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Chapter Notes: Electromagnetic Waves Physics Class 12


Notes for Electromagnetic Waves chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

 

Light may be described as a wave. The wave equation for light propagating in x-direction in vacuum may be written as

E = {E_o}\sin \omega \left( {t - x/c} \right)

where E is the sinusoidally varying electric field at position x at time t. The constant c is the speed of light in vacuum.
There is also a sinusoidally varying magnetic field associated with electric field when light propagates. The magnetic field is perpendicular to the direction of propagation as well as to the electric field E. It is given by

B = {B_o}\sin \omega \left( {t - x/c} \right)

Such a combination of mutually perpendicular electric and magnetic fields is referred to as an electromagnetic wave in vacuum.

MAXWELL'S DISPLACEMENT CURRENT

Maxwell discovered the concept of displacement current. Displacement current is that which results due to the displacement of electrons. The displacement of electrons is caused by the time varying electric field. The generalized form of Ampere's law \oint {B.dl} = {\mu _o}I, which includes both conduction current {I_C} (current flowing through the conducting wire) and displacement current {I_C} as given by Maxwell i.e.

\oint {B.dl} = {\mu _o}\left( {{I_C} + {I_D}} \right)

where {I_C} = {{dq} \over {dt}} ,  where {{dq} \over {dt}} is rate of flow of charge.
and {I_D} = {\varepsilon _o}{{d{\Phi _E}} \over {dt}}
Here  {\Phi _E}   is electric flux.
Therefore modified Ampere's law may be expressed as

\int {\vec B.d\vec l} = {\mu _o}\left( {{I_C} + {\varepsilon _o}{{d{\Phi _E}} \over {dt}}} \right)

Example 1
Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and is equal to 0.15 A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b)  Obtain the displacement current across the plates.
(c)  Is Kirchhoff's first rule valid at each plate of the capacitor? Explain.
Given that {\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}.

Solution:

Here,   d = 5.0 mm = 5.0 x 10-3 m; R = 12 cm = 12 x 10-2 m, I = 0.15 A.
{\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}
Now,   A = \pi R2 = \pi (12 x 10-2)2 = 1.44 \pi x 10-2 m2
(a) The capacitance of parallel plate capacitor is given by

C = {{{\varepsilon _o}A} \over d} = {{8.85 \times {{10}^{ - 12}} \times 1.44\pi \times {{10}^{ - 2}}} \over {5 \times {{10}^{ - 3}}}} = 80.1 \times {10^{ - 12}}V

Now, q = CV
Therefore,  {{dq} \over {dt}} = C{{dV} \over {dt}}   or   I = C{{dV} \over {dt}}
or   {{dV} \over {dt}} = {1 \over C} = {{0.15} \over {80.1 \times {{10}^{ - 12}}}} = 1.873 \times {10^9}V{s^{ - 1}}

(b) Displacement current is equal to conduction current i.e. 0.15 A.
(c) Yes, Kirchhoff's law will be valid at each plate of the capacitor, provided electric current means the sum of the conduction and displacement currents.

Maxwell's Equations

The four basic laws of electricity and magnetism i.e. Gauss's law in electrostatics, Gauss's law in magnetism, Faraday's law of electromagnetic induction and Maxwell-Ampere's circuital law are called Maxwell's equations.

1. Gauss's Law in Electrostatics

\oint {\vec E.d\vec s} = q/{\varepsilon _o}         

 2. Gauss's Law in Magnetism

\oint {\vec B.d\vec s} = 0

3. Faraday's Law of Electromagnetic Induction

                 \varepsilon = - {{d{\Phi _B}} \over {dt}}  or  \oint {\vec E.d\vec l = - {{d{\Phi _B}} \over {dt}}}

4. Maxwell-Ampere's circuital Law               

\oint {\vec B.d\vec l} = {\mu _o}\left( {{I_C} + {\varepsilon _o}{{d{\Phi _E}} \over {dt}}} \right)

Nature of Electromagnetic Waves

Maxwell's concept of displacement current led to the conclusion that an electric field varying with time at a point produces magnetic field at that point. This symmetry in the law of electricity and magnetism leads to the conclusion that a time varying electric field gives rise to a time varying magnetic field and vice versa. Electric and magnetic fields in electromagnetic wave is of transverse nature.

At any instant, the electric and magnetic fields varying sinusoidally with x can be expressed by equation
{E_y} = {E_o}\sin \left( {x - ct} \right)              (1)
and   {B_Z} = {B_o}\sin \left( {x - ct} \right)     (2)
Here {E_o} and {B_o} are the amplitudes of electric and magnetic fields along y-axis and z-axis respectively.
Faraday's law of electromagnetic induction states that

\oint {\vec E.d\vec l} = - {{d{\Phi _B}} \over {dt}}    (3)

From equation (3) it can be proved that for a plane electromagnetic wave propagating along x-axis,

{{\partial {E_y}} \over {\partial x}} = - {{\,\partial {B_z}} \over {\partial t}}     (4)

i.e. a magnetic field varying with time gives rise to an electric field varying in space.
From equation (1), we have

{{\partial {E_y}} \over {\partial x}} = {E_o}\cos \left( {x - ct} \right)

and from equation (2), we have

{{\partial {B_y}} \over {\partial t}} = - c{B_o}\cos \left( {x - ct} \right)

Substituting for {{\partial {E_y}} \over {\partial x}}  and {{\partial {B_z}} \over {\partial t}}  in equation (1)

{E_o} = c{B_o}     or     c = {{{E_o}} \over {{B_o}}}

It shows that the velocity of electromagnetic waves is equal to the ratio of amplitudes of electric and magnetic fields.
Velocity of electromagnetic waves is also given by

c = {1 \over {\sqrt {{\mu _o}{\varepsilon _o}} }}

Example 2
Green light of mercury has a wavelength 5.5 x 10-5 cm.
(a)  What is the frequency in MHz and period in \mu s in vacuum?
(b)  What is the wavelength in glass, if refractive index of glass is 1.5?
Given c = 3 x 108 m/s.

