \] \(\textbf{Superpose the magnetic fields:}\) \[ \vec B_{\mathrm{total}}=\vec B_P+\vec B_Q \] \[ \vec B_{\mathrm{total}}=\frac1c(-80\hat i+60\hat j) \] Its magnitude is \[ B=\frac1c\sqrt{(-80)^2+60^2} \] \[ B=\frac{100}{c} \] \(\textbf{Determine the cross-product direction:}\) \[ \vec E_{\mathrm{total}}\times\vec B_{\mathrm{total}} =\frac1c(60\hat i+80\hat j)\times(-80\hat i+60\hat j) \] Only the cross terms contribute: \[ \vec E_{\mathrm{total}}\times\vec B_{\mathrm{total}} =\frac1c[3600(\hat i\times\hat j)-6400(\hat j\times\hat i)] \] Using \[ \hat i\times\hat j=\hat k,\qquad \hat j\times\hat i=-\hat k, \] we obtain \[ \vec E_{\mathrm{total}}\times\vec B_{\mathrm{total}} =\frac{10000}{c}\hat k \] The resultant energy-flow direction is positive \(z\). \(\textbf{Physical interpretation:}\) The electric fields are perpendicular at the selected event. Their magnitudes therefore combine vectorially, not by direct addition or subtraction. The corresponding magnetic components are also perpendicular and have the same resultant magnitude divided by \(c\). The two waves may have different frequencies, but each propagates in the same direction through vacuum. Their instantaneous superposition remains compatible with the common propagation direction. The relative field orientation need not be fixed over time when the frequencies differ. This comparison requires reconstructing the signed magnetic field before superposing the waves. \(\textbf{Final answer:}\) \[ \boxed{E=100\,\text{V m}^{-1},\qquad B=\frac{100}{c}\,\text{T},\qquad \vec E\times\vec B\parallel+\hat k} \]

Explanation: \(\textbf{Identify the two detection mechanisms:}\) The straight aerial measures the electric-field component along \(y\). The conducting loop responds to changing magnetic flux through its area. Although both devices receive information from the same wave, their measured quantities are different. \(\textbf{Determine the straight-aerial reading:}\) The electric-field component is \[ E_y(t)=E_0\cos(\omega t) \] At \(t=0\), \[ E_y(0)=E_0\cos0 \] \[ E_y(0)=+E_0 \] At \(t=T/4\), \[ \omega t=\frac{\pi}{2} \] Therefore, \[ E_y(T/4)=E_0\cos\frac{\pi}{2} \] \[ E_y(T/4)=0 \] \(\textbf{Construct the magnetic flux through the loop:}\) The magnetic field is directed along the loop's chosen positive normal. Its flux is \[ \Phi_B(t)=AB_z(t) \] Therefore, \[ \Phi_B(t)=\frac{AE_0}{c}\cos(\omega t) \] \(\textbf{Determine the induced emf:}\) By Faraday's law, \[ \mathcal E=-\frac{d\Phi_B}{dt} \] Differentiating, \[ \mathcal E(t)=+\frac{A\omega E_0}{c}\sin(\omega t) \] \(\textbf{Evaluate the induced emf at the two instants:}\) At \(t=0\), \[ \mathcal E(0)=\frac{A\omega E_0}{c}\sin0 \] \[ \mathcal E(0)=0 \] At \(t=T/4\), \[ \mathcal E(T/4)=\frac{A\omega E_0}{c}\sin\frac{\pi}{2} \] \[ \mathcal E(T/4)=+\frac{A\omega E_0}{c} \] \(\textbf{Interpret the phase difference between detector responses:}\) At the initial instant, the electric and magnetic fields are at positive maxima. However, the magnetic flux is instantaneously stationary, so the loop emf vanishes. One-quarter period later, the electric and magnetic fields pass through zero. Their rates of change are then greatest in magnitude. The loop consequently registers a maximum induced emf while the straight electric-field probe reads zero. Thus, two receivers detecting the same electromagnetic wave can produce signals with different phases because they respond to different field quantities. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} t=0:\quad&E_y=+E_0,\quad\mathcal E=0\\ t=T/4:\quad&E_y=0,\quad\mathcal E=+\frac{A\omega E_0}{c} \end{aligned}} \]

