NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves provide detailed answers to textbook theory questions, numerical problems, worksheets and exercises that will help you sort important topics of electromagnetic waves and prepare electromagnetic waves class 12 notes.

NCERT solutions for class 12 Physics Chapter 8 Electromagnetic Waves is an important topic in CBSE Class 12 Examination. In class 12 physics, there are many complicated formulas and equations. In order to score good marks in class 12th examination, it is important to solve NCERT questions provided at the end of each chapter.

Most frequently, questions that are asked in CBSE Class 12 Physics Exams are directly fetched from the NCERT textbook. Electromagnetism is one of the most frequently asked topics in exams.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Electromagnetism is a kind of physical attraction that occurs in electrically charged particles and this chapter consists of various subtopics which is equally important to learn. When a capacitor is charged using an external source, there can be a potential difference between two capacitive plates, we will show how to calculate that along with the displacement. This question will be solved using Kirchhoff’s rules.

Concepts involved in class 12 Physics Chapter 8 Electromagnetic Waves

  1. Introduction
  2. Displacement Current
  3. Electromagnetic Waves
    1. Sources of electromagnetic waves
    2. Nature of electromagnetic waves
  4. Electromagnetic Spectrum
    1. Radio waves
    2. Microwaves
    3. Infrared waves
    4. Visible rays
    5. Ultraviolet rays
    6. X-rays
    7. Gamma rays.

We will determine the RMS value of the conduction current and we will be analysing the similarities between conduction current and displacement current. We will be analysing the similarities among the wavelengths of X-rays, red lights and radio waves. In this solution, you will be seeing questions on the wavelength of electromagnetic waves travelling in a vacuum.

Do you want to know what the frequency of electromagnetic waves produced by the oscillator is? Want to know about the electric field part of the harmonic electromagnetic wave in a vacuum? Check out the answers below. We will be obtaining photo-energy of different parts of the electromagnetic spectrum and perceiving how to obtain different scales of photon energies of electromagnetic radiation.

We will be gaining knowledge on how to prove that the energy density of one field is equal to the average energy density of another field. We know that there are more fundamental forces such as weak and strong nuclear force and gravitational force. You will be finding questions on them in a different chapter. The questions mentioned in this chapter are very common in exams and if prepared thoroughly, will definitely make you understand electromagnetism nice and easy.

Class 12 Physics NCERT Solutions Electromagnetic Waves Important Questions


Q 8.1) Two circular plates having a radius of 12 cm each and separated by 5 cm are used to make a capacitor as shown in Figure 8.6. An external source charges this capacitor. 0.15 A is the charging current which remains constant.

(a) Determine the capacitance and the rate of charge of the potential difference between the two capacitive plates.

(b) Calculate the displacement current across the capacitive plates.

(c) Kirchhoff’s first rule (junction rule) is applicable to each plate of the capacitor. Yes, or No. Give Reasons.

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Answer 8.1:

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 ε0=8.85×1012  C2N1m2\varepsilon_{0} = 8.85\times 10^{-12}\; C^{2}N^{-1}m^{-2}

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

C=ε0AdC = \frac{\varepsilon _{0} A}{d}

where,

A = Area of each plate = πr2\pi r^{2} C=ε0πr2dC = \frac{\varepsilon _{0} \pi r^{2}}{d}

= 8.85×1012×π(0.12)20.05\frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}

= 8.0032×1012  F8.0032\times 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:

dqdt=CdVdt\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}

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

dVdt=IC∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}

=>0.1580.032×1012=1.87×109  V/s\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s

Therefore, the change in potential difference between the plates is 1.87×109  V/s1.87\times 10^{9}\; 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 ofconduction and displacement for current.

 

Q 8.2) Circular plates each of radius 6.0 cm having a capacitance of 100 pF is used to make a parallel plate capacitor (Fig. 8.7). The capacitor is connected to a 230 V ac supply with an (angular) frequency of 300 rad s-1.

(a) Determine RMS value of the conduction current

(b) Is conduction current equivalent to the displacement current?

(c) At a point 3.0 cm find out the amplitude of B from the axis between the plates.

