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NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Edited By Safeer PP | Updated on Sep 14, 2022 01:00 PM IST

NCERT Exemplar Class 12 Physics solutions chapter 8 deals with the concept of accelerating charge which produces electric field as well as magnetic field which collectively form electromagnetic waves. Stationary charges have only electric fields associated with it whereas charges moving with a constant velocity have both electric and magnetic fields associated with it. NCERT Exemplar Class 12 Physics chapter 8 solutions explains different types of electromagnetic waves and their application. .Class 12 Physics NCERT exemplar solutions chapter 8 also exhibits electromagnetic waves as a transverse wave and gives information about Maxwell’s equations of (unification of electricity and magnetism) electromagnetism.
Also check - NCERT Solutions for Class 12 Physics

This Story also Contains
  1. Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves
  2. What will the Students learn from NCERT Exemplar Class 12 Physics Solutions Chapter 8?
  3. NCERT Exemplar Class 12 Physics Chapter Wise Links
  4. Important Topics To Cover For Exams From NCERT Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Exemplar solutions for Class 12 Physics chapter 8 are very useful and detailed from the point of view of board exams as well as competitive exams like NEET and JEE Main. NCERT Exemplar Class 12 Physics solutions chapter 8 PDF download is also available, prepared by experts after thorough study.
Also see - NCERT Solutions for Class 12 Other Subjects

Question:1

One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

  1. microwave region
  2. visible region
  3. ultraviolet region
  4. infrared region

Answer:

The answer is the option (c) ultraviolet region
Solution: E = hf
E = required energy to dissociate a carbon monoxide molecule = 11 eV
h = Planck's constant = 6.62 * 10^{-34} j-s
\\f = frequency; f = E/h = 11 * 1.6 * \frac{10^{-19}}{6.62} * 10^{-34} Hz = 2.65 * 10^{15} Hz.
The frequency radiation belongs to the ultraviolet region.

Question:2

A linearly polarized electromagnetic wave given as E=E_0\hat{i}\cos\left ( kz-\omega t \right ) 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
(a) E_r=E_0\hat{i}\cos\left ( kz+\omega t \right )
(b) E_r=-E_0\hat{i}\cos\left ( kz-\omega t \right )
(c) E_r=-E_0\hat{i}\sin\left ( kz-\omega t \right )
(d) E_r=-E_0\hat{i}\cos\left ( kz+\omega t \right )

Answer:

The answer is the option (a)

E_r=E_0\hat{i}\cos\left ( kz+\omega t \right )
Solution: A wave used to be the same rather its phase changes by 180° or π radian when it is reflected from a denser medium or wall that is perfectly reflecting and made with optically inactive material.

Question:3

Light with an energy flux of 20 wb/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered for 30 minutes is
A. 108 × 104 kg m/s
B. 36 × 10-5 kg m/s
C. 1.08 × 107 kg m/s
D. 36 × 10-4 kg m/s

Answer:

Answer: The answer is the option (d) 36 \times 10^{-4} kg m/s

Solution: \phi = energy\; flux = 20 \;wb/ cm^{2}
\\A = Surface \;area = 30 cm^{2}\\\\t = time = 30 \times 60 seconds = 1800 seconds

U = Energy falls on the surface in that time = \phi At = 20 \times 30 \times 1800 J = 1080000 J

Momentum of the incident light = \frac{U}{c} = 1080000/3 \times 10^{8} = 36 \times 10^{-4} kg m/s

As the momentum of the reflected light is zero, then momentum of the delivered light is 36 \times 10^{-4} kg \;m/s

Question:3

The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is
A. E/2
B. E/\sqrt{2}
C. \sqrt{2}/E
D. 2E

Answer:

I\propto \sqrt{P}

The answer is the option (b) E/\sqrt{2}

Question:5

If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of the electromagnetic wave is along
a. E
b. B
c. E \times B
d. B \times E

Answer:

Answer: The answer is the option (c) E \times B

Question:6

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is
a. c:1
b. \sqrt{c}:1
c. c^{2}:1
d. 1:1

Answer:

The answer is the option (d) 1: 1

Question:7

An EM wave radiates outwards from a dipole antenna, with E_{0} as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as
a. \frac{1}{r^{3}}
b. \frac{1}{r}
c. \frac{1}{r^{2}}
d. remains constant

Answer:

A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector falls intensity inversely as the distance (r) from the antenna

The answer is the option (b) 1/r

Question:8

An electromagnetic wave travels in vacuum along z-direction:

E=\left ( E_1\hat{i}+E_1\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_1\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_1\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), b), and d).

