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## NCERT Class 12 Physics Textbook Chapter 8 With Answer PDF Free Download

### Chapter 8: Electromagnetic Waves

#### 8.1 Introduction

In Chapter 4, we learned that an electric current produces a magnetic field and that two current-carrying wires exert a magnetic force on each other.

Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true?

Does an electric field changing with time give rise to a magnetic field?

James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates a magnetic field.

While applying Ampere’s circuital law to find a magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in Ampere’s circuital law.

He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency.

Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities.

These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism.

The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time-varying electric and magnetic fields that propagate in space.

The speed of the waves, according to these equations, turned out to be very close to the speed of light( 3 ×108 m/s), obtained from optical measurements.

This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism, and light.

Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today.

#### Displacement Current

We have seen in Chapter 4 that an electrical current produces a magnetic field around it. Maxwell showed that for logical consistency, a changing electric field must also produce a magnetic field.

This effect is of great importance because it explains the existence of radio waves, gamma rays and visible light, as well as all other forms of electromagnetic waves.

To see how a changing electric field gives rise to a magnetic field, let us consider the process of

charging of a capacitor and applying Ampere’s circuital the law was given by (Chapter 4)

“B.dl = µ 0 i (t) (8.1)

to find a magnetic field at a point outside the capacitor.

Figure 8.1(a) shows a parallel plate capacitor C which is a part of the circuit through which time-dependent current i (t) flows.

Let us find the magnetic field at a point such as P, in a region outside the parallel plate capacitor.

For this, we consider a plane circular loop of radius r whose plane is perpendicular to the direction

of the current-carrying wire, and which is centered symmetrically with respect to the wire [Fig. 8.1(a)].

From symmetry, the magnetic field is directed along the circumference of the circular loop and is the same in magnitude at all points on the loop so that if B is the magnitude of the field, the left side of Eq. (8.1) is B (2π r).

So we have B (2πr) = µ0 i (t)

Author | NCERT |

Language | English |

No. of Pages | 19 |

PDF Size | 419 KB |

Category | Physics |

Source/Credits | ncert.nic.in |

### NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves

**Q 8.3) What physical quantity is the same for X-rays of wavelength 10 ^{–10}m, the red light of wavelength 6800 Å and radiowaves of wavelength 500m?**

**Answer 8.3:**

The speed of light (3\times 10^{8}3×108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

**Q 8.4) 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 itswavelength?**

**Answer 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\times 10^{6}\;s^{-1}30×106*s*−1

Speed of light in vacuum, C = 3\times 10^{8}3×108 m/s

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

= \frac{3\times 10^{8}}{30\times 10^{6}}30×1063×108 = 10 m

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

**Answer 8.5:**

A radio can tune to minimum frequency, v_{1} = 7.5\; MHz = 7.5\times 10^{6}\; Hz*v*1=7.5*M**H**z*=7.5×106*H**z*

Maximum frequency, v_{2} = 12\; MHz = 12\times 10^{6}\; Hz*v*2=12*M**H**z*=12×106*H**z*

Speed of light, c = 3\times 10^{8}\; m/s3×108*m*/*s*

Corresponding wavelength for v_{1}*v*1 can be calculated as:\lambda_{1} = \frac{c}{v_{1}}*λ*1=*v*1*c*=\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m=7.5×1063×103=40*m*

Corresponding wavelength for v_{2}*v*2 can be calculated as:\lambda_{2} = \frac{c}{v_{2}}*λ*2=*v*2*c*=\frac{3\times 10^{3}}{12\times 10^{6}} = 25\;m=12×1063×103=25*m*

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 10 ^{9} 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., 10^{9} Hz.

**Q 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?**

**Answer 8.7:**

Amplitude of magnetic field of an electromagnetic wave in a vacuum,B_{0} = 510\; nT = 510\times 10^{-9}\; T*B*0=510*n**T*=510×10−9*T*

Speed of light in vacuum, c = 3\times 10^{8}\; m/s3×108*m*/*s*

Amplitude of electric field of an electromagnetic wave is given by the relation,E = cB_{0} = 3\times 10^{8}\times 510\times 10^{-9} = 153\; N/C*E*=*c**B*0=3×108×510×10−9=153*N*/*C*

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

**Q 8.8) Suppose that the electric field amplitude of an electromagnetic wave is E_{0} = 120\; N/C E0=120N/C and that its frequency is v = 50 MHz.(a) Determine B_{0},\; \omega,\; k\;and\; \lambdaB0,ω,kandλ (b) Find expressions for E and B.**

**Answer 8.8:**

Electric field amplitude, E_{0} = 120\; N/C*E*0=120*N*/*C*

Frequency of source, v = 50 MHz = 50\times 10^{6}50×106 Hz

Speed of light, c = 3\times 10^{8}3×108 m/s

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

= \frac{120}{3\times 10^{8}}3×108120

= 40\times 10^{-8}\;=400\times 10^{-9} T = 400\; nT40×10−8=400×10−9*T*=400*n**T*

Angular frequency of source is given by:\omega =2\pi v=2\pi \times 50\times 10^{6}=3.14\times 10^{8}\,rads^{-1}*ω*=2*π**v*=2*π*×50×106=3.14×108*r**a**d**s*−1

= 3.14\times 10^{8}3.14×108 rad/s

Propagation constant is given as:k = \frac{\omega }{c}*k*=*c**ω*

= \frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m3×1083.14×108=1.05*r**a**d*/*m*

Wavelength of wave is given by:\lambda = \frac{c}{v}*λ*=*v**c*

= \frac{3\times 10^{8}}{50\times 10^{6}}50×1063×108 = 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.

Equation of electric field vector is given as:\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}*E*=*E*0*s**i**n*(*k**x*–*ω**t*)*j*

= 120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}120*s**i**n*[1.05*x*–3.14×108*t*]*j*

And, magnetic field vector is given as:\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}*B*=*B*0*sin*(*kx*–*ωt*)*k*\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}*B*=(400×10−9)*sin*[1.05*x*–3.14×108*t*]*k*

NCERT Class 12 Physics Textbook Chapter 8 With Answer PDF Free Download