Electromagnetic Waves MCQs with Answers – Part 5 (NEET Physics)

Explanation: Statement I is correct because the ultraviolet-blocking sheet is initially placed before the fluorescent marker. It can absorb the lamp's ultraviolet radiation before that radiation reaches the material. Consequently, the excitation responsible for visible fluorescence can be strongly reduced. Statement II is correct because moving the sheet beyond the marker changes which process occurs first. The lamp's ultraviolet radiation can then reach and excite the fluorescent material. The marker may emit visible light after absorbing suitable ultraviolet radiation. Because the sheet is stated to transmit this visible light, the visible detector can register radiation emitted by the marker. Statement III is incorrect because thermal infrared emission is not automatically determined by whether the sheet is before or after the marker. The problem explicitly holds the marker's temperature and infrared emissive properties constant. Its own thermal infrared emission therefore need not increase during the change. The infrared detector can continue responding to suitable emitted infrared radiation in either arrangement. Statement IV is correct because fluorescence involves absorption and subsequent emission by matter. The position of the ultraviolet-blocking sheet determines whether ultraviolet excitation can occur before the radiation reaches the visible detector. The resulting change in the visible signal is therefore associated with the order of material interactions. It is not evidence that ultraviolet and visible light have different speeds in vacuum. Both are electromagnetic radiation. The experiment combines source wavelength, material absorption, new visible emission, infrared emission and detector sensitivity. The three signals must be traced according to their distinct physical origins rather than treated as one unchanged beam passing through passive filters. Statements I, II and IV are correct, while Statement III incorrectly links unchanged thermal emission to the sheet's position.

Explanation: \(\textbf{Identify the changing quantity:}\) The electric-field magnitude is constant, but its direction changes. Electric flux through a fixed plane depends on the component of the field normal to that plane. Therefore, constant field magnitude does not guarantee constant electric flux. \(\textbf{Construct the electric flux:}\) For a uniform field, \[ \Phi_E=A\vec E\cdot\hat n \] Substituting the field and surface normal, \[ \Phi_E=AE_0[\hat i\cos(\omega t)+\hat j\sin(\omega t)] \cdot\left(\frac35\hat i+\frac45\hat j\right) \] Therefore, \[ \Phi_E=\frac{AE_0}{5}[3\cos(\omega t)+4\sin(\omega t)] \] \(\textbf{Differentiate to obtain displacement current:}\) \[ I_d=\varepsilon_0\frac{d\Phi_E}{dt} \] Hence, \[ I_d=\frac{\varepsilon_0AE_0\omega}{5} [-3\sin(\omega t)+4\cos(\omega t)] \] The common numerical factor is \[ \varepsilon_0AE_0\omega =(9.0\times10^{-12})(0.020)(2.0\times10^4)(50) \] \[ \varepsilon_0AE_0\omega=1.8\times10^{-7}\,\text{A} \] \(\textbf{Evaluate the first instant:}\) At \(t=0\), \[ I_d(0)=\frac{1.8\times10^{-7}}5[0+4] \] \[ I_d(0)=+1.44\times10^{-7}\,\text{A} \] \(\textbf{Evaluate the second instant:}\) At \[ t=\frac{\pi}{2\omega}, \] the phase is \[ \omega t=\frac{\pi}{2} \] Therefore, \[ I_d\left(\frac{\pi}{2\omega}\right) =\frac{1.8\times10^{-7}}5[-3+0] \] \[ I_d\left(\frac{\pi}{2\omega}\right) =-1.08\times10^{-7}\,\text{A} \] \(\textbf{Interpret the signs:}\) The positive first value indicates increasing flux relative to the selected normal. At the second instant, the normal component of the rotating electric field is decreasing. The signed displacement current consequently reverses. The result depends on the changing projection of the field, even though its full magnitude remains unchanged. This distinguishes vector-field variation from variation of field strength alone. \(\textbf{Final answer:}\) \[ \boxed{I_d(0)=+1.44\times10^{-7}\,\text{A},\qquad I_d\left(\frac{\pi}{2\omega}\right)=-1.08\times10^{-7}\,\text{A}} \]

Explanation: \(\textbf{Express the travelling-wave fields at the observation point:}\) A compatible pair of vacuum fields is \[ \vec E(0,t)=E_0\hat j\cos(\omega t) \] \[ \vec B(0,t)=B_0\hat k\cos(\omega t) \] where \[ B_0=\frac{E_0}{c} \] The electric and magnetic fields oscillate in phase. \(\textbf{Evaluate Observation P:}\) At \[ t=\frac T4, \] the phase is \[ \omega t=\frac{\pi}{2} \] Therefore, \[ \vec E_P=0,\qquad\vec B_P=0 \] The instantaneous total electromagnetic energy density is \[ u=\frac12\varepsilon_0E^2+\frac{B^2}{2\mu_0} \] Consequently, \[ u_P=0 \] \(\textbf{Evaluate Observation Q:}\) At \[ t=\frac T2, \] the phase is \[ \omega t=\pi \] Thus, \[ \vec E_Q=-E_0\hat j \] \[ \vec B_Q=-B_0\hat k \] Their magnitudes equal their initial maximum magnitudes. Using \[ \frac{B_0^2}{\mu_0}=\varepsilon_0E_0^2, \] the energy density is \[ u_Q=\frac12\varepsilon_0E_0^2+\frac12\varepsilon_0E_0^2 \] \[ u_Q=\varepsilon_0E_0^2 \] \(\textbf{Check the direction of propagation:}\) At Q, both field vectors have reversed. However, \[ (-\hat j)\times(-\hat k)=+\hat i \] Therefore, the direction of electromagnetic energy propagation remains positive \(x\). The sign reversal of both oscillating fields does not reverse the direction of the travelling wave. At P, the instantaneous field energy density and energy flow vanish locally, but the electromagnetic disturbance continues propagating. This distinguishes oscillation of the fields from reversal of propagation direction.