Explanation: \(\textbf{Identify the transmission requirement:}\) The total transmitted fraction must satisfy \[ T_{\mathrm{total}}\le0.05 \] For the stated passive materials, transmission fractions multiply when sheets are placed in series. \(\textbf{Check one sheet of each material:}\) For one P sheet followed by one Q sheet, \[ T_{PQ}=T_PT_Q \] Substituting, \[ T_{PQ}=(0.40)(0.25) \] \[ T_{PQ}=0.10=10\% \] This exceeds the \(5\%\) maximum. \(\textbf{Check two P sheets:}\) \[ T_{PP}=(0.40)^2 \] \[ T_{PP}=0.16=16\% \] This also fails. \(\textbf{Check two Q sheets:}\) \[ T_{QQ}=(0.25)^2 \] \[ T_{QQ}=0.0625=6.25\% \] This exceeds the limit and therefore fails. \(\textbf{Check the remaining arrangement:}\) One P sheet followed by two Q sheets gives \[ T_{PQQ}=(0.40)(0.25)^2 \] \[ T_{PQQ}=0.40(0.0625) \] \[ T_{PQQ}=0.025 \] \[ T_{PQQ}=2.5\% \] This satisfies the required maximum. \(\textbf{Determine the minimum sheet count among the given arrangements:}\) All three two-sheet arrangements fail. The specified three-sheet arrangement passes. Therefore, the smallest number of sheets among the proposed successful arrangements is three. \(\textbf{Physical interpretation:}\) Gamma radiation is electromagnetic radiation capable of interacting with matter. Suitable shielding reduces the transmitted intensity. The transmitted fraction depends on the properties and arrangement of the shielding materials. A smaller transmission fraction indicates a greater reduction in the emerging gamma-ray power under the stated measurement conditions. The problem requires checking the complete numerical acceptance condition, not merely selecting the material with the smaller one-sheet transmission. \(\textbf{Final answer:}\) \[ \boxed{T_{PQQ}=2.5\%} \]

Explanation: \(\textbf{Identify the relevant timing conditions:}\) The transmitted pulse begins at \(t=0\) and ends at \(t=2.0\,\mu\text{s}\). Its leading and trailing edges travel towards each reflector. The reflected edges return after the corresponding round-trip propagation time. The receiver becomes active at \(t=2.0\,\mu\text{s}\). \(\textbf{Calculate the nearer reflector's round-trip time:}\) The nearer target distance is \[ R_1=150\,\text{m} \] For radar reflection, \[ \Delta t_1=\frac{2R_1}{c} \] Substituting, \[ \Delta t_1=\frac{2(150)}{3.0\times10^8} \] \[ \Delta t_1=1.0\,\mu\text{s} \] \(\textbf{Determine the nearer echo interval:}\) The leading edge returns at \[ t_{1,\mathrm{start}}=1.0\,\mu\text{s} \] The trailing edge returns one pulse duration later: \[ t_{1,\mathrm{end}}=1.0+2.0 \] \[ t_{1,\mathrm{end}}=3.0\,\mu\text{s} \] Thus, the nearer echo occupies \[ 1.0\,\mu\text{s}\le t\le3.0\,\mu\text{s} \] Its first portion arrives before the receiving circuit becomes active. Only the later portion can be detected. \(\textbf{Calculate the farther reflector's round-trip time:}\) For \[ R_2=450\,\text{m}, \] the round-trip time is \[ \Delta t_2=\frac{2(450)}{3.0\times10^8} \] \[ \Delta t_2=3.0\,\mu\text{s} \] \(\textbf{Determine the farther echo interval:}\) The leading edge returns at \[ t_{2,\mathrm{start}}=3.0\,\mu\text{s} \] The trailing edge returns at \[ t_{2,\mathrm{end}}=3.0+2.0 \] \[ t_{2,\mathrm{end}}=5.0\,\mu\text{s} \] The entire farther echo arrives after the receiver becomes active. \(\textbf{Interpret the practical limitation:}\) A radar cannot necessarily receive its entire echo from a very nearby target while the transmitter occupies the receiving circuit. The nearer echo begins arriving during the transmission interval. Its late portion can still be detected. The farther echo begins after transmission ends and can be received completely under the stated assumptions. The effect arises from finite electromagnetic propagation time and finite pulse duration. It does not mean that the nearer reflected microwaves propagate at a different speed. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} \text{Near echo: }&1.0\text{–}3.0\,\mu\text{s}\quad(\text{partially received})\\ \text{Far echo: }&3.0\text{–}5.0\,\mu\text{s}\quad(\text{fully received}) \end{aligned}} \]

Explanation: Experiment P demonstrates that electromagnetic radiation can travel through vacuum. The transmitting aerial contains oscillating charges that generate electromagnetic fields. These fields propagate away from the source. A separated receiving aerial responds to the arriving electromagnetic wave. Conduction electrons from the transmitting aerial do not have to travel through vacuum to the receiver. Therefore, P's interpretation is incorrect. Experiment Q involves absorption of microwave radiation by matter. The sample receives energy from the incident electromagnetic field. Under the stated conditions, this transferred energy can increase the sample's internal energy and temperature. Q's interpretation is therefore correct. Experiment R involves fluorescence. The material absorbs ultraviolet radiation and subsequently emits visible radiation. This is a material-mediated emission process. The visible radiation need not have the original ultraviolet frequency. Both wavelengths have the same vacuum propagation speed, but this does not require their frequencies to be identical. Thus, R's interpretation is incorrect. Experiment S uses X-ray transmission through matter. Different specimen regions can absorb or transmit different fractions of the incident X-rays. A suitable detector can record these differences and provide information about the specimen. S's interpretation is correct. The four cases require distinguishing wave propagation, energy absorption, material re-emission and transmission-based imaging. The underlying radiation is electromagnetic in every case, but the source, material interaction and detector response must be analysed separately. Only Q and S have correct interpretations.

