NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Question:8

An electromagnetic wave travels in vacuum along z-direction:

$E=\left ( E_1\hat{i}+E_2\hat{j} \right ) \cos \left ( kz-\omega t \right )$.
Choose the correct options from the following:

a) the associated magnetic field is given as $B=\frac{1}{c}\left ( E_1\hat{i}+E_2\hat{j} \right ) \cos \left ( kz-\omega t \right )$

b) the associated magnetic field is given as $B=\frac{1}{c}\left ( E_1\hat{i}-E_2\hat{j} \right ) \cos \left ( kz-\omega t \right )$
c) the given electromagnetic waves is plane polarised
d) the given electromagnetic field is circularly polarised

Answer:

The correct answers are the options a), and c).

In electromagnetic wave, the electric field vector is given as,

$
E=\left(E_1 \hat{i}+E_2 \hat{j}\right) \cos (k z-\omega t)
$
In electromagnetic wave, the associated magnetic field vector,

$
B=\frac{E}{c}=\frac{\left(E_1 \hat{i}+E_2 \hat{j}\right)}{c} \cos (k z-\omega t)
$
As $E$ and $B$ are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to $E$ as well as $B$, so the given electromagnetic wave is plane polarised.

Question:8.9

An electromagnetic wave travelling along $z$-axis is given as: $\mathbf{E}=\mathbf{E}_0 \cos (k z-\omega t$.$) . \text{Choose the correct options from the following;}$
(a) The associated magnetic field is given as $\mathbf{B}=\frac{1}{c} \hat{\mathbf{k}} \times \mathbf{E}=\frac{1}{\omega}(\hat{\mathbf{k}} \times \mathbf{E})$.
(b) The electromagnetic field can be written in terms of the associated magnetic field as $\mathbf{E}=c(\mathbf{B} \times \hat{\mathbf{k}})$.
(c) $\hat{\mathbf{k}} \cdot \mathbf{E}=0, \hat{\mathbf{k}} \cdot \mathbf{B}=0$.
(d) $\hat{\mathbf{k}} \times \mathbf{E}=0, \hat{\mathbf{k}} \times \mathbf{B}=0$.

Answer: The correct answers are the options (a, b and c)

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

$
\mathrm{E}=E_0 \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 $\hat{\mathbf{k}} \times \mathbf{E}$.
As,

$
\begin{aligned}
& B_o=\frac{E_o}{c} \\
& \mathbf{B}=\frac{1}{c}(\hat{\mathbf{k}} \times \mathbf{E})
\end{aligned}
$
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 $\hat{\mathbf{k}} \cdot \mathbf{B}=1 E \cos 90^{\circ}=0$.

Question:8.10

A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B
a) B_x, E_y
b) E_x, B_y
c) E_y, B_z
d) E_z, B_y

Answer: The correct answers are the options,

c) E_y, B_z
d) E_z, B_y
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.

Question:8.11

A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. The electromagnetic waves produced:
a) fall in the region of radio waves
b) will have a wavelength of 0.3 m
c) will have a frequency of 10⁹ Hz
d) will have a frequency of 2 × 10⁹ Hz

Answer: The correct answers are the options,
a. fall in the region of radio waves
b. will have a wavelength of 0.3 m
c. will have a frequency of 10⁹ Hz
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 $v=10^9 \mathrm{~Hz}$.
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.

Question:8.12

The source of electromagnetic waves can be a charge
a) at rest
b) moving with a constant velocity
c) falling in an electric field
d) moving in a circular orbit

Answer: The correct answers are the options,

c) falling in an electric field
d) 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.

