NCERT Solutions for Class 12 Physics Chapter 8 – Electromagnetic Waves

Question 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 0.15 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.

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m463c7cab

Answer:

Radius of each circular plate, r = 12 cm = 0.12 m

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

Charging current, I = 0.15 A

Permittivity of free space, NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m5ad86d09= 8.85 × 10−12 C2 N−1 m−2

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

C NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 2681019d

Where,

A = Area of each plate NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 2b5a206d

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m2e354312

Charge on each plate, q = CV

Where,

V = Potential difference across the plates

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m64a6d5c7

Therefore, the change in the potential difference between the plates is 1.87 ×109 V/s.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.

(c) Yes

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

Page No 286:

Question 8.2:

A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius = 6.0 cm has a capacitance = 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 at a point 3.0 cm from the axis between the plates.

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 24d7fde2

Answer:

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

Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F

Supply voltage, V = 230 V

Angular frequency, ω = 300 rad s−1

(a) Rms value of conduction current, I NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m511e02b6

Where,

XC = Capacitive reactance

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 2c077c2d

∴ I = V × ωC

= 230 × 300 × 100 × 10−12

= 6.9 × 10−6 A

= 6.9 μA

Hence, the rms value of conduction current is 6.9 μA.

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

(c) Magnetic field is given as:

B NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m534a462d

Where,

μ0 = Free space permeability NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m71d48dbe

I0 = Maximum value of current =NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 46384fe9

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

B NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 145f0279

= 1.63 × 10−11 T

Hence, the magnetic field at that point is 1.63 × 10−11 T.

Question 8.3:

What physical quantity are the same for X-rays of wavelength 10−10 m, the red light of wavelength 6800 Å, and radiowaves of wavelength 500 m?

Answer:

The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

Question 8.4:

A plane electromagnetic wave travels in a 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 MHz, what is its wavelength?

Answer:

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

Frequency of the wave, ν = 30 MHz = 30 × 106 s−1

Speed of light in a vacuum, c = 3 × 108 m/s

The wavelength of a wave is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 5537be2

Question 8.5:

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

Answer:

A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz

Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz

Speed of light, c = 3 × 108 m/s

The corresponding wavelength for ν1 can be calculated as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 52a633bd

The corresponding wavelength for ν2 can be calculated as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 68666ca8

Thus, the wavelength band of the radio is 40 m to 25 m.

Question 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?

Answer:

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., 109 Hz.

Question 8.7:

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

Answer:

Amplitude of magnetic field of an electromagnetic wave in a vacuum,

B0 = 510 nT = 510 × 10−9 T

Speed of light in a vacuum, c = 3 × 108 m/s

The amplitude of the electric field of the electromagnetic wave is given by the relation,

E = cB0

= 3 × 10× 510 × 10−9 = 153 N/C

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

Question 8.8:

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0ω, k, and λ. (b) Find expressions for and B.

Answer:

Electric field amplitude, E0 = 120 N/C

Frequency of source, ν = 50.0 MHz = 50 × 106 Hz

Speed of light, c = 3 × 10m/s

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 1a65cebe

Angular frequency of the source is given as:

ω = 2πν

= 2π × 50 × 106

= 3.14 × 108 rad/s

Propagation constant is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 28a96be

The wavelength of wave is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m75d9b67

(b) 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:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m42eff519

And, magnetic field vector is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m5e1b818b

Question 8.9:

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula hν (for the 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?

Answer:

Energy of a photon is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 55ab5a45

Where,

h = Planck’s constant = 6.6 × 10−34 Js

c = Speed of light = 3 × 108 m/s

λ = Wavelength of radiation

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m3045abba

The given table below lists the photon energies for different parts of an electromagnetic spectrum for differentλ.

λ (m)103110−310−610−810−1010−12
(eV)12.375 × 10−1012.375 × 10−712.375 × 10−412.375 × 10−112.375 × 10112.375 × 10312.375 × 105

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 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 field equals the average energy density of the field. [= 3 × 108 m s−1.]

Answer:

Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz

Electric field amplitude, E0 = 48 V m−1

Speed of light, c = 3 × 108 m/s

(a) Wavelength of a wave is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 22983216

(b) Magnetic field strength is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 353ba0ee

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 5c79f20b

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 10b63b33

Where,

0 = Permittivity of free space

μ0 = Permeability of free space

We have the relation connecting E and B as:

E = cB … (1)

Where,

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m4283a516 … (2)

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 35efa045

Squaring both sides, we get

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 60feb102

Page No 287:

Question 8.11:

Suppose that the electric field part of an electromagnetic wave in vacuum is = {(3.1 N/C) cos [(1.8 rad/m) + (5.4 × 106 rad/s)t]}NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m7397950a .

(a) What is the direction of propagation?

(b) What is the wavelength λ?

(c) What is the frequency ν?

(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.

Answer:

(a) From the given electric field vector, it can be inferred that the electric field is directed along the negative x-direction. Hence, the direction of motion is along the negative y-direction i.e., NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 5322208d.

(b) It is given that,

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m7ea219e6

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m219f2a82

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

Electric field amplitude, E0 = 3.1 N/C

Angular frequency, ω = 5.4 × 108 rad/s

Wavenumber, k = 1.8 rad/m

Wavelength, NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m7058dde4 = 3.490 m

(c) Frequency of wave is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m5d1c866f

(d) Magnetic field strength is given as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html c6e50d5

Where,

c = Speed of light = 3 × 108 m/s

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 3dda08f5

(e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z-direction. Hence, the general equation for the magnetic field vector is written as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 7f9d6a8f

Question 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 1 m from the bulb?

(b) at a distance of 10 m?

Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

The power rating of the bulb, P = 100 W

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 4dd19828Power of visible radiation,

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 8e1176

Hence, the power of visible radiation is 5W.

(a) Distance of a point from the bulb, d = 1 m

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 3c4eca7c

(b) Distance of a point from the bulb, d1 = 10 m

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m3b07f590

Question 8.13:

Use the formula λm T= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Answer:

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

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 7ecb12aa

Where,

λm = maximum wavelength

T = temperature

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

For λm = 10−4 cm; NS 10 11 08 Sravana 12 Physics 8 15 NRJ html maa4c391

For λm = 5 ×10−5 cm; NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m31469ff4

For λm = 10−6 cm; NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 16a0f56a 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.

Question 8.14:

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 cm (wavelength emitted by atomic hydrogen in interstellar space).

(b) 1057 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 Å – 5896 Å [double lines of sodium]

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

(Mössbauer spectroscopy)].

Answer:

(a) Radio waves; belongs to the short-wavelength end of the electromagnetic spectrum.

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

(c) Temperature, T = 2.7 °K

λis given by Planck’s law as:

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html 46ffdadf

This wavelength corresponds to microwaves.

(d) This is the yellow light of the visible spectrum.

(e) Transition energy is given by the relation,

E = hν

Where,

= Planck’s constant = 6.6 × 10−34 Js

ν = Frequency of radiation

Energy, = 14.4 K eV

NS 10 11 08 Sravana 12 Physics 8 15 NRJ html m23c2c4f2

This corresponds to X-rays.

Question 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 radio telescopes 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?

Answer:

(a) Long-distance radio broadcasts use shortwave bands because only these bands can be refracted 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 not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long-distance TV transmissions.

(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. 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.

(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching 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 humans to survive.

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


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