Chapter 8 – Electromagnetic Waves Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 8 - Electromagnetic Waves Questions and Answers.

1. Figure Drawn below Shows a Capacitor Made of Two Circular Plates Each of Radius
12 cm12 cm
, and Separated by
5.0 cm5.0 cm
. The capacitor is being Charged by an External Source (not Shown in the Figure). The Charging Current is Constant and Equal to
0.15 A0.15 A

1. Calculate the Capacitance and the Rate of Change of the Potential Difference Between the Plates.

Ans: Radius of each circular plates,
r=12 cm=0.12 mr=12 cm=0.12 m
Distance between the given plates,
d=5 cm=0.05 md=5 cm=0.05 m
Charging current, I=0.15AI=0.15A
The permittivity of free space, ε0=8.85×10−12C2N−1m−2ε0=8.85×10−12C2N−1m−2
Capacitance between the two plates can be given as:
Charge on each plate is given as,
, where,
V = Potential difference across the plates
Differentiating both sides with respect to time (t), we get:
As, dqdt=Idqdt=I, so, the rate of change of potential difference between the plates can be given as:

2. Obtain the Displacement Current Across the Plates.

Ans: The displacement current across the plates would be the same as the conduction current. Hence, the displacement current, IdId would be
0.15 A0.15 A

3. Is Kirchhoff’s First Rule (junction Rule) Valid at Each Plate of the Capacitor? Explain.

Ans: Yes, Kirchhoff’s first rule would be valid at each plate of the capacitor, provided that sum of conduction and displacement currents, i.e., I=Ic+IdI=Ic+Id (junction rule of Kirchhoff’s law).

2) A Parallel Plate Capacitor (figure) Made of Circular Plates Each of Radius
R=6.0 cmR=6.0 cm
has a capacitance
C=100 pFC=100 pF
. The Capacitor is Connected to a
230 V230 V
ac supply with a (angular) frequency of

1. What is the RMS Value of the Conduction Current?

Ans: Radius of each circular plate,
R=6.0 cm=0.06 mR=6.0 cm=0.06 m
Capacitance of a parallel plate capacitor,
C=100 pF=100×10−12FC=100 pF=100×10−12F
Supply voltage,
V=230 VV=230 V
Angular frequency, ω=300rads−1ω=300rads−1
Rms value of conduction current can be given as: I=VXCI=VXC.
Where, XCXC = capacitive currant =1ωC=1ωC
Therefore, the RMS value of conduction current will be 6.9μA6.9μA.

2. Is the Conduction Current Equal to the Displacement Current?

Ans: Yes, the conduction current will be the same as the displacement current. i.e., the conduction current is equal to the displacement current.

3. Determine the Amplitude of B→B→ at a point
3.0 cm3.0 cm
from the Axis Between the Plates.

Ans: Magnetic field is given as: B→=μ0r2πR2I0B→=μ0r2πR2I0
μ0μ0= Free space permeability =4π×10−7NA−2=4π×10−7NA−2
I0I0 = Maximum value of current = 2–√I2I
rr= Distance between the plates from the axis = 3.0cm=0.03cm3.0cm=0.03cm.
Therefore, the magnetic field at that point will be 1.63×10−11T1.63×10−11T.

3. What Physical Quantity is the Same for X-Rays of Wavelength
, red light of wavelength
and radio waves of wavelength
500 m500 m

Ans: The speed of light (
) is independent of the wavelength in the vacuum. Hence, it is the same for all wavelengths in a 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
, what is its wavelength?

Ans: The electromagnetic wave is travelling along the z-direction, in a vacuum. The electric field (E) and the magnetic field (H) will lie in the x-y plane and they will be mutually perpendicular.
Frequency of the wave,
ν=30 Mhz=30×106s−1ν=30 MHz=30×106s−1
Speed of light in a vacuum,
Wavelength, λλ of the wave can be given as: λ=cνλ=cν

5. A Radio Can Tune in to Any Station in the
7.5 Mhz7.5 MHz
12 Mhz12 MHz
Band. What is the corresponding Wavelength Band?

Ans: A radio can tune to minimum frequency,
ν1=7.5 Mhz=7.5×106Hzν1=7.5 MHz=7.5×106Hz
Maximum frequency,
ν2=12 Mhz=12×106Hzν2=12 MHz=12×106Hz
Speed of light,
Wavelength for frequency, ν1ν1 can be calculated as:
Wavelength for frequency, ν2ν2 can be calculated as:
Therefore, the corresponding wavelength band would be between 40m40m to 25m25m.

6. A Charged Particle Oscillates About Its Mean Equilibrium Position With a Frequency of 109Hz109Hz. What is the Frequency of the Electromagnetic Waves Produced by the Oscillator?