Question:8.13

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?
a) radiation pressure is I/c if the wave is totally reflected
b) radiation pressure is I/c if the wave is totally absorbed
c) radiation pressure is in the range I/c < p < 2I/c for real surface
d) radiation pressure is 2I/c if the wave is totally reflected

Answer: The correct answers are the options,

a) radiation pressure is I/c if the wave is totally absorbed
c) radiation pressure is in the range I/c < p < 2I/c for real surface
d) radiation pressure is 2I/c if the wave is totally reflected

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}
$
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 $p=\frac{I}{\mathrm{C}}-\left(-\frac{I}{\mathrm{C}}\right)=\frac{2 I}{\mathrm{C}}$.
Here, $p$ lies between $\frac{I}{c}$ and $\frac{2 I}{c}$.

Question:8.14

Why is the orientation of the portable radio with respect to broadcasting stations important?

Answer:

We know that the electromagnetic waves are plane-polarized. Therefore, the antennas are to be placed in parallel to the vibration of the electric or magnetic field of the wave. That is why the orientation of the portable radio with respect to the broadcasting station is important.

Question:8.15

Why does a microwave oven heat up a food item containing water molecules most efficiently?

Answer:

The frequency of the microwave is exactly the same as that of the resonant frequency of the water molecules in a food item. That is why, the microwave oven heats up a food item containing water molecules most efficiently.

Question:8.16

The charge on a parallel plate capacitor varies as $q = q_0 \cos 2 \pi vt$. The plates are very large and close together. Neglecting the edge effects, find the displacement current through the capacitor?

Answer:

The displacement current through the capacitor is,

Here,

$
I_d=I_c=\frac{d q}{d t}
$
Putting this value in Eq (i), we get

$
\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}
$

Question:8.17

A variable frequency a.c source is connected to a capacitor. How will the displacement current change with a decrease in frequency?

Answer:

Capacitive reactance,

$X_c= \frac{1}{2} \pi fc$

As the frequency is decreasing, therefore capacitive reactance $(X_c)$ will increase with time and the current that is flowing through will become inversely proportional to $X_c$.

Question:8.18

The magnetic field of a beam emerging from a filter facing a floodlight is given by $B_0 = 12 \times 10^{-8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t)T.$ What is the average intensity of the beam?

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

$
\begin{aligned}
I_{\mathrm{av}} & =\frac{1}{2} \frac{B_0^2}{\mu_0} \cdot 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
\end{aligned}
$

Question:8.19

Poynting vectors S is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propagation. Mathematically, it is given by $S=\frac{1}{\mu_0} E \times B$. Show the nature of the S versus t graph.

Answer:

In an electromagnetic waves, Let $\vec{E}$ be varying along y-axis, $\vec{B}$ is along z- axis and propagation of wave be along x-axis. Then $\vec{E} \times \vec{B}$ will tell the direction of propagation of energy flow in electromagnetic wave, along x-axis.
Let
$\\ \vec{E} =E_0 \sin\left ( \omega t -kx \right )\hat{j}\\$

$\vec{B} =B_0 \sin\left ( \omega t -kx \right )\hat{k}\\ $

$S=\frac{1}{\mu_0}\left ( \vec{E} \times \vec{B} \right )=\frac{1}{\mu_0}\;E_0B_0\sin^{2}\left ( \omega t- kx\right )\left ( \hat{j}\times \hat{k} \right )\\$

$\Rightarrow S=\frac{E_0B_0}{\mu_0}\sin^{2}\left ( \omega t-kx \right )\hat{i}\; \; \; \; \; \;$
Since $\sin^{2}\left ( \omega t-kx \right )$ is never negative, $\vec{S}\left ( x,t \right )$ always point in the positive X-direction, i.e, in the direction of wave propagation.
The variation of $\left | S \right |$ with time T will be as given in the figure below:

Question:8.20

Professor C.V. Raman surprised his students by suspending a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property

Answer:

Like other waves, the electromagnetic wave has the same properties. Therefore, an electromagnetic wave has energy and momentum. It exerts pressure called radiation pressure due to momentum.

Question:8.21

Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is $\left ( \frac{\varepsilon _{0}\mu r}{2} \right )\frac{dE}{dt}$.

