Mix Examples- Electromagnetic waves Questions in English

(A) The relationship between frequency $(n)$,velocity $(v)$,and wavelength $(\lambda)$ is given by $n = \frac{v}{\lambda}$.
Since the wavelengths $(\lambda)$ are the same,the ratio of frequencies is equal to the ratio of their velocities: $\frac{n_{MW}}{n_{US}} = \frac{v_{MW}}{v_{US}}$.
The speed of a microwave (electromagnetic wave) is $v_{MW} \approx 3 \times 10^8 \ m/s$.
The speed of an ultrasonic sound wave (mechanical wave) in air is $v_{US} \approx 330 \ m/s \approx 3 \times 10^2 \ m/s$.
Therefore,the ratio of frequencies is $\frac{n_{MW}}{n_{US}} \approx \frac{3 \times 10^8}{3 \times 10^2} = 10^6:1$.

(A) Given: Electric field amplitude $E_0 = 100\,V\,m^{-1}$ and magnetising field amplitude $H_0 = 0.265\,A\,m^{-1}$.
The instantaneous energy flow per unit area is given by the Poynting vector $S = E \times H$.
The maximum rate of energy flow (maximum intensity) is given by $S_{max} = E_0 \times H_0$.
Substituting the values: $S_{max} = 100\,V\,m^{-1} \times 0.265\,A\,m^{-1} = 26.5\,W\,m^{-2}$.

(D) The electric field $E$ inside the wire is given by $E = \frac{V}{l} = \frac{IR}{l}$,where $V$ is the potential difference across the wire.
The magnetic field $B$ at the surface of the wire is given by Ampere's Law as $B = \frac{\mu_0 I}{2\pi a}$,where $a$ is the radius of the wire.
The Poynting vector $S$ is defined as $S = \frac{1}{\mu_0} (E \times B)$. Since $E$ is parallel to the wire and $B$ is tangential,the magnitude of the Poynting vector is $S = \frac{EB}{\mu_0}$.
Substituting the values of $E$ and $B$:
$S = \frac{1}{\mu_0} \left( \frac{IR}{l} \right) \left( \frac{\mu_0 I}{2\pi a} \right) = \frac{I^2R}{2\pi al}$.
This vector is directed radially inward,representing the energy flow into the wire to account for Joule heating.

(B) When an airplane flies over a $TV$ antenna,it acts as a moving reflector for the electromagnetic waves.
The $TV$ antenna receives two signals: one directly from the transmitter and another reflected from the airplane.
Since the airplane is moving,the path length of the reflected signal changes continuously,leading to a time-varying phase difference between the direct and reflected signals.
This results in interference,which causes the flickering or shaking effect on the $TV$ screen.

(C) The intensity of the electromagnetic radiation is given as $I = 10^3 \ W m^{-2}$.
The area of the roof is $A = \text{length} \times \text{width} = 6 \ m \times 30 \ m = 180 \ m^2$.
The power $P$ incident on the surface is given by the product of intensity and area:
$P = I \times A$
$P = 10^3 \ W m^{-2} \times 180 \ m^2$
$P = 1.8 \times 10^5 \ W$.

(B) Surface waves travel along the surface of the Earth. As the distance increases,these waves are absorbed by the Earth's surface and are also affected by the curvature of the Earth. Due to the tilting and curvature of the Earth,the signal strength of surface waves decreases rapidly with distance,causing them to eventually disappear.

(A) We know that the speed of light $c$ is related to the permeability of free space $\mu_0$ and permittivity of free space $\varepsilon_0$ by the relation: $\mu_0 \varepsilon_0 = \frac{1}{c^2}$.
Also,the ratio of the magnitude of the electric field $E$ to the magnetic field $B$ in an electromagnetic wave is equal to the speed of light: $\frac{E}{B} = c$.
Substituting these into the given expression:
$\frac{E^2 \mu_0 \varepsilon_0}{B^2} = \left( \frac{E}{B} \right)^2 (\mu_0 \varepsilon_0) = (c)^2 \left( \frac{1}{c^2} \right) = 1$.
The quantity is dimensionless.
Therefore,the dimensional formula is $[M^0 L^0 T^0]$.

