NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves: Key Topics, Question with Solutions PDF

Q.8.1 Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change 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.8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

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

Capacitance between two plates is given by the relation,

C = $\frac{{ϵ}_{0}A}{d}$  , where A = Area of each plate = $\pi r^2$ = $\pi \times {0.12}^2$

C = $\frac{8.85\mathrm{}\times {10}^{-12}\times \mathrm{}\pi \times {0.12}^2}{0.05}$  = 8.007 $\times {10}^{-12}$

F

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{dq}{dt}$ = C  $\frac{dV}{dt}$

But $\frac{dq}{dt}$  = I, therefore

$\frac{dV}{dt}=\frac{I}{C}$  = $\frac{0.15}{8.007\mathrm{}\times {10}^{-12}}$  = 1.87 $\times {10}^{10}$  V/s $$

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 1.87 $\times {10}^{10}$ V/s

The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

 

Q.8.2 A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad 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 B at a point 3.0 cm from the axis between the plates.

Ans. 8.2 Radius of each circular plate, R = 6.0 cm = 0.06 m

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

Supply voltage, V = 230 V

Angular frequency,  $\omega$ = 300 rad/s

rms value of the conduction current, I =  $\frac{V}{X_c}$ , where $X_c=\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\mathrm{}\frac{1}{\mathrm{\omega }\mathrm{C}}$

Hence, I = V  $\times \omega \times C$ = 230  $\times 300\times$ 100  $\times {10}^{-12}$ = 6.9  $\times {10}^{-6}$ A = 6.9 $\mu A$

Yes, the conduction current is equal to the displacement current.

Magnetic field is given as, B =  $\frac{{\mu }_{0}r}{2\pi R^2}{I}_0$ , where

${\mu }_0$ = Free space permeability = 4  $\pi \times {10}^{-7}$ N  $A^{-2}$

$I_0$ = Maximum value of current =  $\surd 2I$

r = distance between two plates on the axis = 3.0 cm = 0.03 m

then, B =  $\frac{4\pi \times {10}^{-7}\times 0.03}{2\pi \times {0.06}^2}$  $\times \surd 2\times$ 6.9  $\times {10}^{-6}$ = 1.626  $\times {10}^{-11}$ T

 

Q.8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radio waves of wavelength 500m?

Ans.8.3 The speed of light (3  $\times {10}^8$ m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

 

Q.8.4 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 MHz, what is its wavelength?

Ans.8.4 The electromagnetic wave travels in a vacuum along the z- direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave,  $\nu$ = 30 MHz = 30  $\times {10}^6$ /s

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

Wavelength of a wave is given as

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

 

Q.8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Ans.8.5 The lowest tuning frequency  ${\nu }_1$ = 7.5 MHz = 7.5  $\times {10}^6$ Hz

The highest tuning frequency  ${\nu }_2$ = 12 MHz = 12  $\times {10}^6$ Hz

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

The wavelength for lowest tuning frequency,  ${\lambda }_1$ =  $\frac{c}{{\nu }_1}$ =  $\frac{3\mathrm{}\times {10}^8}{7.5\mathrm{}\times {10}^6}$ = 40 m

The wavelength for highest tuning frequency,  ${\lambda }_2$ =  $\frac{c}{{\nu }_2}$ =  $\frac{3\mathrm{}\times {10}^8}{12\mathrm{}\times {10}^6}$ = 25 m

The wavelength of the band is 40m to 25 m

 

Q.8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Ans.8.6 The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position, i.e.  ${10}^9$ Hz.

 

Q.8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?

Ans.8.7 The amplitude of magnetic field of the electromagnetic wave in a vacuum,

$B_0$ = 510 nT = 510  $\times {10}^{-9}$ T

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

Amplitude of electric field of the electromagnetic wave is given by the relation,

E = c  $B_0$ = 3  $\times {10}^8\times$ 510  $\times {10}^{-9}$ N/C = 153 N/C

Hence, the electric field part of the wave is 153 N/C.

 

Q.8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is  $\mathit{\nu }$ = 50.0 MHz (a) Determine, B0,  $\mathit{\omega }$ , k, and  $\mathit{\lambda }$ . (b) Find expressions for E and B.

