## Electromagnetic Waves class 12 ncert solutions: Ncert Class 12 Physics Chapter 8 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 8 |

Chapter Name | Electromagnetic Waves class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 8 Electromagnetic Waves? Now you can download Ncert Class 12 Physics Chapter 8 Exercise Solutions pdf from here.

### Question 8.1: 1 Figure 8.5 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.

**Solution 8.1:**

- 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 ϵ
_{o }= 8.85**×**10^{-12}C^{2}N^{-1}m^{-2}

**(a) **The capacitance between the two plates can be calculated as follows: \(C = \frac{\varepsilon _{0} A}{d}\)

where,

A = Area of each plate = 𝝅r^{2}

\(C = \frac{\varepsilon _{0} \pi r^{2}}{d}\)

= \(\frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}\)

= 8.0032 **×** 10^{-12} F

= 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: \(\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}\)

But,\(\frac{\mathrm{d} q}{\mathrm{d} t}\)

= Current (I) \(∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}\)

= \(\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s\)

Therefore, the change in the potential difference between the plates is \(1.87\times 10^{9}\; V/s\)

**(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.

### Question 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.

**Solution 8.2:**

- Radius of each circular plate, R = 6.0 cm = 0.06 m
- Capacitance of a parallel plate capacitor, C = 100 pF = 100
**×**10^{-12}F - Supply voltage, V = 230 V
- Angular frequency, \(\omega = 300\; rad\;s^{-1}\)

**(a) **Rms value of conduction current, I = \(\frac{V}{X_{c}}\)

Where, X_{c} = Capacitive reactance = \(\frac{1}{\omega C}\)

\(∴ I = V\times \omega C\)

= 230 **×** 300 **×** 100 **×** 10^{-12}

= 6.9 **×** 10^{-6} A

= 6.9 μA

Hence, the rms value of conduction current is 6.9 μA

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

**(c) **Magnetic field is given as: \(B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}\)

Where,

μ_{0} = Permeability of free space

= 4𝝅 **×** 10^{-7} N A^{-2}

I_{0} = Maximum value of current = \(\sqrt{2}\; I\)

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

\(∴B = \frac{4\pi\times 10^{-7}\times 0.03\times \sqrt{2}\times 6.9\times 10^{-6}}{2\pi \times (0.06)^{2}}\)

= 1.63 **×** 10^{-11} T

Hence, the magnetic field at the point is 1.63 **×** 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 radiowaves of wavelength 500m?

**Solution 8.3:** The physical quantity that is the same for X-rays of wavelength 10^{–10} m, red light of wavelength 6800 Å, and radiowaves of wavelength 500m is the **speed of light** in a vacuum.

In a vacuum, all electromagnetic waves (whether X-rays, visible light, or radiowaves) travel at the same speed, which is the speed of light, denoted by ( c ). This value is approximately: **c = 3 × 10 ^{8} m/s**

Thus, despite having different wavelengths and frequencies, the speed of X-rays, visible light, and radiowaves remains constant in a vacuum.

### 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:** The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave, v = 30 MHz = 30 **×** 10^{6} _{8}^{-1}

Speed of light in vacuum, C = 3 **×** 10^{8} m/s

The wavelength of a wave is given as \(\lambda = \frac{c}{v}\)

= \(\frac{3\times 10^{8}}{30\times 10^{6}}\)

= 10 m

E and B in x-y plane and are mutually perpendicular, 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:** To find the wavelength corresponding to a frequency, we use the relationship between the speed of light ( c ), the frequency ( f ), and the wavelength \( \lambda \):

\(\lambda = \frac{c}{f}\)

where:

- c is the speed of light in vacuum, c = 3 × 10
^{8}m/s, - f is the frequency in hertz (Hz),
- \( \lambda \) is the wavelength in meters (m).

**Step 1: Wavelength for 7.5 MHz**

First, we calculate the wavelength corresponding to the lower frequency, 7.5 MHz:

\(f = 7.5 \, \text{MHz} = 7.5 \times 10^6 \, \text{Hz}\)

\(\lambda = \frac{3 \times 10^8 \, \text{m/s}}{7.5 \times 10^6 \, \text{Hz}} = 40 \, \text{m}\)

**Step 2: Wavelength for 12 MHz**

Next, we calculate the wavelength corresponding to the upper frequency, 12 MHz:

\(f = 12 \, \text{MHz} = 12 \times 10^6 \, \text{Hz}\)

\(\lambda = \frac{3 \times 10^8 \, \text{m/s}}{12 \times 10^6 \, \text{Hz}} = 25 \, \text{m}\)

**Conclusion:**

The wavelength band corresponding to the frequency range 7.5 MHz to 12 MHz is **Wavelength band: 40 m – 25 m**.

