Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
1.5k views
in Physics by (3.3k points)
closed by

NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves discussed the topics of an important chapter of physics called the electromagnetic waves. Our NCERT Solutions is most suitable for the students who are preparing for the board exams and also for competitive exams like JEE Mains, JEE Advance, etc. NCERT Solutions Class 12 is very well concise by the expert on the subject.

  • Electromagnetic Waves – electromagnetic waves are formed when the electric and magnetic field comes in contact with each other. Both the waves are perpendicular to each other. They form tough and crest while traveling in any direction and their energy or the particle traverse in two axes and together they travel in only one direction. It is formed by the disturbance of electric and magnetic fields hence it is also called the wave formed by the oscillating magnetic and electric fields.
  • Displacement Current – the displacement current does not exhibit any real movement of electric charge as in the case of conduction current. Displacement current is defined as the rate of change of the electric field of displacement. It is said to a capacitor starts charging when there is no conduction of charge between the plates but as the time passes the plates get charged due to the presence of an electric field near it. The SI unit of displacement current is Ampere (A). It is experienced due to varying electromotive forces.
  • Electromagnetic Spectrum – The electromagnetic spectrum is the entire range of light that exists. It includes a wide range of frequencies of light, from very high-frequency waves such as gamma rays to very low frequency of light rays such as ultraviolet rays. The electromagnetic spectrum also consists of VIBGYOR which stands for seven elementary colors of nature which are Violet, Blue, Green, Yellow, Orange, and Red. VIBGYOR is also called visible light.

NCERT Solutions Class 12 Physics is explained in a detailed manner with the help of diagrams, and a graphs flowchart to make it easier for students to learn and revise all topics.

3 Answers

+2 votes
by (3.3k points)
selected by
 
Best answer

NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves

1. The Figure 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 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.

Answer:

Given Values:

The radius of each circular plate (r) is 12 cm or 0.12 m

The distance between the plates (d) is 5 cm or 0.05 m

The charging current (I) is 0.15 A

The permittivity of free space is ε​= 8.85 × 10−12 CN−1m−2

(a) The capacitance between the two plates can be calculated as follows:

\(C = \frac{\varepsilon _{0} A}{d}\)

where,

A = Area of each plate = \(\pi r^{2} C = \frac{\varepsilon _{0} \pi r^{2}}{d}\)

\(= \frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}\)

= 8.0032 × 10−12 F

= 80.032 pF

The charge on each plate is given by,

q = CV

where,

V is the potential difference across the plates

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

\(\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}\)

But, \(\frac{\mathrm{d} q}{\mathrm{d} t}\)​ = Current (I)

\(∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}\) 

\(⇒\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s\)

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.

2. A parallel plate capacitor 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 an (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.

Answer:

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

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

Supply voltage, V = 230 V

Angular frequency, ω = 300rads−1

(a) Rms value of conduction current, I = \(\frac{V}{X_{c}}\)

Where,

Xc​ = Capacitive reactance = \(\frac{1}{\omega C}\)

∴ 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 equivalent to displacement current.

(c) Magnetic field is given as:

\(B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}\)

Where,

μ0​ = Permeability of free space = 4π × 10−7 N A−2

I0​ = Maximum value of current = √2​I

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

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

= 1.63 × 10−11T

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

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

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.

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 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 x-y plane. They are mutually perpendicular.

Frequency of the wave, v = 30 MHz = 30 × 10s−1

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

Wavelength of a wave is given as:

\(\lambda = \frac{c}{v}\)

\(= \frac{3\times 10^{8}}{30\times 10^{6}}\)​ 

= 10 m

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, v1​ = 7.5 MHz = 7.5 × 106Hz

Maximum frequency, v2​ = 12MHz = 12 × 106Hz

Speed of light, c = 3 × 10m/s

Corresponding wavelength for v1​ can be calculated as:

\(\lambda_{1} = \frac{c}{v_{1}}\)

\(=\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m\) 

Corresponding wavelength for v2​ can be calculated as:

\(\lambda_{2} = \frac{c}{v_{2}}\)

\(=\frac{3\times 10^{3}}{12\times 10^{6}} = 25\:m\)

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

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.

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?

Answer:

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

B0​ = 510nT = 510 × 10−9 T

Speed of light in vacuum, c = 3 × 10m/s

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

E = cB0 ​= 3 × 108 × 510 × 10−9 = 153N/C

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

+1 vote
by (3.3k points)
edited by

8. Suppose that the electric field amplitude of an electromagnetic wave is E0​ = 120N/C and that its frequency is v = 50 MHz. (a) Determine B0​, ω, k and λ (b) Find expressions for E and B.

