Electromagnetic Waves Class 12 Important Questions: CBSE Physics Chapter 8

Very Short Answer Questions (1 Mark Question)

1. The charging current for a capacitor is $0.25\Omega $. What is the displacement current across its plates?

Ans: Displacement current is equivalent to and is the same as charging current $025\;{\text{A}}$.

2. Write the following radiations in a descending order of frequencies: red light, x-rays, microwaves, radio waves

Ans: The descending order of frequencies: red light, x-rays, microwaves, radio waves are given as: X-rays > Red light >Microwaves >Radio Waves

3.  How does the frequency of a beam of ultraviolet light change, when it goes from air into glass?

Ans: We know that the frequency of UV light is unaffected by the frequency of a beam of ultraviolet light change and when it goes from air into glass.

4.  What is the ratio of speed of gamma rays and radio waves in vacuum?

Ans: As we know, the ratio of speed of gamma rays and radio waves in vacuum is one .

5. It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals are not reflected by the ionosphere layer of the atmosphere, TV broadcasts from air earth stations must be reflected back to the ground via an artificial satellite.

6. What is the role of ozone layer in the atmosphere?

Ans: The main role of the ozone layer in the atmosphere is that it absorbs all harmful UV rays and shields us from their detrimental effects.

7. What is the nature of waves used in radar?

Ans: The characteristics of the waves utilized in radar employs microwaves.

8. What physical quantity is the same for $X$ rays of wavelength $10 - 10\;{\text{m}}$, red light of wavelength 6800 A and radio waves of wavelength $500\;{\text{m}}$ ?

Ans: As we know that the speed of light $\left( {3 \times {{10}^8}\;{\text{m}}/{\text{s}}} \right)$ all wavelengths are the same in a vacuum. In the vacuum and they are unaffected by the wavelength.

Short Answer Questions (2 Marks Question)

1. Write the application of Infra-red radiations?

Ans: Following are the application of infra-red radiations :

(1) Photographs taken in hazy situations are taken with infrared radiation.

(2) Infrared radiations are employed to uncover the old walls' secret writings.

2. Which constituent radiation of the electromagnetic spectrum is used?

(1) To photograph internal parts of Human body-

(2) For air aircraft navigation

Ans: As the constituent radiation of the electromagnetic spectrum is -

(1) X – Rays are usually used photograph internal parts of Human body

(2) Microwaves are used for air aircraft navigation

3. Electric field in a plane electromagnetic wave is given by

${E_z} = 60\sin \left( {\dfrac{{{{10}^3}x}}{2} + \left( {{{10}^{11}}} \right)\dfrac{{3t}}{2}} \right)V/m$

(a) Write an expression for the magnetic field

Ans: We know that ,

 $C = \dfrac{{{E_0}}}{{{B_0}}}$

${B_O} = \dfrac{{{E_Q}}}{C} = \dfrac{{60}}{{3 \times {{10}^6}}}$

${B_0} = 2 \times {10^{ - 7}}T$

Since magnetic field and electric field are $ \bot $ to each after

$By = 2 \times {10^{ - 7}}T\sin \left( {\dfrac{{{{10}^3}}}{2}x + \left( {{{10}^{11}}} \right)\dfrac{{3t}}{2}} \right) -  \cdots $

(b) The magnitude of wavelength and frequency of the wave

Ans: Compare e.g. (1) with standard equation

${\text{By}} = {{\text{B}}_0}\sin 2\pi \left( {\dfrac{x}{\lambda } + \dfrac{t}{T}} \right)$

$\lambda  = 4\pi  \times {10^{ - 1}}\;{\text{m}}$

Also $2\pi \dfrac{1}{T} = {(10)^{11}}\dfrac{3}{2}$

$\dfrac{1}{T} = v = \dfrac{{3 \times {{10}^{11}}}}{{2 \times 2\pi }}$

$v = \dfrac{3}{{4\pi }} \times {10^{11}}Hz$

4. If the earth did not have atmosphere would its average surface temperature be higher or lower than what it is now?

Ans: The greenhouse effect traps infrared radiation inside the earth's atmosphere, causing the planet to warm. As a result, the earth's average temperature would have been law.

5. Sky waves are not used in transmitting TV signals, why? Suggest two methods by which range of TV transmission can he increased?

Ans: Sky waves are not transmitting TV signal, because the ionosphere does not reflect sky waves, they are not used to broadcast television signals.

