exemplar-phy-12-electromagnetics-waves

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Ans. (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
$\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 $
<Success>Sol. to be continued …</Success>
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Ques. A linearly polarised electromagnetic wave given as ${E}=E_0 \hat{{i}} \cos (k z-\omega t)$ is incident normally on a perfectly reflecting infinite wall at $z=a$. Assuming that the material of the wall is optically inactive, the reflected wave will be given as
<div class="fragment">
(a)$E_r=-E_0 \hat{i} \cos (k z-\omega t)$
(b)$E_r=E_0 \hat{i} \cos (k z+\omega t)$
(c)$E_r=-E_0 \hat{i} \cos (k z+\omega t)$
(d)$E_r=E_0 \hat{i} \sin (k z-\omega t)$
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Thinking process :
When a wave is reflected from a denser medium, then its phase changes by $180^{\circ}$ or $\pi$
Ans. (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 $\hat{{z}}=-\hat{{z}}, \hat{{i}}=-\hat{{i}}$ and additional phase of $\pi$ in the incident wave.
Given, here the incident electromagnetic wave is,
<Success>Sol. to be continued …</Success>
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Ques. Light with an energy flux of $20 W / cm^{2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30 cm^{2}$, the total momentum delivered (for complete absorption) during 30 minutes is
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(a) $36 \times 10^{-5}$ kg m/s
(b) $36 \times 10^{-4}$ kg m/s
(c) $108 \times 10^{4}$ kg m/s
(d) $1.08 \times 10^{7}$ kg m/s
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Ans. (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
$\frac{E^{\prime}}{E}=\frac{50}{100} \Rightarrow \frac{E^{\prime}}{E}=\frac{1}{2} \Rightarrow E^{\prime}=\frac{E}{2}$
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Ans. (d)
The direction of propagation of electromagnetic wave is perpendicular to both electric field vector ${E}$ and magnetic field vector ${B}$, i.e., in the direction of ${E} \times {B}$.
This can be seen by the diagram given below
<img src="https://cdn.mathpix.com/snip/images/Lw9TGLxtx28rVmBqU54maysGCVk3UTfjxn32YJcxQ6Y.original.fullsize.png" height="120px">
Here, electromagnetic wave is along the $z$-direction which is given by the cross product of ${E}$ and ${B}$.
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Ques. An electromagnetic wave travels in vacuum along $z$-direction ${E}=(E_1 \hat{{i}} E_2 \hat{{j}}) \cos (k z-\omega t)$. Choose the correct options from the following
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(a) The associated magnetic field is given as ${B}=\frac{1}{c}(E_1 \hat{{i}}-E_2 \hat{{j}}) \cos (k z-\omega t)$
(b) The associated magnetic field is given as ${B}=\frac{1}{c}(E_1 \hat{{i}}-E_2 \hat{{j}}) \cos (k z-\omega t)$
(c) The given electromagnetic field is circularly polarised
(d) The given electromagnetic wave is plane polarised
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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
$B_0=\frac{E_0}{c}$
Ans. (d)
Here, in electromagnetic wave, the electric field vector is given as,
${E}=(E_1 \hat{{i}}+E_2 \hat{{j}}) \cos (k z-\omega t)$
<Success>Sol. to be continued …</Success>
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Ques. An electromagnetic wave travelling along $z$-axis is given as ${E}={E} _0 \cos (k z-\omega t)$. Choose the correct options from the following
<div class="fragment">
(a) The associated magnetic field is given as $B=\frac{1}{c} \hat{{k}} \times {E}=\frac{1}{\omega}(\hat{{k}} \times {E})$
(b) The electromagnetic field can be written in terms of the associated magnetic field as ${E}=c({B} \times \hat{{k}})$
(c) $\hat{{k}} . {E}=0, \hat{{k}} . {B}=0$
(d) $\hat{{k}} \times {E}=0, \hat{{k}} \times {B}=0$
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Thinking process :
Given, $E=E_0 \cos (k z-w t)$. Thus, it acts along negative y-direction.