Ans.8.8 Electric field amplitude,  $E_0$ = 120 N/C

Frequency of source,  $\nu$ = 50 MHz = 50  $\times {10}^6$ Hz

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

Magnitude of magnetic field strength is given as

$B_0=$  $\frac{E_0}{c}$ =  $\frac{120}{3\mathrm{}\times {10}^8}$ = 4  $\times {10}^{-7}$ T = 400  $\times {10}^{-9}$ T = 400 nT

$\mathrm{A}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{a}\mathrm{s}\mathrm{}$  $\omega$ = 2  $\pi \nu$ = 2  $\times \pi \times$ 50  $\times {10}^6$

= 3.14  $\times {10}^8$ rad/s

Propagation constant is given as

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

Wavelength of wave is given as

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

Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

$\vec{E}$ =  $E_0\sin (kx-\omega t)\stackrel{̂}{j}$

$=$ 120  $\times \mathrm{s}\mathrm{i}\mathrm{n}(1.05\times x-$ 3.14  $\times {10}^{8}t)\stackrel{̂}{j}$

Equation of magnetic field vector is given as:

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

= (400  $\times {10}^{-9})\sin (1.05\times x-$ 3.14  $\times {10}^{8}t$ )  $\stackrel{̂}{k}$

 

Q.8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = h  $\mathit{\nu }$ (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.8.9 Energy of a photon is given as:

E = h  $\nu$ =  $\frac{hc}{\lambda }$ , where

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

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

$\lambda$ = wavelength of radiation

Hence, E =  $\frac{6.6\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{\lambda }$ J =  $\frac{1.98\times {10}^{-25}}{\lambda }$ J =  $\frac{1.98\times {10}^{-25}}{\lambda \times 1.602\times {10}^{-19}}$ eV =  $\frac{1.236\times {10}^{-6}}{\lambda }$ eV

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

 

Q.8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V 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 × 108 m s–1.]

Ans.8.10 Frequency of the electromagnetic wave, $\nu$  = 2.0 $\times {10}^{10}$  Hz

Electric field amplitude, $E_0$  = 48 V/m

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

Wavelength of a wave is given by,

$\lambda =\frac{c}{\nu }$ = $\frac{3\mathrm{}\times {10}^8}{2.0\mathrm{}\times {10}^{10}}$ = 0.015 m

Magnetic field strength is given by

$B_0=$ $\frac{E_0}{c}$ = $\frac{48}{3\mathrm{}\times {10}^8}$ = 1.6  ${10}^{-12}$ T

Energy density of the electric field is given as,

$U_E$ =  $\frac{1}{2}{ϵ}_0{E}^2$ and energy density of magnetic field is given by,

$U_B$ = $\frac{1}{2}\frac{B^2}{{\mu }_0}$

$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}$

$E_0$  = Permittivity of free space

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

We have the relation connecting E and B as:

E = cB, where c = $\frac{1}{\sqrt[]{{ϵ}_0{\mu }_0}}$

Hence E = $\frac{B}{\sqrt[]{{ϵ}_0{\mu }_0}}$ $\stackrel{̂}{\mathit{i}}$

$E^2$  = $\frac{B^2}{{ϵ}_0{\mu }_0}$

${ϵ}_0{E}^2=$ $\frac{B^2}{{\mu }_0}$

$2U_E=2U_B$  or $U_E$ = $U_B$

 

Q.8.12 About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation?

(a) at a distance of 1m from the bulb?

(b) at a distance of 10 m?

Assume that the radiation is emitted isotropically and neglect reflection.

Ans.8.12 Power rating of the bulb, P = 100 W

Power of visible radiation, $P_v$ = 5% of 100 W = 5 W

Distance from the bulb, d = 1 m

Intensity of radiation at that point is given as:

I = $\frac{P_v}{4\pi d^2}$ = $\frac{5}{4\pi \times 1^2}$ = 0.398 W/ $m^2$

Distance from the bulb, d = 10 m

Intensity of radiation at that point is given as:

I = $\frac{P_v}{4\pi d^2}$ = $\frac{5}{4\pi \times {10}^2}$ = 3.978 $\times {10}^{-3}$  W/ $m^2$

 

Q.8.13 Use the formula ${\mathrm{}\lambda }_m$ T = 0.29 cmK 

to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Ans.8.13 A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Plank’s law. It can be given by the relation,

T = 0.29 cmK

${\mathrm{}\lambda }_m=\frac{0.29}{T}$ cm K, where ${\mathrm{}\lambda }_{\mathit{m}}$

= maximum wavelength and T = temperature

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

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-4}$ cm, T = 2900 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${5\times 10}^{-5}$ cm, T = 5800 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-6}$  cm, T = 290 $\times {10}^3$ $°K$ and so on

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

 

Q.8.15 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 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.8.15 Long distance radio broadcasts use shortwave bands because only these bands can be refracted by ionosphere.

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 not reflected by ionosphere. Hence, satellites are helpful in reflecting TV signals. Also they help in long distance TV transmissions.

With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.

The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiation present in sunlight and prevents from reaching Earth’s surface.

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 decreases rapidly, making it chilly and difficult for human survival.

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 clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light to reach the atmosphere. Also, it will lead to the depletion of the ozone layer.