NCERT Solutions for Class 12 Physics Chapter 8
NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves PDF – eSaral

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves PDF

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Question 1. Figure $8.6$ shows a capacitor made of two circular plates, each of radius $12 \mathrm{~cm}$ and separated by $5.0 \mathrm{~cm}$. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 \mathrm{~A}$.

Solution. (a) Calculate the capacitance and the rate of change of the 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.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves PDF Image 1

Solution. Given that

The magnitude of charging current, $\mathrm{I}=0.15 \mathrm{~A}$

Distance between the plates, $d=5 \mathrm{~cm}=0.05 \mathrm{~m}$

The Radius of each circular plate,

$\mathrm{r}=12 \mathrm{~cm}=0.12 \mathrm{~m}$

The permittivity of free space, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

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

$\mathrm{C}=\frac{\varepsilon_{0} \times \pi \mathrm{r}^{2}}{\mathrm{~d}}=\frac{8.85 \times 10^{-12} \times \pi \times(0.12)^{2}}{0.05}$

$=8.0032 \times 10^{-12} \mathrm{~F}$

$=80.032 \mathrm{pF}$

Charge on each plate, $q=C V$

Where $\mathrm{V}=$ Potential difference across the plates

Differentiate both sides with respect to time (t)

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

But, $\frac{d q}{d t}=$ current $(I)$

$\therefore \frac{d V}{d t}=\frac{I}{C}$

$\Rightarrow \frac{0.15}{80.032 \times 10^{-12}}=1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$

Therefore, the change in the potential difference between the plates with respect to time is $1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$

(b) Here the displacement current in the plates is the same as the conduction current Hence, the displacement current, $i_{d}$ is $0.15 \mathrm{~A}$.

(c) Yes

If we take the sum of conduction and displacement current, then Kirchhoff’s first rule is valid at each plate of the capacitor

Question 2. A parallel plate capacitor (Figure. 8.7) made of circular plates each of radius $\mathrm{R}=$ $6.0 \mathrm{~cm}$ has a capacitance $\mathrm{C}=100 \mathrm{pF}$. The capacitor is connected to a $230 \mathrm{~V}$ ac supply with an (angular) frequency of $300 \mathrm{rad} \mathrm{s}^{-1}$

(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 $\mathrm{B}$ at a point $3.0 \mathrm{~cm}$ from the axis between the plates.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves PDF Image 2

Solution. The capacitance of a parallel plate capacitor, $C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}$

The radius of each circular plate, $R=6.0 \mathrm{~cm}=0.06 \mathrm{~m}$

The capacitance of a parallel plate capacitor, $C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}$

The voltage supply to the capacitor, $V=230 \mathrm{~V}$

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

(a) Rms value of conduction current, $I=\frac{V}{x_{C}}$

Where, $X_{C}=$ Capacitive reactance $=\frac{1}{\omega C}$

$\therefore I=V \times \omega C=230 \times 300 \times 100 \times 10^{-12}$

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

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

(b) Yes, the magnitude of conduction current is equal to displacement current.

(c) The magnetic field is given as $B=\frac{\mu_{0} r}{2 \pi R^{2}} I_{0}$

Where, $\mu_{o}=$ Free space permeability $=4 \pi \times 10^{-7} \mathrm{~N} \mathrm{~A}^{-2}$

$I_{o}=$ Maximum value of current $=\sqrt{2} I$

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

$\therefore \mathrm{B}=\frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi(0.06)^{2}}$

$=1.63 \times 10^{-11} \mathrm{~T}$

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

Question 3. What physical quantity is the same for X-rays of wavelength $10^{-10} \mathrm{~m}$ red light of wavelength 6800 A and radio waves of wavelength $500 \mathrm{~m}$ ?

Solution. The magnitude of the speed of light $\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ in a vacuum is the same for all type of wavelengths. It does not depend on the wavelength in the vacuum.

Question 4. A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is $30 \mathrm{MHz}$, what is its wavelength?

Solution. Given that

The electromagnetic wave travels along the $z$-direction in a vacuum.

The electric field (E) and the magnetic field (B) are in the $x-y$ plane. They are mutually perpendicular.

