WBCHSE Class 12 Physics Notes For Electromagnetic Waves

Class 12 Physics Electromagnetic Waves Introduction

In 1865, James Clerk Maxwell analyzed the nature of light. According to him, light is a progressive wave of an electric and a magnetic field. In other words, light is an electromagnetic wave. At this time, except for visible light, infrared, and ultraviolet, the existence of other electromagnetic waves was unknown.

  • As per Maxwell’s electromagnetic theory, an electric field and a magnetic field exist at each point on the path of propagation of an electromagnetic wave. These two fields undergo periodic vibrations.
  • These vibrations propagate from one point to the next, and thus electromagnetic waves spread. Electric and magnetic fields can spread in all directions even without any material medium (solid, liquid, or gaseous). Hence, electromagnetic waves can propagate in a vacuum also.

Class 12 Physics Electromagnetic Waves Displacement Current

As per Ampere’s circuital law (See chapter ‘Electromagnetism; section 1.5),

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\) → (1)

But the law has a limitation. What would be the form of the above equation for a varying current I, i.e., for a varying electric field inside a conductor, is not mentioned.

For example, during the charging of a capacitor, varying current flows through the circuit though no current flows in between the plates of the capacitor. Equation (1) is not applicable in such cases.

Maxwell corrected this error by adding another term to the right side of equation (1). This additional term is,

⇒ \(\mu_0 I_d=\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\) → (2)

Maxwell mentioned the quantity Jd as displacement current. Note that there is no valid reason for using the word ‘displacement’ in this context But, the word is entrenched in the language of physics.

\(\epsilon_0 \text { and } \phi_E\left(=\int \vec{E} \cdot d \vec{A}\right)\) in equation (2) is the permittivity of free space and electrical flux through a Gaussian surface respectively.

  • Therefore, is the rate of change of electrical flux and clearly, \( I_=\epsilon_0 \frac{d \phi_E}{d t}\) is 1h expression representing varying electrical field.
  • This varying electrical field is equivalent to the die current flow between the plates of the capacitor during charging (known as displacement current). So, now there is no discontinuation in the current flow. Hence, the general form of Ampere’s circuital law is,

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I+\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\) → (3)

which is also known as Ampere-Maxwell law.

The meaning of this expression can be understood with help. The electric flux through surface s2 is \(\phi_E=\int \vec{E} \cdot d \vec{A}=E A\), where A is the area of the capacitor plates and E is the magnitude of the uniform electric field between the plates.

If Q is the charge on the plates at any instant, then \(E=\frac{Q}{\epsilon_0 A}\).

Therefore, \(\phi_E=E A=\frac{Q}{\epsilon_0}\)

Hence, the displacement current through S2 is

∴ \(I_d=\epsilon_0 \frac{d \phi_E}{d t}=\frac{d Q}{d t}\)

Electromagnetic Waves Displacement Current

The displacement current through S2 is exactly equal to the conduction current through S1.

The displacement current can be identified as the source of the magnetic field on the surface S2. The displacement current has its physical origin in the time-varying electric field. The central point of this formalism is that a magnetic field can be produced in two ways—if

  1. Current flows through a conductor or
  2. The electric field varies with time in a region.

∴ The general form of Faraday’s law of induction is, \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\).

  • It can be concluded from this law that,’ a changing magnetic field creates an electric field (that is an emf in a conductor). Similarly, from the above law, we can conclude that a changing electric field sets up a magnetic field.
  • In other words, the concept of displacement current develops an important connection between varying electric fields and magnetic fields. These varying electric and magnetic fields spread out in all directions and are known as electromagnetic waves.

Definition: When varying electric and magnetic fields exist in a place then these varying fields spread out in all directions like a wave which is called an electromagnetic wave.

Class 12 Physics Electromagnetic Waves The Spectrum Of Electromagnetic Waves

Different types of electromagnetic waves, their wavelengths, sources, and uses are given in the following table.

Electromagnetic Waves Different Types Of Electromagnetic Waves Their Wavelengths, Sources And Uses

Today we know about the existence of the electromagnetic spectrum. The whole world is immersed in electromagnetic waves of different wavelengths. The sun is the main source of most of these waves.