Explanation: \(\textbf{Reconstruct the opposing magnetic-field contributions:}\) Let the instantaneous signed electric-field components of P and Q be \(E_P\) and \(E_Q\). Their complete electric field is \[ \vec E=(E_P+E_Q)\hat j \] Wave P travels along positive \(x\). Its magnetic field is therefore \[ \vec B_P=\frac{E_P}{c}\hat k \] Wave Q travels along negative \(x\). Its magnetic field is \[ \vec B_Q=-\frac{E_Q}{c}\hat k \] Consequently, \[ \vec B=\frac{E_P-E_Q}{c}\hat k \] \(\textbf{Determine the instantaneous energy-flow vector:}\) The electromagnetic energy-flow vector is \[ \vec S=\frac1{\mu_0}\vec E\times\vec B \] Substituting, \[ \vec S=\frac{(E_P+E_Q)(E_P-E_Q)}{\mu_0c}\hat i \] Using \[ \frac1{\mu_0c}=c\varepsilon_0, \] we obtain \[ \vec S=c\varepsilon_0(E_P^2-E_Q^2)\hat i \] The mixed product cancels algebraically. \(\textbf{Average the net energy flow:}\) For a sinusoidal wave, \[ \langle E^2\rangle=\frac12E_0^2 \] Therefore, \[ \langle\vec S\rangle =\frac12c\varepsilon_0(E_{0P}^2-E_{0Q}^2)\hat i \] Substituting, \[ \langle\vec S\rangle =\frac12(3.0\times10^8)(8.85\times10^{-12})(80^2-40^2)\hat i \] \[ \langle\vec S\rangle =\frac12(2.655\times10^{-3})(6400-1600)\hat i \] \[ \langle\vec S\rangle=6.372\hat i\,\text{W m}^{-2} \] \(\textbf{Determine the total electromagnetic energy density:}\) For the complete superposed field, \[ u=\frac12\varepsilon_0(E_P+E_Q)^2 +\frac1{2\mu_0}\left(\frac{E_P-E_Q}{c}\right)^2 \] Using \(1/(\mu_0c^2)=\varepsilon_0\), \[ u=\frac12\varepsilon_0[(E_P+E_Q)^2+(E_P-E_Q)^2] \] Expanding and simplifying, \[ u=\varepsilon_0(E_P^2+E_Q^2) \] \(\textbf{Average over a complete oscillation:}\) \[ \langle u\rangle =\frac12\varepsilon_0(E_{0P}^2+E_{0Q}^2) \] Substituting, \[ \langle u\rangle =\frac12(8.85\times10^{-12})(6400+1600) \] \[ \langle u\rangle=3.54\times10^{-8}\,\text{J m}^{-3} \] \(\textbf{Interpret the result:}\) The two waves individually transport energy in opposite directions. Because P has the greater electric-field amplitude, it transports greater cycle-averaged intensity. The net average energy flow is consequently along positive \(x\). The complete electromagnetic energy density contains contributions from both waves. It is not determined by subtracting their energies. The final results are independent of the fixed relative phase because the mixed terms cancel in the complete energy-flow and energy-density expressions. This distinguishes the net transport of electromagnetic energy from the energy present in a region containing oppositely travelling waves. \(\textbf{Final answer:}\) \[ \boxed{\langle\vec S\rangle=+6.372\hat i\,\text{W m}^{-2},\qquad \langle u\rangle=3.54\times10^{-8}\,\text{J m}^{-3}} \]