Methods of increasing range of TV transmission

(1) Tall antenna

(2) Geostationary satellites

6. "Greater the height of a TV transmitting antenna, greater is its coverage." Explain.

Ans: As we know that ,

 ${\text{d}} = \sqrt {2 h R} $

The distance to which TV coverage can be done will rise as height is raised.

7. A plane electromagnetic wave travels in vacuum along $z$ -direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is $30{\text{MHz}}$, what is its wavelength?

Ans: In a vacuum, the electromagnetic wave travels in the -direction. In the x y plane, there is an electric field (E) and a magnetic field (M). They are perpendicular to one other.

Frequency of the wave, $v = 30{\text{MHz}} = 30 \times {10^6}\;{{\text{s}}^{ - 1}}$

Speed of light in a vacuum, $c = 3 \times {10^1}\;{\text{m}}/{\text{s}}$

Wavelength of a wave is given as:

$\lambda  = \dfrac{c}{v}$

$ = \dfrac{{3 \times {{10}^8}}}{{30 \times {{10}^8}}} = 10\;{\text{m}}$

8. $A$ radio can tune in to any station in the $7.5{\text{MHz}}$ to $12{\text{MHz}}$ band. What is the corresponding wavelength band?

Ans: As we have given that,

A radio can tune to minimum frequency, ${v_1} = 7.5{\text{MHz}} = 7.5 \times {10^6}\;{\text{Hz}}$

Maximum frequency, ${v_2} = 12{\text{MHz}} = 12 \times {10^4}\;{\text{Hz}}$

Speed of light, $c = 3 \times {10^4}\;{\text{m}}/{\text{s}}$

Then,

Corresponding wavelength far $v$ can be calculated as:

${\lambda _1} = \dfrac{c}{{{v_i}}}$

$ = \dfrac{{3 \times {{10}^1}}}{{7.5 \times {{10}^2}}} = 40m$

We get ,

Corresponding wavelength far ${v_2}$ can be calculated as:

${\lambda _2} = \dfrac{c}{{v_2^n}}$

$ = \dfrac{{3 \times {{10}^2}}}{{12 \times {{10}^5}}} = 25\;{\text{m}}$

Thus, the wavelength band of the radio is $40\;{\text{m}}$ to $25\;{\text{m}}$.

9. A charged particle oscillates about its mean equilibrium position with a frequency of ${10^9}\;{\text{Hz}}$. What is the frequency of the electromagnetic waves produced by the oscillator?

Ans: The frequency of an electromagnetic wave generated by the oscillator is the same as that of a charged particle osculating about its mean position, i.e. ${10^2}\;{\text{Hz}}$.

The magnetic field portion of a harmonic electromagnetic wave in vacuum has an amplitude of ${B_0} = 510{\text{nT}}$.

10. What is the amplitude of the electric field part of the wave?

Ans: Magnetic field amplitude of an electromagnetic wave in a vacuum,

$B0 = 510{\text{nT}} = 510 \times {10^{ - 7}}\;{\text{T}}$

Speed of light in a vacuum, $c = 3 \times {10^2}m/s$

The relationship, gives the amplitude of the electromagnetic wave's electric field.

$E = {B_0}$

$ = 3 \times 108 \times 510 \times {10^{ - 7}} = 153\;{\text{N}}/{\text{C}}$

Therefore, the electric field part of the wave is $153\;{\text{N}}/{\text{C}}$

Long Answer Questions (3 Marks Question)

1. In a plane electromagnetic wave, the electric field oscillates sinusoid ally with a frequency of $2 \times {10^{16}}\;{\text{Hz}}$ and amplitude $48\;{\text{V}}/{\text{m}}$.

(a) What is the wavelength of the emf wave?

Ans: (a) We have given that ,

 ${\text{V}} = 2 \times {10^{{\text{Ls}}}}{\text{Hz}}$

${{\text{E}}_0} = 48\;{\text{V}}/{\text{m}}$

$\lambda \text{ }\!\!~\!\!\text{ }=\dfrac{C}{V}=\text{ }\dfrac{3\times {{10}^{2}}}{2\times 10{}^\circ }$

$\vec A = 1.5 \times {10^{ - 2}}m$

(b) Calculate the amplitude of the oscillating magnetic field.