  • In the increasing order of frequencies, i.e., decreasing order of wavelengths, a few important em waves are radio waves or Hertzian rays, microwave, infrared, visible light, ultraviolet rays, X-rays, gamma rays, cosmic rays, etc.
  • which has been discussed in the previous table. There is no distinct line of demarcation of different electromagnetic waves which, almost invariably, overlap to some extent.

All electromagnetic waves are generated due to the motion of accelerated charges. German physicist Heinrich Hertz, in 1888, first generated electromagnetic waves in the range of radio waves on an experimental basis.

And also made arrangements for the identification of the wave, ye took the help of an oscillating electric dipole. An oscillating electric dipole of this type is known as a Hertzian oscillator or Hertzian dipole.

  1. Most substances absorb infrared waves very easily. On absorbing infrared radiation, the internal energy of matter increases; hence there is a temperature rise.
  2. The ozone layer in the upper atmosphere, mainly in the stratosphere, absorbs the ultraviolet rays and converts them to infrared rays. Some chemicals reduce the density of the ozone layer by reacting with it. These substances are called ozone-depleting substances or ODS in short. Most of the ODS are made artificially and used in aerosol spray cans and refrigerants. Uses of this type of matter are trying to be banned.

Electromagnetic Waves Electromagnetic Spectrum

Class 12 Physics Electromagnetic Waves Characteristics Of Electromagnetic Wave

1. Electric field \(\vec(E)\) and magnetic field \(\vec(B)\) are perpendicular to the direction of propagation of the wave. Hence electromagnetic waves are transverse.

Electromagnetic Waves Characteristics Of Electromagnetic Wave

2. Magnetic field being a varying field, an electric field is induced (following Faraday’s laws of electromagnetic induction) perpendicular to it.

At the same time, varying electric fields induce a magnetic field (following Maxwell’s law of displacement current).

3. Let an electromagnetic wave propagate along the positive x direction. In that case, \(\vec{E}\) field and \(\vec{B}\) field oscillate parallel to the y and z-axes. Expressions for \(\vec{E}\) and \(\vec{B}\) fields at a distance x from the origin O are given in scalar form, by

E = E0 sin(ωt – kx) → (1)

and B = B0 sin (ωt – kx) → (2)

where E0, and B0, are the amplitudes of the electric field and magnetic field, while ω and k are the angular frequency and the magnitude of the propagation vector, respectively.

It is to be noted that none of the two fields exists independently of the other. A time-varying magnetic field induces an electric field.

Simultaneously, a time-varying electric field induces a magnetic field. Thus, self-sustaining electromagnetic waves travel through space.

Electromagnetic Waves Self-Sustaining Electromagnetic Waves Travel Through Space

This depicts an em wave with its components along the y and z-axes oscillating in the xy-plane and zx-plane and representing equations (1) and (2) respectively. They represent the field vectors in magnitude and direction at different distances from the origin at any time t.

Equations (1) and (2) show that the frequencies of the electric and magnetic fields are equal and these two fields are always in the same phase.

Speed of electromagnetic wave in free space,

⇒ \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\) → (3)

where μ0 and ε0 are respectively permeability and permittivity of free space.

Here, = μ0 x 10-7 N. s2.C-2

ε0 = 8.854 X 10-2 C2. N-2. m-2

Substituting the values in equation (3) we get,

⇒ \(c=\frac{1}{\sqrt{4 \pi \times 10^{-7} \times 8.854 \times 10^{-12}}}=2.998 \times 10^8\)

≈ 3.0 x 102 m. s-1

In addition, the velocity of an electromagnetic wave can also be obtained from the amplitudes of electric and magnetic fields as,

⇒ \(\frac{E_0}{B_0}=c\) → (4)

From equations (1) and (2) we get,

∴ \(\frac{E}{B}=\frac{E_0}{B_0}=c\)

If the frequency and the wavelength of an electromagnetic wave in a medium are respectively f and λ, then in that medium the velocity of the wave will be fλ. If the medium changes, the wavelength as well as the velocity will change, but the frequency will remain the same because frequency is a characteristic of the source.