Answer:

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 $\frac{\varepsilon_0 \mu_0 r}{2} \frac{d E}{d t}$

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}{rlr}
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 \phi_E}{d t} & {\left[\because I_d=\frac{E_0 d \phi_E}{d t}\right]} \\
& =\frac{\mu_0 \varepsilon_0}{2 \pi r} \frac{d}{d t}\left(E \pi r^2\right)=\frac{\mu_0 \varepsilon_0}{2 \pi r} \pi r^2 \frac{d E}{d t} & {\left[\because \phi_E=E \pi r^2\right]} \\
B & =\frac{\mu_0 \varepsilon_0 r}{2} \frac{d E}{d t} &
\end{array}
$

Question:8.22

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 runaways during fog and mist conditions
a) identify and name the part of the electromagnetic spectrum to which these radiations belong
b) arrange these wavelengths in ascending order of their magnitude
c) write one more application of each

Answer:

a)

i) $\lambda _{1}$ is used in satellite communication : a microwave
ii) $\lambda _{2}$ is used to kill germs in water purifier for killing germs : a UV ray
iii) $\lambda _{3}$ is used in improving the visibility: a X-ray
iv) $\lambda _{4}$ is used to improve visibility in runaways during fog and mist conditions: an infrared ray

b) By arranging those wavelengths in ascending order: $\lambda _{3}<\lambda _{2}<\lambda _{4}<\lambda _{1}$

c)

i) Gamma rays: Those are used in nuclear structures.
ii) X rays: Those are used in medical diagnosis.
iii) UV rays: Those are used in the preservation of food.
iv) Infrared rays: Those are used to take photograph

Question:8.23

Show that the average value of radiant flux density S over a single period T is given by
$S= \frac{(\frac{1}{2}c\mu _{0})}{E{_{0}}^{2}}$

Answer:

Radiant flux density $S=\frac{1}{\mu_0}(\mathbf{E} \times \mathbf{B})=c^2 \varepsilon_0(\mathbf{E} \times \mathbf{B})$ $
\left[\because c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right]
$

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

$
\begin{aligned}
\mathbf{E}_0 & =\mathbf{E}_0 \cos (k x-\omega t) \\
\mathbf{B} & =\mathbf{B}_0 \cos (k x-\omega t) \\
\mathbf{E} \times \mathbf{B} & =\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \\
\mathbf{S} & =\mathbf{c}^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\
& =c^2 \varepsilon_0\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t)
\end{aligned}
$

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

$
\begin{aligned}
S_{\mathrm{av}} & =c^2 \varepsilon_0\left|\mathbf{E}_0 \times \mathbf{B}_0\right| \frac{1}{T} \int_0^T \cos ^2(k x-\omega t) d t \\
& =c^2 \varepsilon_0 E_0 B_0 \times \frac{1}{T} \times \frac{T}{2} \ \ \ \ {\left[\because \int_0^T \cos ^2(k x-\omega t) d t=\frac{T}{2}\right]} \\
\Rightarrow \quad S_{\mathrm{av}} & =\frac{c^2}{2} \varepsilon_0 E_0\left(\frac{E_0}{c}\right) \ \ {\left[\text { As, } c=\frac{E_0}{B_0}\right]}\\
& =\frac{c}{2} \varepsilon_0 E_0^2=\frac{c}{2} \times \frac{1}{c^2 \mu_0} E_0^2 \ \ \ {\left[c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{c^2 \mu_0}\right]} \\
\Rightarrow \quad S_{\mathrm{av}} & =\frac{E_0^2}{2 \mu_0 c}
\end{aligned}
$

Hence proved.

Question:8.24

You are given a $2\mu F$ parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?

Answer:

The capacitor has a capacitance of $C=2\mu F$
Displacement current $I_{d}=1\; mA$
Charge in the capacitor,
$q=CV$
or, $I_{d}dt =C\; dV \; \; \; \; \; \; \; \; \; \left [ as,q=it \right ]$
or, $I_{d} =C\; \frac{dV}{dt}$
or, $\frac{dV}{dt}= \frac{(1\times10^{-3})}{(2\times10^{-6})}$
or, $\frac{dV}{dt}= 500\frac{V}{s}$
Therefore, by applying a voltage of $500\frac{V}{s}$, we will get the mentioned displacement current.