Ans: The frequency of an electromagnetic wave produced by the oscillator will be equal to the frequency of charged particle oscillating about its mean position i.e.,

7. The Amplitude of the Magnetic Field Part of a Harmonic Electromagnetic Wave in Vacuum is B0 = 510nT. What is the Amplitude of the Electric Field Part of the Wave?

Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum, is:
Speed of light in a vacuum, c=3×108m/sc=3×108m/s
Amplitude of electric field of the electromagnetic wave can be given as:

8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120N/C and that its frequency is
ν=50.0 Mhzν=50.0 MHz
1. Determine, B0B0, ωω, kk, and λλ.

Ans: Electric field amplitude, E0=120N/CE0=120N/C
Frequency of source,
Speed of light,
Magnetic field strength can be given as:
Angular Frequency of Source Can be Given As:
Propagation Constant Can be Given As:
Wavelength of Wave Can be Given As:

2. Find Expressions for E and B.

Ans: If the wave propagates in the positive x-direction. Then, the electric field vector would be in the positive y-direction and the magnetic field vector will lie in the positive z-direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector can be given as:
And, magnetic field vector can be given as:

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?

Ans: The energy of photon can be given as:
hh = Planck’s constant =6.6×10−34Js=6.6×10−34Js
cc = speed of light =3×108m/s=3×108m/s
λλ = wavelength of radiation
For Different Values of in an Electromagnetic Spectrum, Photon Energies
are Listed in Below Table:

λ(m)λ(m) E(eV)E(eV)
103103 12.375×10−1012.375×10−10
1 12.375×10−712.375×10−7
10−310−3 12.375×10−412.375×10−4
10−610−6 12.375×10−112.375×10−1
10−810−8 12.375×10112.375×101
10−1010−10 12.375×10312.375×103
10−1210−12 12.375×10512.375×105

10. In a Plane Electromagnetic Wave, the Electric Field Oscillates Sinusoidally at a Frequency of 2×1010Hz2×1010Hz and amplitude 48Vm- 1
1. What is the Wavelength of the Wave?

Ans: Frequency of the electromagnetic wave,
Electric field amplitude,
E0=48 Vm−1E0=48 Vm−1
Speed of light,
Wavelength of wave can be given as:

2. What is the Amplitude of the Oscillating Magnetic Field?

Ans: Magnetic field strength can be given as:

3. Show That the Average Energy Density of the E Field Equals the Average Energy Density of the B field. C = 3 x 108m/s

Ans: Energy density of the electric field is given as:
Energy density of the magnetic field can be given as:
∈0∈0 = Permittivity of free space
μ0μ0 = Permeability of free space
The relation between E and B can be given as:
E=cBE=cB ….(1)
c=1∈0μ0√c=1∈0μ0 ….(2)
Substituting equation (2) in equation (1), we get:
Squaring both the sides, we get:

11. Suppose That the Electric Field Part of an Electromagnetic Wave in Vacuum is
1. What is the Direction of Propagation?

Ans: From the given electric field vector, we can say that the electric field will be directed along the negative x direction. Hence, the direction of motion will be along the negative y direction i.e., −j−j.

2. What is the Wavelength λλ?

Ans: The general equation for electric field in the positive x direction is given as:
. By comparing the given equation,
with general equation, we get:
Electric field amplitude, E0=3.1N/CE0=3.1N/C
Wave number, k=1.8rad/mk=1.8rad/m
Angular frequency, ω=5.4×106rad/sω=5.4×106rad/s
Wavelength, λ=2πk=2π1.8=3.490mλ=2πk=2π1.8=3.490m.

3. What is the Frequency νν?

Ans: The frequency, νν of the wave can be given as: ν=ω2πν=ω2π

4. What is the Amplitude of the Magnetic Field Part of the Wave?

Ans: Magnetic field strength can be given as:

5. Write an Expression for the Magnetic Field Part of the Wave.

Ans: We can observe that the magnetic field vector will be directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
, hence,

12. About
of the power of a
100 W100 W
light bulb is converted to visible radiation. What is the average intensity of visible radiation
1. At a Distance of
1 m1 m
from the bulb? Assume That the Radiation Is Emitted Isotropically and Neglect Reflection.

Ans: Power rating of bulb,
P=100 WP=100 W
of its power has been converted into visible radiation.
Therefore, power of visible radiation is given as:
Average intensity at distance, d=1md=1m can be given as:

2. At a Distance of 10m? Assume That the Radiation is Emitted Isotropically and Neglect Reflection.

Ans: Power of visible radiation can be given as:
Average intensity at distance, d=10md=10m can be given as:

13. Use the Formula λm=0.29TcmKλm=0.29TcmK to Obtain the Characteristic Temperature Ranges for Different Parts of the Electromagnetic Spectrum. What Do the Numbers That You Obtain Tell You?