Some Quantities Related To EM Waves

Energy density: We have seen that the electric field, E = E0sin(ωt – kx), and magnetic field, B = B0 sin (ωt – kx).

Consider a volume element dV of the medium in which an electromagnetic wave is propagating. At a certain moment, the energy carried by the volume is U. As the energy stored is due to both electric and magnetic fields, we have

U= UE+UM

Here, UE = stored energy due to the electric field in the volume element dV

= \(\frac{1}{2} \epsilon_0 E^2 \cdot d V\)

UM = stored energy due to magnetic field in the volume element dV

= \(\frac{1}{2} \frac{B^2}{\mu_0} \cdot d V\)

So, at any position x of the electromagnetic wave at time t, the energy density of the electric field,

∴ \(u_E=\frac{U_E}{d V}=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \epsilon_0 E_0^2 \sin ^2(\omega t-k x)\)

The energy stored in a unit volume of the space between two plates of a charged air capacitor, \( u_=\frac {1}{2} c_0 E^2[/atex].

We can prove that the total stored energy of the charged capacitor = uE x volume of the space between two plates of capacitor = [latex]\frac{1}{2} C V_{P d}^2\), where, C = capacitance of the capacitor and VP(t = potential difference between the plates of the capacitor.

and the energy density of the magnetic field,

∴ \(u_M=\frac{U_M}{d V}=\frac{1}{2 \mu_0} B^2=\frac{1}{2 \mu_0} B_0^2 \sin ^2(\omega t-k x)\)

The above equation is used to calculate the energy, stored in an inductor.

Average value of sin2# or cos2# in a complete cycle = \(\frac{1}{2}\)

Here, \(\left\langle\sin ^2(\omega t-k x)\right\rangle=\frac{1}{2}\)

Therefore, the average energy density,

⇒ \(\bar{u}_E=\frac{1}{2} \epsilon_0 E_0^2 \times \frac{1}{2}=\frac{1}{4} \epsilon_0 E_0^2\)

and \(\bar{u}_M=\frac{1}{2 \mu_0} B_0^2 \times \frac{1}{2}=\frac{1}{4 \mu_0} B^2=\frac{1}{4} c^2 \epsilon_0\left(\frac{E_0}{c}\right)^2\)

= \(\frac{1}{4} \epsilon_0 E_0^2\) [∵\(c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \text { or } \frac{1}{\mu_0}=c^2 \epsilon_0\)]

so \(\bar{u}_E=\bar{u}_M\)

It implies that the energy is equally distributed between electric and magnetic fields during the propagation of electromagnetic waves.

The average energy density of electromagnetic waves,

∴ \(\bar{u}=\bar{u}_E+\bar{u}_M=\frac{1}{4} \epsilon_0 E_0^2+\frac{1}{4} \epsilon_0 E_0^2=\frac{1}{2} \epsilon_0 E^2\)

For a complete cycle, the average value can be written as u instead of \(\bar{u}\).

Hence energy density,

⇒ \(u=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2 \mu_0} B_0^2\) → (1)

In general, for an isotropic medium of permittivity ε, the energy density,\(u=\frac{1}{2} \epsilon E_0^2\), where ε = Kε0, K is the dielectric constant of the medium.

Intensity: In an electromagnetic field, if we consider a unit area around a point, perpendicular to the direction of radiation, then the electromagnetic energy incident on that area per second is called the intensity of electromagnetic radiation at that point.

Electromagnetic Waves Some Quantities Related To EM Waves

Let P is a point in an electromagnetic field. The wave propagates toward the point P with velocity c. A unit area is considered around P, which is perpendicular to the direction of propagation of radiation.

In time dt, the wave travels a distance equal to cdt in the direction of propagation. Imagine a cylinder of length cdt and unit cross-sectional area on the path of the wave such that it crosses the cylinder normally.

The energy of the electromagnetic wave incident at point P is equal to the energy stored in the volume of the imaginary cylinder (cdt X 1 = cdt).

