Answer

(c) Given, energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms $E=11 eV$

We know that, $E=h \nu$, where $h=6.62 \times 10^{-34} J$-s

$$ \begin{aligned} \Rightarrow \quad 11 eV & =h \nu \\ v & =\frac{11 \times 1.6 \times 10^{-19}}{h} J \\ & =\frac{11 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}} J \\ & =2.65 \times 10^{15} Hz \end{aligned} $$

$$ \nu=\text { frequency } $$

This frequency radiation belongs to ultraviolet region.

Thinking Process

When a wave is reflected from a denser medium, then its phase changes by $180^{\circ}$ or $\pi$

Answer

(b) When a wave is reflected from denser medium, then the type of wave doesn’t change but only its phase changes by $180^{\circ}$ or $\pi$ radian.

Thus, for the reflected wave $\hat{\mathbf{z}}=-\hat{\mathbf{z}}, \hat{\mathbf{i}}=-\hat{\mathbf{i}}$ and additional phase of $\pi$ in the incident wave.

Given, here the incident electromagnetic wave is,

$$ \mathbf{E}=E_0 \hat{\mathbf{i}} \cos (k z-\omega t) $$

The reflected electromagnetic wave is given by

$$ \begin{aligned} r & =E_0(\hat{i}) \cos [k(-z)-\omega t+\pi] \\ & =-E_0 \hat{i} \cos [-(k z+\omega t)+\pi] \\ & =E_0 \hat{i} \cos [-(k_{z}+\omega t)=E_0 \hat{\mathbf{i}} \cos (k z+\omega t)] \end{aligned} $$

Answer

(b) Given, energy flux $\phi=20 W / cm^{2}$

Area, $\quad A=30 cm^{2}$

Time, $\quad t=30 min=30 \times 60 s$

Now, total energy falling on the surface in time $t$ is, $U=\phi A t=20 \times 30 \times(30 \times 60) J$

Momentum of the incident light $=\frac{U}{C}$

$$ =\frac{20 \times 30 \times(30 \times 60)}{3 \times 10^{8}} \Rightarrow=36 \times 10^{-4} kg-ms^{-1} $$

Momentum of the reflected light $=0$

$\therefore$ Momentum delivered to the surface

$$ =36 \times 10^{-4}-0=36 \times 10^{-4} kg-ms^{-1} $$

Thinking Process

Electric field intensity on a surface due to incident radiation is,

$$ \begin{aligned} I_{av} \propto E_o^2 \\ \frac{P_{av .}}{A} \propto E_o^2 \\ \end{aligned} $$

Here,

$$ P_{a v} \propto E_0^{2} \quad[\because A \text { is same in bothcases }] $$

Answer

(c) We know that,

$$ E_0 \propto \sqrt{P_{av}} $$

$\frac{(E_O)_1}{(E_O)_2} = \sqrt{\frac{(P_a{av})_1} {(P_a{av})_2} \Rightarrow \frac{E}{E_O}_2} = \sqrt{\frac{1000}{5}} \\ $

according to question, P=50 W, P=100 W

$\therefore \quad$ Putting these value in Eq.(i), we get

$$ \frac{E^{\prime}}{E}=\frac{50}{100} \Rightarrow \frac{E^{\prime}}{E}=\frac{1}{2} \Rightarrow E^{\prime}=\frac{E}{2} $$

Answer

(d) The direction of propagation of electromagnetic wave is perpendicular to both electric field vector $\mathbf{E}$ and magnetic field vector $\mathbf{B}$, i.e., in the direction of $\mathbf{E} \times \mathbf{B}$.

This can be seen by the diagram given below

Here, electromagnetic wave is along the $z$-direction which is given by the cross product of $\mathbf{E}$ and $\mathbf{B}$.