Question:9

An electromagnetic wave travelling along the z-axis is given as: E = E_0 \cos (kz -\omega t). Choose the correct options from the following
(a) \hat{k}\times E=0,\hat{k}\times B=0
(b) \hat{k}. E=0,\hat{k}\times B=0
(c) the associated magnetic field is given as
B= \frac{1}{c}\hat{k} \times E=\frac{1}{\omega }\left ( \hat{k}\times E \right )
(d) the electromagnetic field can be written in terms of the associated magnetic field as
E=c\left ( B\times \hat{k} \right )

Answer:

Answer: The correct answers are the options (b, c and d)
The direction of propagation of an electromagnetic wave is always along the direction of vector product \vec{E}\times \vec{B}. Refer to figure
(a)
\\\vec{B}=B\hat{j}=B\left ( \hat{k}\times \hat{i} \right )=\frac{E}{c}\left ( \hat{k}\times \hat{i} \right )\\\\ =\frac{1}{c}\left [ k\times E\hat{i} \right ]=\frac{1}{c}\left [ \hat{k} \times \vec{E} \right ]\: \: \: As\frac{E}{B}=c
54623

(b)
\\\vec{E}=E\hat{i}=cB\left ( \hat{j} \times\hat{k}\right )=c\left (B \hat{j} \times\hat{k}\right )=c\left (\vec{B}\times \hat{k} \right )
(c)
\\ \hat{k}\cdot \vec{E}=\hat{k}\cdot \left ( E\hat{i} \right )= 0,\vec{k}.\vec{B}=\hat{k}\cdot \left ( B\hat{j} \right )=0
(d)
\\ \hat{k}\times \vec{E}=\hat{k}\cdot \left ( E\hat{i} \right )= E\left ( \hat{k} \times \hat{i}\right )=E\hat{j}\\\\and\: \: \hat{k}\times \vec{B}= \hat{k}\times \left ( B\hat{j} \right )=B\left ( \hat{k}\times \hat{j} \right )=-B\hat{i}

Question:10

A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B
a) Bx, Ey
b) Ex, By
c) Ey, Bz
d) Ez, By

Answer:

Answer: The correct answers are the options,
c) Ey, Bz
d) Ez, By
Solution: The direction of propagation of electromagnetic waves is perpendicular with the magnetic field vector(B) and also with the electric field vector(E). Therefore, the electromagnetic wave that is propagating along the x-direction, will have electric and magnetic field vectors either in Y-direction or 2-direction.

Question: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:

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
Here, the particle that produces electromagnetic waves of frequency is equal to the frequency at which it oscillates about its mean equilibrium position.
So, the frequency of electromagnetic waves produced by the charged particle is v= 109 Hz.
Wavelength (\lambda ) = c/v = 3 \times 108 /109 = 0.3 m
The 109 Hz frequency falls under the radio wave region.

Question: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:

Answer: The correct answers are the options,
c) falling in an electric field
d) moving in a circular orbit

Question: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:

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

Question: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: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: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:

Here, the value of displacement current through the capacitor is Id

Therefore,

I_d = I_c = \frac{dq}{dt}.......a)

Given,

q = q_0 \cos 2 \pi vt

Substituting the values in the equation a), we get

I_d = I_c = \left ( \frac{dq}{dt} \right )\left ( q_0 \cos 2 \pi vt \right )= -2 \pi v \;q_0 \sin2 \pi vt

Question: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 (Xc) will increase with time and the current that is flowing through will become inversely proportional to Xc.

Question: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:

Here, given equation,

B_0 = 12 \times 10^{-8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t)T.