Explanation: \(\textbf{Distinguish the energy-density period from the wave period:}\) At a fixed position, a sinusoidal plane wave can have \[ E(t)=E_0\cos(\omega t) \] Its instantaneous total energy density is \[ u(t)=\varepsilon_0E_0^2\cos^2(\omega t) \] The squared cosine reaches a maximum twice during one complete field-oscillation cycle. Therefore, \[ T_u=\frac T2 \] \(\textbf{Recover the wave period:}\) The measured interval between consecutive energy-density maxima is \[ T_u=1.0\,\text{ns} \] Hence, \[ T=2T_u \] \[ T=2.0\,\text{ns} \] \(\textbf{Determine frequency and wavelength:}\) \[ f=\frac1T \] \[ f=\frac1{2.0\times10^{-9}} \] \[ f=5.0\times10^8\,\text{Hz} \] The vacuum wavelength is \[ \lambda=\frac cf \] \[ \lambda=\frac{3.0\times10^8}{5.0\times10^8} \] \[ \lambda=0.60\,\text{m} \] \(\textbf{Recover the electric-field amplitude:}\) At an energy-density maximum, \[ u_{\max}=\varepsilon_0E_0^2 \] Therefore, \[ E_0=\sqrt{\frac{u_{\max}}{\varepsilon_0}} \] Substituting, \[ E_0=\sqrt{\frac{8.85\times10^{-8}}{8.85\times10^{-12}}} \] \[ E_0=\sqrt{10^4} \] \[ E_0=100\,\text{V m}^{-1} \] \(\textbf{Determine the magnetic-field orientation:}\) At the first maximum, the electric field points along positive \(y\). The wave travels along positive \(x\). Therefore, \[ \vec B=\frac1c(\hat i\times\vec E) \] The magnetic field points along positive \(z\), and its magnitude is \[ B_0=\frac{100}{3.0\times10^8} \] \[ B_0=3.33\times10^{-7}\,\text{T} \] \(\textbf{Physical interpretation:}\) The energy-density graph repeats more frequently than the signed electric-field oscillation. A detector measuring only field energy cannot distinguish positive from negative field maxima. Consequently, treating the interval between energy-density maxima as the full electric-field period would produce an incorrect wavelength. The field direction at the specified positive electric maximum supplies the separate information needed to reconstruct magnetic-field orientation. \(\textbf{Final answer:}\) \[ \boxed{E_0=100\,\text{V m}^{-1},\qquad\lambda=0.60\,\text{m},\qquad\vec B=+3.33\times10^{-7}\hat k\,\text{T}} \]

Explanation: Statement I is correct because reflection and thermal emission are distinct radiation processes. The reflected microwave pulse retains the characteristic frequency of the incident microwave radiation under the stated stationary-target conditions. The absorbed portion transfers energy to the object. The warmed object can subsequently emit infrared radiation, which occupies a different part of the electromagnetic spectrum. The emitted infrared radiation is not required to have the original microwave wavelength. Statement II is correct because the two signals need not be generated at the same instant. A promptly reflected microwave pulse can begin its return journey without waiting for substantial heating. Infrared emission from the warmed object may arise after the absorption and heating process. A difference in emission times can therefore produce a difference in reception times. Statement III is incorrect because ordinary electromagnetic propagation through vacuum does not automatically convert microwaves into infrared radiation. The intervening material absorbs energy and can subsequently emit new radiation. The two-stage material process must not be confused with a freely travelling wave changing frequency spontaneously. Statement IV is correct because the returned microwave signal depends on how the object reflects the incident radiation. The infrared signal depends on radiation emitted by the object and can reflect its thermal state. The two detectors therefore examine different consequences of electromagnetic interaction with the same object. The signals may differ in wavelength, timing and intensity while sharing the same vacuum propagation speed. The correct combination is Statements I, II and IV.

Explanation: \(\textbf{Determine enclosed displacement current:}\) The current density is uniform over the circular plate projection. For a concentric path of radius \(r\le R\), \[ I_{\text{enc}}=I_d\frac{\pi r^2}{\pi R^2} \] Therefore, \[ I_{\text{enc}}=I_d\frac{r^2}{R^2} \] \(\textbf{Calculate the current enclosed at two-thirds radius:}\) At \[ r=\frac{2R}{3}, \] the enclosed fraction is \[ \frac{I_{\text{enc}}}{I_d}=\left(\frac23\right)^2 \] \[ \frac{I_{\text{enc}}}{I_d}=\frac49 \] Therefore, \[ I_{\text{enc}}=\frac49(12\,\text{mA}) \] \[ I_{\text{enc}}=\frac{16}{3}\,\text{mA} \] \(\textbf{Relate magnetic field to enclosed current:}\) For the concentric circular path, \[ B(2\pi r)=\mu_0I_{\text{enc}} \] Substituting the enclosed-current expression, \[ B(2\pi r)=\mu_0I_d\frac{r^2}{R^2} \] Hence, \[ B=\frac{\mu_0I_dr}{2\pi R^2} \] \(\textbf{Compare the three radii:}\) For fixed total current and plate radius, \[ B\propto r \] Therefore, \[ B_{R/3}:B_{2R/3}:B_R =\frac R3:\frac{2R}{3}:R \] Dividing by \(R/3\), \[ B_{R/3}:B_{2R/3}:B_R=1:2:3 \] \(\textbf{Physical interpretation:}\) The enclosed displacement current increases with the square of the measuring radius. However, magnetic-field circulation is distributed along a circular path whose circumference increases linearly with radius. The resulting magnetic-field magnitude therefore increases linearly inside the uniform plate projection. The enclosed current at \(2R/3\) is not two-thirds of the total current. It is four-ninths of the total current because current is distributed over area. This distinguishes radial current-enclosure reasoning from the final magnetic-field scaling. \(\textbf{Final answer:}\) \[ \boxed{B_{R/3}:B_{2R/3}:B_R=1:2:3,\qquad I_{\text{enc}}(2R/3)=\frac{16}{3}\,\text{mA}} \]