Ans: For calculating the amplitude of the oscillating magnetic field,

We have ,

${{\text{E}}_0} = c{B_0}$

${B_0} = \dfrac{{{E_Q}}}{C} = \dfrac{{48}}{{3 \times {{10}^2}}}$

${B_0} = 16 \times {10^{ - 1}}$ Tesla

(c) Calculate average energy density of the electromagnetic field at the wave

Ans: Now,

 For  Energy density

$U = \dfrac{1}{2}8.85 \times {10^{ - 12}} \times 48 \times 48$

$U = 1 \times {10^{ - 2}}J/{w^3}$

2. Find the wavelength of electromagnetic waves of frequency $6 \times {10^2}\;{\text{Hz}}$ in free space. Give two applications of the type of wave.

Ans: We have given that ,

 ${\text{V}} = 6 \times {10^{12}}\;{\text{Hz}}$

${{\text{U}}_{{\text{sing }}}}\lambda  = \dfrac{c}{v}$

$\lambda  = \dfrac{{3 \times {{10}^2}}}{{6 \times {{10}^{ - 2}}}}$

$\lambda  = 5 \times {10^{ - 1}}\;{\text{m}}$

Following are the applications for infra-red radiations 

(1) It keeps the earth warm.

(2) Infra-red lamps are used to treat muscular strains.

3. A plane monochromatic wave lies in the visible region. It is represented by the sinusoidal variation with time by the following components of electric field

$Ex = 0,Ey = 4\sin \left[ {\dfrac{{2\pi }}{\lambda }(x - 1t)} \right],Fz = 0$

Where $v = 5 \times {10^{14}}{\text{Hg}}$ And $\lambda $ is the wavelength of light.

(a) What is the direction of propagation of the wave?

Ans: We know that,

The direction of propagation of the wave is along the $ + x = $ axis.

(b) What is its amplitude?

Ans: Amplitude= 4 units

(c) Compute the component of magnetic field?

Ans: Component of magnetic of field

$Bz = \dfrac{{Eo}}{C} = \dfrac{4}{{3 \times {{10}^2}}}$

$Bz = 1.33 \times {10^{ - 6}}$ Tesla

4. Write the characteristics of emf waves? Write the expression for velocity of electromagnetic waves in terms of permittivity and permeability of the medium?

Ans: The Characteristics of emf waves are as follows :

(1) It travels in free space with speed of light $c = 3 \times {10^6}\;{\text{m}}/{\text{s}}$,

(2) Electromagnetic waves are transverse in nature.

Velocity of emf waves in vacuum C $ = \dfrac{1}{{\sqrt {{\mu _0}{ \in _6}} }}$

5. The electric field of a plane electromagnetic wave in vacuum is represented ${\text{b}}y$ -

$Ex = 0,Ey = 0.5\cos \left( {2\pi  \times {{10}^8}(t - x/c)} \right)$ and ${E_z} = 0$

a) What is the direction of propagation of electromagnetic wave?

Ans: For the given equation 

It represents the wave is propagating along $ + \bar x$ - axis

b) Determine the wavelength of the wave?

Ans: Now, by Comparing equation with the standard one

${\text{Ey}} = {{\text{E}}_0}\cos w(t - x/c)$

$w = 2\pi  \times {10^2}$

${z^\prime }\vec t{\text{t}} = {2^\prime }z \times {10^8}$

$v = {10^8}$

$ \Rightarrow \lambda  = \dfrac{c}{v} = \dfrac{{3 \times {{10}^8}}}{{{{10}^8}}}$

$\lambda  = 3m$

c) Compute the component of associated magnetic field

Ans: As it is associated magnetic field to electric field and the direction of propagation. Since wave is propagating along ${\text{x}} = $ axis, electric field is along, ${\text{y}} = $ axis

Thus, magnetic field is along $z = $ axis

$ \Rightarrow {B_x} = 0,{B_y} = 0$

$B_z = {B_0}\cos \left( {2\pi  \times {{10}^1}(t - x/x)} \right)$

$B_z = {\dfrac {{E}_{0}}{C}}\cos \left( {2\pi  \times {{10}^1}(t - x/x)} \right)$

6. Find the wavelength of electromagnetic waves of frequency $5 \times {10^{18}}\;{\text{H}}{{\text{z}}^{{\text{in free space}}{\text{. }}}}$ Give its two applications.

Ans: By Using ${\text{C}} = \lambda v$

$v = 5 \times {10^{15}}\;{\text{Hz}}$

$\lambda  = \dfrac{{3 \times {{10}^1}}}{{5 \times {{10}^{^\circ 6}}}}$

$\lambda  = 0.6 \times {10^{ - 11}} = 6 \times {10^{ - 12}}\;{\text{m}}$

These are Gamma Rays.