Standard equation,

B_0 = 12 \times 10^{-8 } \;T

By comparing the given equation with the standard equation, we get:

I_{av} = \frac{1}{2}\;\left ( \frac{B_0^{2}}{u_0} \right )c = 1.71 \; \;19 \;wb/m^{2}

Question:19

Pointing 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 Twill be as given in the figure below:
54620

Question: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: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:

21
Here, I_{d} is the displacement current between the two plates of a parallel plate capacitor.
r = distance between the two plates
Therefore,
B=\mu _{0}*\frac{2I_{d}}{4\pi r}
=\frac{(\mu _{0}I_{d})}{2\pi r}
=\frac{\mu _{0}}{2\pi r} \varepsilon _{0}\left ( \frac{d\phi _{E}}{dt} \right )\; \; \; \; \; \; \; \; \left [ as, I_{d}=\varepsilon _{0} \left ( \frac{d\phi _{E}}{dt} \right )\right ]
=\left ( \frac{\mu _{0}\varepsilon _{0}}{2\pi r} \right ) \frac{d(E\pi r^{2})}{dt}\; \; \; \; \; \; \; \; \; \; \; \; \left [ as,\phi _{E}=E\pi r^{2} \right ]
=\left ( \frac{\mu _{0}\varepsilon _{0}r}{2} \right )\left ( \frac{dE}{dt} \right )

Question: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: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:

We know that radiant flux density is:
\overrightarrow{S}=\frac{1}{\mu _{0}}(\overrightarrow{E}\times \overrightarrow{B})=C^{2}\epsilon _{0} (\overrightarrow{E}\times \overrightarrow{B}) [as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}]
Let, along the X-axis, electromagnetic waves are propagating. If along the Y-axis, the electric field vector of electromagnetic waves is there. In the same way, along the z-axis, magnetic field vectors will be there. Hence,
E=E_{0}\cos (kx-\omega t)
B=B_{0}\cos (kx-\omega t)
E\times B=\left ( E_{0}B_{0} \right )\cos^{2} (kx-\omega t)
S=C^{2}\varepsilon _{0}\left ( E\times B \right )=C^{2}\varepsilon _{0}\left ( E_{0}B_{0} \right )\cos ^{2}\left ( kx-\omega t \right )
The average value of the radiant flux density is
S_{av}=C^{2}\varepsilon _{0}\left | E_{0}\times B_{0} \right |\frac{1}{T}\int_{0}^{T} \cos ^{2}\left ( kx-\omega t \right )dt=c^{2}\varepsilon _{0}E_{0}B_{0}\left ( \frac{1}{T} \right )\left ( \frac{T}{2} \right )=\left ( c^{2}\frac{\varepsilon _{0}E_{0}}{2} \right ) \left ( \frac{E_{0}}{c} \right ) \; \; \; \; \;
S_{av}=\left ( \frac{\varepsilon _{0}c E{_{0}}^{2}}{2} \right )\; \; \; \; \; \; \; \left [ as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}} \right ]
Therefore,
S_{av}=\left ( \frac{E{_{0}}^{2}}{2c\mu _{0}} \right )

Question: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: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: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: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: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 ).
28


Answer:

The electric field in an infinitely long thin wire is
\overrightarrow{E}=\frac{\lambda \hat{e}_{s}}{2\pi \epsilon _{0}a}\hat{j}
Magnetic field due to the wire is
\overrightarrow{B}=\frac{\mu _{0}\lambda v}{2\pi a}\hat{i}
The equivalent current flowing through the wire is
\overrightarrow{S}=\frac{\lambda ^{2}v}{4\pi^{2}\epsilon _{0} a^{2}}\hat{k}

Question: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:

The separation between the plates of the capacitor is
V(t)=V_{0}\; \sin (2\pi vt)
Ohm’s law for the conduction of current density
J_{0c}=\frac{v}{\rho d}
The displacement current density is
J_{0d}=\frac{2\pi v\varepsilon V_{0}}{d}
The fraction of the conduction of current density and the displacement density
=\frac{J_{0}d}{J_{0}c}=\frac{4}{9}

Question: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_{0} with the displacement current I_{0}d