Answer:

Electric field amplitude, E0​ = 120N/C

Frequency of source, v = 50 MHz = 50 × 106 Hz

Speed of light, c = 3 × 108 m/s

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

\(B_{0} = \frac{E_{0}}{c}\)

\(= \frac{120}{3\times 10^{8}}\)

= 40 × 10−8 = 400 × 10−9 T = 400 nT

Angular frequency of source is given by:

ω = 2πv = 2π × 50 × 106 = 3.14 × 108 rads−1

= 3.14 × 108 rad/s

Propagation constant is given as:

\(k = \frac{\omega }{c}\)

\(= \frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m\)

Wavelength of wave is given by:

\(\lambda = \frac{c}{v}\)

\( \frac{3\times 10^{8}}{50\times 10^{6}}\)​ = 6.0 m

(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:

\(\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}\)

\(= 120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}\)

And, magnetic field vector is given as:

\(\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}\)

\(\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}\) 

9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = 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:

The energy of a photon is given as:

E = hv = \(\frac{hc}{\lambda}\)

Where,

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

c = Speed of light = 3 × 10m/s

If the wavelength λ is in metre and the energy is in Joule, then by dividing E by 1.6 × 10-19 will convert the energy into eV.

\(E=\frac{hc}{\lambda \times 1.6\times 10^{-19}}\,eV\) 

a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m, therefore the photon energy can be calculated as follows:

\(E=\frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-10}\times 1.6\times 10^{-19}}=12.4\times 10^3\approx 10^4\,eV\) 

Therefore,

λ = 10−10 m, energy = 104 eV

λ = 10−14 m, energy = 10eV

The energy for Gamma rays ranges from 104 to 108 eV.

b) The wavelength for X-rays ranges between 10-8 m to 10-13 m

For λ = 10-8,

\(Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-8}\times 1.6\times 10^{-19}}=124\approx 10^2\,eV\)

For λ = 10-13 m, energy = 107 eV

c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.

For 4 × 10-7 m,

\(Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}\times 1.6\times 10^{-19}}=3.1\approx 10^{10}\,eV\)

For 6 × 10-7 m, the energy is equal to 103 eV.

The energy of the ultraviolet radiation varies between 1010 to 103 eV.

d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.

For 4 × 10-7, the energy is the same as above, that is 1010 eV

For 7 × 10-7 m, the energy is 100 eV

e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.

The energy for 7 × 10-7 m is 100 eV

The energy for 7 × 10-14 m is 10-3 eV

f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.

For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.

g) For radio waves, the wavelength ranges from 1 m to few km.

For 1 m, the energy is 10-6 eV.

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

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 × 10ms−1]

Answer:

Frequency of the electromagnetic wave, v = 2 × 1010 Hz

Electric field amplitude, E0​ = 48 V m−1

Speed of light, c = 3 × 10m/s

(a) Wavelength of a wave is given as:

\(\lambda = \frac{c}{v}\)

\(= \frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m\) 

(b) Magnetic field strength is given as:

\(B_{0} = \frac{E_{0}}{c}\)

\(= \frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T\)

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

\(U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}\)

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

\(U_{B} = \frac{1}{2\mu_{0}}B^{2}\) 

Where,

ϵ0​ = Permittivity of free space

μ0​ = Permeability of free space

E = cB  …(1)

Where,

\(c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\)​  …(2)

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

\(E = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B\)

Squaring on both sides, we get

\(E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}\)

\(\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}\)

\(\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}\)

⇒ UE ​= U

+1 vote
by (3.3k points)

11. Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} \(\hat i\) .

(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) The direction of motion is along the negative y-direction. i.e., along -j.

(b) The given equation is compared with the equation,

E = E0 cos (ky + ωt)

⇒ k = 1.8 rad/s

ω = 5.4 x 10rad/s

λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m

(c) Frequency, ν = ω/2π 

=  5.4 x 106/(2 x 3.14) 

= 0.859  x 10Hz

(d) Amplitude of the magnetic field, B0 = E0/c

= 3.1/(3 x 108

= 1.03 x 10-8 T 

= 10.3 x 10-9 T 

= 10.3 nT

(e) Bz = B0 cos (ky + ωt) \(\hat k\)

{(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} \(\hat k\) 

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.

Answer:

(a) Average intensity of the visible radiation, I = P’/4πd2

Here, the power of the visible radiation, P’ = (5/100) x 100 = 5 W

At d = 1 m

I = P’/4πd2 = 5/(4 x 3.14 x 12) = 5/12.56 = 0.39 W/m2

(b) At d = 10 m

I = P’/4πd2 = 5/(4 x 3.14 x 102)  = 5/1256 = 0.39 x 10-2 W/m2

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:

We have the equation,  λm T = 0.29 cm K

⇒ T = (0.29/λm)cm K

Here, T is the temperature

λm is the maximum wavelength of the wave

For λ m = 10-4 cm

T = (0.29/10-4)cm K = 2900 K

For the visible light, λ= 5 x 10-5 cm

T = (0.29/ 5 x 10-5 )cm K ≈ 6000 K

a lower temperature will also produce wavelength but not with maximum intensity.

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 (short-wavelength end)

(b) Radio waves (short-wavelength end)

(c) Microwave

(d) Visible light (Yellow)

(e) X-rays (or soft γ-rays) region

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) Ionosphere reflects waves in the shortwave bands.

(b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.

(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.

(d) Ozone layer absorbs the ultraviolet radiations from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.

(e) If the atmosphere is not present, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.

(f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...