Applications for gamma rays :

(1) These rays are used to get information regarding atomic structure.

(2) They have very high penetrating power so they are used for detection purpose

7. 

(1) State the condition under which a microwave oven heats up food items containing water molecules most efficiently?

Ans: (1) Microwaves must have a frequency that is equal to the resonant frequency of the water molecules in the food item.

(2) Name the radiations which are next to these radiations in emf. Spectrum having (a) Shorter wavelength (b) Longer wavelength

(a) visible light is the radiation are the radiation next to shorter wavelength.

(b) Microwaves light is the radiation are the radiation next to longer wavelength.

8. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = {h_v}$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Ans: It is given that ,

 Energy of a photon is given as:

$E = hv = \dfrac{{hc}}{\lambda }$

Where,

$h = $ Planck's constant $ = 6.6 \times {10^{ - 34}}Js$

$c = $ Speed of light $ = 3 \times {10^8}\;{\text{m}}/{\text{s}}$

$\lambda $ = Wavelength of radiation

$\therefore E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } = \dfrac{{19.8 \times {{10}^{ - 26}}}}{\lambda }$

$ = \dfrac{{19.8 \times {{10}^{ - 26}}}}{{\lambda  \times 1.6 \times {{10}^{ - 19}}}} = \dfrac{{12.375t{{10}^{ - 7}}}}{\lambda }{\mathbf{eV}}$

The given table lists show the photon energies for different parts of an electromagnetic spectrum far different $\lambda $.

The spacing of the relevant energy levels of a source is determined by the photon energies for different portions of its spectrum.

9. About $5\% $ of the power of a $100\;{\text{W}}$ light bulb is converted to visible radiation. What is the average intensity of visible radiation?

(a) at a distance of $1\;{\text{m}}$ from the bulb?

(b) at a distance of $10\;{\text{m}}$ ?

Assume that the radiation is emitted isotropic ally and neglect reflection.

Ans: The power rating of bulb, $P = 100\;{\text{W}}$

It is given that about $S\% $ of its power is converted into visible radiation.

$\therefore $Power of visible radiation,

${P^\prime } = \dfrac{5}{{100}} \times 100 = 5{\pi ^\prime }$

Hence, the power of visible radiation is $5\;{\text{W}}$.

(a)Now, the Distance of a point from the bulb, $d = 1\;{\text{m}}$ Hence, intensity of radiation at that point is

given as

$1 = \dfrac{{{P^\prime }}}{{4\pi {d^2}}}$

$ = \dfrac{5}{{4\pi {{(1)}^2}}} = 0.398{\text{BF}}/{{\text{m}}^2}$

(b) Distance of a point from the bulb, ${d_1} = 10\;{\text{m}}$ Hence, intensity of radiation at that point is given as,

$1 = \dfrac{{{P^\prime }}}{{4\pi {{\left( {{d_1}} \right)}^2}}}$

$ = \dfrac{5}{{4\pi {{(10)}^2}}} = 0.00398\;{\text{W}}/{{\text{m}}^2}$

10. Use the formula ${\lambda _x}T = 0.29\;{\text{cm}}\;{\text{K}}$ to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Ans: A continuous spectrum of wavelengths is produced by a body at a specific temperature. Planck's law determines the wavelength corresponding to the greatest intensity of radiation in the case of a black substance. It can be determined by the relationship.

We know that,

${\lambda _\gamma } = \dfrac{{0.29}}{T}cmK$

Where,

${\lambda _k} = $ maximum wavelength

$T = $ temperature

Thus, the temperature for different wavelengths can be obtained as

For ${\lambda _{{\text{s }}}} = {10^{ - 4}}\;{\text{cm}}$;

$T = \dfrac{{0.29}}{{10}^{-4}} = {{2900}^{0}}K$

For ${\lambda _n} = 5 \times {10^{ - 5}}\;{\text{c}}{{\text{m}}^\prime }$

$T=\dfrac{0.29}{5\times {{10}^{-5}}}=5800{}^\circ \text{K}$

For ${\lambda _{\text{m}}} = {10^{ - 6}}\;{\text{cm}}$;

$T=\dfrac{0.29}{{{10}^{-5}}}=290000{}^\circ \text{K}$ and so on.

Temperature ranges are required for avoiding radiations in different sections of the electromagnetic spectrum, according to the results. As the wavelength gets shorter, the temperature gets hotter.

Very Long Answer Questions (5 Marks Question)

1. Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.

(a) $21\;{\text{cm}}$ (wavelength emitted by atomic hydrogen in interstellar space).