Frequency of the wave, $v=30 \mathrm{MHz}=30 \times 10^{6} \mathrm{~Hz}$

Speed of light in a vacuum, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

We know that expression for the wavelength of a wave is given as:

$\lambda=\frac{c}{v}$

$\lambda=\frac{3 \times 10^{8}}{30 \times 10^{6}}=10 \mathrm{~m}$

Hence the wavelength of the electromagnetic wave is $10 \mathrm{~m}$

Question 5. A radio can tune in to any station in the $7.5 \mathrm{MHz}$ to $12 \mathrm{MHz}$ bands. What is the corresponding wavelength band?

Solution. Given that,

The minimum frequency of radio wave, $v_{1}=7.5 \mathrm{MHz}=7.5 \times 10^{6} \mathrm{~Hz}$

Maximum frequency, $v_{2}=12 \mathrm{MHz}=12 \times 10^{6} \mathrm{~Hz}$

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

Wavelength for $v_{1}$ can be calculated as:

$\lambda=\frac{c}{v_{1}}$

$=\frac{3 \times 10^{8}}{7.5 \times 10^{6}}=40 \mathrm{~m}$

Wavelength for $v_{2}$ can be calculated as:

$\lambda=\frac{c}{v_{1}}$

$=\frac{3 \times 10^{8}}{12 \times 10^{6}}=25 \mathrm{~m}$

Hence the wavelength range of the radio is $40 \mathrm{~m}$ to $25 \mathrm{~m}$.

Question 6. A charged particle oscillates about its mean equilibrium position with a frequency of $10^{9} \mathrm{~Hz}$. What is the frequency of the electromagnetic waves produced by the oscillator?

Solution. About its mean position, the frequency of an electromagnetic wave produced by the oscillator is the same as the frequency of a charged particle oscillating, i.e., $10^{9} \mathrm{~Hz}$.

Question 7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $\mathrm{B}_{0}=510 \mathrm{nT}$. What is the amplitude of the electric field part of the wave?

Solution. Given that,

The amplitude of the magnetic field of an electromagnetic wave in a vacuum,

$\mathrm{B}_{\mathrm{o}}=510 \mathrm{nT}=510 \times 10^{-9} \mathrm{~T}$

Speed of light in a vacuum, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

The relation between the amplitude of the electric field and the magnetic field is

$c=\frac{E_{0}}{B_{0}}$

$E=c B_{o}=3 \times 10^{8} \times 510 \times 10^{-9}=153 \mathrm{~N} / \mathrm{C}$

Therefore, the amplitude of the electric field part of the wave is $153 \mathrm{~N} / \mathrm{C}$.

Question 8. Suppose that the electric field amplitude of an electromagnetic wave is $\mathrm{E}_{0}=120 \mathrm{~N} / \mathrm{C}$ and that its frequency is $v=50.0 \mathrm{MHz}$.

(a) Determine, $\mathrm{B}_{0} \omega, k$, and $\lambda$.

(b) Find expressions for $\mathrm{E}$ and $\mathrm{B}$.

Solution. Given that,

The amplitude of Electric field, $E_{o}=120 \mathrm{~N} / \mathrm{C}$

Frequency of electromagnetic wave, $v=50.0 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}$

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

(a) The magnitude of magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

$=\frac{120}{3 \times 10^{8}}$

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

Angular frequency of the electromagnetic wave is given as:

$=\omega=2 \pi v=2 \pi \times 50 \times 10^{6}=3.14 \times 10^{8} \mathrm{rad} / \mathrm{s}$

Propagation constant is given as:

$k=\frac{\omega}{c}$

$=\frac{3.14 \times 10^{8}}{3 \times 10^{8}}=1.05 \mathrm{rad} / \mathrm{m}$

The wavelength of the wave is given as:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{50 \times 10^{6}}=6.0 \mathrm{~m}$

(b) Suppose that the wave is propagating in the positive $x$ direction. Then, the direction of the electric field vector will be in the positive $y$-direction, and the magnetic field vector will be in the positive $z$ direction because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

$\vec{E}=E_{0} \sin (k x-\omega t) \hat{\jmath}$

$=120 \sin \left[1.05 x-3.14 \times 10^{8} t\right] \hat{\jmath}$

And, magnetic field vector is given as:

$\vec{B}=B_{0} \sin (k x-\omega t) \hat{k}$ $\vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] \hat{k}$

Question 9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $\mathrm{E}=\mathrm{hv}$ (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of $\mathrm{eV}$ 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?