Answer:

a) The displacement current density is given as
\overrightarrow{J}_{d}=\frac{2\pi I_{0}}{\lambda ^{2}}ln\frac{a}{s}\sin \; 2\pi vt\hat{k}
b) Total displacement current
I^{d}=\int J_{d}2\pi sds
30
I^{d}=\left ( \frac{\pi a}{\lambda } \right )2I_{0}\; \sin \; 2\pi vt
c) The displacement current is
\frac{I_{0}d}{I_{0}}=\left ( \frac{a\pi }{\lambda } \right )^{2}

Question: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
31
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) 31i
\oint \overrightarrow{E}.\overrightarrow{dl}=E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]
b) 31ii
\oint \overrightarrow{B}.\overrightarrow{ds}=-\frac{B_{0}h}{k}\left [ \cos (kz_{2}-\omega t)-\cos (kz_{1}-\omega t) \right ]
c) Substituting the above equations in the following equation we get
\oint E.dl=-\frac{d\phi _{B}}{dt}=-\frac{d}{dt}\oint B.ds
\\E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]=\frac{B_0hw}{k} [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]\\

\frac{E_{0}}{B_{0}}=C
d) 31iv
\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt}
\\B_0 h[\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]=\frac{E_0h}{kw} [\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]
\\B_0 h=\frac{E_0h}{kw}
We get
c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}

Question: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

Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

  • Introduction
  • Displacement Current
  • Electromagnetic Waves
  • Sources of electromagnetic waves
  • Nature of electromagnetic waves
  • Electromagnetic Spectrum
  • Radio waves
  • Microwaves
  • Infrared waves
  • Visible rays
  • Ultraviolet rays
  • X-rays
  • Gamma rays

What will the Students learn from NCERT Exemplar Class 12 Physics Solutions Chapter 8?

From Class 12 Physics NCERT Exemplar solutions chapter 8, the students will be able to learn a lot about various properties, and facts of different kinds of rays present in the electromagnetic spectrum and how they are used differently in different places and situations. They will also get an insight into application of these concepts in various daily activities, and also about how the electromagnetic waves work in real life to provide various facilities such as the process of carrying of radio and TV signals from broadcasting stations, how such waves are used for these purposes, how light carries energy from sun to earth and much more. From NCERT Exemplar Solutions For Class 12 Physics chapter 8, the learners will also be able to verify the nature of electromagnetic waves as polarised light with its use in transfer of Radio and TV signals which place the antenna at a desired alignment, and in a particular direction to catch the signals from broadcasting stations.

NCERT Exemplar Class 12 Physics Chapter Wise Links

Important Topics To Cover For Exams From NCERT Class 12 Physics Chapter 8 Electromagnetic Waves

· NCERT Exemplar Class 12 Physics solutions chapter 8 also discuss the nature and source of electromagnetic field and defines electric and magnetic field vector which vibrate perpendicular to each other i.e. the electric field along x-axis, and magnetic field along y-axis and also shows the direction of which can travel through vacuum wave propagation perpendicular to both fields i.e. along the z-axis.

· The electromagnetic waves is featured with a variety of qualities explained in NCERT Exemplar solutions for Class 12 Physics chapter 8 along with various phenomena attached to it such as the radiation pressure, the energy and momentum carrier along with formula to calculate the amount of energy transported by such waves on any surface with the help of pointing vector.

· NCERT Exemplar Class 12 Physics solutions chapter 8 also includes questions related to the electromagnetic spectrum containing different kinds of waves such as UV rays, Gamma rays, X-rays, Infrared rays, etc. having different frequencies and wavelengths.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. What all things one will learn from this chapter?

 One will learn all the basics of the electromagnetic waves and their properties like radio waves, microwaves, IR rays, UV rays, gamma rays etc.

2. Are these questions helpful in entrance exams?

 Yes, these questions and NCERT Exemplar Class 12 Physics solutions chapter 8 are highly useful in preparation of the entrance exams of both engineering and medical science.  

3. Are these questions as per the marking scheme of CBSE?

  Yes, each and every question and Class 12 Physics NCERT Exemplar solutions chapter 8 is solved in a way that is deemed fit by CBSE involving each step and diagram that gets separate marks.

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