Explanation: The capacitor has constant capacitance, so its charge follows \(Q=CV\). At the initial instant, the voltage is \(+V_0\). Therefore, the charge on the specified positive-reference plate is \(+CV_0\). At the final instant, the voltage is \(-V_0\). Its charge is consequently \(-CV_0\). Statement I is correct because the charge magnitudes are equal but the signs are opposite. The stored electrical energy is \[ U=\frac12CV^2 \] Since the endpoint voltage magnitudes are equal, the endpoint stored energies are equal. Statement II is correct. For a constant-capacitance capacitor, the displacement current is \[ I_d=C\frac{dV}{dt} \] Differentiating the specified voltage gives \[ \frac{dV}{dt}=-\frac{2V_0}{\tau} \] Therefore, \[ I_d=-\frac{2CV_0}{\tau} \] throughout the stated interval. The current does not reverse merely because the instantaneous voltage changes sign. Statement III is incorrect. At the midpoint, the voltage and stored electric energy are zero. However, the voltage continues changing at a nonzero rate. The enclosed electric flux therefore has a nonzero rate of change. According to the Ampere–Maxwell law, this changing flux can produce nonzero magnetic-field circulation. Statement IV is correct. The example separates the sign of stored charge, the energy associated with its magnitude squared, and the rate at which the charge changes. These quantities have different behaviours even in the same capacitor experiment.

Explanation: \(\textbf{Separate source timing from propagation time:}\) The radio and microwave pulses leave the transmitter at different instants. However, both travel through vacuum and cover the same distance. Their propagation speeds are equal to \(c\), irrespective of their frequencies. \(\textbf{Determine the common travel time:}\) The path length is \[ L=900\,\text{m} \] The propagation time for either pulse is \[ t_{\text{travel}}=\frac Lc \] Substituting, \[ t_{\text{travel}}=\frac{900}{3.0\times10^8} \] \[ t_{\text{travel}}=3.0\times10^{-6}\,\text{s} \] \[ t_{\text{travel}}=3.0\,\mu\text{s} \] \(\textbf{Calculate the two arrival times:}\) The radio pulse is emitted at \(t=0\). Therefore, \[ t_{R,\text{arrival}}=0+3.0 \] \[ t_{R,\text{arrival}}=3.0\,\mu\text{s} \] The microwave pulse is emitted at \(2.0\,\mu\text{s}\). Therefore, \[ t_{M,\text{arrival}}=2.0+3.0 \] \[ t_{M,\text{arrival}}=5.0\,\mu\text{s} \] The arrival-time difference is \[ \Delta t=5.0-3.0 \] \[ \Delta t=2.0\,\mu\text{s} \] \(\textbf{Determine the radio wavelength:}\) The radio frequency is \[ f_R=100\,\text{MHz}=1.0\times10^8\,\text{Hz} \] Using \[ \lambda=\frac cf, \] we obtain \[ \lambda_R=\frac{3.0\times10^8}{1.0\times10^8} \] \[ \lambda_R=3.0\,\text{m} \] \(\textbf{Determine the microwave wavelength:}\) The microwave frequency is \[ f_M=10\,\text{GHz}=1.0\times10^{10}\,\text{Hz} \] Therefore, \[ \lambda_M=\frac{3.0\times10^8}{1.0\times10^{10}} \] \[ \lambda_M=0.030\,\text{m} \] The wavelength ratio is \[ \frac{\lambda_R}{\lambda_M}=\frac{3.0}{0.030} \] \[ \frac{\lambda_R}{\lambda_M}=100 \] \(\textbf{Physical interpretation:}\) The radio pulse reaches the receiver first solely because it was transmitted earlier. The microwave's greater frequency does not give it a greater vacuum propagation speed. The carrier wavelengths nevertheless differ by a factor of one hundred. This separates pulse-emission timing, electromagnetic-wave propagation speed and carrier wavelength. \(\textbf{Final answer:}\) \[ \boxed{t_R=3.0\,\mu\text{s},\quad t_M=5.0\,\mu\text{s},\quad \lambda_R:\lambda_M=100:1} \]

Explanation: \(\textbf{Distinguish fixed voltage from fixed capacitance:}\) The two capacitors remain at the same constant voltage because they are connected in parallel to the ideal source. However, their capacitances change with time. For either capacitor, \[ Q=CV \] When \(V=V_0\) remains constant, \[ I=\frac{dQ}{dt}=V_0\frac{dC}{dt} \] Therefore, a capacitor can carry current despite a constant applied voltage when its capacitance changes. \(\textbf{Determine the current in Branch A:}\) Differentiating the first capacitance, \[ \frac{dC_A}{dt}=\frac{C_0}{\tau} \] Consequently, \[ I_A=V_0\frac{dC_A}{dt} \] \[ I_A=+\frac{C_0V_0}{\tau} \] The positive sign indicates charge entering A's specified positive plate. \(\textbf{Determine the current in Branch B:}\) Differentiating, \[ \frac{dC_B}{dt}=-\frac{C_0}{\tau} \] Therefore, \[ I_B=V_0\frac{dC_B}{dt} \] \[ I_B=-\frac{C_0V_0}{\tau} \] B loses charge from its specified positive plate. \(\textbf{Apply current continuity at the common junction:}\) The algebraic source current is \[ I_s=I_A+I_B \] Substituting, \[ I_s=\frac{C_0V_0}{\tau}-\frac{C_0V_0}{\tau} \] \[ I_s=0 \] \(\textbf{Check total capacitance and charge:}\) The equivalent capacitance is \[ C_{\mathrm{eq}}=C_A+C_B \] \[ C_{\mathrm{eq}}=3C_0 \] It is constant. Accordingly, \[ Q_{\mathrm{total}}=3C_0V_0 \] also remains constant. \(\textbf{Physical interpretation:}\) A zero current in the common source lead does not imply zero current in every branch. Charge can be transferred between the two capacitors through the connecting circuit. Each changing capacitor has its own electric-flux variation and corresponding signed displacement-current contribution. The branch contributions cancel algebraically at the common junction. This distinguishes total-circuit current from the individual currents associated with changing electric flux in separate capacitor gaps. \(\textbf{Final answer:}\) \[ \boxed{I_A=+\frac{C_0V_0}{\tau},\qquad I_B=-\frac{C_0V_0}{\tau},\qquad I_s=0} \]