(b) $1057{\text{MHz}}$ (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).

(c) 2.7 K (temperature associated with the isotropic radiation filling all space-thought to be a relic of the 'big-bang' origin of the universe).

(d) \[5890\text{A}{}^\circ \text{ }\!\!~\!\!\text{ }-5896\text{A}{}^\circ \text{ }\!\!~\!\!\text{ }\] (double lines of sodium)

(e) 14.4 keV [energy of a particular transition in 57 Fe nucleus associated with a famous high resolution spectroscopic method

(Mossbauer spectroscopy)

Ans: By using the question we know that ,

(a) Radio waves; they belong to the electromagnetic spectrum's short wavelength end.

(b) Radio waves; they belong to the short wavelength end of the spectrum. 

(c) Now, Temperature, $T = {2.7^ \circ }K$

By ${\lambda _m}$ is given by Planck's law as:

${\lambda _m} = \dfrac{{0.29}}{{2.7}} = 0.11\;{\text{cm}}$

Microwaves relate to this wavelength.

(d) This is the visible spectrum's yellow light.

(e) The relation gives the transition energy:

$E = h{\text{v}}$

Where,

$h = $ Planck's constant $ = 6.6 \times {10^{ - 14}}{J_5}$

$v = $ Frequency of radiation

Energy, $E = 14.4\;{\text{KeV}}$

$\therefore v = \dfrac{E}{h}$

$ = \dfrac{{14.4 \times {{10}^3} \times 1.6 \times {{10}^{ - 18}}}}{{6.6 \times {{10}^{ - 34}}}}$

$ = 3.4 \times {10^{18}}\;{\text{Hz}}$

This corresponds to ${\text{X}}$ -rays.

2. Answer the following questions:

(a) Long distance radio broadcasts use short-wave bands. Why?

Ans: Shortwave bands are used for long-distance radio broadcasts because only these bands can be refracted by the ionosphere.

(b) It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals have high frequencies and energy, satellites are required for long-distance TV transmissions. As a result, the ionosphere does not reflect these signals. As a result, satellites aid in the reflection of television transmissions.

(c) Optical and radio telescopes are built on the ground but   -ray astronomy is possible only from satellites orbiting the earth. Why? 

Ans: As far as -ray astronomy is concerned, -rays are absorbed by the atmosphere. Visible and radio waves, on the other hand, can pass through it. As a result, optical and radio telescopes are constructed on the ground, but X-ray astronomy is only conceivable thanks to satellites orbiting the Earth. 

(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Ans: The ozone layer at the top of the atmosphere is critical for human survival because it absorbs damaging ultraviolet radiation from the sun and keeps it from reaching the Earth's surface.

(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? 

Ans: There would be no greenhouse effect on the Earth's surface if it did not have an atmosphere. As a result, the Earth's temperature would swiftly drop, making it cold and uncomfortable to live on and difficult for human survival.

(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with a devastating effect on life on earth. What might be the basis of this prediction?

Ans: On the surface of the Earth, a global nuclear war would be catastrophic. The Earth will experience a harsh winter following a nuclear war because the war will produce clouds of smoke that will cover the majority of the sky, preventing solar light from reaching the atmosphere. It will also contribute to the ozone layer's depletion.

3. Suppose that the electric field part of an electromagnetic wave in vacuum is $E = \{ (3.1$ $\left. {{\text{N}}/{\text{C}})\cos \left[ {(1.8{\text{rad}}/{\text{m}})y + \left( {5.4 \times {{10}^6}{\text{rad}}/{\text{s}}} \right)t} \right]} \right\}_i^*$

(a) What is the direction of propagation?

(b) What is the wavelength$\lambda $?

(c) What is the frequency $v$ ?

(d) What is the amplitude of the magnetic field part of the wave?

(e) Write an expression for the magnetic field part of the wave.

Ans: (a) It may be deduced from the supplied electric field vector that the electric field is directed in a negative direction. As a result, the movement is in the negative direction .i.e., $ - j$.

