NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

Edited By Vishal kumar | Updated on Sep 11, 2023 09:08 AM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 8 –Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves- If you're a Class 12 student concerned about NCERT solutions electromagnetic waves, you're in the right place on Careers360. On this NCERT solutions page, you'll find answers to all the exercises and additional exercise questions from one to fifteen. These class 12 chapter 8 physics NCERT solutions, authored by subject experts, are detailed and presented in easy-to-understand language

NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves
NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

The Electromagnetic Waves NCERT solutions cover the question related to the concept of displacement current. Problems related to the Electromagnetic Spectrum are also discussed in NCERT Solutions for Class 12 Physics chapter 8 Electromagnetic Waves. The class 12 physics ch 8 NCERT solutions help in testing the knowledge about the concepts studied in a chapter. Students must refer to these solutions of NCERT to prepare for the examination effectively. The NCERT Solutions for Class 12 Physics Chapter 8 PDF download also helps in preparing for various entrance exams like JEE, NEET, etc.

The NCERT textbook is usually explicitly referenced in the questions that are asked in the CBSE EMW class 12 exams. One of the subjects that comes up frequently on the board test is electromagnetism. Therefore, it is advised that students use the electromagnetic waves NCERT solutions to get a firm understanding of both the chapter and the subject.

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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves: Exercise Question and Answer

1.(a) 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.

Calculate the capacitance and the rate of change of potential
difference between the plates.

1594195598206

Answer:

Radius of the discs(r) = 12cm

Area of the discs(A) = \pi r^{2}=\pi (0.12)^{2}=.045m^{2}

Permittivity, \epsilon _{0} = 8.85\times 10^{-12}C^{2}N^{-1}m^{2}

Distance between the two discs = 5cm=0.05m

Capacitance= \frac{\varepsilon _{0}A}{d}

= \frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}

= 8.003\times 10^{-12}F

=8.003 pF

Rate\ of \ change\ of \ potential =\frac{dv}{dt}

But

V =\frac{Q}{C}

Therefore,

Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C}

=\frac{0.15A}{8.003pF}

= 1.87\times 10^{10}Vs^{-1}

1(b). 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.

Obtain the displacement current across the plates.

1594195608735

Answer:

The value of the displacement current will be the same as that of the conduction current which is given to be 0.15A. Therefore displacement current is also 0.15A.

1.(c) 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.

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

1594195616698

Answer:

Yes, Kirchoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first but once we take into consideration both the conduction and displacement current the Kirchoff's first rule will hold good.

2.(a) 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} .

What is the rms value of the conduction current?

1594195911624

Answer:

Capacitance(C) of the parallel plate capacitor = 100 pF

Voltage(V) = 230 V

Angular Frequency (ω) = 300\ rad\ s^{-1}

Rms\ Current(I) =\frac{Voltage}{Capacitive\ Reactance}

Capacitive \ Reactance(X_{c})= \frac{1}{C\omega }=\frac{1}{100\ pF\times 300\ rad\ s^{-1}}

X_{c}= 3.33\times 10^{7} \Omega

I=\frac{V}{X_{c}}=\frac{230V}{3.33\times 10^{7}\Omega }= 6.9\times 10^{-6}A

RMS value of conduction current is 6.9\mu A

2.(b) 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}

Is the conduction current equal to the displacement current?

1594195919158


Answer:

Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere – Maxwell Law.

2.(c) 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}

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

1594195925850

Answer:

We know the Ampere - Maxwells Law,

\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})

Between the plates conduction current i_{c}=0 .

For a loop of radius r smaller than the radius of the discs,

\mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}

B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}

B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}

Since we have to find the amplitude of the magnetic field we won't use the RMS value but the maximum value of current.

\\i_{max}=\sqrt{2}\times i_{rms}\\ i_{max}=\sqrt{2}\times6.9\mu A\\ i_{max}=9.76\mu A

\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\ B_{amp}=1.63\times 10^{-11}T

The amplitude of B at a point 3.0 cm from the axis between the plates is 1.63\times 10^{-11}T .

3. What physical quantity is the same for X-rays of wavelength 10 ^{-10} m, red light of wavelength 6800 \dot{A} and radiowaves of wavelength 500m?

Answer:

The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to 3\times 10^{8}ms^{-1} (speed of light in vacuum).

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?

Answer:

Since the electromagnetic wave travels along the z-direction its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.

Frequency of wave = 30 MHz

Wavelength=

\frac{Speed\ of\ light}{Frequency}=\frac{3\times 10^{8}}{30\times 10^{6}}

=10m .