Solution. The energy of a photon is given as:

$E=h v=\frac{h c}{\lambda}$

Where,

$h=$ Planck’s constant $=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$

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

$\lambda=$ Wavelength of radiation

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

$=\frac{19.8 \times 10^{-26}}{\lambda \times 1.6 \times 10^{-16}}=\frac{12.375 \times 10^{-7}}{\lambda} \mathrm{eV}$

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

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves PDF Image 3

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

Question 10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \mathrm{~Hz}$ and amplitude $48 \mathrm{~V} \mathrm{~m}^{-1}$

(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 $\mathrm{E}$ field equals the average energy density of the B field. $\left[\mathrm{c}=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right]$

Solution. Given that,

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

The amplitude of the electric field, $E_{o}=48 \mathrm{~V} \mathrm{~m}^{-1}$

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

(a) The wavelength of a wave is given as:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{2 \times 10^{10}}=0.015 \mathrm{~m}$

(b) Magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

$=\frac{48}{3 \times 10^{8}}=1.6 \times 10^{-7} \mathrm{~T}$

(c) The energy density of the electric field is given as:

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

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

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

Where,

$\varepsilon_{0}=$ Permittivity of free space

$\mu_{0}=$ Permeability of free space

We have the relation connecting $\mathrm{E}$ and $\mathrm{B}$ as:

$E=c B \ldots$… 1 )

Where,

$c=\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}} \ldots(2)$

Putting equation (2) in equation (1), we get

$E=\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}} B$

$\varepsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$

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

$\Rightarrow U_{E}=U_{B}$

Hence the average energy density of the $\mathrm{E}$ field equals the average energy density of the B field

Additional Exercises

Question 11. Suppose that the electric field part of an electromagnetic wave in vacuum is

$\mathrm{E}=\left\{(3.1 \mathrm{~N} / \mathrm{C}) \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(5.4 \times 10^{6} \mathrm{rad} / \mathrm{s}\right) \mathrm{t}\right]\right\} \hat{\mathbf{i}}$

(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.

Solution. (a) From the given electric field vector, we can say that the electric field is directed along the positive $x$ direction. Hence, the direction of motion is along the negative y-direction i.e. $-\hat{\jmath}$.

(b) It is given that,

$\vec{E}=3.1 \mathrm{~N} / \mathrm{C} \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}\right) t\right] \hat{\imath} \ldots$

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

$\vec{E}=E_{0} \sin (k x-\omega t) \hat{\imath} \ldots(2)$

On comparing equations (1) and (2), ‘

we can find

Electric field amplitude, $E_{o}=3.1 \mathrm{~N} / \mathrm{C}$\

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

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

The wavelength, $\lambda=\frac{2 \pi}{1.8}=3.490 \mathrm{~m}$

(c) Frequency of wave is given as:

$B_{0}=\frac{E_{0}}{c}$

Where $c=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

$\therefore B_{0}=\frac{3.1}{3 \times 10^{8}}=1.03 \times 10^{-7} \mathrm{~T}$

(e) After seeing the given vector field, it can be observed that the magnetic field vector is directed along the positive $z$ direction. Hence, the general equation for the magnetic field vector is written as:

$\vec{B}=B_{0} \cos (k y+\omega t) \hat{k}$

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

Question 12. About $5 \%$ of the power of a $100 \mathrm{~W}$ light bulb is converted to visible radiation. What is the average intensity of visible radiation?

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

(b) at a distance of $10 \mathrm{~m}$ ? Assume that the radiation is emitted isotopically and neglect reflection.

Solution. Given that,

The power rating of the bulb, $P=100 \mathrm{~W}$

It is given that about $5 \%$ of its power is converted into visible radiation.

Power of visible radiation, $P^{\prime}=\frac{5}{100} \times 100=5 \mathrm{~W}$

Hence, the power of visible radiation is $5 \mathrm{~W}$.

