Ncert Solution Class 12th Physics Chapter – 8 Electromagnetic Waves Question & Answer

June 24, 2022

NCERT Solutions Class 12th Physics Chapter – 8 Electromagnetic Waves

Textbook	NCERT
class	Class – 12th
Subject	Physics
Chapter	Chapter -8
Chapter Name	Electromagnetic Waves
Category	Class 12th Physics Question & Answer
Medium	English
Source	last doubt

NCERT Solutions Class 12th Physics Chapter – 8 Electromagnetic Waves

?Chapter – 8?

✍Electromagnetic Waves✍

?Question & Answer?

Q 1. The 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 figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change of the 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 Solution Class 12th Physics Chapter - 8 Electromagnetic Waves Question & Answer

Answer:

(a) $C = ε_{0} A / d = 80.1 p F$
$d t d Q = C d t d V$
$d t d V = 80.1 \times 10 ^{-12} 0.15 = 1.87 \times 1 0^{9} V s^{-1}$

(b)The displacement current across the plate is same as the conduction current.Hence, the displacement current , $I_{d} = 0.15 A$

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

Q 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 an (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 Solution Class 12th Physics Chapter - 7 Alternating Current Question & Answer

Answer:

$(a)$ RMS value of the conduction current,

I = V/ X ^{c}

X_{c} = 1 ω C

\Rightarrow I = V ω C

\Rightarrow I = 230 \times 300 \times 100 \times 1 0^{-12}

\Rightarrow I = 6.9 μ A

$(b)$ Yes. Conduction current is equal to the displacement current.

$(c)$ $B = μ ^{o} r I ^{o /} 2 π R ^{2}$

$I_{o} =\sqrt 2 I$

r is distance between plates from the axis, $r = 0.03 m$

putting values

$B = 4 π \times 10 ^{-7} \times 0.03 \times\sqrt 2 \times 6.9 \times 10 ^{-6 /} 2 π \times 0.06 ^{2}$

$\Rightarrow B = 1.63 \times 1 0^{-11} T$

Q 3. What physical quantity is the same for X-rays of wavelength 10^–10m, the red light of wavelength 6800 Å and radiowaves of wavelength 500m?

Answer: The speed in vacuum is the same for all, $c = 3 \times 1 0^{8} m / s$

Q 4. A plane electromagnetic wave travels in vacuum along the 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: Wave travel along z-direction.

Electric and magnetic field are in XY plane and perpendicular to each other.

λ = c / ν = 10 m

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

Answer:

$f_{1} = 7.5 \times 1 0^{6} H z$

f_{2} = 12 \times 1 0^{6} H z

λ_{1} = c / f_{1} = 40 m

λ_{2} = c / f_{2} = 25 m

So the range is 40m to 25m

Q 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?

Answer: 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 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?

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

B₀ = 510 nT = 510 × 10⁻⁹ T

Speed of light in a vacuum, c = 3 × 10⁸ m/s

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

E = cB₀ = 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C

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

Q 8. Suppose that the electric field amplitude of an electromagnetic wave is $E_{0} = 120 N / C$ and that its frequency is $ν = 50.0 M H z$ .

(a) Determine, $B_{0}, ω, k, a n d λ$

(b) Find the expression for E and B.

Answer: (a)

$B_{o} = E_{o} / c = 400 n T$

ω = 2 π ν = 3.14 \times 1 0^{8} r a d / s .

k = ω / c = 1.05

rad/m

λ = c / ν = 6 m

(b) Suppose wave is moving along x-direction, Electric field be along y-direction and magnetic field along the z-direction.

$E = E_{o} s i n (k x - ω t) j^= 120 s i n (1.05 x - 3.14 \times 1 0^{8} t) j^$

Similarly,

$B = 4 \times 1 0^{-7} s i n (1.05 x - 3.14 \times 1 0^{8} t) k^$

Q 9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the 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:

Energy of a photon is given as:

Ncert Solution Class 12th Physics Chapter - 8 Electromagnetic Waves Question & Answer

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

λ = Wavelength of radiation

Ncert Solution Class 12th Physics Chapter - 8 Electromagnetic Waves Question & Answer

The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.

λ (m)	10³	1	10⁻³	10⁻⁶	10⁻⁸	10⁻¹⁰	10⁻¹²
E (eV)	12.375 × 10⁻¹⁰	12.375 × 10⁻⁷	12.375 × 10⁻⁴	12.375 × 10⁻¹	12.375 × 10¹	12.375 × 10³	12.375 × 10⁵

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 10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 1 0^{10} H z$ and amplitude $48 V m^{-1}$ .