Answer 8.2:

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

Capacitance of a parallel plate capacitor, C = 100 pF = 100×1012  F100\times 10^{-12}\; F

Supply voltage, V = 230 V

Angular frequency, ω=300  rad  s1\omega = 300\; rad\;s^{-1}

 

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

Where,

XcX_{c} = Capacitive reactance

= 1ωC\frac{1}{\omega C} I=V×ωC∴ I = V\times \omega C

= 230×300×100×1012230\times 300\times 100\times 10^{-12}

= 6.9×106A6.9\times 10^{-6} A

= 6.9  μA6.9\; \mu A

Hence, the rms value of conduction current is 6.9  μA6.9\; \mu A.

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

(c) Magnetic field is given as:

B=μ0r2πR2I0B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}

Where,

μ0\mu_{0} = Permeability of free space = 4π×107  N  A24\pi \times 10^{-7}\; N\;A^{-2} I0I_{0} = Maximum value of current = 2  I\sqrt{2}\; I

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

B=4π×107×0.03×2×6.9×1062π×(0.06)2∴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}}

= 1.63×1011  T1.63\times 10^{-11}\; T

Hence, the magnetic field at that point is 1.63×1011  T1.63\times 10^{-11}\; T.

 

 

Q 8.3) For X-rays of wavelength 101010^{-10}m, the red light of wavelength6800 Å and radiowaves of wavelength 500 m, what physical quantity could be the same?

Answer 8.3:

The speed of light (3×1083\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) What can be understood about the directions of magnetic and electric field vectors of a plane electromagnetic wave travelling in vacuum along the z-direction. What is the wavelength of the electromagnetic wave when its frequency is 30 MHz?

Answer 8.4:

The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) andthe magnetic field (H) are in the x-y plane. They are mutually perpendicular. Frequency of the wave, v = 30 MHz = 30×106  s130\times 10^{6}\;s^{-1}

Speed of light in vacuum, C = 3×1083\times 10^{8} m/s

Wavelength of a wave is given a:

λ=cv\lambda = \frac{c}{v}

= 3×10830×106\frac{3\times 10^{8}}{30\times 10^{6}} = 10 m

 

 

Q 8.5) What is the wavelength band of a radio that can tune in to any station in the 7.5 MHz to 12 MHz bands?

Answer 8.5:

A radio can tune to minimum frequency, v1=7.5  MHz=7.5×106  Hzv_{1} = 7.5\; MHz = 7.5\times 10^{6}\; Hz

Maximum frequency, v2=12  MHz=12×106  Hzv_{2} = 12\; MHz = 12\times 10^{6}\; Hz

Speed of light, c = 3×108  m/s3\times 10^{8}\; m/s

Corresponding wavelength for v1v_{1} can be calculated as:

λ1=cv1\lambda_{1} = \frac{c}{v_{1}} 3×1037.5×106=40  m\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m

Corresponding wavelength for v2v_{2} can be calculated as:

λ2=cv2\lambda_{2} = \frac{c}{v_{2}} 3×10312×106=25  m\frac{3\times 10^{3}}{12\times 10^{6}} = 25\;m

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

 

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

Answer 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., 109 Hz.

 

Q 8.7) What is the amplitude of the electric field part of the harmonic electromagnetic wave whose amplitude of the magnetic field part in a vacuum is B0=510  nTB_{0} = 510\; nT?

Answer 8.7:

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

B0=510  nT=510×109  TB_{0} = 510\; nT = 510\times 10^{-9}\; T

Speed of light in vacuum, c = 3×108  m/s3\times 10^{8}\; m/s

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

E=cB0=3×108×510×109=153  N/CE = cB_{0} = 3\times 10^{8}\times 510\times 10^{-9} = 153\; N/C

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

 

 

Q 8.8)Determine, (a) B0,  ω,  k  and  λB_{0},\; \omega,\; k\;and\; \lambda supposing that the electric field amplitude of an electromagnetic wave is E0=120  N/CE_{0} = 120\; N/C and that its frequency is v = 50 MHz. (b) Also find expressions for E and B.