Thinking Process

Intensity of electromagnetic wave, $I=U_{a v} c$

where, $\quad U_{a v}=$ Average energy

and $\quad c=$ speed to light

Answer

(c) Intensity in terms of electric field $U_{av}=\frac{1}{2} \varepsilon_0 E_0^{2}$

Intensity in terms of magnetic field $U_{av}=\frac{1}{2} \frac{B_0^{2}}{\mu_0}$

Now taking the intensity in terms of electric field.

$$ \quad(U_{\text{av}}) \text{ electric field } =\frac{1}{2} E_0^{2} $$ $$=\frac{1}{2} (c B_0)^{2} \quad(\because E_0=c B_0) =\frac{1}{2} \times c^{2} B^{2} $$

$$\text { But, } \quad =\frac{1}{\sqrt{\mu_0 }} \therefore\quad(U_{\text{av}})_{\text {Electric field }}$$

Thus, the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector

Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is $1: 1$.

Answer

(c) From a diode antenna, the electromagnetic waves are radiated outwards.

The amplitude of electric field vector $(E_0)$ which transports significant energy from the source falls off intensity inversely as the distance $(r)$ from the antenna, i.e., $E_0 \propto \frac{1}{r}$.

Thinking Process

From Maxwell’s equations, it is seen that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as

$$ B_0=\frac{E_0}{c} $$

Answer

(d) Here, in electromagnetic wave, the electric field vector is given as,

$$ \mathbf{E}=(E_1 \hat{\mathbf{i}}+E_2 \hat{\mathbf{j}}) \cos (k z-\omega t) $$

In electromagnetic wave, the associated magnetic field vector,

$$ \mathbf{B}=\frac{E}{C}=\frac{E_1 \hat{i}+E_2 \hat{j}}{c} \cos (k z-\omega t) $$

Also, $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to $\mathbf{E}$ as well as $\mathbf{B}$, so the given electromagnetic wave is plane polarised.

Thinking Process

Given, $E=E_{0} \cos (k z-w t)$. Thus, it acts along negative y-direction.

Answer

$(a, b, c)$

Suppose an electromagnetic wave is travelling along negative z-direction. Its electric field is given by

$$ \mathbf{E}=E_{O} \cos (k z-\omega t) $$

which is perpendicular to $z$-axis. It acts along negative $y$-direction.

The associated magnetic field $\mathbf{B}$ in electromagnetic wave is along $x$-axis i.e., along $\hat{\mathbf{k}} \times \mathbf{E}$.

$$ \begin{array}{ll} \text { As, } & B_{0}=\frac{E_{0}}{C} \\ \therefore & B=\frac{1}{C}(\hat{\mathbf{k}} \times \mathbf{E}) \end{array} $$

The associated electric field can be written in terms of magnetic field as

$$ \mathbf{E}=c(\mathbf{B} \times \hat{\mathbf{k}}) $$

Angle between $\hat{\mathbf{k}}$ and $\mathbf{E}$ is $90^{\circ}$ between $\hat{\mathbf{k}}$ and $\mathbf{B}$ is $90^{\circ}$. Therefore, $\mathbf{E}=1 E \cos 90^{\circ}=0$ and $\hat{\mathbf{k}} \cdot \mathbf{B}=1 E \cos 90^{\circ}=0$

Answer

$(b, d)$ $\newline$

(a) $E_{x}, B_{y}$ $\newline$

(b) $E_{y}, B_{z}$ $\newline$

(c) $B_{x}, E_{y}$ $\newline$

(d) $E_{z}, B_{y}$ $\newline$

As electric and magnetic field vectors $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other as well as perpendicular to the direction of propagation of electromagnetic wave.

Here in the question electromagnetic wave is propagating along $x$-direction. So, electric and magnetic field vectors should have either $y$-direction or $z$-direction. $\newline$

Thinking Process

The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.

Answer

$(a, c, d)$

Given, frequency by which the charged particles oscillates about its mean equilibrium position $=10^{9} \mathrm{~Hz}$.

So, frequency of electromagnetic waves produced by the charged particle is $v=10^{9} \mathrm{~Hz}$.

$$ \text { Wavelength } \lambda=\frac{c}{v}=\frac{3 \times 10^{8}}{10^{9}}=0.3 \mathrm{~m} $$

Also, frequency of $10^{9} \mathrm{~Hz}$ fall in the region of radiowaves.

