NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 8
Chapter Name : Electromagnetic Waves

Q8.1 Figure 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 gure). 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 rst rule (junction rule)
valid at each plate of the capacitor? Explain.

Answer. Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε = 8.85 × 10 C N m 0
−12 2 −1 −2

(a) Capacitance between the two plates is given by the relation,
ε A
0
=
c d

A = Area of each plate = πr 2

2
E0 πr
C =
d
2
8.85 × 10π × 0.84
=
0.05
−12
= 8.0032 × 10 F = 80.032pF

Charge on each plate, q = CV
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
dq dV
= C
dt dt
dq
But, = current (I )
dt
dV I
∴ =
dt C

0.15 9
⇒ = 1.87 × 10 V/s
−12
80.032×10

Therefore, the change in potential difference between the plates is 1.87 × 10 V/s.
9

(b) The displacement current across the plates is the same as the conduction current. Hence, the
displacement current, I˙ is 0.15 A.
d

(c) Yes ,Kirchhoff's rst rule is valid at each plate of the capacitor provided that we take the sum of
conduction and displacement for current.

Page : 285 , Block Name : Exercise

Q8.2 A parallel plate capacitor 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

Answer. Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100pF = 100 × 10
−12
F

Supply voltage, V = 230 V
Angular frequency, ω = 300rads −1

(a) Rms value of conduction current, =
V
I
XC

Where
XC = Capacitive resistance
1
=
ωC

∴ I = v × ωC
−12
= 230 × 300 × 100 × 10

−6
= 6.9 × 10 A

= 6.9μA

Hence, the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic eld is given as:
μ0 r
= I0
2πR2

μ = Free space permeability = 4π × 10
−7 −2
0 NA

I = Maximum value of current = √2I
0

r = Distance between the plates from the axis = 3.0 cm = 0.03 m
−7 −6
4π×10 ×0.03×√2×6.9×10
= 2
2π ×(06)

−11
= 1.63 × 10 T

Hence, the magnetic eld at that point is 1.63 × 10 −11
T .

Page : 286 , Block Name : Exercise

Q8.3 What physical quantity is the same for X-rays of wavelength 10 red light of wavelength
−10
m

6800 Å and radiowaves of wavelength 500m?

Answer. The speed of light (3 × 10 m/s) in a vacuum is the same for all wavelengths. It is
3

independent of the wavelength in the vacuum.

Page : 286 , Block Name : Exercise

Q8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about
the directions of its electric and magnetic eld vectors? If the frequency of the wave is 30 MHz,
what is its wavelength?

Answer. The electromagnetic wave travels in a vacuum along the z-direction. The electric eld (E)
and the magnetic eld (H) are in the x-y plane. They are mutually perpendicular.
Frequency of the wave, v = 30 MHz = 30 × 10 s
6 −1

Speed of light in a vacuum, c = 3 × 10 m/s 8

Wavelength of a wave is given as:
c
λ =
v
s
3×10
= 6
= 10m
30×10

Page : 286 , Block Name : Exercise

Q8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding
wavelength band?

Answer. A radio can tune to minimum frequency, v = 7.5MHz = 7.5 × 10 Hz
1
6

Maximum frequency, v = 12MHz = 12 × 10 Hz
2
6

Speed of light, c = 3 × 10 m/s 8

Corresponding wavelength for can be calculated as:
c
λ1 =
v1
8
3 × 10
= = 40m
6
7.5 × 10

Corresponding wavelength for V2 can be calculated as:
c
λ2 =
v2
8
3 × 10
= = 25m
6
12 × 10

Thus, the wavelength band of the radio is 40 m to 25 m.
The photon energies for the different parts of the spectrum of a source indicate the spacing of the
relevant energy levels of the source.

Page : 286 , Block Name : Exercise

Q8.10 In a plane electromagnetic wave, the electric eld oscillates sinusoidally at a frequency of
Hz and amplitude 48 V m
10 −1
2.0 × 10

(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic eld?
(c) Show that the average energy density of the E eld equals the average energy density of the B
eld. [c = 3 × 10 ms ] 8 −1

Answer. Frequency of the electromagnetic wave, v = v = 2.0 × 10 10
Hz

Electric eld amplitude, E = 48Vm 0
−1

Speed of light, c = 3 × 10 m/s 8

(a) Wavelength of a wave is given as:
c
λ =
v
8
3 × 10
= = 0.015m
10
2 × 10

(b) Magnetic eld strength is given as:
E0
B0 =
c
48
−7
= = 1.6 × 10 T
8
3 × 10

(c) Energy density of the electric eld is given as:
1 2
UE = ϵ0 E
2

And, energy density of the magnetic eld is given as:
1 2
Un = B
2μ0

ϵ = Permittivity of free space
0

μ = Permeability of free space
0

We have the relation connecting E and B as:
E = cВ ... (1)
c = ....(2)
1

√ϵ0 μ0

Putting equation (2) in equation (1), we get
1
E = B
√ϵ0 μ0

Squaring both sides, we get
2 1 2
E = B
ϵ0 μ0

2
2 B
ϵ0 E =
μ0

2
1 2 1 B
ϵ0 E =
2 2 μ0

⇒ UE = UB

Page : 286 , Block Name : Exercise

Q8.11 Suppose that the electric eld part of an electromagnetic wave in vacuum is
E = {(3.1N/C) cos[(1.8rad/m)y + (5.4 × 10 rad/s) t]} i .
^ 6

(a) What is the direction of propagation?
(b) What is the wavelength λ ?
(c) What is the frequency ν ?
(d) What is the amplitude of the magnetic eld part of the wave?
(e) Write an expression for the magnetic eld part of the wave.

