Displacement Current and Properties of EM Waves — NEET Physics PYQ MCQs with Solutions

Solution: Given, $V=V_0\sin(\omega t)$ and $Q=CV$. Differentiating w.r.t. time, $\dfrac{dQ}{dt}=C\dfrac{dV}{dt}=C\dfrac{d}{dt}(V_0\sin\omega t)=CV_0\,\omega\cos(\omega t)$. Since $I_d=\dfrac{dQ}{dt}$, we get $I_d=V_0\,\omega C\cos(\omega t)$.

Solution: Comparing with $B_y=B_0\sin(kx\pm\omega t)$, we get $k=\pi\times10^{3}\,\text{m}^{-1}$. Since $\lambda=\dfrac{2\pi}{k}$, $\lambda=\dfrac{2\pi}{\pi\times10^{3}}=2\times10^{-3}\,\text{m}$.

Solution: We know that $\dfrac{E_0}{B_0}=c$, where $E_0$ and $B_0$ are the peak values of the electric and magnetic fields respectively. So $E_0:B_0=c:1$. Hence, the correct option is (d).

Solution: Given, average flux $=20\,\text{W/cm}^2$, surface area $=20\,\text{cm}^2$, time $=1\,\text{min}=60\,\text{s}$. For a non-reflecting surface, energy received $=$ average flux $\times$ surface area $\times$ time $=20\times20\times60=24\times10^{3}\,\text{J}$.

Solution: The displacement current $I_d=\varepsilon_0\dfrac{d\phi}{dt}=C\dfrac{dV}{dt}$ (using $C=\dfrac{\varepsilon_0A}{d}$). Given $C=20\,\mu\text{F}$ and $\dfrac{dV}{dt}=3\,\text{V/s}$, so $I_d=20\times10^{-6}\times3=60\times10^{-6}\,\text{A}=60\,\mu\text{A}$. Since the displacement current between the plates always equals the conduction current in the connecting wires, $I_c=I_d=60\,\mu\text{A}$.

Solution: Here, velocity of the EM wave, $\vec{v}=v\hat{i}$, and the instantaneous oscillating electric field, $\vec{E}=E\hat{j}$. In an electromagnetic wave, the electric and magnetic field vectors are mutually perpendicular to each other and to the direction of propagation, i.e. $\vec{E}\times\vec{B}\parallel\vec{v}$. Comparing, the direction of the oscillating magnetic field of the wave is along the $-z$ direction.

Solution: A charge at rest has only an electric field around it, no magnetic field. A charge moving with constant velocity produces both electric and magnetic fields, but they do not change with time, so no EM wave is produced. Only a charge moving with non-zero acceleration produces electric and magnetic fields that change with space and time, generating a propagating electromagnetic wave.

Solution: Initial momentum of the radiation, $P_i=\dfrac{E}{c}$. Since the surface is perfectly reflecting, the reflected momentum is $P_r=-\dfrac{E}{c}$. The change in momentum of light is $\Delta P_{\text{light}}=P_r-P_i=-\dfrac{2E}{c}$. Thus, by Newton's third law, the momentum transferred to the surface is $\dfrac{2E}{c}$.

Solution: For a perfectly reflecting surface, $F_{\text{average}}=\dfrac{2IA}{c}$, where $I=25\times10^{4}\,\text{W/m}^2$ is the energy flux, $A=15\times10^{-4}\,\text{m}^2$ is the surface area, and $c=3\times10^{8}\,\text{m/s}$. So $F_{\text{average}}=\dfrac{2\times25\times10^{4}\times15\times10^{-4}}{3\times10^{8}}=2.50\times10^{-6}\,\text{N}$.

Solution: The electromagnetic wave equation is $E=E_0\cos(kz-\omega t)$. Speed of the electromagnetic wave, $v=\dfrac{\omega}{k}=c$. Comparing the given equation $E=\hat{i}\,40\cos(kz-6\times10^{8}t)$, we get $\omega=6\times10^{8}\,\text{rad/s}$ and $E_0=40\hat{i}$. So $k=\dfrac{\omega}{c}=\dfrac{6\times10^{8}}{3\times10^{8}}=2\,\text{m}^{-1}$.

Solution: We know that $\vec{E}\times\vec{B}$ points in the direction of wave propagation. For $\vec{E}=E_0\hat{i}$ and $\vec{B}=B_0\hat{j}$, $\vec{E}\times\vec{B}=E_0B_0(\hat{i}\times\hat{j})=E_0B_0\hat{k}$, which is along $+z$, matching the given propagation direction.

Solution: The time-varying electric and magnetic fields of an electromagnetic wave are mutually perpendicular to each other, and both are also perpendicular to the direction of propagation. So statement (c), which says they are parallel to each other, is false.

Solution: Comparing the given equation with $E_y=E_0\cos(\omega t-kx)$, we get $\omega=2\pi\times10^{6}\,\text{rad/s}$ and $k=\pi\times10^{-2}\,\text{rad/m}$. Frequency, $f=\dfrac{\omega}{2\pi}=10^{6}\,\text{Hz}$. Wavelength, $\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{\pi\times10^{-2}}=200\,\text{m}$. Since the field is $E_y$ and depends on $\omega t-kx$, the wave moves along the $+x$-direction with frequency $10^{6}\,\text{Hz}$ and wavelength $200\,\text{m}$.

Solution: An electromagnetic wave is composed of oscillating electric and magnetic fields in mutually perpendicular planes, and these oscillations are perpendicular to the direction of propagation. The direction of propagation of the electromagnetic wave is given by the Poynting vector, $\vec{S}=\vec{E}\times\vec{H}=\dfrac{\vec{E}\times\vec{B}}{\mu_0}$, which is parallel to $\vec{E}\times\vec{B}$.

Solution: For the electron to pass undeflected, the electric force on it must balance the magnetic force, i.e. $eE=evB$, so $v=\dfrac{E}{B}$, or $v=\dfrac{|\vec{E}|}{|\vec{B}|}$.

Solution: A stationary charge produces only an electric field; a uniformly moving charge produces a localised electromagnetic field around it. An accelerated charge produces changing electric and magnetic fields that regenerate each other, mutually perpendicular and perpendicular to the direction of propagation — this radiating disturbance is the electromagnetic wave. Hence, electromagnetic radiations are caused by accelerated charges.

Solution: The relation between velocity of light $c$, frequency $f$ and wavelength $\lambda$ is $c=f\lambda$, so $\lambda=\dfrac{c}{f}$. Given $c=3\times10^{8}\,\text{m/s}$, $f=100\,\text{Hz}$, so $\lambda=\dfrac{3\times10^{8}}{100}=3\times10^{6}\,\text{m}$.

Solution: Refractive index of a medium is given by $n=\sqrt{\varepsilon_r\mu_r}$, where $\varepsilon_r=\dfrac{\varepsilon}{\varepsilon_0}$ and $\mu_r=\dfrac{\mu}{\mu_0}$. So $n=\sqrt{\dfrac{\varepsilon}{\varepsilon_0}\cdot\dfrac{\mu}{\mu_0}}=\sqrt{\dfrac{\varepsilon\mu}{\varepsilon_0\mu_0}}$.

Solution: According to Maxwell, electromagnetic waves have sinusoidal variations of the electric and magnetic field vectors at right angles to each other, as well as at right angles to the direction of wave propagation. Both fields vary with time and space and have the same frequency, reaching their maxima and minima at the same place and the same time — i.e. they are in phase, while being mutually perpendicular.