Thinking Process

An electromagnetic wave can be produced by accelerated or oscillating charge.

Answer

( $b, d)$

Here, in option (b) charge is moving in a circular orbit.

In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.

Also, we know that a charge starts accelerating when it falls in an electric field.

Answer

$(a, c, d)$

Radiation pressure $(p)$ is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.

Momentum per unit time per unit area

$$ =\frac{\text { Intensity }}{\text { Speed of wave }}=\frac{I}{\mathrm{c}} $$

Change in momentum per unit time per unit area $=\frac{\Delta I}{C}=$ radiation pressure $(p)$ i.e.,

$$ p=\frac{\Delta I}{c} $$

Momentum of incident wave per unit time per unit area $=\frac{I}{c}$

When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area $=0$.

Radiation pressure $(p)=$ change in momentum per unit time per unit area $=\frac{\Delta I}{c}=\frac{I}{c}-0=\frac{I}{c}$.

When wave is totally reflected, then momentum of the reflected wave per unit time per unit area $=-\frac{I}{\mathrm{c}}$, Radiation pressure $p=\frac{I}{\mathrm{c}}–\frac{I}{\mathrm{c}}=\frac{2 I}{\mathrm{c}}$.

Here, $p$ lies between $\frac{I}{c}$ and $\frac{2 I}{c}$.

Answer

The orientation of the portable radio with respect to broadcasting station is important because the electromagnetic waves are plane polarised, so the receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave.

Answer

Microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.

Answer

The displacement current through the capacitor is,

Here,

Putting this value in $\mathrm{Eq}$ (i), we get

$$ I_{d}=I_{C}=\frac{d q}{d t} $$

$$ \begin{aligned} & I_{d}=I_{c}=-q_{0} \sin 2 \pi v t \times 2 \pi v \\ & I_{d}=I_{c}=-2 \pi v q_{0} \sin 2 \pi v t \end{aligned} $$

Thinking Process

Capacities reactance $X_{c}$ is inversely proportional to the displacement current i.e., $X_{c} \propto \frac{1}{I}$.

Answer

Capacitive reaction $X_{C}=\frac{1}{2 \pi f C}$,

$\therefore \quad X_{c} \propto \frac{1}{f}$

As frequency decreases, $X_{C}$ increases and the conduction current is inversely proportional to $X_{c} \because I \propto \frac{1}{X_{c}}$

So, displacement current decreases as the conduction current is equal to the displacement current.

Answer

Magnetic field $\mathbf{B}=B_{0}$ sin $\omega t$

Given, equation $B=12 \times 10^{-8} \sin \left(1.20 \times 10^{7} z-3.60 \times 10^{15} t\right) \mathrm{T}$.

On comparing this equation with standard equation, we get

$$ B_{0}=12 \times 10^{-8} $$

The average intensity of the beam $I_{\mathrm{av}}=\frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}} \cdot \mathrm{C}=\frac{1}{2} \times \frac{\left(12 \times 10^{-8}\right)^{2} \times 3 \times 10^{8}}{4 \pi \times 10^{-7}}$

$$ =1.71 \mathrm{~W} / \mathrm{m}^{2} $$

Answer

Consider and electromagnetic waves, let $\mathbf{E}$ be varying along $y$-axis, $\mathbf{B}$ is along z-axis and propagation of wave be along $x$-axis. Then $\mathbf{E} \times \mathbf{B}$ will tell the direction of propagation of energy flow in electromegnetic wave, along $x$-axis.

Let

$$ \begin{aligned} \mathbf{E} & =E_{0} \sin (\omega t-k x) \hat{\mathbf{j}} \\ \mathbf{B} & =B_{0} \sin (\omega t-k x) \hat{\mathbf{k}} \\ \mathbf{S} & =\frac{1}{\mu_{0}}(\mathbf{E} \times \mathbf{B})=\frac{1}{\mu_{0}} E_{0} B_{0} \sin ^{2}(\omega t-k x)[\hat{\mathbf{j}} \times \hat{\mathbf{k}}] \\ & =\frac{E_{0} B_{0}}{\mu_{0}} \sin ^{2}(\omega t-k x) \hat{\mathbf{i}} \end{aligned} $$

The variation of $|\mathbf{S}|$ with time $t$ will be as given in the figure below

Answer

An electromagnetic wave carries energy and momentum like other waves.

Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped professor CV Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. The tails of the camets are also due to radiation pressure.

Answer

Consider the figure ginen below to prove that the magneti field $\mathrm{B}$ at a point in between the plater of a paravel- plate copocior during charging is $\frac{\varepsilon_{0} \mu_{0} r}{2} \frac{d E}{d t}$

Let $I_{d}$ be the displacement current in the region between two plates of parallel plate capacitor, in the figure.

The magnetic field induction at a point in a region between two plates of capacitor at a perpendicular distance $r$ from the axis of plates is

$$ \begin{array}{rlrl} B & =\frac{\mu_{0} 2 I_{d}}{4 \pi r}=\frac{\mu_{0}}{2 \pi r} I_{d}=\frac{\mu_{0}}{2 \pi r} \times \varepsilon_{0} \frac{d \varphi_{E}}{d t} & & \because I_{d}=\frac{E_{0} d \varphi_{E}}{d t} \\ & =\frac{\mu_{0} \varepsilon_{0}}{2 \pi r} \frac{d}{d t}\left(E r^{2}\right)=\frac{\mu_{0} \varepsilon_{0}}{2 \pi r} \pi^{2} \frac{d E}{d t} & \\ B & =\frac{\mu_{0} \varepsilon_{0} r}{2} \frac{d E}{d t} & & {\left[\because \varphi_{E}=E r^{2}\right]} \end{array} $$

Answer

(a) (i) Microwave is used in satellite communications.

So, $\lambda_{1}$ is the wavelength of microwave.

(ii) Ultraviolet rays are used to kill germs in water purifier. So, $\lambda_{2}$ is the wavelength of UV rays.

(iii) $X$-rays are used to detect leakage of oil in underground pipelines. So, $\lambda_{3}$ is the wavelength of X-rays.

(iv) Infrared is used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.

(b) Wavelength of X-rays $<$ wavelength of UV $<$ wavelength of infrared $<$ wavelength of microwave.

$$ \Rightarrow \quad \lambda_{3}<\lambda_{2}<\lambda_{4}<\lambda_{1} $$

(c) (i) Microwave is used in radar.

(ii) UV is used in LASIK eye surgery.

(iii) $X$-ray is used to detect a fracture in bones.

(iv) Infrared is used in optical communication.

Answer

Radiant flux density $S=\frac{1}{\mu_{0}}(\mathbf{E} \times \mathbf{B})=c^{2} \varepsilon_{0}(\mathbf{E} \times \mathbf{B})$

$$ \because c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} $$

Suppose electromagnetic waves be propagating along $x$-axis. The electric field vector of electromagnetic wave be along $y$-axis and magnetic field vector be along $z$-axis. Therefore,

and $$ E_0 =E_0 \cos(k x-\omega t) $$ $$ {B} =B_0 \cos(k x-\omega t) \\ $$

$$E \times B =\left(E_0 \times B_0\right) \cos^{2}(k x-\omega t) \\ $$ $$S =c^{2}0(E \times B) \\ =c^{2}0\left(E_0 \times B_0\right) \cos^{2}(k x-\omega t) $$

Average value of the magnitude of radiant flux density over complete cycle is

$$ \begin{array}{rlrl} S_{av} =c^{2} 0\left|E_0 \times B_0\right| \frac{1}{T} \int_{0}^{T} \cos ^{2}(k x-\omega t) d t \newline \\ =c^{2} 0 E_{0} B_{0} \times \frac{1}{T} \times \frac{T}{2} \because \int_{0}^{T} \cos ^{2}(k x-\omega t) d t=\frac{T}{2} \newline \\ \Rightarrow S_{av}=\frac{c^{2}}{2} 0 E_{0} \frac{E_{0}}{c} \text { As, } c=\frac{E_{0}}{B_{0}} \newline \\ =\frac{c}{2} 0 E_{0}^{2}=\frac{c}{2} \times \frac{1}{c^{2} \mu_{0}} E_{0}^{2} c=\frac{1}{\sqrt{\mu_{0} 0}} \text { or } 0=\frac{1}{c^{2} \mu_{0}} \newline \\ \Rightarrow \quad S_{av} =\frac{E_{0}^{2}}{2 \mu_{0} c} \newline\text { Hence proved. } \end{array} $$