Answer. (a) from the given electric eld vector, it can be inferred that the elkctric eld is directed
along the negative x direction. Hence , the direction of the motion is along the negative y direction
i.e. −^j

(b) It is given that,
→
E = 3.1N/C cos[(1.8rad/m)y + (5.4 × 10 rad/s) t] i .. (1)
^ s

The general equation for the electric eld vector in the positive x direction can be written as:
→ = E sin(kx − ωt)^i ..... (2)
E 0

On comparing equations (1) and (2), we get
Electric eld amplitude,E = 3.1N/C 0

Angular frequency, ω = 5.4 × 10 rad/s
8

Wave number, k = 1.8 rad/m
Wavelength,λ = = 3.490m
2π

1.8

(c) Frequency of wave is given as:
ω
v =
2π
8
5.4 × 10
7
= = 8.6 × 10 Hz
2π

(d) Magnetic eld strength is given as:
E0
B0 =
c

C = Speed of light = 3 × 10 m/s 8

3.1 −7
∴ B0 = 8
= 1.03 × 10 T
3×10

(e) On observing the given vector eld, it can be observed that the magnetic eld vector is directed
along the negative z direction. Hence, the general equation for the magnetic eld vector is written
as:
→ ^
B = B0 cos(ky + ωt)k
−7 6 ^
= {(1.03 × 10 T) cos[(1.8rad/m)y + (5.4 × 10 rad/s) t]} k

Page : 287 , Block Name : Additional Exercise

Q8.12 About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the
average intensity of visible radiation
(a) at a distance of 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect re ection.

Answer. Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
Power of visible radiation, P = ′ 5
× 100 = 5w
100

Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
′
P
I =
2
4πd
5
2
= = 0.398W/m
2
4π(1)

(b) Distance of a point from the bulb, d = 10 m 1

Hence, intensity of radiation at that point is given as:
′
P
I =
2
4π(d1 )

5
2
= = 0.00398W/m
2
4π(10)

Page : 287 , Block Name : Additional Exercise

Q8.13 Use the formula λ T = 0.29 cm K to obtain the characteristic temperature ranges for
m

different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Answer. A body at a particular temperature produces a continous spectrum of wavelengths. In case
of a black body, the wavelength corresponding to maximum intensity of radiation is given
according to Planck's law. It can be given by the relation,
0.29
λm = cmK
T

λm = maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
−4 T 0.29 ∘
λm = 10 cm; = −4
= 2900 K
10
−5 0.29 ∘
λm = 5 × 10 cmi T = −5
= 5800 K
5×10
0.29
T=

and so on.
−6 −6 ∘
10
λm = 10 cm = 290000 K
i

The numbers obtained tell us that temperature ranges are required for obtaining radiations in
different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding

temperature increases.

Page : 287 , Block Name : Additional Exercise

Q8.14 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.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as
Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation lling all space-thought to be a relic
of the ‘big-bang’ origin of the universe].
(d) 5890 Å - 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high
resolution spectroscopic method (Mössbauer spectroscopy)].

Answer. (a) Radio waves: it belongs to the short wavelength end of the electromagnetic spectrum.
(b) Radio waves; it belongs to the short wavelength end.
∘
(c) Temperature, T = 2.7 Kλm is given by Planck's law as:

0.29
λm = = 0.11cm
2.7

This wavelength corresponds to microwaves.

(d) This is the yellow light of the visible spectrum.

(e) Transition energy is given by the relation,
−34
E = hv Where, h = Planck's constant = 6.6 × 10 Js

v = Frequency of radiation

Energy, E = 14.4KeV

E
∴ v =
h
3 −19
14.4×10 ×1.6×10
= −34
6.6×10

18
= 3.4 × 10 Hz

This corresponds to X -rays.

Q8.15 Answer the following questions:
(a) Long distance radio broadcasts use shortwave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from
satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the
earth did not have an atmosphere, would its average surface temperature be higher or lower than
what it is now?
(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. (a) Long distance radio broadcasts use shortwave bands because only these bands can be
refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals
are of high frequencies and high energies. Thus, these signals are not re ected by the ionosphere.
Hence, satellites are helpful in re ecting TV signals. Also, they help in long distance TV
transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible
and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground,
while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it
absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the
Earth's surface.
(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the
Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and
dif cult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-
nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that
would cover maximum parts of the sky, thereby preventing solar light form reaching the
atmosphere. Also, it will lead to the depletion of the ozone layer.

Page : 287 , Block Name : Additional Exercise