Question:8.25

Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is $\frac{I}{c}$.

Answer:

Energy received by the surface per second, $E = IA$
No.of photons received per second by the surface
$=N=\frac{E}{E_{photon}}=\frac{E\lambda }{hc}=\frac{IA\lambda }{hc}$
For a perfectly absorbing surface, is
$\Delta P=\frac{h}{\lambda }$
Force, $F=N.\Delta P=\left ( \frac{IA\lambda }{hc} \right )\left ( \frac{h}{\lambda } \right )=\left ( \frac{IA}{c} \right )$
Pressure,
$P=\frac{F}{A}=\frac{I}{c}$

Question:8.26

What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant. What geometrical characteristics of the LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

Answer:

When the distance is doubled, the intensity of light becomes one-fourth.
Following are the geometrical characteristics of the LASER those missing in the case of normal light from the bulb:
a) unidirectional
b) monochromatic
c) coherent light
The above characteristics are missing in the case of light from the bulb

Question:8.27

Even though an electric field E exerts a force qE on a charged particle, yet the electric field of an EM wave does not contribute to the radiation pressure. Explain.

Answer:

Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure because radiation pressure is the result of the action of the magnetic field of the wave on the electric currents which is induced by the electric field.

Question:8.28

An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the z-axis. The wire is set into motion along its length with a uniform velocity $v=v\hat{k}_{z}$ Calculate the pointing vectors $S=\frac{1}{\mu _{0}}\left ( E \times B \right )$.

Answer:

$\begin{aligned} \mathbf{E} & =\frac{\lambda \hat{\mathbf{e}}_s}{2 \pi \varepsilon_0 a} \hat{\mathbf{j}} \\ \mathbf{B} & =\frac{\mu_0 \mathrm{i}}{2 \pi a} \hat{\mathbf{i}}=\frac{\mu_0 \lambda v}{2 \pi a} \hat{\mathbf{i}} \\ \mathbf{S} & =\frac{1}{\mu_0}[\mathbf{E} \times \mathbf{B}]=\frac{1}{\mu_0}\left[\frac{\lambda_{\hat{\mathbf{j}}}}{2 \pi \varepsilon_0 a} \times \frac{\mu_0}{2 \pi a} \lambda \sqrt{\hat{\mathbf{i}}}\right] \\ & =\frac{\lambda^2 v}{4 \pi^2 \varepsilon_0 a^2}(\hat{\mathbf{j}} \times \hat{\mathbf{i}})=-\frac{\lambda^2 v}{4 \pi^2 \varepsilon_0 a^2} \hat{\mathbf{k}}\end{aligned}$

Question:8.29

Seawater at frequency v = 4 x 10⁸ Hz has permittivity $\varepsilon =80\; \varepsilon _{0},$ permeability $\mu =\mu _{0}$ and resistivity $\rho =0.25\; \Omega m.$ Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source $V(t)=V_{0}\; \sin (2\pi vt)$. What fraction of the conduction current density is the displacement current density?

Answer:

Suppose distance between the parallel plates is $d$ and applied voltage $V(t)=V_o2\pi vt$ Thus, electric field

$
\begin{array}{ll}
\text { Now using Ohm's law, } & \begin{aligned}
E & =\frac{V_0}{d} \sin (2 \pi v t) \\
J_c & =\frac{1}{\rho} \frac{V_0}{d} \sin (2 \pi v t) \\
\Rightarrow &
\end{aligned}=\frac{V_0}{\rho d} \sin (2 \pi v t)=J_0^c \sin 2 \pi v t \\
\text { Here, } & J_0^c=\frac{V_0}{\rho d}
\end{array}
$