(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 $\times 1 0^{8} m s^{-1}$ ]

Answer:

(a)

The wavelength is given by

λ = c = 3 \times 10 ^{8 /} 2 \times 10 ^{10} = 1.5 \times 1 0^{-2} m

(b)

B_{o} = E_{o} / c = 1.6 \times 1 0^{-7} T

(c)

Energy density due to the electric field,

E_{E} = 1/2 ϵ_{o} E^{2}

Energy density due to the magnetic field,

E_{B} = 1/ 2 B^{2} / μ_{o}

also,

c = 1/\sqrt ϵ ^{0} μ ^{o}

on solving above equations,

E_{E} = E_{B}

Q 11. Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]} ˆi .
(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 field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Answer: (a) Electric field is directed along negative x-direction.

hence the direction of motion is along the negative y-direction.

(b)

From the given equation, $k = 1.8 r a d / m$

$λ = 2 π / k = 3.49$

(c)

Frequency of wave is $ν = ω / 2 π = 8.6 \times 1 0^{7} = 86 M h z$

(d)

For the given equation, $E_{o} = 3.1 N / C$

$B_{o} = E_{o} / c = 1.03 \times 1 0^{-7} = 103 n T$

(e)

Magnetic wave is directed along negative z-direction.

Thus, $B = B_{o} c o s (k y + ω t) k^$

$\Rightarrow B = 1.03 \times 1 0^{-7} c o s (1.8 y + 5.4 \times 1 0^{6} t) k^$

Q 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 reflection.

Answer: Given: The power rating of bulb is 100 W, 5 % of the power is converted into visible radiation.

(a)

The power of visible radiation is given as,

P v = 5 100 ×P

Where, P v is the power of visible radiation and P is the power rating of bulb.

By substituting the values in the above expression, we get

P v = 5 100 ×100 W =5 W

The intensity of radiation at d=1 m is given as,

I= P v 4π d 2

Where, I is the intensity of radiation and d is the distance at which intensity is measured.

By substituting the values in the above expression, we get

I= 5 4π ( 1 ) 2 =0.398 W/ m 2

Thus, the intensity of radiation at distance of 1 m from the bulb is 0.398 W/ m 2

(b)

The intensity of radiation at d=10 m is given as,

I= P v 4π d 2

Where, I is the intensity of radiation and d is the distance at which intensity is measured.

By substituting the values in the above expression, we get

I= 5 4π ( 10 ) 2 =0.00398 W/ m 2

Thus, the intensity of radiation at distance of 10 m from the bulb is 0.00398 W/ m 2 .

Q 13. Use the formula λ_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:

We have the equation, λ_m T = 0.29 cm K

⇒ T = (0.29/λ_m )cm K

Here, T is the temperature

λ_m is the maximum wavelength of the wave

For λ_m= 10^-4 cm

T = (0.29/10^-4)cm K = 2900 K

For the visible light, λ_m= 5 x 10^-5cm

T = (0.29/ 5 x 10^-5)cm K ≈ 6000 K

Note: a lower temperature will also produce wavelength but not with maximum intensity.

Q 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 filling 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 (short-wavelength end)
(b) Radio waves (short-wavelength end)
(c) Microwave
(d) Visible light (Yellow)
(e) X-rays (or soft γ-rays) region

Q 15. Answer the following questions:
(a) Long-distance radio broadcasts use short-wave 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) Ionosphere reflects waves in the shortwave bands.
(b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.
(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
(d) Ozone layer absorbs the ultraviolet radiations from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
(e) If the atmosphere is not present, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.
(f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.

NCERT Solutions Class 12th Physics Chapter – 8 Electromagnetic Waves

NCERT Solutions Class 12th Physics Chapter – 8 Electromagnetic Waves

(a) C = ε0​A/d=80.1 pF dtdQ​=CdtdV​ dtdV​=80.1×10−120.15​=1.87×109Vs−1

f1​=7.5×106 Hz

(a) $C = ε_{0} A / d = 80.1 p F$
$d t d Q = C d t d V$
$d t d V = 80.1 \times 10 ^{-12} 0.15 = 1.87 \times 1 0^{9} V s^{-1}$

$f_{1} = 7.5 \times 1 0^{6} H z$