(D) $1$. The quantity $x = E/B$ represents the speed of an electromagnetic wave,which has dimensions $[LT^{-1}]$.
$2$. The quantity $y = \sqrt{1/(\mu_0 \varepsilon_0)}$ is the speed of light in vacuum,which also has dimensions $[LT^{-1}]$.
$3$. The quantity $z = l/(RC)$. Since the time constant $\tau = RC$ has dimensions of time $[T]$,and $l$ is length $[L]$,the dimensions of $z$ are $[L/T] = [LT^{-1}]$.
$4$. Since $x$,$y$,and $z$ all have dimensions of speed $[LT^{-1}]$,they all have the same dimensions.

(B) To determine the position of a device in $3D$ space (latitude,longitude,and altitude),we need $3$ satellites to solve for these $3$ variables. However,because the receiver's clock is not perfectly synchronized with the satellite clocks,a $4^{th}$ satellite is required to calculate the time offset (clock bias). Therefore,a minimum of $4$ satellites is necessary for accurate $GPS$ positioning.

(D) The radiation power $P$ of a linear antenna is given by $P \propto \frac{l^2}{\lambda^2}$.
From the graph,the slope of the line is $\frac{P}{l^2} \propto \frac{1}{\lambda^2}$.
Therefore,$\text{Slope} \propto \frac{1}{\lambda^2}$,which implies $\lambda \propto \frac{1}{\sqrt{\text{Slope}}}$.
Thus,$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{\text{Slope}_B}{\text{Slope}_A}} = \sqrt{\frac{\tan(53^\circ)}{\tan(37^\circ)}}$.
Using $\tan(53^\circ) \approx \frac{4}{3}$ and $\tan(37^\circ) \approx \frac{3}{4}$,we get:
$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4/3}{3/4}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$.

(A) The effective refractive index of the ionosphere is given by the formula:
$n_{eff} = \sqrt{1 - \frac{80.5N}{f^2}}$
where $N$ is the electron density and $f$ is the frequency of the $e-m$ wave.
From this formula,it is clear that the refractive index of the ionosphere depends on the frequency $f$ of the $e-m$ waves. Therefore,Statement $-2$ is false.
Short wave communication (sky wave propagation) relies on the refraction of $e-m$ waves by the ionosphere,which acts like a medium with a variable refractive index. When the frequency is appropriate,the waves are bent back towards the Earth,which is often described as total internal reflection. Thus,Statement $-1$ is true.

(D) The Assertion is incorrect because environmental damage,specifically the release of chlorofluorocarbons $(CFCs)$,leads to the depletion of the ozone layer in the stratosphere,not an increase.
The Reason is also incorrect because the ozone layer acts as a shield that absorbs harmful ultraviolet $(UV)$ radiation. Therefore,a decrease (depletion) of ozone leads to an increase in $UV$ radiation reaching the earth,not the other way around.
Since both the Assertion and the Reason are false,the correct option is $D$.

(D) The Assertion is incorrect because optical waves (like those used in optical fibres) have a much higher frequency and bandwidth compared to microwaves,making them superior for signal transmission.
The Reason is also incorrect because all electromagnetic waves,including microwaves and optical waves,travel at the same speed $(c = 3 \times 10^8 \ m/s)$ in a vacuum.
Therefore,both the Assertion and the Reason are incorrect.

(C) The Lorentz force on a charged particle is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
At $t=0$ and $z = \frac{\pi}{k}$,the electric field is $\overrightarrow{E} = E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \cos(\pi) = -E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$.
For an electromagnetic wave,the direction of propagation is $\hat{k}_{prop} = \frac{\overrightarrow{E} \times \overrightarrow{B}}{|E||B|}$. Here,the wave propagates in the $-z$ direction (since the argument is $kz + \omega t$),so $\hat{k}_{prop} = -\hat{k}$.
Thus,$\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times \overrightarrow{B} = -\hat{k} \cdot \frac{E_{0}}{c}$. Solving this,we get $\overrightarrow{B} = -\frac{E_{0}}{c} \left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$.
The magnetic force is $\overrightarrow{F}_{m} = q(\overrightarrow{v} \times \overrightarrow{B}) = q(v_{0}\hat{k} \times [-\frac{E_{0}}{c} \frac{\hat{i}-\hat{j}}{\sqrt{2}}]) = -q \frac{v_{0}E_{0}}{c} \left(\frac{\hat{j}+\hat{i}}{\sqrt{2}}\right)$.
Since $\frac{v_{0}}{c} \ll 1$,the magnetic force is negligible compared to the electric force.
Therefore,$\overrightarrow{F} \approx q\overrightarrow{E} = -q E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$,which is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.