### Question 8.6: A charged particle oscillates about its mean equilibrium position with a frequency of 10^{9} Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

**Solution 8.6:** The frequency of the electromagnetic waves produced by a charged particle oscillating about its mean equilibrium position is **equal to the frequency of the oscillation** of the charged particle itself.

In this case, the particle oscillates with a frequency of 10^{9} Hz (or 1 GHz), which means the frequency of the electromagnetic waves produced will also be 10^{9} Hz.

**Conclusion:**

The frequency of the electromagnetic waves produced by the oscillator is 10^{9} Hz (1 GHz).

### Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B_{0} = 510 nT. What is the amplitude of the electric field part of the wave?

**Solution 8.7:** The amplitude of the magnetic field of an electromagnetic wave in a vacuum,

B_{0} = 510 nT = 510 **×** 10-9 T

Speed of light in vacuum, c = 3 **×** 10^{8} m/s

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

E = cB_{0} = 3 **×** 10^{8} **×** 510 **×** 10^{-9} = 153 N/C

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

### Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E_{0} = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B_{0} ,ω, k, and \( \lambda \). (b) Find expressions for E and B.

**Solution 8.8:**

- Electric field amplitude, E
_{0}= 120 N/C - Frequency of source, v = 50 MHz = 50
**×**10^{6}Hz - Speed of light, c = 3
**×**10^{8}m/s

**(a) **Magnitude of magnetic field strength is given as: \(B_{0} = \frac{E_{0}}{c} \)

= \(\frac{120}{3\times 10^{8}} \)

= 40 **×** 10^{-8} = 400 **×** 10^{-9} T = 400nT

The angular frequency of the source is given by:

\(\omega =2\pi v=2\pi \times 50\times 10^{6} \)

\(=3.14\times 10^{8}\,rads^{-1} \)

= 3.14 **×** 10^{8} rad/s

The propagation constant is given as: \(k = \frac{\omega }{c} \)

= \(\frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m \)

The wavelength of the wave is given by: \(\lambda = \frac{c}{v} \)

= \(\frac{3\times 10^{8}}{50\times 10^{6}} \)

= 6.0 m

**(b) **Suppose the wave is propagating in the positive x-direction. Then, the electric field vector will be in the positive y-direction, and the magnetic field vector will be in the positive z-direction. This is because all three vectors are mutually perpendicular.

The equation of the electric field vector is given as:

\(\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j} \)

= \(120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j} \)

And the magnetic field vector is given as:

\(\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k} \)

\(\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k} \)

### Question 8.9: 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?

**Solution 8.9:** Energy of a photon is given as: E = hv = \(\frac{hc}{\lambda}\)

Where,

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

c = Speed of light = 3 × 10^{8} m/s

λ = Wavelength of radiation

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.

### Question 8.10: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 **×** 10^{10} Hz 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 × 10
^{8 }m s^{–1}.]

**Solution 8.10:**

- Frequency of the electromagnetic wave, v = 2.0
**×**10^{10}Hz - Electric field amplitude, E
_{0}= 48 V m^{-1} - Speed of light, c = 3
**×**10^{8}m/s

**(a)** Wavelength of a wave is given as: \(\lambda = \frac{c}{v}\)

= \(\frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m\)

**(b)** Magnetic field strength is given as: \(B_{0} = \frac{E_{0}}{c}\)

= \(\frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T\)

**(c) **Energy density of the electric field is given as: \(U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}\)

And the energy density of the magnetic field is given as: \(U_{B} = \frac{1}{2\mu_{0}}B^{2}\)

Where,

ϵ_{o} = Permittivity of free space

μ_{0} = Permeability of free space

E = cB …. (1)

Where,

\(c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\) ……(2)

Putting equation (2) in equation (1), we get \(E = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B\)

Squaring on both sides, we get\(E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}\)

\(\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}\)

= \(\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}\)

= \(U_{E} = U_{B}\)