(a) The distance of a point from the bulb, $d=1 \mathrm{~m}$

Hence, the intensity of radiation at that point is given as:

$I=\frac{P^{\prime}}{4 \pi d^{2}}$

$=\frac{5}{4 \pi(I)^{2}}=0.398 \mathrm{~W} / \mathrm{m}^{2}$

(b) The distance of a point from the bulb, $d_{1}=10 \mathrm{~m}$

Hence, the intensity of radiation at that point is given as:

$I=\frac{p^{\prime}}{4 \pi\left(d_{1}\right)^{2}}$

$=\frac{5}{4 \pi(10)^{2}}=0.00398 \mathrm{~W} / \mathrm{m}^{2}$

Question 13. Use the formula $\lambda_{\mathrm{m}} \mathrm{T}=0.29 \mathrm{~cm} \mathrm{~K}$ to obtains the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Solution. At a particular temperature, a body produces a continuous spectrum of wavelengths. In the case of a black body, we can find the wavelength corresponding to the maximum intensity of radiation by Planck’s law. It can be given by the relation,

$\lambda_{m}=\frac{0.29}{T} \mathrm{~cm}$

Where, $\lambda_{m}=$ maximum wavelength and $T=$ temperature.

Thus, the temperature for different wavelengths can be obtained as:

For $\lambda_{m}=10^{-4} \mathrm{~cm} ; T=\frac{0.29}{10^{-4}}=2900^{\circ} \mathrm{K}$

For $\lambda_{m}=5 \times 10^{-5} \mathrm{~cm} ; \mathrm{T}=\frac{0.29}{5 \times 10^{-4}}=5800^{\circ} \mathrm{K}$

For $\lambda_{\mathrm{m}}=10^{-6} \mathrm{~cm} ; \mathrm{T}=\frac{0.29}{10^{-6}}=290000^{\circ} \mathrm{K}$

The obtained values tell us that temperature ranges are required for obtaining cadiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.

Question 14. 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) $21 \mathrm{~cm}$ (wavelength emitted by atomic hydrogen in interstellar space).

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

(c) $2.7 \mathrm{~K}$ [temperature associated with the isotropic radiation filling all spacethought to be a relic of the ‘big-bang’ origin of the universe].

(d) 5890 A – 5896 A [double lines of sodium]

(e) $14.4 \mathrm{keV}$ [energy of a particular transition in ${ }^{57} \mathrm{Fe}$ nucleus associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].

Solution. (a) $21 \mathrm{~cm}$ come under Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.

(b) Radio waves; it belongs to the short wavelength end

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

$\lambda_{\mathrm{m}}$ is given by Planck’s law as:

$\lambda_{m}=\frac{0.29}{2.7}=0.11 \mathrm{~cm}$ This wavelength corresponds to microwaves.

(d) wavelength range comes under the yellow light of the visible spectrum.

(e) Transition energy is given by the relation,

$E=h v$ Where, $h=$ Planck’s constant $=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$

$v=$ Frequency of radiation

Energy, $E=14.4 \mathrm{keV}$

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

$=\frac{14.4 \times 10^{3} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$

$=3.4 \times 10^{18} \mathrm{~Hz}$

$3.4 \times 10^{18} \mathrm{~Hz}$ comes under X-rays.

Question 15. Answer the following questions:

(a) Long-distance radio broadcasts use short-wave bands. Why? (b) It is necessary to use satellites for long-distance TV transmission. Why?

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

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

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

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

Solution. (a) Shortwave frequencies are capable of reaching any location on the Earth because they can be reflected by the ionosphere.

(b) It is necessary to use satellites for long-distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are mainly not reflected by the ionosphere. Hence, satellites are very helpful in reflecting TV signals in a very big area on the earth.

(c) The earth’s atmosphere cannot absorb visible light and radio waves. But $\mathrm{X}$ rays being of much smaller wavelength are absorbed by the atmosphere. Therefore, X-ray astronomy is possible only from satellites orbiting the earth at the height of $36000 \mathrm{~km}$ above the earth’s surface. At such a height, the atmosphere is very thin, and X-rays are not absorbed.

(d) Ozone layer on the top of the atmosphere is very important for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface. In the presence of ultraviolet, it would be very difficult for anything to survive on the surface. Plants cannot live and grow in heavy uitraviolet radiation, nor can the plankton that serves as food for most of the ocean life. The ozone layer acts as a shield to absorb the UV rays, and keep them from doing damage at the Earth’s surface.

(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.

(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce smoke that would cover most of the part of the sky, thereby preventing solar light from reaching the atmosphere. Also, it will cause the depletion of the ozone layer.

Also Read,

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