(B) The intensity of radiation $I$ at a distance $r$ from a source of power $P$ is given by $I = \frac{P}{4 \pi r^2}$.
Also,the intensity is related to the electric field amplitude $E$ by the relation $I = \frac{1}{2} c \epsilon_0 E^2$.
For the $100 \, W$ bulb: $\frac{1}{2} c \epsilon_0 E^2 = \frac{100}{4 \pi (3)^2}$.
For the $60 \, W$ bulb: $\frac{1}{2} c \epsilon_0 (\sqrt{\frac{x}{5}} E)^2 = \frac{60}{4 \pi (3)^2}$.
Dividing the second equation by the first,we get: $(\sqrt{\frac{x}{5}})^2 = \frac{60}{100}$.
$\frac{x}{5} = \frac{6}{10} = \frac{3}{5}$.
Therefore,$x = 3$.

(D) The power $P$ of the transmitter is $100\,kW = 100 \times 10^{3}\,W = 10^{5}\,W$.
The time $t$ is $1\,hour = 3600\,s$.
The energy $E$ radiated is given by the formula $E = P \times t$.
Substituting the values, we get $E = 10^{5}\,W \times 3600\,s$.
$E = 3600 \times 10^{5}\,J = 36 \times 10^{7}\,J$.

(B) The velocity of the electromagnetic wave in the medium is given by $v = \frac{E_{0}}{B_{0}}$.
Substituting the given values: $v = \frac{2.25}{1.5 \times 10^{-8}} = 1.5 \times 10^{8}\,m/s$.
The total distance traveled by the signal (to the target and back) is $d_{total} = 2 \times 3\,km = 6 \times 10^{3}\,m$.
The time taken for the echo to reach the radar is $t = \frac{d_{total}}{v}$.
$t = \frac{6 \times 10^{3}}{1.5 \times 10^{8}} = 4 \times 10^{-5}\,s$.
Thus,the time is $4 \times 10^{-5}\,s$.

(B) The total power of the bulb is $P_{total} = 100 \, W$.
Only $5 \%$ of this power is converted into visible radiation.
Therefore,the power of visible radiation is $P_{vis} = \frac{5}{100} \times 100 \, W = 5 \, W$.
The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P_{vis}}{A}$,where $A = 4 \pi r^2$ is the surface area of a sphere.
Given $r = 10 \, m$,the area is $A = 4 \pi (10)^2 \, m^2$.
Thus,the intensity is $I = \frac{5}{4 \pi (10)^2} \, W/m^2$.

(D) The power radiated $P$ by a short linear antenna of length $l$ is given by the relation $P \propto \left(\frac{l}{\lambda}\right)^2$.
This indicates that the radiated power is directly proportional to the square of the ratio of the antenna length to the wavelength of the electromagnetic wave.

(D) Given: Capacitance $C = 200 \ pF = 200 \times 10^{-12} \ F$,Voltage $V_{rms} = 230 \ V$,Angular frequency $\omega = 300 \ rad/s$.
The capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{300 \times 200 \times 10^{-12}} = \frac{1}{6 \times 10^{-8}} = \frac{10^8}{6} \ \Omega$.
The $rms$ conduction current $I_c$ is $I_c = \frac{V_{rms}}{X_C} = 230 \times 300 \times 200 \times 10^{-12} \ A$.
$I_c = 230 \times 6 \times 10^{-8} \ A = 1380 \times 10^{-8} \ A = 13.8 \times 10^{-6} \ A = 13.8 \ \mu A$.
According to the Maxwell-Ampere law,the displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ in the connecting wires. Therefore,$I_d = I_c = 13.8 \ \mu A$.

