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NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 8: Electromagnetic Waves

May 31, 2017 15:00 IST

    NCERT Solutions for Class 12 Physics, Chapter 6 - Electromagnetic Induction are available here. Some important topics of this chapter are displacement current, Maxwell’s equations, electromagnetic waves, sources of electromagnetic waves, nature of electromagnetic waves, electromagnetic spectrum etc. Questions from these topics are frequently asked in CBSE Class 12 Physics board exam and competitive. You can also download these solutions (in PDF format) with the help of download link given at the end of this article.    

    NCERT Solutions for Class 12 Physics ‒ Chapter 8: Electromagnetic Waves are given below

    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 0.15A.

    (a) Calculate the capacitance and the rate of change 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.

    NCERT Solutions Class 12th Physics Chapter 8 Electromagnetc Waves Question 8.1

    Solution 8.1:

    Here,

    r = 0.12 m

    d = 0.05 m

    I = 0.15 A

    (a) We know that,

    C = ϵo(A/d)

    Calculating area,

    A = 3.14 × 0.12 × 0.12

    A = 0.045 sq. m

    Substituting in above equation,

    C = 80.1 pF

    Now,

    Rate of change of potential difference

    dQ / dt = C.(dV/dt)

    Or

    dV/dt = I / C

    Thus, dV / dt = 0.15 / (80.1pF)

    Or dV / dt = 1.87 x 109 V / s.

    (b) Here, displacement current is equal to conduction current i.e. 0.15 Ampere.

    (c) It will be valid if both the conduction current and displacement current are taken into account.

    NCERT Exemplar Class 12 Physics Chapter 8  - Electromagnetic Waves

    Question 8.2: 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.

    (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.

    NCERT Solutions Class 12th Physics Chapter 8 Electromagnetc Waves Question 8.2

    Solution 8.2:

    Given:

    R = 0.06 m

    C = 100 pF

    V = 230 V

    ω = 300 rad / s

    (a) Irms = V. ω C

    Substitution values, we have Irms = 6.9 micro-amperes

    (b) Yes

    (c) For oscillating B and i

    Bo = (μo r io)/(2πR2)

    io = 1.414 × Irms

    Putting required value, we have

    Bo = 1.06 × 10‒11 T.

    Question 8.3: What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radio waves of wavelength 500m?

    Solution 8.3:

    Speed. All electromagnetic waves travel with same speed.

    Question 8.4: 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?

    Solution 8.4:

    E and B will be mutually perpendicular and perpendicular to direction of propagation in x-y plane.

    Wavelength = c / f

    = (3 x 108)/(30 x 106)

    = 10 m.

    Question 8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

    Solution 8.5:

    λ1 = c/f1 = 40 cm

    λ2 = c/f2 = 25 cm

    Wavelength band = 40 m – 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?

    Solution 8.6:

    Same, i.e., 109 Hz.

    Question 8.7: 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?

    Solution 8.7:

    Given,Bo = 510 nT

    Using the relation

    ⇒Eo / Bo = c

    ⇒Eo = c.Bo

    ⇒Eo = (3 × 108).(510 × 10-9 )

    ⇒Eo = 153 N / C

    Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E = 120 N/C and that its frequency is ν = 50.0 MHz.

    (a) Determine, B,ω, k, and λ.

    (b) Find expressions for E and B.

    Solution 8.8:

    Given:

    Eo = 120 N / C

    v = 50 MHz

    (a) Bo = Eo / c

    Or Bo = 400 nT

    λ = c/v

    ⇒ λ = 6 m.

    k = 2 π/ λ

    Or k = 1.05 rad /m

    ⇒ ω = c k or ω = 3.14 × 108 rad/s.

    (b) E = Eo sin (kx  ‒ ωt)

    E = 120 sin [1.05 x‒ (3.14×108) t] j

    B = Bo sin (kx  ‒ ωt)

    B = 400 × 10‒9 sin [1.05x ‒ (3.14×108) t] k

    Class 12 NCERT Solutions: All Subjects

    Question 8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = (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?

    Solution 8.9:

    We know

    E = hv = hc /λ

    Solving for different parts of electromagnetic spectrum

    (a) Long Radio Waves, 105 m

    E = 1.24 x 10-11 eV = 12.4 peV

    (b) AM Radio

    λ = 100 m

    E = 1.24 x 10-8 eV = 12.4 neV

    (c) Television and FM Radio

    λ = 10 m

    E = 1.24 x 10-7 eV = 0.124 μeV

    (d) Short Radio Waves

    λ = 10-1 m

    E = 1.24 x 10-5 eV = 12.4 μeV

    (e) Microwaves

    λ = 10-2 m

    E = 1.24 x 10-4 eV = 0.124 meV

    (f) Infrared

    λ = 10-4 m

    E = 1.24 x 10-2 eV = 12.4 meV

    (g) Visible

    λ = 10-6 m

    E = 1.24 eV

    (h) Ultraviolet

    λ = 10-7 m

    E = 12.4 eV

    (i) X-rays

    λ = 10-10 m

    E = 1.24 x 104 eV = 12.4 keV

    Energy of a photon indicates the amount of energy it needs to be emitted.

    Question 8.10: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 Vm–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 × 108 m s–1.]

    Solution 8.10:

    Given:

    Frequency = 2 x 1010 Hz

    Eo = 48 V / m

    (a) Wavelength is given by

    λ = c /f

    ⇒ λ = 300000000 / 20000000000 = 0.015 m

    (b) Using the relation

    Eo / Bo = c

    Bo = Eo / c

    or

    Bo = 48 / 300000000 = 1.6 x 10-7 T.

    (c) Average energy of electric field, UE = (ϵoE2)/2

    Average energy of magnetic field, UB = B2/(2μo)

    Now, E and B are related by the expression, so, E = cB

    [(2UE)/ϵo]1/2 = [1/μoϵo] × [2μoUB]1/2

    Squaring both sides and cancelling the common terms, we get, UE = UB

    Hence prove.

    Download NCERT Solutions for Class 12 Physics - Chapter 8 Electromagnetic Waves in PDF format

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