The intensity of radiation at P is equal to the energy incident per second i.e., the energy stored in the volume, c x 1 = c.

So, the energy contained in that cylinder =cu, where u = energy density.

Therefore, the intensity of electromagnetic wave at the point P, by definition

I = cu

By using the equation (l), we get,

⇒ \(I=c u=\frac{1}{2} c \epsilon_0 E_0^2=\frac{1}{2 \mu_0} c B_0^2\) → (2)

Incidentally, unlike in mechanical waves, this energy flow per unit area, per unit time, called energy flux is denoted by S, not I.

Hence, \(S=\frac{d u}{A d t}=c u\)

Average value S is given by \(\bar{S}=\frac{1}{2 \mu_0} E_0 B_0=\frac{1}{2} \epsilon_0 c E_0^2\)

\(\vec{S}\) called Poynting vector, is given by \(\vec{S}=\frac{1}{\mu_0}(\vec{E} \times \vec{B})=\vec{E} \times \vec{H}\)

The direction of \(\vec{S}\) is given by the direction of propagation of the wave, as shown.

If the point P is considered in an isotropic medium of electric permittivity ε and permeability μ instead of free space, then the equation (2) will be

∴ \(I=v u=\frac{1}{2} v \epsilon E_0^2=\frac{1}{2 \mu} v B_0^2\)

Where, \(v=\frac{1}{\sqrt{\epsilon \mu}}\) = velocity of the electromagnetic radiation in tha medium.

Unit of the intensity of electromagnetic radiation

= J.m-2.s-1 = W.m-2

Its dimension = \(\frac{M L^2 T^{-2}}{L^2 T}=M T^{-3}\)

Radiation Pressure: EM wave has linear momentum as well as energy. This means that radiation can exert force and hence pressure on any surface on which it is incident during Its propagation through a medium.

The force exerted on a unit area of the surface, perpendicular to the direction of propagation is called radiation pressure.

From thermodynamical analysis, we can show that (this analysis is beyond our syllabus),

Radiation pressure,\(p=\frac{1}{3} u\). where u = energy density of electromagnetic radiation [see the equation (1)].

The unit of pressure (p) = N.m-2; the unit of energy density

(u) = J.m-3.

Generally, both the units are the same as N.m = J.

The dimension of both p and u = ML-1T-2.

The force due to the radiation pressure is too small to detect under everyday circumstances. In subjects, like astronomy or astrodynamics-related works, its importance cannot be ignored.

For example, if the radiation pressure of the sun had been ignored, the space crafts, Viking-1 and Viking-2 (sent by NASA to Mars’s orbit to collect information about Mars) would have missed Mars’s orbit by about 15000 km.

Class 12 Physics Electromagnetic Waves Numerical Examples

Example 1. If the velocity of the em wave in a vacuum is 3 x 108 m.s-1 and the -magnetic permeability of the vacuum is 4π x 10-7 N.C-2. s2, find the electric permittivity of the vacuum.

Solution:

If permittivity and permeability of free space are ε0 and μ0, respectively,

∴ \(c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \quad \text { or, } c^2=\frac{1}{\epsilon_0 \mu_0}\)

∴ \(\epsilon_0=\frac{1}{\mu_0 c^2}=\frac{1}{4 \pi \times 10^{-7} \times 9 \times 10^{16}}\)

[∵ μ0 = 4π x 10-7 N.C-2. s2 and c = 3 x 108 m.s-1]

= 8.842 x 10-12 C2.N-1 m-2

Example 2. The amplitude of the electric field of a plane electromagnetic wave is 48 V.m-1. What is the amplitude of the magnetic field of the wave?

Solution:

In this case, the amplitude of the electric field

E0 = 48 V.m-1

As \(\frac{E_0}{B_0}=c\) amplitude of magnetic field, \(B_0=\frac{E_0}{c}\)

∴ \(B_0=\frac{48}{3 \times 10^8}\) [∵ c = 3 x 108 m.s-1]

= 16 x 10-8 Wb.m-2

Example 3. The electric field of an electromagnetic wave is, E = 10-5 sin(12 x 1015 t-4 x 107x). Find the frequency, velocity, and wavelength of the wave. Write the equation of the magnetic field corresponding to this wave. Assume all quantities in SI unit.