Explanation: \(\textbf{Separate the stationary and oscillating magnetic fields:}\) The probe records the vector sum of a background field and the travelling-wave magnetic field. Let the signed background component be \(B_s\). Let the wave component at the positive electric-field maximum be \(B_w\). Because the wave fields are in phase, half a period later the wave magnetic field reverses sign. The static background does not reverse. \(\textbf{Construct the two measurement equations:}\) At \(t=0\), \[ B_s+B_w=5.0\times10^{-7} \] At \(t=T/2\), \[ B_s-B_w=1.0\times10^{-7} \] The values are expressed in tesla. \(\textbf{Recover the background field:}\) Adding the equations, \[ 2B_s=6.0\times10^{-7} \] Therefore, \[ B_s=3.0\times10^{-7}\,\text{T} \] The background magnetic field points along positive \(z\). \(\textbf{Recover the wave magnetic field:}\) Subtracting the second equation from the first, \[ 2B_w=4.0\times10^{-7} \] Therefore, \[ B_w=+2.0\times10^{-7}\,\text{T} \] At \(t=0\), the wave magnetic field points along positive \(z\). \(\textbf{Verify the vacuum field relation:}\) The electric-field amplitude is \[ E_0=60\,\text{V m}^{-1} \] Therefore, \[ \frac{E_0}{B_0}=\frac{60}{2.0\times10^{-7}} \] \[ \frac{E_0}{B_0}=3.0\times10^8\,\text{m s}^{-1} \] This agrees with the vacuum propagation speed. \(\textbf{Determine the propagation direction:}\) At the positive electric maximum, \[ \vec E=+60\hat j \] and the wave magnetic field is directed along \(+\hat k\). Their ordered vector product is \[ \hat j\times\hat k=+\hat i \] Therefore, the wave propagates along positive \(x\). \(\textbf{Physical interpretation:}\) The total measured magnetic field remains positive at both observations. Nevertheless, the travelling-wave component reverses between the two instants. The independent background field conceals that reversal in the raw measurements. Subtracting the background is necessary before applying the plane-wave amplitude and directional relations. The ratio \(E_0/B_0=c\) does not apply to the total magnetic-field reading when it contains an independent static contribution. \(\textbf{Final answer:}\) \[ \boxed{\vec B_{\mathrm{static}}=+3.0\times10^{-7}\hat k\,\text{T}, \qquad\hat n=+\hat i} \]

Explanation: \(\textbf{Compare the capacitor voltages:}\) The phases at the two instants are \[ \omega t_1=\frac{\pi}{4} \] and \[ \omega t_2=\frac{3\pi}{4} \] Therefore, \[ V_1=V_0\cos\frac{\pi}{4}=+\frac{V_0}{\sqrt2} \] \[ V_2=V_0\cos\frac{3\pi}{4}=-\frac{V_0}{\sqrt2} \] The voltages have equal magnitudes but opposite signs. \(\textbf{Determine the displacement currents:}\) For a fixed-capacitance capacitor, \[ I_d=C\frac{dV}{dt} \] Differentiating, \[ I_d=-C\omega V_0\sin(\omega t) \] At the first instant, \[ I_{d1}=-C\omega V_0\sin\frac{\pi}{4} \] \[ I_{d1}=-\frac{C\omega V_0}{\sqrt2} \] At the second instant, \[ I_{d2}=-C\omega V_0\sin\frac{3\pi}{4} \] \[ I_{d2}=-\frac{C\omega V_0}{\sqrt2} \] Thus, the signed currents are identical and negative. \(\textbf{Compare stored energies:}\) The energy is \[ U=\frac12CV^2 \] Because \(V_1^2=V_2^2\), \[ U_1=U_2=\frac14CV_0^2 \] However, equal instantaneous energies need not imply equal rates of energy change. \(\textbf{Determine the energy-change rates:}\) Differentiating the energy relation, \[ \frac{dU}{dt}=CV\frac{dV}{dt} \] Since \(I_d=C\,dV/dt\), \[ \frac{dU}{dt}=VI_d \] At \(t_1\), \[ \left(\frac{dU}{dt}\right)_1 =\frac{V_0}{\sqrt2}\left(-\frac{C\omega V_0}{\sqrt2}\right) \] \[ \left(\frac{dU}{dt}\right)_1=-\frac12C\omega V_0^2 \] At \(t_2\), \[ \left(\frac{dU}{dt}\right)_2 =-\frac{V_0}{\sqrt2}\left(-\frac{C\omega V_0}{\sqrt2}\right) \] \[ \left(\frac{dU}{dt}\right)_2=+\frac12C\omega V_0^2 \] \(\textbf{Interpret the contrasting behaviours:}\) The plate charge decreases at both instants because the signed current is negative. At the first instant, the charge magnitude decreases towards zero, so stored energy decreases. At the second instant, the negative charge magnitude increases, so stored energy increases. The magnetic-field direction associated with the signed displacement current is the same at both instants. However, the direction of change of stored electrical energy is opposite. This distinguishes the sign of current from the sign of electrical-energy transfer. \(\textbf{Final answer:}\) \[ \boxed{I_{d1}=I_{d2}=-\frac{C\omega V_0}{\sqrt2},\quad \left(\frac{dU}{dt}\right)_1=-\frac12C\omega V_0^2,\quad \left(\frac{dU}{dt}\right)_2=+\frac12C\omega V_0^2} \]

