NCERT Solutions for Class 12 Physics Chapter 8 – Electromagnetic Waves

NCERT Solutions for Class 12 Electromagnetic Waves Chapter 8 are here by the subject experts at SimplyAcad to help students score well in their upcoming physics board examination.

 

The 12th board is the foundation of the student’s career, Hence, students must make sure to regularly practise and revise. Electromagnetic Waves Chapter 8 focuses on various phenomena that will enhance students’ creativity. The physics enthusiasts will love the chapter and hope these solutions provided for the Electromagnetic Waves Chapter 8 will benefit the most.

 

NCERT Solutions for Class 12 Electromagnetic Waves Chapter 8

Q 8.1) 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.

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

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

 

= 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:

Therefore, the change in the potential difference between the plates is

(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 of conduction and displacement for current.

Q 8.2) 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.

Answer 8.2:

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

Capacitance of a parallel plate capacitor, C = 100 pF =

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

(c) Magnetic field is given as:

Q 8.3)What physical quantity is the same for X-rays of wavelength $10^{-10} \, \text{m}$, red light of wavelength $6800 \, \text{Å}$, and radio waves of wavelength $500 \, \text{m}$?

Answer:

X-rays, red light, and radio waves are all electromagnetic waves. The wavelengths and frequencies of these waves are different, but they all travel at the same speed in a vacuum, which is the speed of light $c$.

$$c = 3 \times 10^8 \, \text{m/s}$$

Final answer:

The speed in a vacuum is the same for all electromagnetic waves.

 

Q 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?

Answer 8.4

The wave travels along the $z$-direction. The electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ are in the $XY$ plane and are perpendicular to each other.

The wavelength $\lambda$ can be calculated using the formula:

$$\lambda = \frac{c}{\nu}$$

where $c$ is the speed of light and $\nu$ is the frequency. Given the frequency is 30 MHz ($30 \times 10^6 \, \text{Hz}$), we can find:

$$\lambda = \frac{3 \times 10^8 \, \text{m/s}}{30 \times 10^6 \, \text{Hz}} = 10 \, \text{m}$$

Thus, the wavelength is $10 \, \text{m}$.

 

Q 8.5) A radio can tune into any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?

Answer 8.5:

A radio can tune to a minimum frequency,

Maximum frequency,

Speed of light, c =

Corresponding wavelength for

𝑣1

can be calculated as:

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

 

Q 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?

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) The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave?

Answer 8.7:

The amplitude of the magnetic field of an electromagnetic wave in a vacuum,

Speed of light in vacuum, c =

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

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

 

Q 8.8) Question:

Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 \, \text{N/C}$ and that its frequency is $\nu = 50.0 \, \text{MHz}$.

(a) Determine $B_0$, $\omega$, $k$, and $\lambda$.

(b) Find the expression for $\mathbf{E}$ and $\mathbf{B}$.

Solution:

(a)

The magnetic field amplitude $B_0$ is given by:

$$B_0 = \frac{E_0}{c} = \frac{120 \, \text{N/C}}{3 \times 10^8 \, \text{m/s}} = 400 \, \text{nT}$$

The angular frequency $\omega$ is:

$$\omega = 2 \pi \nu = 2 \pi \times 50.0 \times 10^6 \, \text{Hz} = 3.14 \times 10^8 \, \text{rad/s}$$

The wave number $k$ is:

$$k = \frac{\omega}{c} = \frac{3.14 \times 10^8 \, \text{rad/s}}{3 \times 10^8 \, \text{m/s}} = 1.05 \, \text{rad/m}$$

The wavelength $\lambda$ is:

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8 \, \text{m/s}}{50.0 \times 10^6 \, \text{Hz}} = 6 \, \text{m}$$

(b)

Suppose the wave is moving along the $x$-direction, with the electric field along the $y$-direction and the magnetic field along the $z$-direction.

The electric field $\mathbf{E}$ is:

$$\mathbf{E} = E_0 \sin(kx – \omega t) \hat{j} = 120 \sin(1.05 x – 3.14 \times 10^8 t) \hat{j}$$

Similarly, the magnetic field $\mathbf{B}$ is:

$$\mathbf{B} = B_0 \sin(kx – \omega t) \hat{k} = 4 \times 10^{-7} \sin(1.05 x – 3.14 \times 10^8 t) \hat{k}$$

 

Q 8.9)Question:

Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 \, \text{N/C}$ and that its frequency is $\nu = 50.0 \, \text{MHz}$.

(a) Determine $B_0$, $\omega$, $k$, and $\lambda$.

(b) Find the expression for $\mathbf{E}$ and $\mathbf{B}$.

Solution:

(a)

The magnetic field amplitude $B_0$ is:

$$B_0 = \frac{E_0}{c} = \frac{120 \, \text{N/C}}{3 \times 10^8 \, \text{m/s}} = 400 \, \text{nT}$$

The angular frequency $\omega$ is:

$$\omega = 2 \pi \nu = 2 \pi \times 50.0 \times 10^6 \, \text{Hz} = 3.14 \times 10^8 \, \text{rad/s}$$

The wave number $k$ is:

$$k = \frac{\omega}{c} = \frac{3.14 \times 10^8 \, \text{rad/s}}{3 \times 10^8 \, \text{m/s}} = 1.05 \, \text{rad/m}$$

The wavelength $\lambda$ is:

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8 \, \text{m/s}}{50.0 \times 10^6 \, \text{Hz}} = 6 \, \text{m}$$

(b)

Suppose the wave is moving along the $x$-direction, the electric field is along the $y$-direction, and the magnetic field is along the $z$-direction.

The electric field $\mathbf{E}$ is:

$$\mathbf{E} = E_0 \sin(kx – \omega t) \hat{j} = 120 \sin(1.05 x – 3.14 \times 10^8 t) \hat{j}$$

Similarly, the magnetic field $\mathbf{B}$ is:

$$\mathbf{B} = B_0 \sin(kx – \omega t) \hat{k} = 4 \times 10^{-7} \sin(1.05 x – 3.14 \times 10^8 t) \hat{k}$$

 

Q 8.10) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude of 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 =

Answer 8.10:

Frequency of the electromagnetic wave, v =

Electric field amplitude,

Speed of light, c =

(a) Wavelength of a wave is given as:

(b) Magnetic field strength is given as:

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

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

Q 8.11) 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) The direction of motion is along the negative y-direction, i.e. along -j.

(b) The given equation is compared with the equation,

E = E0 cos (ky + ωt)

⇒ k = 1.8 rad/s

ω = 5.4 x 106 rad/s

λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m

(c) Frequency, ν = ω/2π =  5.4 x 106/(2 x 3.14) = 0.859  x 106 Hz.

(d) Amplitude of the magnetic field, B0 = E0/c

= 3.1/(3 x 108) = 1.03 x 10-8 T= 10.3 x 10-9 T= 10.3 nT.

(e) Bz = B0 cos (ky + ωt)ˆk ={(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ

Q 8. 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 neglects reflection.

Answer:

(a) Average intensity of the visible radiation, I = P’/4πd2

Here, the power of the visible radiation, P’ = (5/100) x 100 = 5 W

At d = 1 m

I = P’/4πd2 = 5/(4 x 3.14 x 12) = 5/12.56 = 0.39 W/m2

(b) At d = 10 m

I = P’/4πd2 = 5/(4 x 3.14 x 102)  = 5/1256 = 0.39 x 10-2 W/m2

Q 8.13) 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?

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 10-5 cm

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 8. 14) Given below are some famous numbers associated with electromagnetic radiation 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 8.15) 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 radiation from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
(e) If the earth did not have an atmosphere, 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’.