Solution:

The electric field of the given electromagnetic wave,

E = 10-5 sin(12 x 1015 t – 4 x 107x)

Compared with the expression for the electric field,

E = E0sin(ωt – kx) we get,

ω = 12 x 1015 rad.s-1

∴ 2πf = 12 x 1015

or \(f=\frac{12 \times 10^{15}}{2 \pi}=1.9 \times 10^{15} \mathrm{~Hz}\)

Hence, the frequency of the wave, f = 1.9 x 1015 Hz.

The velocity of the wave,

⇒ \(v=\frac{\omega}{k}=\frac{12 \times 10^{15}}{4 \times 10^7}=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Wavelength \(\lambda=\frac{c}{f}=\frac{3 \times 10^8}{1.9 \times 10^{15}}=1.58 \times 10^{-7} \mathrm{~m}\)

Equation of the corresponding magnetic field,

⇒ \(B=B_0 \sin (\omega t-k x)=\frac{E_0}{c} \sin (\omega t-k x)\) [∵\(\frac{E_0}{B_0}=c\)]

⇒ \(\frac{10^{-5}}{3 \times 10^8} \sin \left(12 \times 10^5 t-4 \times 10^7 x\right)\)

∴ 3.33 x 10-14 sin(12 x 1015 t-4 x 107x)T

Example 4. In an electromagnetic Held, the amplitude of the electric field at a point is 3 V m-1. Calculate the energy density and Intensity of the wave at that point. Given μ0 = 4π x 10-7H. m-1

Solution:

The amplitude of the electric field, E0 = 3 V. m-1

⇒ \(B_0=\frac{E_0}{c}=\frac{3}{3 \times 10^8}=10^{-8} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

So, the energy density,

⇒ \(u=\frac{1}{2 \mu_0} B_0^2=\frac{1}{2 \times 4 \pi \times 10^{-7}} \times\left(10^{-8}\right)^2\)

 3.98 x 10-11J . m-3

Intensity, I = cu = (3 x 108) x (3.98 x 10-11)

= 0.012 W. m-2

Example 5. A rectangular parallel plate capacitor of dimension 5 cm x 4 cm is charged in such a way that the rate of change of electric field between the two plates is 5.65 x 1011V. m-1 . s-1. Calculate the displacement current for this capacitor.

Given ε0 = 8.85 x 10-12 F. m-1

Solution:

Area of a rectangular plate,

A = (5 x 4) cm2 = 20 cm2 = 0.002 m2

If E is the electric field between two plates, then the electric flux,

⇒ \(\phi_E=E A ; \frac{d \phi_E}{d t}=A \frac{d E}{d t}\)

Therefore, the displacement current,

⇒ \(I_d=\epsilon_0 \frac{d \phi_E}{d t}=\epsilon_0 A \frac{d E}{d t}\)

= (8.85 x 10-12) x 0.002 x (5.65 x 1011)

= 0.01 A = 10 mA

Example 6. Calculate the intensity and the rms value of an electric field of an electromagnetic wave at a distance of 10 m from a 100 W electric bulb.

Given ε0 = 8.85 x 10-12 F. m-1

Solution:

Area of a spherical surface of radius 10 m

A = 4π (10)2 = 400 m2

∴ Intensity, I, \(\frac{100 \mathrm{~W}}{400 \pi \mathrm{m}^2}=\frac{1}{4 \pi} \mathrm{W} \cdot \mathrm{m}^{-2}=0.08 \mathrm{~W} \cdot \mathrm{m}^{-2}\)

Here, \(I=\frac{1}{2} c \epsilon_0 E_0^2=c \epsilon_0 E_{\mathrm{rms}}^2\) [latex]E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}[/latex]

∴ \(E_{\mathrm{rms}}=\sqrt{\frac{I}{c \epsilon_0}}=\sqrt{\frac{0.08}{\left(3 \times 10^8\right) \times\left(8.85 \times 10^{-12}\right)}}\)

= 5.5 V. m-1

 

Class 12 Physics Electromagnetic Waves Very Short Answer Type Questions And Answers

Displacement Current

Question 1. The current which comes into play in a region where the electric flux is changing with time is called___________.