Answer

Given, capacitance of capacitor $C=2 \mu \mathrm{F}$,

Displacement current $I_{d}=1 \mathrm{~mA}$ Charge

or

$$ \begin{aligned} q & =C V \\ I_{d} d t & =C d V \\ I_{d} & =C \frac{d V}{d t} \\ 1 \times 10^{-3} & =2 \times 10^{-6} \times \frac{d V}{d t} \\ \frac{d V}{d t} & =\frac{1}{2} \times 10^{+3}=500 \mathrm{~V} / \mathrm{s} \end{aligned} \quad[\because q=i t] $$

$$ \text { or } $$

So, by applying a varying potential difference of $500 \mathrm{~V} / \mathrm{s}$, we would produce a displacement current of desired value.

Answer

Pressure $=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}$

Force is the rate of change of momentum

i.e.,

$$ F=\frac{d p}{d t} $$

Energy in time $d t$,

$$ U=p \cdot \operatorname{Cor} p=\frac{U}{C} $$

$\therefore \quad$ Pressure $=\frac{1}{A} \cdot \frac{U}{C \cdot d t}$

$$ \text { Pressure }=\frac{I}{C} \quad \because I=\text { Intensity }=\frac{U}{A \cdot d t} $$

Answer

As the distance is doubled, the area of spherical region $\left(4 \pi^{2}\right)$ will become four times, so the intensity becomes one fourth the initial value $\because I \propto \frac{1}{r^{2}}$ but in case of laser it does not spread, so its intensity remain same.

Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following

(i) Unidirection

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristic are missing in the case of light from the bulb.

Answer

Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle.

Hence, electric field is not responsible for radiation pressure.

Answer

Given,

$$ \begin{aligned} B & =\frac{\mu_{0} i}{2 \pi a} \hat{i}=\frac{\mu_{0} \lambda v}{2 \pi a} \hat{i} \newline\\ \therefore \quad S & =\frac{1}{\mu_{0}}[E \times B]=\frac{1}{\mu_{0}} \frac{\lambda_{\hat{j}}}{2 \pi \varepsilon_{0} a} \times \frac{\mu_{0}}{2 \pi a} \lambda \sqrt{\hat{i}} \newline\\ & =\frac{\lambda^{2} v}{4 \pi^{2} \varepsilon_{0} a^{2}}(\hat{j} \times \hat{i})=-\frac{\lambda^{2} V}{4 \pi^{2} \varepsilon_{0} a^{2}} \hat{k} \end{aligned} \quad\left[\because I=\lambda_{V}\right] $$

$$ E=\frac{\lambda \hat{e}_{s}}{2 \pi 0 a} \hat{j} $$

Thinking Process

The conduction current density is given by the Ohm’s law = Electric field between the plates.

Answer

Suppose distance between the parallel plates is $d$ and applied voltage $V_{(t)}=V_{0} 2 \pi v t$.

Thus, electric field

Now using Ohm’s law,

$$ \begin{aligned} E=\frac{V_{0}}{d} \sin (2 \pi v t) \newline \\ J_{c} =\frac{1}{\rho} \frac{V_{0}}{d} \sin (2 \pi v t) \newline \\ =\frac{V_{0}}{\rho d} \sin (2 \pi v t)= \newline \\ J_{0}^{c} =\frac{V_{0}}{\rho d} \end{aligned} $$

$$ \Rightarrow \quad=\frac{V_{0}}{\rho d} \sin (2 \pi v t)=J_{0}^{c} \sin 2 \pi v t $$

