Electromagnetic Waves — NEET Physics PYQ MCQs with Solutions

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: 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: 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: 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: 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: 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: 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.

Solution: $\gamma$-rays have the shortest wavelength because they have higher frequency than UV-rays, X-rays and microwaves.

Solution: It is an electromagnetic wave. The frequency of the microwave oven must match the resonant frequency of the water molecules so that the water molecules oscillate and large heat is developed.

Solution: The decreasing order of wavelength of various rays is Microwave $>$ Infrared $>$ Ultraviolet $>$ Gamma rays.

Solution: The velocity of electromagnetic waves in free space is given by $c = \dfrac{1}{\sqrt{\varepsilon_0\mu_0}}$.

Solution: A green house's glass transmits short-wavelength infrared radiation and visible light in, but absorbs the long-wavelength infrared radiation emitted back by the plants at night, trapping heat inside. This is called the green house effect.

Solution: The ozone layer extends from about 30 km to 50 km above the earth's surface and absorbs the major part of ultraviolet radiations coming from the sun. The range of ultraviolet radiation is $100$–$4000\,\text{\AA}$, so it blocks radiation of wavelength less than $3\times10^{-7}\,\text{m}$ (3000 \AA).

Solution: Space communication refers to sending, receiving and processing information through space, which occurs via ground (surface) wave propagation and sky wave propagation (among others), so a signal can reach another point as either a sky wave or a ground wave.

Solution: Due to their high penetrating power, X-rays are used for investigation of the structure of solids. The Laue spot method and rotating crystal method are used for this purpose: X-rays fall on the solid under investigation and their structure is recorded on a photographic plate.

Solution: Gamma rays have wavelength range $6\times10^{-14}\,\text{m}$ to $1\times10^{-10}\,\text{m}$, X-rays have wavelength range $1\times10^{-13}\,\text{m}$ to $3\times10^{-8}\,\text{m}$. Blue light and red light lie in the visible range, which extends from $4000\,\text{\AA}$ to $7800\,\text{\AA}$. Hence, red light has the longest wavelength.

Solution: Wavelength ranges: Gamma rays $6\times10^{-14}$ to $1\times10^{-10}\,\text{m}$, X-rays $1\times10^{-13}$ to $3\times10^{-8}\,\text{m}$, microwaves $10^{-3}$ to $0.3\,\text{m}$, radio waves greater than $0.1\,\text{m}$. So, radiowaves are the longest waves.

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: 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: 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: 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: 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: 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: 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: $E = h\nu = \dfrac{hc}{\lambda} \Rightarrow \lambda = \dfrac{hc}{E} = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{15\times10^{3}\times1.6\times10^{-19}} \approx 0.828\times10^{-10}\,\text{m} = 0.828\,\text{\AA}$. This wavelength lies in the X-ray region.