(b) It is given that,

$\vec E = 3.1\;{\text{N}}/{\text{C}}\cos \left[ {(1.8{\text{rad}}/{\text{m}})y + \left( {5.4 \times {{10}^8}{\text{rad}}/{\text{s}}} \right)t} \right]_i^* \ldots .$ (i)

The general equation for the electric field vector in the positive $x$ direction can be written as:

$\vec E = {E_0}\sin (kx - (i)t) = {j^{ -  \ldots (2)}}$

On comparing equations (1) and (2), we get

Electric field amplitude, ${E_0} = 3.1\;{\text{N}}/{\text{C}}$

Angular frequency, $\omega  = 5.4 \times {10^8}{\text{rad}}/{\text{s}}$

Wave number, $k = 1.8{\text{rad}}/{\text{m}}$

Wavelength, $\lambda  = \dfrac{{2\pi }}{{1.8}} = 3.490\;{\text{m}}$

(c) Now, Frequency of wave is given as:

$\lambda  = \dfrac{{2\pi }}{{1.8}}$

$ = \dfrac{{5.4 \times {{10}^3}}}{{2\pi }} = 8.6 \times {10^7}\;{\text{Hz}}$

(d) We know that ,

 Magnetic field strength is given as:

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

Where,

$c = $ Speed of light $ = 3 \times {10^3}\;{\text{m}}/{\text{s}}$

$\therefore {B_0} = \dfrac{{3.1}}{{3 \times {{10}^8}}} = 1.03 \times {10^{ - 1}}T$

(e) When looking at the given vector field, it's clear that the magnetic field vector is pointing in the wrong direction. As a result, the magnetic field vector's general equation is expressed as:

$\vec B = {B_0}\cos (ky + (i)t)\hat k$

$ = \left\{ {\left( {1.03 \times {{10}^{ - 7}}T} \right)\cos \left[ {(1.8{\text{rad}}/{\text{m}})y + \left( {5.4\;{\text{K}}t{{10}^6}{\text{rad}}/{\text{s}}} \right)t} \right]} \right\}\hat k$

4. In a plane electromagnetic wave, the electric field oscillates sinusoid ally at a frequency of $2.0 \times {10^{10}}{\text{H}}z$ and amplitude $48{\text{v}}_{\text{m}}^{ - 1}$.

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c $ = 3 \times {10^8}\;{\text{m}}\;{{\text{s}}^{ - 1}}$. )

Ans: As it is given that,

Frequency of the electromagnetic wave, $v = 2.0 \times {10^{10}}\;{\text{Hz}}$

Electric field amplitude, ${E_0} = 48\;{\text{V}}\;{{\text{m}}^{ - 1}}$

Speed of light, $c = 3 \times {10^3}\;{\text{m}}/{\text{s}}$

(a) What is the wavelength of the wave?

Ans: Now, Wavelength of a wave is given as:

$\lambda  = \dfrac{c}{v}$

$ = \dfrac{{3 \times {{10}^8}}}{{2 \times {{10}^{12}}}} = 0.015m$

(b) What is the amplitude of the oscillating magnetic field?

Ans: Where,

Magnetic field strength is given as:

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

$ = \dfrac{{48}}{{3 \times {{10}^2}}} = 1.6 \times {10^{ - 7}}T$

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c $ = 3 \times {10^8}\;{\text{m}}\;{{\text{s}}^{ - 1}}$. )

Ans: Then, 

Energy density of the electric field is given as:

${U_E} = \dfrac{1}{2}{ \in _0}{E^2}\ldots (1)$

And, energy density of the magnetic field is given as:

${U_B} = \dfrac{1}{{2{\mu _0}}}{B^2}\ldots (2)$

Where,

\[\varepsilon_{0}\] - Permittivity of free space

${\mu _0} = $ Permeability of free space

We have the relation connecting $E$ and $B$ as:

$E = cB \ldots (3)$

Where,

$c=\dfrac{1}{\sqrt{{\mu_{0}}{\varepsilon_{0}}}}$

Now, Substituting equation (3) in equation (1), we get

Squaring both sides, we get

$ \Rightarrow {U_E} = {U_B}$

5. Suppose that the electric field amplitude of an electromagnetic wave is ${E_0} = 120\;{\text{N}}/{\text{C}}$ and that its frequency is $v = 50.0{\text{MHz}}$. (a) Determine, ${B_0},\omega ,k$, and $\lambda $. (b) Find expressions for ${\mathbf{E}}$ and ${\mathbf{B}}$.