Ans. (a, b, c)
Suppose an electromagnetic wave is travelling along negative z-direction. Its electric field is given by
${E}=E_O \cos (k z-\omega t)$
which is perpendicular to $z$-axis. It acts along negative $y$-direction.
<Success>Sol. to be continued …</Success>
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The associated magnetic field ${B}$ in electromagnetic wave is along $x$-axis i.e., along $\hat{{k}} \times {E}$.
As, $B_0=\frac{E_0}{C} $
$\therefore B=\frac{1}{c}(\hat{{k}} \times {E}) $
The associated electric field can be written in terms of magnetic field as
${E}=c({B} \times \hat{{k}})$
Angle between $\hat{{k}}$ and ${E}$ is $90^{\circ}$ between $\hat{{k}}$ and ${B}$ is $90^{\circ}$.
Therefore, ${E}=1 E \cos 90^{\circ}=0$ and $\hat{{k}} . {B}=1 E \cos 90^{\circ}=0$
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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.
Ans. (a, c, d)
Given, frequency by which the charged particles oscillates about its mean equilibrium position $=10^{9}$ Hz.
So, frequency of electromagnetic waves produced by the charged particle is $v=10^{9}$ Hz.
Wavelength $\lambda=\frac{c}{v}=\frac{3 \times 10^{8}}{10^{9}}$ = 0.3 m
Also, frequency of $10^{9}$ Hz fall in the region of radiowaves.
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Thinking process :
An electromagnetic wave can be produced by accelerated or oscillating charge.
Ans. (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.
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Ques. An EM wave of intensity $I$ falls on a surface kept in vacuum and exerts radiation pressure $p$ on it. Which of the following are true?
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(a) Radiation pressure is $\frac{I}{c}$ if the wave is totally absorbed
(b) Radiation pressure is $\frac{I}{c}$ if the wave is totally reflected
(c) Radiation pressure is $\frac{2 I}{c}$ if the wave is totally reflected
(d) Radiation pressure is in the range $\frac{I}{c}<p<\frac{2 I}{c}$ for real surfaces
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Ans. (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
$=\frac{\text{ Intensity }}{\text{ Speed of wave }}=\frac{I}{c}$
Change in momentum per unit time per unit area $=\frac{\Delta I}{c}=$ radiation pressure $(p)$ i.e.,
$p=\frac{\Delta I}{c}$
<Success>Sol. to be continued …</Success>
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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
$=-\frac{I}{c}$, Radiation pressure $p=\frac{I}{c}–\frac{I}{c}=\frac{2 I}{c}$.
Here, $p$ lies between $\frac{I}{c}$ and $\frac{2 I}{c}$.
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Thinking process :
Capacities reactance $X_c$ is inversely proportional to the displacement current i.e., $X_c \propto \frac{1}{I}$.
Ans.
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.
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Ans.
Magnetic field ${B}=B_0$ sin $\omega t$
Given, equation $B=12 \times 10^{-8} \sin (1.20 \times 10^{7} z-3.60 \times 10^{15} t) T$.
On comparing this equation with standard equation, we get
$B_0=12 \times 10^{-8}$
The average intensity of the beam $I_{av}=\frac{1}{2} \frac{B_0^{2}}{\mu_0} . C=\frac{1}{2} \times \frac{(12 \times 10^{-8})^{2} \times 3 \times 10^{8}}{4 \pi \times 10^{-7}}$
= 1.71 W/m$^{2}$
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Ans.
Consider and electromagnetic waves, let ${E}$ be varying along $y$-axis, ${B}$ is along z-axis and propagation of wave be along $x$-axis. Then ${E} \times {B}$ will tell the direction of propagation of energy flow in electromegnetic wave, along $x$-axis.