(D) The vector $\vec{S}$ is known as the Poynting vector,which represents the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field.
The $SI$ unit of $\vec{S}$ is $\text{W/m}^2$ (Watts per square meter).
Dimensional analysis:
$1$. $\text{Power} = \text{Energy} / \text{Time}$,so $\text{Power} / \text{Area} = \text{Energy} / (\text{Area} \times \text{Time})$. This matches the definition of $\vec{S}$. Thus,$(D)$ is correct.
$2$. $\text{Force} / (\text{Length} \times \text{Time}) = (\text{Force} \times \text{Length}) / (\text{Length}^2 \times \text{Time}) = \text{Energy} / (\text{Area} \times \text{Time})$. This also matches the dimensions of $\vec{S}$. Thus,$(B)$ is correct.
$3$. $\text{Energy} / (\text{charge} \times \text{current}) = \text{Energy} / (\text{charge} \times \text{charge} / \text{time}) = \text{Energy} \times \text{time} / \text{charge}^2$. This does not match.
$4$. $\text{Energy} / \text{Volume}$ has dimensions of $\text{Pressure}$ or $\text{Energy density}$,which is $\text{J/m}^3$. This does not match.
Therefore,the correct options are $(B)$ and $(D)$.

(A) For a linear antenna of length $l$ (where $l \ll \lambda$),the power radiated $P$ is given by the relation $P \propto \frac{1}{\lambda^2}$.
This implies that the power radiated is inversely proportional to the square of the wavelength.
Therefore,as the wavelength $\lambda$ increases,the power radiated decreases.
Thus,the correct proportionality is $\lambda^{-2}$.

(D) The power radiated by a linear antenna is given by the formula $P \propto (L/\lambda)^2$, where $L$ is the length of the antenna and $\lambda$ is the wavelength of the radiation.
Given the ratio of lengths $L_1 : L_2 = 2 : 3$ and the ratio of wavelengths $\lambda_1 : \lambda_2 = 8 : 9$.
The ratio of powers is given by:
$\frac{P_1}{P_2} = \left( \frac{L_1}{L_2} \right)^2 \times \left( \frac{\lambda_2}{\lambda_1} \right)^2$
Substituting the given values:
$\frac{P_1}{P_2} = \left( \frac{2}{3} \right)^2 \times \left( \frac{9}{8} \right)^2$
$\frac{P_1}{P_2} = \frac{4}{9} \times \frac{81}{64}$
$\frac{P_1}{P_2} = \frac{1}{1} \times \frac{9}{16} = \frac{9}{16}$
Therefore, the ratio of effective powers radiated is $9: 16$.

(A) The energy density of an electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$. The total energy is $E_E = u_E \times V$. Given $V = (10 \ cm)^3 = (0.1 \ m)^3 = 10^{-3} \ m^3$.
$E_E = \frac{1}{2} \times 8.9 \times 10^{-12} \times (10^7)^2 \times 10^{-3} = 0.445 \ J$.
The energy density of a magnetic field is $u_B = \frac{B^2}{2\mu_0}$. The total energy is $E_B = u_B \times V$.
$E_B = \frac{(0.25)^2}{2 \times 4\pi \times 10^{-7}} \times 10^{-3} = \frac{0.0625 \times 10^{-3}}{25.12 \times 10^{-7}} \approx 24.88 \ J \approx 25 \ J$.
Thus,the energies are $0.445 \ J$ and $25 \ J$.

$(C)$ Gauss's law relates the total electric flux through a closed surface to the enclosed charge, which is $(V)$ Total electric flux.
$(B)$ Faraday's law states that an induced electromotive force is proportional to the $(III)$ Change in magnetic flux.
$(C)$ Ampere's law relates the magnetic field around a closed loop to the electric current passing through the loop. In the context of Maxwell's modification, it involves the $(IV)$ Change in electric flux (displacement current).
$(D)$ Kirchhoff's laws are based on $(II)$ Electric charge conservation (Junction rule) and energy conservation (Loop rule).
Therefore, the correct matching is $A-V, B-III, C-IV, D-II$.

(C) The generation,propagation,and detection of electromagnetic waves are the fundamental principles behind wireless communication systems. Radio and television broadcasting rely entirely on the transmission of electromagnetic waves through space,which are then detected by receivers to convert them back into audio and visual signals. Therefore,radio and television are based on these principles.