Here,

Now the displacement current density is given as

$$ \begin{array}{rlrl} J_{d} =\varepsilon \frac{\delta E}{d t}=\frac{\varepsilon \delta}{d t} \newline \\ =\frac{\varepsilon 2 \pi v V_{0}}{d} \cos (2 \pi v t) \newline \\ \Rightarrow =J_{0}^{d} \cos (2 \pi v t) \newline \\ \text { where, } J_{0}^{d} =\frac{2 \pi V \varepsilon V_{0}}{d} \newline \\ \Rightarrow \frac{J_{0}^{d}}{J_{0}^{c}} =\frac{2 \pi v \varepsilon V_{0}}{d} \cdot \frac{\rho d}{V_{0}}=2 \pi v \varepsilon \rho \newline \\ =2 \pi \times 80 \varepsilon_{0} v \times 0.25=4 \pi \varepsilon_{0} v \times 10 \newline \\ =\frac{10 v}{9 \times 10^{9}}=\frac{4}{9} \end{array} $$

Thinking Process

Displacement current density

$$ J_{d}=\varepsilon_{0} \frac{dE}{d t} $$

Answer

(i) Given, the induced electric field at a distance $r$ from the wire inside the cable is

$$ \mathbf{E}(s, t)=\mu_{0} I_{0} v \cos (2 \pi v t) \ln \frac sa \hat{\mathbf{k}} $$

Now, displacement current density,

$$ Jd=\varepsilon_{0} \frac{d \mathbf{E}}{d t}=\varepsilon_{0} \frac{d}{d t} \mu_{0} I_{0} v \cos (2 \pi v t) \ln \frac{s}{a} \hat{\mathbf{k}} $$

(ii)

$\begin{aligned} I _d & =\int J _d s d s d \theta=\int _{s=0}^a J _d s d s \int _0^{2 \pi} d \theta=\int _{s=0}^a J _d s d s \\ & =\int _{s=0}^a[\frac{2 \pi}{\lambda^2} I_0 \log _e(\frac{a}{s}) s d s \sin 2 \pi v t] \times 2 \pi \\ & =(\frac{2 \pi}{\lambda})^2 I _0 \int _{s=0}^a(\frac{a}{s}) s d s \sin 2 \pi v t \\ \Rightarrow \qquad & =(\frac{2 \pi}{\lambda})^2 I _0 \int _{s=0}^a \ln (\frac{a}{s}) \frac{1}{2} d(s^2) \cdot \sin 2 \pi v t \end{aligned}$

$\begin{aligned} & =\frac{a^2}{2}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right) \cdot d\left(\frac{s}{a}\right)^2 \\ & =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right)^2 \cdot d\left(\frac{s}{a}\right)^2 \\ & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{s}{a}\right)^2 \cdot d\left(\frac{s}{a}\right)^2 \end{aligned}$

$\begin{array}{rlrl} & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \times(-1) & {\left[\because \int_{s=0}^a \ln \left(\frac{s}{a}\right)^2 d\left(\frac{s}{a}\right)^2=-1\right]} \\ \therefore & I_d =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \\ \Rightarrow \qquad & =\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0 \sin 2 \pi v t & \end{array}$

(iii) The displacement current,

$$ \begin{aligned} & I_{d}=\frac{2 \pi a^{2}}{2 \lambda} I_{0} \sin 2 \pi v t=I_{0 d} \sin 2 \pi v t \\ \text { Here, } & I_{0 d}=\frac{2 \pi a}{2 \lambda}^{2} I_{0}=\frac{a \pi^{2}}{\lambda} I_{0} \\ \therefore & \frac{I_{0 d}}{I_{0}}=\frac{a \pi^{2}}{\lambda} \end{aligned} $$

Answer

(i) Consider the figure given below

During the propagation of electromagnetic wave a long $z$-axis, let electric field vector $(\mathbf{E})$ be along $x$-axis and magnetic field vector $\mathbf{B}$ along $y$-axis, i.e., $\mathbf{E}=E_{0} \hat{\mathbf{i}}$ and $\mathbf{B}=B_{0} \hat{\mathbf{j}}$.