Explanation: \(\textbf{Recover the displacement current from magnetic circulation:}\) The measuring path encloses the complete changing electric flux between the plates. No conduction current crosses the insulating capacitor gap. Therefore, the Ampere–Maxwell law gives \[ \oint\vec B\cdot d\vec l=\mu_0I_d \] Rearranging, \[ I_d=\frac{\displaystyle\oint\vec B\cdot d\vec l}{\mu_0} \] Substituting, \[ I_d=\frac{4\pi\times10^{-9}}{4\pi\times10^{-7}} \] \[ I_d=+1.0\times10^{-2}\,\text{A} \] \[ I_d=+10\,\text{mA} \] The positive sign is fixed by the specified loop orientation. \(\textbf{Relate current to the changing plate charge:}\) For the fixed-capacitance capacitor, \[ Q=CV \] Therefore, \[ \frac{dQ}{dt}=C\frac{dV}{dt}=I_d \] The plate charge is increasing in the chosen signed direction. \(\textbf{Determine the instantaneous energy-change rate:}\) The stored electrical energy is \[ U=\frac12CV^2 \] Differentiating, \[ \frac{dU}{dt}=CV\frac{dV}{dt} \] Since \[ C\frac{dV}{dt}=I_d, \] we obtain \[ \frac{dU}{dt}=VI_d \] \(\textbf{Substitute the signed voltage and current:}\) At the specified instant, \[ V=-10\,\text{V} \] and \[ I_d=+0.010\,\text{A} \] Therefore, \[ \frac{dU}{dt}=(-10)(+0.010) \] \[ \frac{dU}{dt}=-0.10\,\text{W} \] \(\textbf{Check the physical meaning:}\) The capacitor voltage is negative, but it is increasing because the signed current is positive. Its magnitude is therefore decreasing at that instant. The stored energy, which depends on \(V^2\), must consequently decrease. A positive displacement current does not automatically imply increasing capacitor energy. The energy-change rate depends on the product of signed voltage and signed current. This experiment combines magnetic-field circulation, displacement-current inference and the capacitor's instantaneous energy balance. \(\textbf{Final answer:}\) \[ \boxed{I_d=+10\,\text{mA},\qquad\frac{dU}{dt}=-0.10\,\text{W}} \]

Explanation: \(\textbf{Express the additional propagation delay:}\) The slab replaces an equal length of vacuum along the propagation path. The time required inside the slab is \[ t_m=\frac{L}{v_m} \] The time required through the corresponding vacuum distance would be \[ t_0=\frac Lc \] Therefore, the additional delay is \[ \Delta t=t_m-t_0 \] \[ \Delta t=L\left(\frac1{v_m}-\frac1c\right) \] \(\textbf{Recover the slab thickness:}\) The given speeds are \[ v_m=1.5\times10^8\,\text{m s}^{-1} \] \[ c=3.0\times10^8\,\text{m s}^{-1} \] Substituting, \[ 4.0\times10^{-9} =L\left(\frac1{1.5\times10^8}-\frac1{3.0\times10^8}\right) \] The difference in reciprocal speeds is \[ \frac1{1.5\times10^8}-\frac1{3.0\times10^8} =\frac1{3.0\times10^8} \] Therefore, \[ L=(4.0\times10^{-9})(3.0\times10^8) \] \[ L=1.2\,\text{m} \] \(\textbf{Determine the wavelength inside the slab:}\) The carrier frequency remains unchanged at a stationary interface: \[ f=5.0\times10^8\,\text{Hz} \] The wavelength in the slab is \[ \lambda_m=\frac{v_m}{f} \] Substituting, \[ \lambda_m=\frac{1.5\times10^8}{5.0\times10^8} \] \[ \lambda_m=0.30\,\text{m} \] \(\textbf{Determine the magnetic-field amplitude:}\) The specified plane-wave relation is \[ E_0=v_mB_0 \] Therefore, \[ B_0=\frac{E_0}{v_m} \] Substituting, \[ B_0=\frac{60}{1.5\times10^8} \] \[ B_0=4.0\times10^{-7}\,\text{T} \] \(\textbf{Verify the propagation delay:}\) The slab travel time is \[ t_m=\frac{1.2}{1.5\times10^8}=8.0\,\text{ns} \] The corresponding vacuum travel time is \[ t_0=\frac{1.2}{3.0\times10^8}=4.0\,\text{ns} \] Their difference is indeed \(4.0\,\text{ns}\). \(\textbf{Physical interpretation:}\) The delay determines the slab thickness because both propagation speeds are known. The frequency determines the medium wavelength once the medium speed is specified. The local electric-field amplitude determines the magnetic amplitude through the same medium speed. These are distinct deductions from propagation timing, wave periodicity and field relationships. \(\textbf{Final answer:}\) \[ \boxed{L=1.2\,\text{m},\qquad\lambda_m=0.30\,\text{m},\qquad B_0=4.0\times10^{-7}\,\text{T}} \]