Ans: Given that ,

Electric field amplitude, ${E_0} = 120\;{\text{N}}/{\text{C}}$

Frequency of source, $v = 50.0{\text{MHz}} = 50 \times {10^5}\;{\text{Hz}}$

Speed of light, $c = 3 \times {10^8}\;{\text{m}}/{\text{s}}$

(a) Magnitude of magnetic field strength is given as:

${B_0}\dfrac{{{E_0}}}{c}$

$ = \dfrac{{120}}{{3 \times {{10}^2}}}$

$ = 4 \times {10^{ - 7}}T = 400nT$

Angular frequency of source is given as:

$\omega  = 2m = 2\pi  \times 50 \times {10^6}$

$ = 3.14 \times 108{\text{rad}}/{\text{s}}$

Propagation constant is given as:

$k = \dfrac{{(i)}}{c}$

$ = \dfrac{{3.14 \times {{10}^8}}}{{3 \times {{10}^8}}} = 1.05{\text{rad}}/{\text{m}}$

Now,

Wavelength of wave is given as:

$\lambda  = \dfrac{c}{v}$

$ = \dfrac{{3 \times {{10}^8}}}{{50 \times {{10}^2}}} = 6.0\;{\text{m}}$

(b) Assume the wave is travelling in a positive direction. The electric field vector will then point in the positive direction, and the magnetic field vector will also point in the positive direction. This is due to the fact that all three vectors are perpendicular to one another.

Equation of electric field vector is given as:

$\bar E = {E_0}\sin (kc - (i)t)j$

$ = 120\sin \left[ {1.05x - 3.14 \times {{10}^1}t} \right]j$

And, magnetic field vector is given as:

$\bar B = {B_0}\sin (kx - (t)t)\hat k$

$\vec B = \left( {4 \times {{10}^{ - 1}}} \right)\sin \left[ {1.05x - 3.14 \times {{10}^9}t} \right]\hat k$

6. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius $R = 6.0\;{\text{cm}}$ has a capacitance $C = 100{\text{pF}}$. The capacitor is connected to a $230\;{\text{V}}$ ac supply with a (angular) frequency of $300{\text{rad}}{{\text{s}}^{ - 1}}$

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of ${\mathbf{B}}$ at a point $3.0\;{\text{cm}}$ from the axis between the plates.

Ans: Given that , 

Radius of each circular plate, $R = 6.0\;{\text{cm}} = 0.06\;{\text{m}}$

Capacitance of a parallel plate capacitor, $C = 100{\text{pF}} = 100 \times {10^{ - 12}}\;{\text{F}}$

Supply voltage, $V = 230\;{\text{V}}$

Angular frequency, $\omega  = 300{\text{rad}}{{\text{s}}^{ - 1}}$

(a) Rms value of conduction current, $I = \dfrac{V}{{{X_\ell }}}$

Where,

$ = \dfrac{1}{{\omega C}}$

$\therefore I = V \times \omega C$

$ = 230 \times 300 \times 100 \times {10^{ - 12}}$

$ = 6.9 \times {10^{ - 6}}A$

$ = 6.9{\text{HA}}$

Hence, the rms value of conduction current is 6.9 $\mu {\text{A}}$

(b) Yes, the displacement current equal to conduction current 

(c) Where Magnetic field is given as:

$B = \dfrac{{{\mu _e}{r^r}}}{{2\pi {R^2}}}{I_e}$

Where,

${\mu _0} = $ Free space permeability $ = 4\pi  \times {10^{ - 7}}\;{\text{N}}{A^{ - 1}}$

$10 = $ Maximum value of current $ = \sqrt 2 l$

$r = $ Distance between the plates from the axis $ = 3.0\;{\text{cm}} = 0.03\;{\text{m}}$

$\therefore B = \dfrac{{4\pi  \times {{10}^{ - 7}} \times 0.03 \times \sqrt 2  \times 6.9 \times {{10}^{ - 5}}}}{{2\pi  > 0{{(06)}^2}}}$

$ = 1.63 \times {10^{ - 11}}T$ Hence, the magnetic field at that point is $1.63 \times {10^{ - 11}}T$.

7. Figure 8.6 shows a capacitor made of two circular plates each of radius $12\;{\text{cm}}$, and separated by $5.0\;{\text{cm}}$. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15\;{\text{A}}$

(a) Calculate the capacitance and the rate of charge of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.

Ans: Given data,

Radius of each circular plate, $r = 12\;{\text{cm}} = 0.12\;{\text{m}}$

Distance between the plates, $d = 5\;{\text{cm}} = 0.05\;{\text{m}}$

Charging current, $I = 0.15\;{\text{A}}$

Permittivity of free space, ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{\text{C}}2\;{{\text{N}}^{ - 1}}\;{{\text{m}}^{ - 2}}$

(a) Capacitance between the two plates is given by the relation.