Now the displacement current density is given as

$
\begin{aligned}
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) \\
& =J_0^d \cos (2 \pi V t) \\
\text { where, } \quad & \quad J_0^d=\frac{2 \pi V \varepsilon V_0}{d} \\
\Rightarrow \quad \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 \\
& =2 \pi \times 80 \varepsilon_0 V \times 0.25=4 \pi \varepsilon_0 V \times 10 \\
& =\frac{10 \mathrm{~V}}{9 \times 10^9}=\frac{4}{9}
\end{aligned}
$

Question:8.30

A long straight cable of length l is placed symmetrically along the z-axis and has radius a. The cable consists of a thin wire and a coaxial conducting tube. An alternating current $I(t)=I_{0}\; \sin \left ( 2\pi vt \right )$ flows down the central thin wire and returns along the coaxial conducting tube. The induced electric field at a distance s from the wire inside the cable is $E(s,t)=\mu _{0}I_{0}v\; coz\; \left ( 2\pi vt \right ).$In $\left ( \frac{s}{a} \right )\hat{k}$
a) calculate the displacement current density inside the cable
b) integrate the displacement current density across the cross-section of the cable to find the total displacement current I
c) compare the conduction current $I_{o}$ with the displacement current $I_{o}d$

Answer:

(a) 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 \left(\frac{s}{a}\right) \hat{\mathbf{k}}
$
Now, displacement current density,

$
\mathbf{J}_d=\varepsilon_0 \frac{d \mathbf{E}}{d t}=\varepsilon_0 \frac{d}{d t}\left[\mu_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}\right]
$

$\begin{aligned} & =\varepsilon_0 \mu_0 I_0 v \frac{d}{d t}[\cos 2 \pi v t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \\ & =\frac{1}{c^2} I_0 v^2 2 \pi[-\sin 2 \pi v t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \\ & =\frac{v^2}{c^2} 2 \pi I_0 \sin 2 \pi v t \ln \left(\frac{a}{s}\right) \hat{\mathbf{k}} \\ & =\frac{1}{\lambda^2} 2 \pi I_0 \ln \left(\frac{a}{s}\right) \sin 2 \pi v t \hat{\mathbf{k}} \\ & =\frac{2 \pi I_0}{\lambda_2} \ln \frac{a}{s} \sin 2 \pi v t \hat{\mathbf{k}}\end{aligned}$

(b)

$\begin{aligned} I_d & =\int J_d s d s d \theta=\int_{s=0}^a J_d s d s \int_0^{2 \pi} d \theta=\int_{s=0}^a J_d s d s \times 2 \pi \\ & =\int_{s=0}^a\left[\frac{2 \pi}{\lambda^2} I_0 \log _e\left(\frac{a}{s}\right) s d s \sin 2 \pi v t\right] \times 2 \pi \\ & =\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \int_{s=0}^a\left(\frac{a}{s}\right) s d s \sin 2 \pi v t \\ & =\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \int_{s=0}^a \ln \left(\frac{a}{s}\right) \frac{1}{2} d\left(s^2\right) \cdot \sin 2 \pi v t \\ & =\frac{a^2}{2}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right) \cdot d\left(\frac{s}{a}\right)^2 \\ & =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{a}{s}\right)^2 \cdot d\left(\frac{s}{a}\right)^2 \\ & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \int_{s=0}^a \ln \left(\frac{s}{a}\right)^2 \cdot d\left(\frac{s}{a}\right)^2\end{aligned}$

$\begin{aligned} & =-\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t \times(-1) \\ \therefore & I_d =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi v t =\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0 \sin 2 \pi v t\end{aligned}$

(c)

The displacement current,

$
\begin{aligned}
I_d =\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0 \sin 2 \pi v t=I_{0 d} \sin 2 \pi v t \\
\text { Here, } I_{0 d} =\left(\frac{2 \pi a}{2 \lambda}\right)^2 I_0=\left(\frac{a \pi}{\lambda}\right)^2 I_0 \\
\therefore \frac{I_{0 d}}{I_0} =\left(\frac{a \pi}{\lambda}\right)^2
\end{aligned}
$