Answer 8.8:

Electric field amplitude, E0=120  N/CE_{0} = 120\; N/C

Frequency of source, v = 50 MHz = 50×10650\times 10^{6} Hz

Speed of light, c = 3×1083\times 10^{8} m/s

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

B0=E0cB_{0} = \frac{E_{0}}{c}

= 1203×108\frac{120}{3\times 10^{8}}

= 4×107  T=400  nT4\times 10^{-7}\; T = 400\; nT

Angular frequency of source is given by:

ω=2nv=2n×50×106\omega = 2nv = 2n\times 50\times 10^{6}

= 3.14×1083.14\times 10^{8} rad/s

Propagation constant is given as:

k=ωck = \frac{\omega }{c}

= 3.14×1083×108=1.05  rad/m\frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m

Wavelength of wave is given by:

λ=cv\lambda = \frac{c}{v}

= 3×10850×106\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:

E=E0  sin(kxωt)  j^\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}

= 120  sin[1.05x3.14×108t]  j^120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}

And, magnetic field vector is given as:

B=B0  sin(kxωt)  k^\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k} B=(4×107)  sin[1.05x3.14×108t]  k^\overline{B} = (4\times 10^{-7})\; sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}

 

Q 8.9) Obtain the photon energy in units of eV for different parts of the electromagnetic spectrum using the formula E = hν (for the energy of a quantum of radiation: photon). How are the obtained different scales of photon energies related to the sources of electromagnetic radiation?

Answer 8.9:

Energy of a photon is given as:

E = hv = hcλ\frac{hc}{\lambda}

Where,

h = Planck’s constant = 6.6×1034  Js6.6\times 10^{-34}\;Js

c = Speed of light = 3×108  m/s3\times 10^{8}\;m/s λ\lambda = Wavelength of radiation

E=6.6×1034×3×108λ=19.8×1026λ  J∴ E = \frac{6.6\times 10^{-34}\times 3\times 10^{8}}{\lambda} = \frac{19.8\times 10^{-26}}{\lambda}\; J

= 19.8×1026λ×1.6×1019=12.375×107λ  eV\frac{19.8\times 10^{-26}}{\lambda \times 1.6\times 10^{-19}} = \frac{12.375\times 10^{-7}}{\lambda}\; eV

The given table lists the photon energies for different parts of an electromagnet spectrum for different λ\lambda.

λ\lambda

(m)

103 1 10310^{-3} 10610^{-6} 10810^{-8} 101010^{-10} 101210^{-12}
E

(eV)

12.375×101012.375\times 10^{-10} 12.375×10712.375\times 10^{-7} 12.375×10412.375\times 10^{-4} 12.375×10112.375\times 10^{-1} 12.375×10112.375\times 10^{1} 12.375×10312.375\times 10^{3} 12.375×10512.375\times 10^{5}

 

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

 

Q 8.10) (a) What is the wavelength of the electromagnetic wave in which the electric field oscillates sinusoidally at a frequency of 2×10102\times 10^{10} Hz and amplitude 48 V m-1. (b) Find the amplitude of the oscillating magnetic field and (c) Prove that the average energy density of the E field equals the average energy density of the B field.           [ c = 3×108  m  s13\times 10^{8}\;m\;s^{-1} ]

Answer 8.10:

Frequency of the electromagnetic wave, v = 2×1010  Hz2\times 10^{10}\;Hz

Electric field amplitude, E0=48  V  m1E_{0} = 48\;V\;m^{-1}

Speed of light, c = 3×108  m/s3\times 10^{8}\;m/s

 

(a) Wavelength of a wave is given as:

λ=cv\lambda = \frac{c}{v}

= 3×1082×1010=0.015  m\frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m

 

(b) Magnetic field strength is given as:

B0=E0cB_{0} = \frac{E_{0}}{c}

= 483×108=1.6×107  T\frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T

 

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

UE=12  ϵ0  E2U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}

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

UB=12μ0B2U_{B} = \frac{1}{2\mu_{0}}B^{2}

Where,

ϵ0\epsilon _{0} = Permittivity of free space

μ0\mu_{0} = Permeability of free space

E = cB  …(1)

Where,

c=1ϵ0  μ0c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}  …(2)

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

E=1ϵ0  μ0  BE = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B

Squaring on both sides, we get

E2=1ϵ0  μ0  B2E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2} ϵ0  E2=B2μ0\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}} 12  ϵ0  E2=12  B2μ0\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}

=> UE=UBU_{E} = U_{B}

 

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