Ans: $C = \dfrac{{{\varepsilon _0}A}}{{{d^\prime }}}\lambda  = \dfrac{c}{v} = \dfrac{{3 \times {{10}^8}}}{{30 \times {{10}^6}}} = 10m$

Where,

$A = $ Area of each plate $ = \pi {r^2}$

$C = \dfrac{{{\varepsilon _0}\pi {r^2}}}{d}$

$ = \dfrac{{8.85 \times 10{\pi ^{12}} \times 0k2{{()}^2}}}{{0.05}}$

$ - 8.0032 \times {10^{ - 12}}F = 80.032pF$

Charge on each plate, $q = CV$

Where,

${\text{V}}$ = Potential difference across the plates

Differentiation on both sides with respect to time $(t)$ gives:

$\dfrac{{dq}}{{dt}} = C\dfrac{{dV}}{{dt}}$

But $\dfrac{{dq}}{{at}} = $ current $(I)$

$\therefore \dfrac{{dV}}{{dt}} = \dfrac{1}{C}$

$ \Rightarrow \dfrac{{0.15}}{{80.032 \times {{10}^{ - 1.2}}}} = 1.87 \times {10^9}\;{\text{V}}/{\text{s}}$

Therefore, the change in potential difference between the plates is $1.87 \times {{10}^{9} }{\text{V}}/{\text{s}}$.

(b) Obtain the displacement current across the plates.

Ans: The displacement current \[\left ( {i_d} \right )\] will be same as the conduction current or equal to the conduction current \[\left ( {i_c} \right )\].

\[\therefore \left ( {i_d} \right )\:=\left ( {i_c} \right )\:=\:0.15\:amp\]

(c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.

Ans: Yes, Kirchhoff’s first law of electrical circuits will be valid at each plate of the given capacitor, unless we consider the total sum of the both conduction and displacement currents.

Important Formulas from Class 12 Physics Chapter 8 Electromagnetic Waves

• Speed of Light:
  $c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}c=μ0​ϵ0​​1​$
  where μ0\mu_0μ0​ is the permeability of free space and ϵ0\epsilon_0ϵ0​ is the permittivity of free space.

• Electromagnetic Wave Equation:
  $E=E0sin⁡(kx−ωt)andB=B0sin⁡(kx−ωt)E = E_0 \sin(kx - \omega t) \quad \text{and} \quad B = B_0 \sin(kx - \omega t)E=E0​sin(kx−ωt)andB=B0​sin(kx−ωt)$

• Ratio of Electric and Magnetic Fields:
  $EB=c\frac{E}{B} = cBE​=c$

• Displacement Current:
  $Id=ϵ0dΦEdtI_d = \epsilon_0 \frac{d\Phi_E}{dt}Id​=ϵ0​dtdΦE​​$

• Energy Density in Electromagnetic Waves:
  $u=12ϵ0E2+12B2μ0u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}u=21​ϵ0​E2+21​μ0​B2​$

• Poynting Vector:
  $\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$

Benefits of Class 12 Physics Chapter 8 Electromagnetic Waves      

• Builds a foundation for advanced topics like optics, quantum mechanics, and wireless communication.

• Explains the behavior and applications of waves in everyday life, such as in radio, television, and mobile communications.

• Enhances problem-solving and analytical skills for board and competitive exams.

• Helps understand the electromagnetic spectrum and its uses in different fields like medicine, astronomy, and technology.

• Prepares students for topics in engineering and applied physics courses.

Tips to Study Class 12 Physics Chapter 8 Electromagnetic Waves Important Questions 

1. Memorise and understand key formulas and their derivations.  

2. Focus on the electromagnetic spectrum and applications of different waves.

3. Solve NCERT textbook problems and additional numerical problems.

4. Use diagrams to visualise the propagation of electromagnetic waves.

5. Practice past year board questions to identify common patterns.

6. Pay attention to the displacement current concept and its implications in Maxwell's equations.

Related Study Materials for CBSE Class 12 Physics Chapter 8

Conclusion

Electromagnetic Waves is an important chapter that bridges physics theory with real-world applications. Mastering its concepts not only boosts exam performance but also lays a strong foundation for advanced studies in science and engineering. Use the FREE PDF of important questions, practice regularly, and achieve excellence in chapter 8 to achieve a good score in the exam as these important questions are created by Vedantu experts and master teachers.

Download CBSE Class 12 Physics Important Questions 2024-25 PDF

CBSE Class 12 Physics Chapter-wise Important Questions and Answers cover topics from all other chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Additional Study Materials for Class 12 Physics