Answer: Displacement Current

Question 2. Write down the mathematical form of Ampere-Maxwell’s law.

Answers: \(\oint \overrightarrow{B \cdot} \cdot d \vec{l}=\mu_0 I+\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\)

Electromagnetic Spectrum

Question 3. The wavelength of a radio wave is 10 m. Is this statement true or false?

Answer: True

Question 4. Wavelengths of visible light, X-rays, and infrared rays are λ1, λ2, and λ3, respectively. Arrange them in ascending order of their magnitudes.

Answer: λ2, λ1, λ3

Question 5. Of X-rays, γ-rays, and ultraviolet rays, which one has the maximum frequency?

Answer: γ – ray

Question 6. Which part of the electromagnetic spectrum has the largest penetrating power?

Answer: Gamma rays

Question 7. Name the electromagnetic waves that have frequencies greater than those of ultraviolet light but less than those of gamma rays.

Answer: x-rays

Question 8. Find the wavelength, in angstrom unit, of electromagnetic waves of frequency 5 x 1019Hz in free space.

Answer: 0.06 A

Production And Propagation Of Electromagnetic Waves

Question 9. An electromagnetic wave is propagating along the positive z-axis. At any moment, if the direction of the magnetic field at a point is along the positive x-axis, what will be the direction of the electric field at that point?

Answer: Along the negative y-axis

Question 10. What is the velocity of a radio wave in a vacuum?

Answer: 3 x 108 m. s-1

Question 11. Light is a progressive wave of _______ and _______ fields.

Answer: Electric, magnetic

Question 12. An electromagnetic wave is a _______ progressive wave

Answer: Transverse

Question 13. Write down the relation among the velocity of electromagnetic waves, the magnetic permeability of vacuum, and the electric permittivity of vacuum.

Answer: \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

Question 14. A Physical quantity is expressed by the ratio of the amplitudes of electric and magnetic fields for an electromagnetic wave.

Answer: velocity of the wave

Question 15. What is the ratio of velocities of light rays of wavelengths 4000A and 8000A in a vacuum?

Answer: 1:1

Question 16. What is the ratio of speeds of infrared rays and ultraviolet rays in a vacuum?

Answer: 1:1

Question 17. Which physical quantity, if any, has the same value waves belonging to the different parts of the electromagnetic spectrum?

Answer: Speed in vacuum

 

Class 12 Physics Electromagnetic Waves Synopsis

Electromagnetic waves exhibit wave wave-like nature as they travel through space.

Electromagnetic wave has both oscillating electric and magnetic field components.

  • Electromagnetic waves are transverse waves.
  • Electromagnetic waves were first experimentally produced by Hertz and the waves were radio waves.
  • When varying electric and magnetic fields exist in a place then these varying fields spread out in all directions like a wave, which is called an electromagnetic wave.
  • The Velocity of electromagnetic waves in a vacuum is equal to that of light.
  • Some electromagnetic waves according to increasing order of frequency (i.e., decreasing order of wavelength):
  • Radio waves (range 104m-0.1m), microwaves (range 0.3 m – 10-4 m ), infrared waves (range 10-3 m – 7 x 10-7 m), visible light (range 7 x 10-7 m – 4 x 10-7 m), ultraviolet rays (range 4 x 10-7 m – 6 x 10-8 m), X-rays (range 10-8 m- 10-11 m), gamma rays (range 10-11 m – 10-14 m).

An LC circuit connected with an AC source is known as an LC oscillator. With the help of an LC Oscillator, electromagnetic waves can be created.

The energy is equally distributed between electric and magnetic fields at the time of propagation of electromagnetic waves in a medium.