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

Answer:

Frequency range = 7.5 MHz to 12 MHz

Speed of light = 3\times 10^{8}ms^{-1}

Wavelength corresponding to the frequency of 7.5 MHz =

\frac{3\times 10^{8}ms^{-1}}{7.5\times 10^{6}Hz}

=40m

Wavelength corresponding to the frequency of 12 MHz =

\frac{3\times 10^{8}ms^{-1}}{12\times 10^{6}Hz}

=25m

The corresponding wavelength band is 25m to 40m.

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

Answer:

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore electromagnetic waves produced will have a frequency of 10 9 Hz.

7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B_0 = 510 n T . What is the amplitude of the electric field part of the wave?

Answer:

Magnetic Field (B 0 )=510 nT = 510 \times 10 -9 T

Speed of light(c) = 3 \times 10 8 ms -1

Electric Field = B 0 \times c

= 510 \times 10 -9 T \times 3 \times 10 8 ms -1

= 153 NC -1

8. Suppose that the electric field amplitude of an electromagnetic wave is E_0 = 120 N/C and that its frequency is \nu =50.0\ MHz (a) Determine, B_0 , \omega , \ k , \lambda (b) Find expressions for E and B.

Answer:

E 0 = 120 NC -1

\nu =50.0\ MHz

(a)

Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}

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

=400 nT

Angular frequency ( \omega ) = 2 \pi \nu

=2 \times \pi \times 50 \times 10^{6}

=3.14 \times 10 8 rad s -1

Propagation constant(k)

= \frac{2\pi }{\lambda }

= \frac{2\pi \nu }{\lambda\nu }

= \frac{\omega }{c}

= \frac{3.14\times 10^{8} }{3\times 10^{8}}

=1.05 rad m -1

Wavelength( \lambda ) = \frac{c}{\nu }

= \frac{3\times 10^{8}}{50\times 10^{6}} = 6 m

Assuming the electromagnetic wave propagates in the positive z-direction the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction as they are mutually perpendicular and E_{0}\times B_{0} gives the direction of propagation of the wave.

\vec{E}=E_{0}sin(kx-\omega t)\hat{i}

= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}


\vec{B}=B_{0}sin(kx-\omega t)\hat{j}

= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}

9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (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?

Answer:

E=h\nu=\frac{hc}{\lambda}

E=h\nu=\frac{6.6\times 10^{-34}\times3\times10^8}{\lambda \times1.6\times10^{-19}}eV=\frac{12.375\times 10^{-7}}{\lambda}eV

Now substitute the different range of wavelength in the electromagnetic spectrum to obtain the energy

EM wave one wavelength is taken from the range Energy in eV
Radio 1 m 1.2375\times10^{-6}
Microwave 1 mm 1.2375\times10^{-3}
Infra-red 1000 nm 1.2375
Light 500 nm 2.475
Ultraviolet 1nm 1237.5
X-rays 0.01 nm 123750
Gamma rays 0.0001 nm 12375000

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

Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 \times 10^8 m s^{-1}. ]

Answer:

The average energy density of the Electric field(U E )

= \frac{1}{2}\epsilon E^{2}

= \frac{1}{2}\epsilon (Bc)^{2} (as E=Bc)

= \frac{1}{2}\epsilon \frac{B^{2}}{\mu \epsilon } (c=\frac{1}{\sqrt{\mu \epsilon }})

=U B

Therefore the average energy density of the electric field is equal to the average energy density of the Magnetic field.

11 (a) Suppose that the electric field part of an electromagnetic wave in vacuum is E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i
What is the direction of propagation?

Answer:

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

E = \left \{ ( 3.1 N/C )\sin [ (\frac{\pi }{2}-(( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t ))] \right \} \hat i

The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.

11 (b) Suppose that the electric field part of an electromagnetic wave in vacuum is

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

What is the wavelength?

Answer:

From the equation of the wave given we can infer k = 1.8 rad m -1

Wavelength(\lambda )=\frac{2\pi }{k}

=\frac{2\pi }{1.8}

=3.49 m

11 (c) Suppose that the electric field part of an electromagnetic wave in vacuum is

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

What is the frequency n?

Answer:

From the given equation of the electric field we can infer angular frequency( \omega ) = 5.4 \times 10 8 rad s -1

Frequency( \nu ) = \frac{\omega }{2\pi }

= \frac{5.4\times10^{8}}{2\pi }

=8.6 \times 10 7 Hz

=86 MHz

11 (d) Suppose that the electric field part of an electromagnetic wave in vacuum is

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i
What is the amplitude of the magnetic field part of the wave?