Ans: At a particular temperature, a body produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given by Planck’s law. It can be represented as:
λmλm = maximum wavelength
TT = temperature
Therefore, the temperature for different wavelengths can be given as:
For, λm=10−4cmλm=10−4cm, T=0.2910−4=2900∘KT=0.2910−4=2900∘K
For, λm=5×10−5cmλm=5×10−5cm, T=0.295×10−5=5800∘KT=0.295×10−5=5800∘K
For, λm=10−6cmλm=10−6cm, T=0.2910−6=290000∘KT=0.2910−6=290000∘K
We can see that, as the maximum wavelength decreases, the corresponding temperature increases. At lower temperature, wavelength produced will not have 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.
1. 21 cm21 cm
(Wavelength Emitted by Atomic Hydrogen in Interstellar Space).

Ans: Wavelength of
21 cm21 cm
belongs to a short radio wave,
which is present at the end of the electromagnetic spectrum.

2. 1057MHz1057MHz
(Frequency of Radiation Arising from Two Close Energy Levels in Hydrogen; Known as Lamb Shift).

Ans: Frequency of
belongs to radio waves. As, radio waves generally belong to the frequency range of 500kHz500kHz to
. The range of wavelength for radio waves is, >0.1m>0.1m.

3. 2.7K2.7K

Ans: Given, temperature,
According to Planck’s law,
Substituting the value of temperature, we get: λ=0.292.7=0.11cmλ=0.292.7=0.11cm.
belongs to the Microwave region of electromagnetic spectrum.

4. 5890A∘−5896A∘5890A∘−5896A∘ doublelinesofsodiumdoublelinesofsodium

Ans: This wavelength belongs to yellow light of visible spectrum.

5. 14.4 keV14.4 keV
[energy of a particular transition in
nucleus Associated With a Famous High Resolution Spectroscopic Method (Mossbauer Spectroscopy)].

Ans: Given: 14.4keV14.4keV
The transition energy can be given as: E=hνE=hν.
hh is Planck’s constant ,
νν is frequency of radiation.
Substituting the values, we get:
Now, 3.4×1018Hz3.4×1018Hz belongs to range of X-ray frequencies. As, range of frequency for X-rays lies between

15. Answer the Following Questions:
1. Long-Distance Radio Broadcasts Use Short-Wave Bands. Why?

Ans: Short wave bands are used for long distance broadcasting because, after getting reflected from ionosphere present in our environment, these waves can reach to longer distances.

2. It Is Necessary to Use Satellites for Long Distance Tv Transmission. Why?

Ans: Yes, for long-distance TV transmission, it is compulsory to use satellites, because TV signal contains very high frequency of range
, and this much high frequency does not get reflected by ionosphere, and communication does not get stablished. Hence, we need to use satellites for TV transmission, over long distance.

3. Optical and Radio Telescopes Are Built on the Ground but X-Ray Astronomy Is Possible Only from Satellites Orbiting the Earth. Why?

Ans: Light and radio waves used in optical and radio telescopes respectively, penetrate through the atmosphere but X-rays used in X-ray astronomy, are absorbed by the atmosphere. Hence, optical and radio telescopes are built on earth’s surface and for X-ray astronomy, satellites orbiting the earth are necessary.

4. The Small Ozone Layer on Top of the Stratosphere is Crucial for Human Survival. Why?

Ans: The ozone layer, present on the top of the stratosphere layer is essential for human survival because it prevents the harmful radiations, (like, ultraviolet rays, γγ rays, cosmic rays) coming from the sun to earth’s atmosphere, which can cause many diseases like cancer and genetic damage in cells and plants. The ozone layer also traps the infrared radiations and maintains earth’s warmth.

5. If the Earth Did Not Have an Atmosphere, Would Its Average Surface Temperature Be Higher or Lower Than What it Is Now?

Ans: If the earth would not have any atmosphere, then there will not
be any greenhouse effect. And if the greenhouse effect would not be present, then infrared radiations and CO2CO2would not retain in the atmosphere. The infrared radiations and CO2CO2 gets reflected from the atmosphere and keep the earth’s surface warm. But in the absence
of an atmosphere, this would not be possible and the average surface
temperature of the earth would become lower.

6. 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?

Ans: A global nuclear war on the earth would be followed by a severe ‘nuclear winter’, as the nuclear weapons would create clouds of smoke and dust, which would stop the solar light to reach the earth’s surface and would stop the greenhouse effect. This would cause extreme winter conditions.