NCERT Solutions For Class 12 Physics Chapter 8 Electromagnetic Waves

 

NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves: Maxwell introduced the term displacement current to remove inconsistency in Ampers law. The solutions of NCERT class 12 physics chapter 8 electromagnetic waves starts with the question related to the concept of displacement current. Other topics covered in the chapter are Maxwell’s equations and electromagnetic waves and their properties and applications, the equation of electromagnetic wave and the relation between the magnetic field, electric field, and speed of light. Problems related to the electromagnetic spectrum are also discussed in CBSE NCERT solutions for class 12 physics chapter 8 electromagnetic waves. The solutions of NCERT helps in testing the knowledge about the concepts studied in the chapter. Different electromagnetic waves and their wavelength are listed in the below table. Questions based on this table are also discussed in the NCERT solutions for class 12 physics chapter 8 electromagnetic waves.

Range of wavelength:

 

Electromagnetic wave type

Wavelength range

Radio wave

Greater than 0.1 m

Microwave

0.1m to 1 mm

Infra-red wave

1mm to 700 nm

Light

700 nm to 400 nm

Ultraviolet

400 nm to 1nm

X-rays

1nm to 0.001nm

Gamma rays

Less than 0.001nm

 

NCERT solutions for class 12 physics chapter 8 electromagnetic waves exercise:

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.

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.

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?

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?

 

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.

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  109 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 (B0)=510 nT = 510 \times 10-9 T

Speed of light(c) = 3 \times 108 ms-1

Electric Field = B \times c

= 510 \times 10-9 T   \times 3 \times 108 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:

E0 = 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\times10^{6}

=3.14\times10rad 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(UE

\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

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\times10rad s-1

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

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

=8.6\times10Hz

=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 (E0) =3.1 NC-1

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

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

=1.03\times10-7 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\times10-3 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.

NCERT solutions for class 12 physics chapter wise:

NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields

Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance

CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity

NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism

Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter

CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction

NCERT solutions for class 12 physics chapter 7 Alternating Current

Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves

CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments

NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions

Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter

CBSE NCERT solutions for class 12 physics chapter 12 Atoms

NCERT solutions for class 12 physics chapter 13 Nuclei

Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Subject wise solutions-

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

  • For CBSE board exam 4% questions are expected from this chapter.
  • CBSE NCERT solutions for class 12 physics chapter 8 electromagnetic waves will give a clear idea about how to use the formulas studied in the chapter.
  • For exams like NEET and JEE main one or two questions are asked from the chapter.
  • Questions based on wave equation are expected from the chapter. The relation between electric field, magnetic field and speed of light is also important. There are questions based on these concepts in the NCERT solutions for class 12 physics chapter 8 electromagnetic waves.
 

Recently Asked Questions

 

Related Articles

Exams
Articles
Questions