Explanation: \(\textbf{Construct the travelling-wave expression:}\) The wave travels along positive \(x\). The first probe records a positive maximum at the spatial and temporal origin. Therefore, a compatible electric-field expression is \[ \vec E(x,t)=80\hat k\cos(kx-\omega t)\,\text{V m}^{-1} \] Here \(k\) denotes the wave number, while \(\hat k\) is the unit vector along positive \(z\). \(\textbf{Determine the temporal phase shift:}\) The frequency is \[ f=250\,\text{MHz}=2.50\times10^8\,\text{Hz} \] The angular frequency is \[ \omega=2\pi f \] Therefore, over \(1.0\,\text{ns}\), \[ \omega\Delta t=2\pi(2.50\times10^8)(1.0\times10^{-9}) \] \[ \omega\Delta t=\frac{\pi}{2} \] \(\textbf{Use the negative maximum at the second event:}\) The second electric field equals \(-E_0\). Therefore, \[ \cos(k\Delta x-\omega\Delta t)=-1 \] The wavelength restriction selects the solution \[ k(0.40)-\frac{\pi}{2}=\pi \] Hence, \[ 0.40k=\frac{3\pi}{2} \] \[ k=\frac{3\pi}{0.80}\,\text{rad m}^{-1} \] \(\textbf{Recover the medium wavelength:}\) Using \[ k=\frac{2\pi}{\lambda}, \] we obtain \[ \lambda=\frac{2\pi}{k} \] \[ \lambda=\frac{2\pi(0.80)}{3\pi} \] \[ \lambda=0.533\,\text{m} \] This lies within the specified wavelength interval. \(\textbf{Determine the propagation speed:}\) \[ v=f\lambda \] Substituting, \[ v=(2.50\times10^8)(0.5333) \] \[ v=1.33\times10^8\,\text{m s}^{-1} \] \(\textbf{Calculate the magnetic-field magnitude:}\) The electric-field amplitude is \[ E_0=80\,\text{V m}^{-1} \] From the specified medium relation, \[ B_0=\frac{E_0}{v} \] \[ B_0=\frac{80}{1.333\times10^8} \] \[ B_0=6.0\times10^{-7}\,\text{T} \] \(\textbf{Determine the magnetic-field direction:}\) At the second event, the electric field points along negative \(z\): \[ \vec E=-80\hat k\,\text{V m}^{-1} \] The propagation direction is positive \(x\): \[ \hat n=\hat i \] Therefore, \[ \vec B=\frac1v(\hat n\times\vec E) \] \[ \vec B=\frac1v[\hat i\times(-80\hat k)] \] Since \[ \hat i\times\hat k=-\hat j, \] we obtain \[ \vec B=+\frac{80}{v}\hat j \] \[ \vec B=+6.0\times10^{-7}\hat j\,\text{T} \] \(\textbf{Check the wave geometry:}\) The electric field points along negative \(z\) and the magnetic field along positive \(y\). Their ordered vector product points along positive \(x\), consistent with the wave's direction of travel. The known wavelength interval also resolves the possible ambiguity from adding complete phase cycles. The result combines temporal phase, spatial phase, propagation speed and transverse-field orientation. \(\textbf{Final answer:}\) \[ \boxed{\lambda=0.533\,\text{m},\quad v=1.33\times10^8\,\text{m s}^{-1},\quad \vec B=+6.0\times10^{-7}\hat j\,\text{T}} \]

Explanation: \(\textbf{Use the measured field gradients to infer propagation:}\) For a wave travelling along positive \(x\), the field depends on \(x\) and \(t\) through a combination such as \(\omega t-kx\). A compatible expression is \[ E_y(x,t)=E_0\sin(\omega t-kx) \] This expression gives \[ E_y(0,0)=0 \] Its time derivative is \[ \frac{\partial E_y}{\partial t} =\omega E_0\cos(\omega t-kx) \] Therefore, \[ \left.\frac{\partial E_y}{\partial t}\right|_{0,0}=+\omega E_0 \] Its spatial derivative is \[ \frac{\partial E_y}{\partial x} =-kE_0\cos(\omega t-kx) \] Hence, \[ \left.\frac{\partial E_y}{\partial x}\right|_{0,0}=-kE_0 \] Both signs agree with the observations. \(\textbf{Reject propagation along negative x:}\) A wave travelling along negative \(x\) would have the opposite relationship between the spatial and temporal slopes. The observed positive time slope and negative spatial slope are therefore consistent with positive-\(x\) propagation. \(\textbf{Evaluate the electric field one-quarter period later:}\) At \[ x=0,\qquad t=\frac T4, \] the phase is \[ \omega t=\frac{\pi}{2} \] Therefore, \[ E_y=E_0\sin\frac{\pi}{2} \] \[ E_y=+E_0 \] \(\textbf{Recover the magnetic-field direction:}\) For a vacuum plane electromagnetic wave, \[ \vec B=\frac1c(\hat n\times\vec E) \] Here, \[ \hat n=\hat i,\qquad \vec E=E_0\hat j \] Thus, \[ \vec B=\frac{E_0}{c}(\hat i\times\hat j) \] \[ \vec B=+\frac{E_0}{c}\hat k \] \(\textbf{Physical interpretation:}\) The electric field vanishes at the initial event, so its instantaneous value alone cannot reveal its oscillation sign or propagation direction. The spatial and temporal slopes supply the missing directional information. The subsequent positive electric-field maximum fixes the corresponding magnetic-field direction. The result illustrates how local field variation, travelling-wave phase and transverse vector geometry must be interpreted together. \(\textbf{Final answer:}\) \[ \boxed{\hat n=+\hat i,\qquad \vec B(0,T/4)=+\frac{E_0}{c}\hat k} \]