Line integral of $E$ over the closed rectangular path 1234 in $x$ - $z$ plane of the figure

$$ \begin{aligned} g \mathbf{E} \cdot \mathbf{d l} & =\int_{1}^{2} E \cdot d l+\sqrt{2} \cdot \frac{3}{E} \cdot d l+\int_{3}^{4} \cdot d l+\int_{4}^{1} E \cdot d l \newline \\ & =\int_{1}^{2} E \cdot d l \cos 90+\int_{2}^{3} E \cdot d l \cos 0+\int_{3}^{4} E \cdot d l \cos 90+\int_{4}^{1} E \cdot d l \cos 180^{\circ} \newline \\ & =E_{0} h\left[\sin \left(k z_{2}-\omega t\right)-\sin \left(k z_{1}-g \omega t\right)\right] \end{aligned} $$

(ii) For evaluating $\boldsymbol{\beta} \mathbf{B} \cdot \mathbf{d s}$, let us consider the rectangle 1234 to be made of strips of are $d s=h d z$ each.

$$ \begin{aligned} \int \mathbf{B} \cdot \mathbf{d} \mathbf{s}=\int \mathbf{B} \cdot \mathbf{d} \mathbf{s} \cos 0 & =\int \mathbf{B} \cdot \mathbf{d} \mathbf{s}=\int_{z_{1}}^{z_{2}} B_{0} \sin (k z-\omega t) h d z \\ & =\frac{-B_{0} h}{k}\left[\cos \left(k z_{2}-\omega t\right)-\cos \left(k z_{1}-\omega t\right)\right] \end{aligned} $$

(iii) Given, $g \mathbf{E} \cdot dl=\frac{-d \varphi_{B}}{d t}=-\frac{d}{d t} g \mathbf{B} \cdot \mathbf{d s}$

Putting the values from Eqs. (i) and (ii), we get

$$ E_{0} h [\sin k z_{2}-\omega t-\sin k z_{1}-\omega t] \\ $$

$$= \frac{-d}{d t} \frac{B_{y} h}{k}t\cos k z_{2}-\omega t-\cos k z_{1}-\omega t. \\ $$ $$= \frac{B_y h}{k} \omega \sin k z_2-\omega t-\sin k z_{1}-\omega t $$

$$\Rightarrow \quad E_{0}=\frac{B_{0} \omega}{k}=B_{y} c \because \frac{\omega}{k}=c$$ $$\Rightarrow \quad \frac{E_0}{B_0}= C$$

(iv) For evaluating $g^{\mathbf{B}} \cdot \mathbf{d l}$, let us consider a loop 1234 in $y$-z plane as shown in figure given below

$$ \begin{aligned} \int \mathbf{B} \cdot \mathbf{d l} & =\int_{1}^{2} \mathbf{B} \cdot \mathbf{d l}+\int_{2}^{3} \mathbf{B} \cdot \mathbf{d l}+\int_{3}^{4} \mathbf{B} \cdot \mathbf{d l}+\int_{4}^{1} \mathbf{B} \cdot \mathbf{d l}\newline \\ & =\int_{1}^{2} \mathbf{B} \cdot \mathbf{d l} \cos 0+\int_{2}^{3} \mathbf{B} \cdot \mathbf{d l} \cos 90^{\circ}+\int_{3}^{4} \mathbf{B} \cdot \mathbf{d l} \cos 180^{\circ}+\int_{4}^{1} \mathbf{B} \cdot \mathbf{d l} \cos 90^{\circ}\newline \\ & =B_{0} h\left[\sin \left(k z_{1}-\omega t\right)-\sin \left(k z_{2}-\omega t\right)\right. \end{aligned} $$

Now to evaluate $\varphi_{E}=\int \mathbf{E} \cdot \mathbf{d s}$, let us consider the rectangle 1234 to be made of strips of area $h d_{2}$ each.