Let
${E} =E_0 \sin (\omega t-k x) \hat{{j}} $
${B} =B_0 \sin (\omega t-k x) \hat{{k}} $
<Success>Sol. to be continued …</Success>
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Ans.
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 C.V 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.
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Ans.
Consider the figure ginen below to prove that the magneti field $B$ at a point in between the plater of a paravel- plate copocior during charging is $\frac{\varepsilon_0 \mu_0 r}{2} \frac{d E}{d t}$
<img src="https://cdn.mathpix.com/snip/images/uOmOu6YMFDhE1PoOvudn3UPONXt42tSHhQsVlnqwMJM.original.fullsize.png" height="100px">
Let $I_d$ be the displacement current in the region between two plates of parallel plate capacitor, in the figure.
<Success>Sol. to be continued …</Success>
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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
$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}(E r^{2})=\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} $
$ {[\because \varphi_E=E r^{2}]} $
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Ques. Electromagnetic waves with wavelength
(i) $\lambda_1$ is used in satellite communication.
(ii) $\lambda_2$ is used to kill germs in water purifies.
(iii) $\lambda_3$ is used to detect leakage of oil in underground pipelines.
(iv) $\lambda_4$ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.
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Ans.
(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.
<Success>Sol. to be continued …</Success>
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Ans.
Radiant flux density $S=\frac{1}{\mu_0}({E} \times {B})=c^{2} \varepsilon_0({E} \times {B})$
$\because c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
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) $
<Success>Sol. to be continued …</Success>
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$ {B} =B_0 \cos(k x-\omega t) $
$E \times B =(E_0 \times B_0) \cos^{2}(k x-\omega t) $
$S =c^{2}0(E \times B) $
$=c^{2}0(E_0 \times B_0) \cos^{2}(k x-\omega t) $
Average value of the magnitude of radiant flux density over complete cycle is
$ S _{av} =c^{2} 0|E_0 \times B_0| \frac{1}{T} \int_0^{T} \cos ^{2}(k x-\omega t) d t $
<Success>Sol. to be continued …</Success>
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$=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} $
$\Rightarrow S _{av}=\frac{c^{2}}{2} 0 E_0 \frac{E_0}{c} \text{ As, } c=\frac{E_0}{B_0} $
$=\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} $
$\Rightarrow \quad S _{av} =\frac{E_0^{2}}{2 \mu_0 c}$ Hence proved.
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Ans.
As the distance is doubled, the area of spherical region $(4 \pi^{2})$ 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
 
<Success>Sol. to be continued …</Success>
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Ques. An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the $z$-axis (figure). The wire is set into motion along its length with a uniform velocity $v=v \hat {k_z}$. Calculate the pointing vector $S=\frac{1}{\mu_0}(E \times B)$.
<img src="https://cdn.mathpix.com/snip/images/jpggZFusSg4D-itqLibwkotUprbiWkUYHjGTaNpM6ck.original.fullsize.png" height="200px">
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Ans.
Given,
$B =\frac{\mu_0 i}{2 \pi a} \hat{i}=\frac{\mu_0 \lambda v}{2 \pi a} \hat{i} $
$\therefore 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}} $
$ =\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}$
$[\because I=\lambda_V]$
$E=\frac{\lambda \hat {e_s}}{2 \pi 0 a} \hat{j}$
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Now the displacement current density is given as 
$J_d =\varepsilon \frac{\delta E}{d t}=\frac{\varepsilon \delta}{d t} $
$ =\frac{\varepsilon 2 \pi v V_0}{d} \cos (2 \pi v t) $
$\Rightarrow =J_0^{d} \cos (2 \pi v t) $
where, $J_0^{d} =\frac{2 \pi V \varepsilon V_0}{d} $
$\Rightarrow \frac{J_0^{d}}{J_0^{c}} =\frac{2 \pi v \varepsilon V_0}{d} . \frac{\rho d}{V_0}=2 \pi v \varepsilon \rho $
$ =2 \pi \times 80 \varepsilon_0 v \times 0.25=4 \pi \varepsilon_0 v \times 10 $
$ =\frac{10 v}{9 \times 10^{9}}=\frac{4}{9} $
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Ques. A long straight cable of length $l$ is placed symmetrically along $z$-axis and has radius $a(«l)$. The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I(t)=I_0 \sin (2 \pi v t)$ flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance $s$ from the wire inside the cable is
$E(s, t)=\mu_0 I_0 v \cos (2 \pi v t) \ln \frac sa \hat{k}$
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current $I^{d}$.