Explanation: \(\textbf{Relate the measured total energy density to the instantaneous field:}\) For a freely propagating plane electromagnetic wave in vacuum, the instantaneous electric and magnetic energy densities are equal. Therefore, \[ u=\frac12\varepsilon_0E^2+\frac{B^2}{2\mu_0} \] Using the plane-wave field relation, \[ \frac{B^2}{\mu_0}=\varepsilon_0E^2 \] we obtain \[ u=\varepsilon_0E^2 \] \(\textbf{Recover the instantaneous electric-field magnitude:}\) The measured energy density is \[ u=3.6\times10^{-8}\,\text{J m}^{-3} \] Therefore, \[ E=\sqrt{\frac{u}{\varepsilon_0}} \] Substituting, \[ E=\sqrt{\frac{3.6\times10^{-8}}{1.0\times10^{-11}}} \] \[ E=\sqrt{3600} \] \[ E=60\,\text{V m}^{-1} \] \(\textbf{Determine the full electric-field amplitude:}\) The observation is made when \[ E=\frac35E_0 \] Therefore, \[ E_0=\frac53E \] \[ E_0=\frac53(60) \] \[ E_0=100\,\text{V m}^{-1} \] \(\textbf{Calculate the magnetic-field amplitude:}\) For the vacuum wave, \[ E_0=cB_0 \] Thus, \[ B_0=\frac{E_0}{c} \] \[ B_0=\frac{100}{3.0\times10^8} \] \[ B_0=3.33\times10^{-7}\,\text{T} \] \(\textbf{Determine the time-averaged energy density:}\) For a sinusoidal plane wave, \[ \langle u\rangle=\frac12\varepsilon_0E_0^2 \] Substituting, \[ \langle u\rangle=\frac12(1.0\times10^{-11})(100)^2 \] \[ \langle u\rangle=5.0\times10^{-8}\,\text{J m}^{-3} \] \(\textbf{Interpret the distinction:}\) The measured instantaneous energy density is smaller than the time-averaged value because the observation occurs away from a field maximum. The instantaneous field magnitude must therefore be recovered before determining the wave amplitude. Using the measured instantaneous value directly as the amplitude would underestimate the magnetic amplitude and the average energy density. The problem distinguishes a local observation from the full oscillation amplitude and from the time average. \(\textbf{Final answer:}\) \[ \boxed{E_0=100\,\text{V m}^{-1},\quad B_0=3.33\times10^{-7}\,\text{T},\quad \langle u\rangle=5.0\times10^{-8}\,\text{J m}^{-3}} \]

Explanation: \(\textbf{Avoid assuming that the two positive maxima are consecutive:}\) A positive maximum repeats after every complete period. Therefore, the interval \(4.0\,\text{ns}\) may contain an integer number of cycles. Let this number be \(n\), where \(n\) is a positive integer. Then, \[ 4.0\,\text{ns}=nT \] \(\textbf{Construct the possible frequencies:}\) Since \[ f=\frac1T, \] we obtain \[ f=\frac{n}{4.0\times10^{-9}} \] \[ f=0.25n\,\text{GHz} \] The frequency restriction is \[ 0.40\lt f\lt 1.00\quad\text{GHz} \] Therefore, the possible integer values are \[ n=2\quad\text{or}\quad n=3 \] These correspond to \[ f=0.50\,\text{GHz} \] or \[ f=0.75\,\text{GHz} \] \(\textbf{Use the intermediate negative maximum:}\) The probe records a negative maximum at \(t=2.0\,\text{ns}\). This instant is halfway between the two recorded positive maxima. Starting from the positive maximum at \(t=0\), the electric field may be expressed as \[ E(t)=E_0\cos(2\pi ft) \] At \(t=2.0\,\text{ns}\), \[ 2\pi ft =2\pi(0.25n\times10^9)(2.0\times10^{-9}) \] \[ 2\pi ft=n\pi \] Thus, \[ E(2.0\,\text{ns})=E_0\cos(n\pi) \] For an even \(n\), this is a positive maximum. For an odd \(n\), it is a negative maximum. The observation therefore requires odd \(n\). Among the allowed candidates, only \[ n=3 \] satisfies this condition. \(\textbf{Determine the frequency and wavelength:}\) \[ f=0.25(3) \] \[ f=0.75\,\text{GHz} \] Using \[ \lambda=\frac cf, \] we obtain \[ \lambda=\frac{3.0\times10^8}{0.75\times10^9} \] \[ \lambda=0.40\,\text{m} \] \(\textbf{Physical interpretation:}\) Two repeated positive maxima alone do not identify the fundamental period unless they are known to be consecutive. The independent frequency range narrows the possible number of intervening cycles. The intermediate negative maximum then eliminates the remaining even-cycle possibility. This example shows how an additional signed field observation resolves an ambiguity in temporal wave measurements. \(\textbf{Final answer:}\) \[ \boxed{f=0.75\,\text{GHz},\qquad\lambda=0.40\,\text{m}} \]

Explanation: \(\textbf{Interpret the spatial energy-density graph:}\) For a sinusoidal plane electromagnetic wave, the instantaneous energy density varies as the square of the oscillating field. At a fixed time, its spatial dependence therefore contains a factor of the form \[ u(x)\propto\cos^2(kx+\phi) \] The squared cosine repeats after a spatial phase change of \(\pi\), rather than \(2\pi\). Consequently, the distance between successive energy-density maxima is \[ \Delta x_u=\frac{\lambda}{2} \] \(\textbf{Recover the wavelength:}\) The measured energy-density maximum spacing is \[ \Delta x_u=0.20\,\text{m} \] Therefore, \[ \lambda=2\Delta x_u \] \[ \lambda=2(0.20) \] \[ \lambda=0.40\,\text{m} \] \(\textbf{Interpret the signed electric-field graph:}\) Successive positive electric-field maxima at one fixed position are separated by a complete period. Therefore, \[ T=1.6\,\text{ns} \] \[ T=1.6\times10^{-9}\,\text{s} \] The frequency is \[ f=\frac1T \] \[ f=\frac1{1.6\times10^{-9}} \] \[ f=6.25\times10^8\,\text{Hz} \] \(\textbf{Determine the medium propagation speed:}\) Using \[ v=f\lambda, \] we obtain \[ v=(6.25\times10^8)(0.40) \] \[ v=2.5\times10^8\,\text{m s}^{-1} \] \(\textbf{Recover the magnetic-field amplitude:}\) At the selected event, \[ E_0=75\,\text{V m}^{-1} \] Using the supplied medium relation, \[ B_0=\frac{E_0}{v} \] \[ B_0=\frac{75}{2.5\times10^8} \] \[ B_0=3.0\times10^{-7}\,\text{T} \] \(\textbf{Determine the vector direction:}\) The wave propagates along positive \(x\), and its electric field points along positive \(y\). Therefore, \[ \vec B=\frac1v(\hat i\times75\hat j) \] \[ \vec B=+3.0\times10^{-7}\hat k\,\text{T} \] \(\textbf{Physical interpretation:}\) The spatial energy-density graph repeats twice as often as the signed field pattern. The temporal electric-field graph directly supplies the carrier period. Combining these distinct periodic measurements gives the propagation speed. The field direction and speed then determine the magnetic-field vector. Confusing energy-density maxima with consecutive same-sign electric-field maxima would underestimate the wavelength and propagation speed. \(\textbf{Final answer:}\) \[ \boxed{\lambda=0.40\,\text{m},\quad v=2.5\times10^8\,\text{m s}^{-1},\quad \vec B=+3.0\times10^{-7}\hat k\,\text{T}} \]