  • When an electromagnetic wave propagates along the positive X-axis, then electric field \(\vec{E}\) and magnetic field \(\vec{B}\) oscillate parallel to the y and z -axis respectively.
  • Expression of electric field and magnetic field at a distance x from the origin at the time t are given by
  • E = E0sin(ωt – kx) and B = B0sin(ωt – kx)

The velocity of electromagnetic waves in free space,

⇒ \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

where μ0 and ε0 are permeability and permittivity of free space respectively.

If μ0 = 4π x 10-7N.C-2. s2;

and ε0 = 8.854 x 10-12 C2.N-1.m-2 then, c = 3.0 x 108m.s-1

If the amplitudes of electric and magnetic fields are E0 and B0 respectively, then

⇒ \(\frac{E}{B}=\frac{E_0}{B_0}=c\)

The energy density of an electromagnetic field,

⇒ \(u=\frac{1}{2} \epsilon_0 E_0^2=\frac{1}{2 \mu_0} B_0^2\)

The intensity of radiation at an OOOpoint in an electromagnetic field,

I = cu

and radiation pressure,

∴ \(p=\frac{1}{3} u\)

Class 12 Physics Electromagnetic Waves Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
  2. Statement 1 is true, and statement 2 is true; statement is not a correct explanation for statement 1.
  3. Statement 1 is true, statement 2 is false.
  4. Statement 1 is false, and statement 2 is true.

Question 1. Statement 1: The electric field 1 and the magnetic field 3 are mutually perpendicular at a point in the electromagnetic wave.

Statement 2: Electromagnetic waves are transverse waves.

Answer: 2. Statement 1 is true, and statement 2 is true; statement 1 is not a correct explanation for statement 1.

Question 2. Statement 1: Electromagnetic waves are transverse waves.

Statement 2: Electromagnetic waves have the property of polarisation.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3. Statement 1: The ratio of the amplitudes of electric and magnetic fields at a point in the electromagnetic wave is the same as the velocity of the wave.

Statement 2: If a medium, μ, and ε are the magnetic permeability and the electric permittivity respectively, then 1/ √με is the velocity of an electromagnetic wave in that medium.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4. Statement 1: During the propagation of visible light and X-rays as electromagnetic waves, X-rays carry more energy than visible light despite both of them having the same amplitudes of electric fields at a point.

Statement 1: The frequency of X-rays is much higher than that of visible light.

Answer: 2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5. Statement 1: During the propagation of electromagnetic wave along the z-axis, if the electric field \(\vec{E}\) at a point is along the x-axis, then the magnetic field B at that point will be along the y-axis.

Statement 2: In the direction of propagation of the electromagnetic wave, the electric field \(\vec{E}\) and the magnetic field \(\vec{B}\) both form a right-handed cartesian coordinate system.

Answer: 3. Statement 1 is true, and statement 2 is false.

Class 12 Physics Electromagnetic Waves Match The Columns

Question 1. Some types of electromagnetic waves and their corresponding frequencies are given in column 1 and column 2 respectively.

Electromagnetic Waves Match The Column Question 1

Answer: 1-C, 2-B, 3-D, 4-A

Question 2. Some quantities and their corresponding dimensions are given in Column 1 and Column 2 respectively.

Electromagnetic Waves Match The Column Question 2

Answer: 1-D, 2-C, 3-A, 4-B

Class 12 Physics Electromagnetic Waves Comprehension Type Questions And Answers

Question 1. Electromagnetic waves propagate through free space or a medium as transverse waves. The electric and magnetic fields are perpendicular to each other as well as perpendicular to the direction of propagation of waves at each point. In the direction of wave propagation, electric field \(\vec{E}\) and magnetic field \(\vec{B}\) form a right-handed cartesian coordinate system. During the propagation of electromagnetic waves, the total energy of electromagnetic waves is distributed equally between electric and magnetic fields. Since e0 and are permittivity and permeability of free space, the velocity of electromagnetic wave, c = (ε0μ0)-1/2. Energy density i.e., energy volume due to electric field at any point, \(u_E=\frac{1}{2} \epsilon_0 E^2\); Similarly, energy density due to magnetic field, \(u_M=\frac{1}{2 \mu_0} B^2\). If the electromagnetic wave propagates along x -direction, then the equations of electric and magnetic field are respectively,
E = E0sin(ωt- kx) and B = B0sin(ωt-kx)