Explanation: \(\textbf{Identify the carrier and amplitude envelope:}\) The electric-field expression contains a rapidly oscillating carrier, \[ \cos(\omega t) \] multiplied by the slowly varying factor \[ A(t)=1+\frac12\cos(\Omega t) \] Because \(\omega\gg\Omega\), the amplitude factor changes little during one carrier cycle. The detector's averaging interval is therefore suitable for comparing carrier-cycle-averaged intensities at different values of the envelope. \(\textbf{Determine the maximum envelope amplitude:}\) When \[ \cos(\Omega t)=+1, \] the factor becomes \[ A_{\max}=1+\frac12 \] \[ A_{\max}=\frac32 \] The corresponding carrier electric-field amplitude is \[ E_{\max}=\frac32E_c \] \(\textbf{Determine the minimum envelope amplitude:}\) When \[ \cos(\Omega t)=-1, \] the factor becomes \[ A_{\min}=1-\frac12 \] \[ A_{\min}=\frac12 \] Therefore, \[ E_{\min}=\frac12E_c \] \(\textbf{Relate cycle-averaged intensity to amplitude:}\) For a sinusoidal plane electromagnetic wave in vacuum, \[ I=\frac12c\varepsilon_0E_0^2 \] Over the stated short averaging intervals, the envelope is approximately constant. Thus, the measured intensities are proportional to the squares of the corresponding local carrier amplitudes. \(\textbf{Calculate the required ratio:}\) \[ \frac{I_{\max}}{I_{\min}} =\frac{E_{\max}^2}{E_{\min}^2} \] Substituting, \[ \frac{I_{\max}}{I_{\min}} =\frac{(3E_c/2)^2}{(E_c/2)^2} \] \[ \frac{I_{\max}}{I_{\min}}=\frac{9/4}{1/4} \] \[ \frac{I_{\max}}{I_{\min}}=9 \] \(\textbf{Interpret the communication mechanism:}\) The transmitter varies the electric-field amplitude envelope in a controlled manner. A suitable receiver can detect variations in the received signal associated with the information being transmitted. The carrier's specified oscillation factor remains \(\cos(\omega t)\). Changing its amplitude does not require electromagnetic propagation through vacuum to occur at a different speed. Nor does an instantaneous increase in electric-field amplitude imply that the carrier wavelength has decreased. The radiation remains electromagnetic and transverse. The intensity changes because energy transport depends on the square of the field amplitude. The experiment connects electromagnetic-wave amplitude, transported energy and radio communication without confusing amplitude variation with a change in vacuum wave speed. \(\textbf{Final answer:}\) \[ \boxed{I_{\max}:I_{\min}=9:1,\qquad E_{\max}:E_{\min}=3:1} \]

Explanation: \(\textbf{Identify the series-current constraint:}\) Both capacitors belong to the same series charging path. Therefore, their total displacement currents have equal magnitudes: \[ I_{dP}=I_{dQ}=8.0\,\text{mA} \] However, equal total currents do not require equal displacement-current densities. The plate areas are different. \(\textbf{Determine the enclosed current at each probe:}\) For a uniformly changing electric field within a circular plate projection, \[ I_{\mathrm{enc}}=I_d\frac{r^2}{R^2} \] At equal interior measuring radii, the enclosed current varies inversely with the square of the plate radius. Therefore, \[ \frac{I_{\mathrm{enc},P}}{I_{\mathrm{enc},Q}} =\frac{R_Q^2}{R_P^2} \] Substituting, \[ \frac{I_{\mathrm{enc},P}}{I_{\mathrm{enc},Q}} =\left(\frac{4.0}{2.0}\right)^2 \] \[ \frac{I_{\mathrm{enc},P}}{I_{\mathrm{enc},Q}}=4 \] \(\textbf{Compare the local magnetic fields:}\) For either concentric path, \[ B(2\pi r)=\mu_0I_{\mathrm{enc}} \] Both probes have the same radius. Consequently, \[ \frac{B_P}{B_Q} =\frac{I_{\mathrm{enc},P}}{I_{\mathrm{enc},Q}} \] \[ \frac{B_P}{B_Q}=4 \] \(\textbf{Determine the charge delivered to each capacitor:}\) The charging interval is \[ \Delta t=1.0\times10^{-6}\,\text{s} \] Therefore, \[ \Delta Q=I\Delta t \] \[ \Delta Q=(8.0\times10^{-3})(1.0\times10^{-6}) \] \[ \Delta Q=8.0\times10^{-9}\,\text{C} \] Each capacitor acquires charge magnitude \(8.0\,\text{nC}\). \(\textbf{Recover the final voltages:}\) Using \[ V=\frac QC, \] for capacitor P, \[ V_P=\frac{8.0\,\text{nC}}{2.0\,\text{nF}} \] \[ V_P=4.0\,\text{V} \] For capacitor Q, \[ V_Q=\frac{8.0\,\text{nC}}{8.0\,\text{nF}} \] \[ V_Q=1.0\,\text{V} \] \(\textbf{Physical interpretation:}\) The capacitors have equal series currents and acquire equal charge magnitudes. The smaller capacitance consequently develops the greater voltage. The smaller plate area also produces the greater changing electric-flux density for the same total displacement current. Thus, a probe enclosing the same small area in both capacitor gaps measures a greater magnetic field in P. The problem separates total series current, locally enclosed displacement current, charge and voltage. \(\textbf{Final answer:}\) \[ \boxed{B_P:B_Q=4:1,\qquad V_P=4.0\,\text{V},\qquad V_Q=1.0\,\text{V}} \]