(D) The wave theory of light was proposed by Christian Huygens, not $C. V. Raman$. $C. V. Raman$ is famous for the discovery of the Raman effect (inelastic scattering of light). Therefore, the pair $C. V. Raman - \text{Wave theory of light}$ is incorrectly matched.

(A) Given: Power of the bulb $P = 200 \ W$,distance $r = 100 \ cm = 1 \ m$,and efficiency $\eta = 11 \% = 0.11$.
The power converted into visible radiation is $P_{vis} = \eta \times P = 0.11 \times 200 \ W = 22 \ W$.
Assuming the bulb acts as a point source,the light spreads uniformly over a sphere of radius $r$.
The intensity $I$ at a distance $r$ is given by $I = \frac{P_{vis}}{4 \pi r^2}$.
Substituting the values: $I = \frac{22}{4 \times 3.14 \times (1)^2} = \frac{22}{12.56} \approx 1.75 \ W \ m^{-2}$.

(C) The maximum rate of energy flow per unit area is given by the magnitude of the Poynting vector,$S = E \times H$.
Given the amplitudes of the electric field $E_0 = 300 \ V m^{-1}$ and the magnetic field $H_0 = 7.9 \ A m^{-1}$.
The maximum rate of energy flow is $S = E_0 \times H_0$.
Substituting the values,we get $S = 300 \times 7.9 = 2370 \ W m^{-2}$.

(C) The intensity $I$ of electromagnetic radiation at a distance $r$ from a point source is given by $I = \frac{P_{out}}{4 \pi r^2}$, where $P_{out} = P_{in} \times \eta$ and $\eta$ is the efficiency.
Also, the intensity in terms of the electric field amplitude $E_0$ is $I = \frac{1}{2} \epsilon_0 E_0^2 c$.
Given: $P_{in} = 100 \ W$, $E_0 = 2 \ V \ m^{-1}$, $r = 10 \ m$, $\epsilon_0 = 8.854 \times 10^{-12} \ F \ m^{-1}$, $c = 3 \times 10^8 \ m \ s^{-1}$.
Calculating intensity: $I = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (2)^2 \times (3 \times 10^8) = 0.0106 \ W \ m^{-2}$.
Now, equating the two expressions for intensity: $0.0106 = \frac{100 \times \eta}{4 \times \pi \times (10)^2}$.
$0.0106 = \frac{100 \times \eta}{4 \times 3.1416 \times 100} = \frac{\eta}{12.566}$.
$\eta = 0.0106 \times 12.566 \approx 0.133$.
Thus, the efficiency is $13.3 \%$.

(B) The power of the laser beam is $P = 100 \,mW = 100 \times 10^{-3} \,W = 0.1 \,W$.
The length of the beam segment is $l = 90 \,cm = 0.9 \,m$.
The speed of light is $c = 3 \times 10^8 \,m/s$.
The time taken for this length of the beam to pass a point is $t = \frac{l}{c} = \frac{0.9}{3 \times 10^8} = 0.3 \times 10^{-8} \,s = 3 \times 10^{-9} \,s$.
The energy stored in this length is $E = P \times t$.
$E = (0.1 \,W) \times (3 \times 10^{-9} \,s) = 3 \times 10^{-10} \,J$.

(B) The given electric field equation is $E = E_0 \sin(\omega t - kx)$,where $E_0 = \sqrt{377} \text{ V/m}$.
The intensity (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} c \varepsilon_0 E_0^2$.
Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we can write $I = \frac{1}{2} \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \varepsilon_0 E_0^2 = \frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} E_0^2$.
Given $\sqrt{\frac{\mu_0}{\varepsilon_0}} = 377$,then $\sqrt{\frac{\varepsilon_0}{\mu_0}} = \frac{1}{377}$.
Substituting the values: $I = \frac{1}{2} \times \frac{1}{377} \times (\sqrt{377})^2 = \frac{1}{2} \times \frac{1}{377} \times 377 = \frac{1}{2} \text{ W/m}^2$.
Comparing this with $\frac{1}{\alpha} \text{ W/m}^2$,we get $\alpha = 2$.