Here, the frequency and the wavelength of oscillating electric and magnetic fields are \(f=\frac{\omega}{2 \pi}\) and \(\lambda=\frac{2 \pi}{k}\) respectively. Thus \(E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}\) and \(B_{\mathrm{rms}}=\frac{B_0}{\sqrt{2}}\), where \(\frac{E_0}{B_0}=c\). Therefore, average energy density, \(\bar{u}_E=\frac{1}{2} \epsilon_0 E_{\mathrm{rms}}^2\) and \(\bar{u}_M=\frac{1}{2 \mu_0} B_{\mathrm{rms}}^2\). The intensity of the electromagnetic wave at a point, \(I=c \bar{u}=c\left(\bar{u}_E+\bar{u}_B\right)\).

To answer the following questions, we assume that in the case of propagation of an electromagnetic wave through free space, c = 3 x 108 m. s-1 and μ0 = 4 π x 10-7 H. m-1

Question 1. If the electromagnetic wave propagates along the x-axis, then the electric field \(\vec{E}\) will be

  1. Along y-axis
  2. Along z-axis
  3. On xy – plane
  4. On yz – plane

Answer: 4. Along y-axis

Question 2. If the peak value of the electric field at a point in an electromagnetic wave is 15 V.m-1, then the average electrical energy density (in J.m-3)

  1. 4.98 x l0-9
  2. 9.95 x 10-9
  3. 4.98 x 10-10
  4. 9.95 x 10-10

Answer: 3. 4.98 x 10-10

Question 3. The peak value of the magnetic field (in Wb.m-2) at that point

  1. 5 x 10-8
  2. 45 x 10-8
  3. 5 x 108
  4. 45 x 108

Answer: 1. 5 x 10-8

Question 4. The average energy density (in J.m-3 )of the electromagnetic wave at that point

  1. 4.98 x 10-9
  2. 9.95 x 10-9
  3. 4.98 x 10-10
  4. 9.95 x 10-10

Answer: 4. 9.95 x 10-10

Question 5. The intensity (in W.m-2) of the electromagnetic wave at that point is almost

  1. 0.15
  2. 0.3
  3. 0.45
  4. 0.6

Answer: 2. 0.3

Question 6. If the wavelength is 1000A, then the frequency (in Hz)

  1. 1013
  2. 3 x 1013
  3. 1015
  4. 3 x 1015

Answer: 4. 3 x 1015

Question 7. Relation between ω and k

  1. \(\omega=c k\)
  2. \(\omega=c k\)
  3. \(\omega=\frac{c}{k}\)
  4. \(\omega=\frac{c}{2 \pi k}\)

Answer: 1. \(\omega=c k\)

Class 12 Physics Electromagnetic Waves Integer Type Questions And Answers

In this type, the answer to each of the equations is a single-digit integer ranging from 0 to 9.

Question 1. If the dielectric constant and relative magnetic permeability of a medium are 4 and 2.25 receptively, then the velocity of an electromagnetic wave in that medium is n x 108m.s-1. Find the value of n.

Answer: 1.

Question 2. When the frequency of an electromagnetic wave and the amplitude of its associated magnetic field at a point are 108HZ and 10-10 T respectively, then the amplitude of the electric field is n x 10-2 V . m-1. Find the value of n.

Answer: 3.

Question 3. Due to the sunlight, if the amplitude of the electric field at the earth’s surface is 900 V m-1, then the amplitude of the magnetic field is n x 10-6 T. Find the value of n.

Answer: 3.

Question 4. What is the wavelength (in meters) of a radio wave of frequency 1.5 x 108 HZ?

Answer: 2.

Question 5. If the intensity of an electromagnetic wave at a point is 0.085 W m-2, then find the average amplitude (in V.m-1) of the electric field at that point.

Answer: 8.

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