$$ \begin{aligned} \varphi_{E} & =\int E \cdot d s=\int E d s \cos 0=\int E d s=\int_{z_{1}}^{z_{0}^{2}} E_{0} \sin \left(k z_{1}-\omega t\right) h d z \\ & =-\frac{E_{0} h}{k}\left[\cos \left(k z_{2}-\omega t\right)-\cos \left(k z_{1}-\omega t\right)\right] \\ \therefore \quad \frac{d \varphi_{E}}{d t} & =\frac{E_{0} h \omega}{k}\left[\sin \left(k z_{1}-\omega t\right)-\sin \left(k z_{2}-\omega t\right)\right] \end{aligned} $$

$$ \begin{aligned} \text { In } \quad g \mathbf{B} \cdot \mathbf{d l} & =\mu_{0} I+\frac{\varepsilon_{0} d \varphi_{E}}{d t}, I=\text { conduction current } \\ & =0 \text { in vacuum } \\ \therefore \quad \quad \quad g \mathbf{B} \cdot \mathrm{dl} & =\mu_{0} \varepsilon \frac{d \varphi_{E}}{d t} \end{aligned} $$

Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get

$$ \begin{array}{rlrl} B_{0} =E_{0} \frac{\omega \mu_{0} \varepsilon_{0}}{k} \newline \\ \Rightarrow \frac{E_{0}}{B_{0}} \frac{\omega}{k} =\frac{1}{\mu_{0} \varepsilon_{0}} \newline \\ \text { But } \frac{E_{0}}{B_{0}} =c \text { and } \omega=c k \newline \\ \Rightarrow c \cdot c =\frac{1}{\mu_{0} \varepsilon_{0}}, \text { therefore } c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \end{array} $$

Answer

(i) The electromagnetic wave carry energy which is due to electric field vector and magnetic field vector. In electromagnetic wave, $E$ and $B$ vary from point to point and from moment to moment. Let $E$ and $B$ be their time averages.

The energy density due to electric field $E$ is

$$ u_{E}=\frac{1}{2} \varepsilon_{0} E^{2} $$

The energy density due to magnetic field $B$ is

$$ u_{B}=\frac{1}{2} \frac{B^{2}}{\mu_{0}} $$

Total average energy density of electromagnetic wave

$$ u_{a v}=u_{E}+u_{B}=\frac{1}{2} \varepsilon_{0} E^{2}+\frac{1}{2} \frac{B^{2}}{\mu_{0}} $$

Let the EM wave be propagating along z-direction. The electric field vector and magnetic field vector be represented by

$$ \begin{aligned} & E=E_{0} \sin (k z-\omega t) \\ & B=B_{0} \sin (k z-\omega t) \end{aligned} $$

The time average value of $E^{2}$ over complete cycle $=\frac{E_{0}^{2}}{2}$

and time average value of $B^{2}$ over complete cycle $=\frac{B_{0}^{2}}{2}$

$$ \begin{aligned} u_{\mathrm{av}} & =\frac{1}{2} \frac{\varepsilon_{0} E_{0}^{2}}{2}+\frac{1}{2} \mu_{0} \frac{B_{0}^{2}}{2} \\ & =\frac{1}{4} \varepsilon_{0} E_{0}^{2}+\frac{B_{0}^{2}}{4 \mu_{0}} \end{aligned} $$

(ii) We know that $E_{0}=c B_{0}$ and $c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$

$$ \begin{array}{rlrl} \therefore \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}} =\frac{1}{4} \frac{E_{0}^{2} / c^{2}}{\mu_{0}}=\frac{E_{0}^{2}}{4 \mu_{0}} \times \mu_{0} \varepsilon_{0}=\frac{1}{4} \varepsilon_{0} E_{0}^{2} \newline\\ \therefore u_{B} =u_{E} \newline\\ \text { Hence, } u_{\mathrm{av}} =\frac{1}{4} \varepsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}} \newline\\ =\frac{1}{4} \varepsilon_{0} E_{0}^{2}+\frac{1}{4} \varepsilon_{0} E_{0}^{2} \newline\\ =\frac{1}{2} \varepsilon_{0} E_{0}^{2}=\frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}} \end{array} $$

Time average intensity of the wave

$$ I_{\mathrm{av}}=u_{\mathrm{av}} \mathrm{c}=\frac{1}{2} \varepsilon_{0} E_{0}^{2} \mathrm{C}=\frac{1}{2} \varepsilon_{0} E_{0}^{2} $$