Explanation: \(\textbf{Separate the two displacement harmonics:}\) The first displacement component is \[ x_1=A\cos(\omega t) \] The second is \[ x_2=\frac A4\cos(2\omega t) \] Their displacement amplitudes are different. Their angular frequencies are also different. \(\textbf{Determine the first acceleration component:}\) Acceleration is the second time derivative of displacement. Therefore, \[ a_1=\frac{d^2x_1}{dt^2} \] \[ a_1=-A\omega^2\cos(\omega t) \] Its amplitude is \[ a_{01}=A\omega^2 \] \(\textbf{Determine the second acceleration component:}\) Differentiating twice, \[ a_2=-\frac A4(2\omega)^2\cos(2\omega t) \] Simplifying, \[ a_2=-A\omega^2\cos(2\omega t) \] Thus, \[ a_{02}=A\omega^2 \] The two acceleration-component amplitudes are equal. \(\textbf{Compare the associated radiation frequencies:}\) Under the stated harmonic-source assumption, \[ f_1=\frac{\omega}{2\pi} \] and \[ f_2=\frac{2\omega}{2\pi}=2f_1 \] \(\textbf{Determine their vacuum wavelength ratio:}\) Using \[ \lambda=\frac cf, \] we obtain \[ \frac{\lambda_1}{\lambda_2}=\frac{f_2}{f_1} \] \[ \frac{\lambda_1}{\lambda_2}=2 \] \(\textbf{Identify the limit of the inference:}\) The calculated acceleration amplitudes describe two components of the source motion. They do not by themselves specify the electric-field amplitudes received at an arbitrary observation point. Those amplitudes also depend on how the radiation is produced and observed. Therefore, equal harmonic acceleration amplitudes cannot be used as sufficient evidence of equal received field amplitudes. The student's error is ignoring the squared angular-frequency factor when differentiating displacement twice. The wavelength conclusion follows from the harmonic frequencies, not from the displacement-amplitude ratio. Source displacement, source acceleration and the properties of detected electromagnetic waves must therefore be distinguished.

Explanation: \(\textbf{Relate magnetic circulation to the rate of electric-flux change:}\) The selected surface has no conduction-current contribution. Therefore, the Ampere–Maxwell law gives \[ C_B=\mu_0\varepsilon_0\frac{d\Phi_E}{dt} \] Rearranging, \[ \frac{d\Phi_E}{dt}=\frac{C_B}{\mu_0\varepsilon_0} \] \(\textbf{Recover the total flux change from the graph:}\) Integrating over the measurement interval, \[ \Delta\Phi_E=\frac1{\mu_0\varepsilon_0}\int_0^{4.0\,\mathrm{ns}}C_B(t)\,dt \] The circulation–time graph is triangular. Its signed area is \[ \int C_B\,dt=\frac12(\text{base})(\text{height}) \] Substituting, \[ \int C_B\,dt =\frac12(4.0\times10^{-9})(2.0\times10^{-8}) \] \[ \int C_B\,dt=4.0\times10^{-17}\,\text{T m s} \] \(\textbf{Calculate the corresponding electric-flux change:}\) Using \[ \frac1{\mu_0\varepsilon_0}=c^2, \] we obtain \[ \Delta\Phi_E=c^2(4.0\times10^{-17}) \] \[ \Delta\Phi_E=(3.0\times10^8)^2(4.0\times10^{-17}) \] \[ \Delta\Phi_E=3.6\,\text{V m} \] \(\textbf{Translate flux change into electric-field change:}\) For the stationary surface with uniform perpendicular field, \[ \Phi_E=EA \] Therefore, \[ \Delta E=\frac{\Delta\Phi_E}{A} \] \[ \Delta E=\frac{3.6}{0.020} \] \[ \Delta E=+180\,\text{V m}^{-1} \] \(\textbf{Determine the final field component:}\) The initial component is \[ E_i=+50\,\text{V m}^{-1} \] Hence, \[ E_f=E_i+\Delta E \] \[ E_f=50+180 \] \[ E_f=+230\,\text{V m}^{-1} \] \(\textbf{Physical interpretation:}\) Magnetic-field circulation gives the instantaneous rate of change of enclosed electric flux. The area under its time graph gives the accumulated flux change after division by \(\mu_0\varepsilon_0\). That flux change must then be converted into an electric-field change using the stationary surface area. The graph does not directly provide the instantaneous electric-field magnitude. The initial field value is needed to determine the final field rather than only its change. \(\textbf{Final answer:}\) \[ \boxed{E_f=+230\,\text{V m}^{-1}} \]