(iii) Compare the conduction current $I_0$ with the displacement current $I_0^{d}$.
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Now, displacement current density,
$Jd=\varepsilon_0 \frac{d {E}}{d t}=\varepsilon_0 \frac{d}{d t} \mu_0 I_0 v \cos (2 \pi v t) \ln \frac{s}{a} \hat{{k}}$
(ii) $ 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 (\frac{2 \pi}{\lambda})^2 I _0 \int _{s=0}^a \ln (\frac{a}{s}) \frac{1}{2} d(s^2) . \sin 2 \pi v t $
<Success>Sol. to be continued …</Success>
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$ =\frac{a^2}{2}(\frac{2 \pi}{\lambda})^2 I_0 \sin 2 \pi v t \int _{s=0}^a \ln (\frac{a}{s}) \cdot d(\frac{s}{a})^2 $
$ =\frac{a^2}{4}(\frac{2 \pi}{\lambda})^2 I_0 \sin 2 \pi v t \int _{s=0}^a \ln (\frac{a}{s})^2 \cdot d(\frac{s}{a})^2 $
$ =-\frac{a^2}{4}(\frac{2 \pi}{\lambda})^2 I_0 \sin 2 \pi v t \int _{s=0}^a \ln (\frac{s}{a})^2 \cdot d(\frac{s}{a})^2 $
$ =-\frac{a^2}{4}(\frac{2 \pi}{\lambda})^2 I_0 \sin 2 \pi v t \times(-1) $
$ {[\because \int _{s=0}^a \ln (\frac{s}{a})^2 d(\frac{s}{a})^2=-1]} $
$\therefore I_d =\frac{a^2}{4}(\frac{2 \pi}{\lambda})^2 I_0 \sin 2 \pi v t $
<Success>Sol. to be continued …</Success>
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Ques. A plane EM wave travelling in vacuum along $z$-direction is given by ${E}=E_0 \sin (k z-\omega t) \hat{{i}}$ and ${B}=B_0 \sin (k z-\omega t) \hat{{j}}$. 
(i) Evaluate $\int {E} . {d l}$ over the rectangular loop 1234 shown in figure. 
(ii) Evaluate $\int {B} . {d s}$ over the surface bounded by loop 1234. 
(iii) Use equation $\int {E} . {d} {l}=\frac{-d \varphi_B}{d t}$ to prove $\frac{E_0}{B_0}=c$. 
(iv) By using similar process and the equation $\int {B} . {d l}=\mu_0 I+\varepsilon_0 \frac{d \varphi_E}{d t}, $ prove that $ c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
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During the propagation of electromagnetic wave a long $z$-axis, let electric field vector $({E})$ be along $x$-axis and magnetic field vector ${B}$ along $y$-axis, i.e., ${E}=E_0 \hat{{i}}$ and ${B}=B_0 \hat{{j}}$.
Line integral of $E$ over the closed rectangular path 1234 in $x$ - $z$ plane of the figure
$ {E} . {d l} =\int_1^{2} E . d l+\sqrt{2} . \frac{3}{E} . d l+\int_3^{4} . d l+\int_4^{1} E . d l $
$ =\int_1^{2} E . d l \cos 90+\int_2^{3} E . d l \cos 0+\int_3^{4} E . d l \cos 90+\int_4^{1} E . d l \cos 180^{\circ} $
$ =E_0 h[\sin (k z_2-\omega t)-\sin (k z_1-g \omega t)] $ 
(ii) For evaluating $\boldsymbol{{}\beta} {B} . {d s}$, let us consider the rectangle 1234 to be made of strips of are $d s=h d z$ each.