Question:8.31

A plane EM wave travelling in vacuum along z-direction is given by
$E=E_{0}\sin (kz-\omega t)\hat{i}\; and \; B=B_{0}\sin (kz-\omega t)\hat{j}$
a) evaluate $\oint E.dl$ over the rectangular loop 1234 shown in the figure

b) evaluate $\int B.ds$ over the surface bounded by loop 1234
c) use equation $\oint E.dl=\frac{-d\phi _{B}}{dt}$ to prove $\frac{E_{0}}{B_{0}}=c$
d) by using a similar process and the equation
$\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt}$ prove that $c=\frac{1}{\sqrt{\mu _{0}\epsilon _{0}}}$

Answer:

a)

During the propagation of electromagnetic wave a long $z$-axis, let electric field vector (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}
\oint \mathbf{E} \cdot \mathbf{d} \mathbf{l} & =\int_1^2 \mathbf{E} \cdot \mathbf{d} \mathbf{l}+\int_2^3 \mathbf{E} \cdot \mathbf{d} \mathbf{l}+\int_3^4 \mathbf{E} \cdot \mathbf{d} \mathbf{l}+\int_4^1 \mathbf{E} \cdot \mathbf{d} \mathbf{l} \\
& =\int_1^2 \mathbf{E} \cdot \mathbf{d} \operatorname{los} 90+\int_2^3 \mathbf{E} \cdot \mathbf{d} \operatorname{lcos} 0+\int_3^4 \mathbf{E} \cdot \mathbf{d} \operatorname{lcos} 90+\int_4^1 \mathbf{E} \cdot \mathbf{d} \operatorname{lcos} 180^{\circ} \\
& =E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-g \omega t\right)\right]
\end{aligned}
$

b)

For evaluating $\int \mathbf{B} \cdot \mathbf{d s}$, 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 s}=\int \mathbf{B} \cdot \mathbf{d s} \cos 0 & =\int \mathbf{B} \cdot \mathbf{d 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}
$

c)

Given, $\oint \mathbf{E} \cdot \mathbf{d l}=\frac{-d \phi_B}{d t}=-\frac{d}{d t} \oint \mathbf{B} \cdot \mathbf{d s}$
Putting the values from Eqs. (i) and (ii), we get

$
\begin{array}{ll}
& E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\
= & \frac{-d}{d t}\left[\frac{B_y h}{k}\left\{\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right]\right. \\
= & \frac{B_y h}{k} \omega\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\
\Rightarrow & E_0=\frac{B_0 \omega}{k}=B_y c \\
\Rightarrow & \frac{E_0}{B_0}=c
\end{array}
$

d)

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

$
\begin{aligned}
\oint \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} \\
& =\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} \\
& =B_0 h\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right.
\end{aligned}
$

Question:8.32

A plane EM wave travelling along z-direction is described by $E=E_{0}\sin (kz-\omega t)\hat{i}\; and\; B=B_{0}\sin (kz-\omega t)\hat{j}$ show that,
i) the average energy density of the wave is given by
$u_{av}=\frac{(u_{E}+u_{B})}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$
ii) the time - averaged intensity of the wave is given by
$I_{av}=u_{av}c=\frac{1}{2} (\varepsilon _{0}E^{2})c$

Answer:

i) The energy density due to electric field E is
$u_{\varepsilon }=\frac{1}{2}(\varepsilon _{0}E^{2})$
The energy density due to magnetic field B is
$u_{B}=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )$
The average energy density of the wave is given by :
$u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$
ii) $u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$
Assuming $u_{E}=u_{B},$
$u_{av}=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )$
We know that $c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}=\left ( \frac{E_{0}}{B_{0}} \right )$
The time -averaged intensity of the wave is given as
$I_{av}=u_{av}c=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )c$