Answer:

From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1 NC -1

Magnetic field amplitude (B_0) =\frac{E_{0}}{c}

=\frac{3.1}{3\times 10^{8}}

=1.03 \times 10 -7 T

11(e) Suppose that the electric field part of an electromagnetic wave in vacuum is

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

Write an expression for the magnetic field part of the wave.

Answer:

As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction the magnetic field vector must be directed along the negative z-direction. ( -\hat{i}\times -\hat{k}=-\hat{j} )

Therefore, \vec{B}= \left \{ B_{0}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat k

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

12.(a) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 1m from the bulb?

Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 1m

=\frac{5}{4\pi (1)^{2}}

=0.398 Wm -2

12.(b) About 5% of the power of a 100 W light bulb is converted to visible
radiation. What is the average intensity of visible radiation at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect
reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 10 m

=\frac{5}{4\pi (10)^{2}}

=3.98 \times 10 -3 Wm -2

13) Use the formula \lambda _m 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:

EM wave one wavelength is taken from the range Temperature T=\frac{0.29}{\lambda}
Radio 100 cm 2.9\times 10^{-3}K
Microwave 0.1cm 2.9 K
Infra-red 100000ncm 2900K
Light 50000 ncm 5800K
Ultraviolet 100ncm 2.9\times10^{6}K
X-rays 1 ncm 2.9\times10^8K
Gamma rays 0.01 nm 2.9\times10^{10}K


These numbers indicate the temperature ranges required for obtaining radiations in different parts of the spectrum

Answer the following questions

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

Answer:

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

Answer the following questions

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

Answer:

As TV signals are of high frequencies they are not reflected by the ionosphere and therefore satellites are to be used to reflect them.

Answer the following questions

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

Answer:

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

Answer the following questions

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

Answer:

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiations coming from the sun which are very harmful to humans.

Answer the following questions

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

Answer:

If the earth did not have an atmosphere its average surface temperature be lower than what it is now as in the absence of atmosphere there would be no greenhouse effect.

Answer the following questions

15 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:

The use of nuclear weapons would cause the formation of smoke clouds preventing the light from the sun reaching earth surface and it will also deplete the atmosphere and therefore stopping the greenhouse effect and thus doubling the cooling effect.

Class 12 physics ch 8 NCERT solutions is an essential resource for students preparing for both board exams and competitive exams like JEE. This chapter can be considered moderately challenging but also offers scoring potential. These solutions provide a clear understanding of electromagnetic waves, making complex concepts manageable, and play a crucial role in helping students excel in both board exams and JEE, contributing significantly to their success. There are 15 questions in the Electromagnetic Waves NCERT solutions. The NCERT solution for Class 12 Physics chapter 8 covers theoretical as well as numerical problems.

NCERT solutions for class 12 physics chapter-wise

NCERT Solutions for Class 12 Physics Chapter 8: Important Formulas and Diagrams

  • Gauss’ law of electricity:

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{"code":"$$\\oint_{}^{}\\vec{E.}d\\vec{s}\\,=\\,\\frac{q}{\\epsilon_{0}}$$","backgroundColor":"#ffffff","backgroundColorModified":null,"font":{"color":"#000000","family":"Arial","size":11},"id":"3","aid":null,"type":"$$","ts":1637054812868,"cs":"J5PxCQICONS3+SPn+w+DFQ==","size":{"width":104,"height":37}}

  • Gauss’s law of magnetism:

{"aid":null,"id":"4","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColor":"#ffffff","type":"$$","code":"$$\\oint_{}^{}\\vec{B}.d\\vec{S}\\,=\\,0$$","ts":1637055016683,"cs":"15C3V/oS4nBin7YLhgWUDg==","size":{"width":97,"height":37}}

  • Faraday’s laws of Electromagnetic induction:

{"id":"5","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"type":"$$","backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$e\\,=\\,-\\,\\frac{d\\phi_{B}}{dt}$$","ts":1637055464878,"cs":"numdiV7y2ABV1sdwjXgcNQ==","size":{"width":93,"height":34}}

Since emf can be defined as the line integral of the electric field, the above relation can be expressed as

{"id":"6","font":{"family":"Arial","color":"#000000","size":11},"type":"$$","backgroundColor":"#ffffff","aid":null,"backgroundColorModified":false,"code":"$$\\oint_{}^{}\\vec{E}.d\\vec{l}\\,=\\,-\\,\\frac{d\\phi_{B}}{dt}$$","ts":1637055605991,"cs":"2It52s47V8EVzWxmp+TIRw==","size":{"width":146,"height":37}}

  • Ampere-Maxwell’s Circuital law:

{"code":"$$\\oint_{}^{}\\vec{B.}d\\vec{l}\\,=\\,\\mu_{0}\\left(i\\,+\\,i_{d}\\right)$$","aid":null,"backgroundColor":"#ffffff","id":"7","type":"$$","backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"ts":1637056068499,"cs":"OGGYEsZmvZcrjuCOSYuwfg==","size":{"width":160,"height":37}}

Where

{"font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$i_{d}\\,=\\,\\epsilon_{0}A\\frac{dE}{dt}$$","type":"$$","aid":null,"id":"8","ts":1637056118397,"cs":"UTVFqDR1iuL9gpEFg4Xrxg==","size":{"width":96,"height":34}}

Electromagnetic Spectrum

  • Gamma Rays:

Wavelength range: 1 x 10¹⁴ to 1 x 10⁻¹⁰m

Frequency range: 3 x 10²² to 3 x 10¹⁸Hz

Production: By transition of atomic nuclei and decay of certain elementary particles.

  • X- rays:

Wavelength range: 1 x 10⁻¹¹ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁹ to 1 x 10¹⁶Hz

Production: By sudden deceleration of high-speed electrons at a high-atomic number target, and also by the electronic transition among the innermost orbits of atoms.

  • Ultraviolet- Rays:

Wavelength range: 1 x 10⁻⁸ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁶ to 1 x 10¹⁴Hz

Production: By sun, arc, vacuum spark, and ionized gases.

  • Visible Light:

Wavelength range: 1 x 10⁻⁷ to 1 x 10⁻⁷m

Frequency range: 7 x 10¹⁴ to 4 x 10¹⁴Hz

Production: Radiated by excited atoms in gases and incandescent bodies.

  • Infrared Radiation

Wavelength range: 8 x 10⁻⁷ to 5 x 10⁻³m

Frequency range: 4x 10¹⁴ to 6 x 10¹⁰ Hz

Production: From hot bodies, and by rotational and vibrational transitions in molecules

  • Radio waves

Wavelength range: 1 x 10⁻¹ to 1 x 10⁴m

Frequency range: 3 x 10⁹ to 3 x 10⁴ Hz

Production: By oscillating electric circuits.

Also, check

Electromagnetic Wave Class 12 Physics NCERT: Topics

The following are the main headings covered in the Chapter 8 Physics Class 12-

  • Concepts of displacement current and Maxwell's equation
  • Basics of electromagnetic waves
  • Electromagnetic Spectrum

Significance of NCERT solutions for class 12 physics chapter 8 electromagnetic induction:

  • For the CBSE board exam, 4% of questions are expected from chapter 8 physics class 12.

  • NCERT solutions for class 12 will give a clear idea about how to use the formulas studied in the NCERT chapter 8 Physics Class 12.

  • For exams like NEET and JEE Main one or two questions are asked from the Electromagnetic Waves NCERT.

  • Questions based on the wave equation are expected from the chapter for the CBSE board exam. The relation between electric field, magnetic field and speed of light is also important.

  • The electromagnetic Waves Class 12 pdf download button is available for the ease of students to prepare offline using Electromagnetic Waves Class 12 NCERT solutions.

Key Features of Electromagnetic Waves Class 12 NCERT SolutionsPDF

  1. Comprehensive Coverage: These ncert solutions electromagnetic waves comprehensively cover all the topics and questions presented in the Class 12 Physics chapter on electromagnetic waves.

  2. Detailed Explanations: Each class 12 chapter 8 physics ncert solutions provides detailed step-by-step explanations, simplifying complex concepts for students.

  3. Clarity and Simplicity: The solutions are presented in clear and simple language, ensuring ease of understanding.

  4. Exam Preparation: These solutions are crucial for board exam preparation and provide valuable support for competitive exams like JEE.

  5. Balanced Complexity: The chapter balances complexity with scoring potential, making it accessible for students aiming to score well.

  6. Free Access: These solutions are available for free, ensuring equal access for all students.

Excluded Content:

Certain portions have been omitted, which include:

  • Example 8.1.
  • Section 8.3.2 discusses the nature of electromagnetic waves (excluding references to ether and content on page 277).
  • Examples 8.4 and 8.5.
  • Exercises 8.11 to 8.15.

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Questions (FAQs)

1. What is the prerequisite of the NCERT chapter electromagnetic waves?

Before starting the preparation of NCERT class 12 chapter 8 it is better to have an idea of chapters 1 to 7. Chapter 1 to 7 is interconnected.

2. How many questions are expected from Class 12 Physics Chapter 8 NCERT solutions for NEET?

One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.

3. What is the weightage of the NCERT class 12 chapter electromagnetic wave for JEE Main?

From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.

4. How many marks are allotted to electromagnetic wave for CBSE board exams?

3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.

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Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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