<Success>Sol. to be continued …</Success>
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(iii) Given, $g {E} . dl=\frac{-d \varphi_B}{d t}=-\frac{d}{d t} g {B} . {d s}$
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 $
$\Rightarrow \quad E_0=\frac{B_0 \omega}{k}=B_y c \because \frac{\omega}{k}=c$
<Success>Sol. to be continued …</Success>
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$\int {B} . {d l} =\int_1^{2} {B} . {d l}+\int_2^{3} {B} . {d l}+\int_3^{4} {B} . {d l}+\int_4^{1} {B} . {d l}$
$ =\int_1^{2} {B} . {d l} \cos 0+\int_2^{3} {B} . {d l} \cos 90^{\circ}+\int_3^{4} {B} . {d l} \cos 180^{\circ}+\int_4^{1} {B} . {d l} \cos 90^{\circ}$
$ =B_0 h[\sin (k z_1-\omega t)-\sin (k z_2-\omega t). $
Now to evaluate $\varphi_E=\int {E} . {d s}$, let us consider the rectangle 1234 to be made of strips of area $h d_2$ each.
 
<Success>Sol. to be continued …</Success>
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$\therefore \frac{d \varphi_E}{d t} =\frac{E_0 h \omega}{k}[\sin (k z_1-\omega t)-\sin (k z_2-\omega t)] $
$\text{ In } g {B} . {d l} =\mu_0 I+\frac{\varepsilon_0 d \varphi_E}{d t}, I=\text{ conduction current } $
$ =0 \text{ in vacuum } $
$\therefore g {B} . dl =\mu_0 \varepsilon \frac{d \varphi_E}{d t} $
Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get
$ B_0 =E_0 \frac{\omega \mu_0 \varepsilon_0}{k} $
<Success>Sol. to be continued …</Success>
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Ques. A plane EM wave travelling along $z$-direction is described by ${E}=E_0 \sin (k z-\omega t) \hat{{i}}$ and ${B}=B_0 \sin (k z-\omega t) \hat{{j}}$. Show that
(i) the average energy density of the wave is given by $u _{av}=\frac{1}{4} \varepsilon_0 E_0^{2}+\frac{1}{4} \frac{B_0^{2}}{\mu_0}$
(ii) the time averaged intensity of the wave is given by $I _{av}=\frac{1}{2} c \varepsilon_0 E_0^{2}$
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Ans.
(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
<Success>Sol. to be continued …</Success>
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$u_B=\frac{1}{2} \frac{B^{2}}{\mu_0}$
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
$ E=E_0 \sin (k z-\omega t) $
$ B=B_0 \sin (k z-\omega t) $
<Success>Sol. to be continued …</Success>
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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 $=\frac{B_0^{2}}{2}$
$u _{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} $
(ii) We know that $E_0=c B_0$ and $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
<Success>Sol. to be continued …</Success>
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$\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} $
$\therefore u_B =u_E $
Hence, $ u _{av} =\frac{1}{4} \varepsilon_0 E_0^{2}+\frac{1}{4} \frac{B_0^{2}}{\mu_0} $
$ =\frac{1}{4} \varepsilon_0 E_0^{2}+\frac{1}{4} \varepsilon_0 E_0^{2} $
$ =\frac{1}{2} \varepsilon_0 E_0^{2}=\frac{1}{2} \frac{B_0^{2}}{\mu_0} $
Time average intensity of the wave
$I _{av}=u _{av} c=\frac{1}{2} \varepsilon_0 E_0^{2} C=\frac{1}{2} \varepsilon_0 E_0^{2}$
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