Solution:

Here, wavelength \lambda = 5.5 x 10-5 cm = 5.5 x 10-7 m
Velocity of light, c = 3 x 108 m/s
(a)  If \nu is the frequency, then

\nu = {c \over \lambda } = {{3 \times {{10}^8}} \over {5.5 \times {{10}^{ - 7}}}} Hz = {{3 \times {{10}^8}} \over {5.5 \times {{10}^{ - 7}} \times {{10}^6}}} = 5.45 \times {10^8} MHz

Time period,

T = {1 \over \mu } = {\lambda \over c} = {{5.5 \times {{10}^{ - 7}}} \over {3 \times {{10}^8}}} = 1.8 \times {10^{ - 15}} \times {10^6} = 1.8 \times {10^{ - 19}}\mu s

(b) Now, refractive index \mu = {{{\rm{veloicty of light in vacuum}}} \over {{\rm{velocity}}{\mkern 1mu} {\mkern 1mu} {\rm{of light in glass}}}} = {c \over \nu }
Therefore, velocity of light in glass \nu = {c \over \mu } = {{3 \times {{10}^8}} \over {1.5}} = 2 \times {10^8}m/s
The wavelength of light in glass, \lambda = {v \over \nu } = vt
 = 2 \times {10^8} \times 1.8 \times {10^{ - 15}} = 3.6 \times {10^{ - 7}}m

Energy Density of Electromagnetic Waves

(a)  The average energy density of electric field is

{u_E} = {1 \over 2}{\varepsilon _o}{E^2} = {1 \over 2}{\varepsilon _o}{\left( {{E_o}/\sqrt 2 } \right)^2} = {1 \over 4}{\varepsilon _o}E_o^2

(b)  The average energy density of magnetic field is,

{u_B} = {{{B^2}} \over {2{\mu _o}}} = {{{{\left( {{B_o}/\sqrt 2 } \right)}^2}} \over {2{\mu _o}}} = {{B_o^2} \over {4{\mu _o}}}

(c) {u_E} = {u_B}
(d) Total average energy density

{u_E} + {u_B} = 2{u_E} = 2{u_B} = {1 \over 2}{\varepsilon _o}E_o^2 = {1 \over 2}{{B_o^2} \over {{\mu _o}}}

The units of {u_E} and {u_B} are J m-3.

Intensity of Electromagnetic Wave

The energy of electromagnetic wave crossing per unit time per unit area perpendicular to the direction of propagation of wave is called the intensity of electromagnetic wave.
The intensity of electromagnetic wave is

I = {{{P_{av}}} \over {4\pi {r^2}}} = {u_{av}} \times c = {1 \over 2}{\varepsilon _o}E_o^2c = {1 \over {2{\mu _o}}}B_o^2c = {1 \over {2{\mu _o}}}.{{E_o^2} \over c}

Momentum of Electromagnetic Wave

The electromagnetic wave during its propagation has linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by

p = U/c

If the electromagnetic wave incident on a material surface is completely absorbed, it will deliver energy U and momentum p = U/c to the surface.
If the incident wave is totally reflected from the surface, the momentum delivered to the surface  = U/c - \left( { - U/c} \right) = 2U/c. Due to which the electromagnetic waves incident on a surface exert a force on the surface.

Radiant flux of electromagnetic wave

According to Maxwell, the accelerated charged particles produce electromagnetic waves. The total radiant flux emitted at any instant is given by
P = {q^2}{a^2}/(6\pi {\varepsilon _0}{c^3}), where q is the charge on the particle and a is its instantaneous acceleration.

Poynting Vector

When an electromagnetic wave advances, the electromagnetic energy flows in the direction of \vec E \times \vec B. The total energy flowing perpendicularly per second per unit area into the surface in free space is called a Poynting vector \vec S, where \vec S = {c^2}{\varepsilon _o}\left( {\vec E \times \vec B} \right) = \left( {\vec E \times \vec B} \right)/{\mu _o}.
The S. I. unit of S is watt/(metre)2.

Radiant Flux Density

The average value of pointing vector (\vec S) over a convenient time interval in the propagation of electromagnetic wave is known as radiant flux density. When energy of electromagnetic wave is incident on a surface, the flux density is called intensity of wave (denoted by I). Thus I = S
A harmonic electromagnetic wave traveling along X-axis in free space can be described by periodic variation of electric and magnetic field along y-axis and z-axis with the equations

\vec E = {\vec E_o}\cos \left( {kx - \omega t} \right)      and     \vec B = {\vec B_o}\cos \left( {kx - \omega t} \right)

Then radiant flux density \vec S is given by

\vec S = {c^2}{\varepsilon _o}\left( {\vec E \times \vec B} \right) = {c^2}{\varepsilon _o}{\vec E_o} \times {\vec B_o}{\cos ^2}\left( {kx - \omega t} \right)

Hence |\vec S| = {c^2}{\varepsilon _o}|{\vec E_o} \times {\vec B_o}|{\cos ^2}\left( {kx - \omega t} \right).
The average value of \vec S over a single period T is given by

S = I = {c^2}{\varepsilon _o}|{\vec E_o} \times {\vec B_o}|{1 \over T}\int\limits_0^T {{{\cos }^2}\left( {kx - \omega t} \right)} dt
= {c^2}{\varepsilon _o}{E_o}{B_o}\sin {90^o}\left[ {{1 \over 2}} \right]                     [since {1 \over T}\int\limits_0^T {{{\cos }^2}\left( {kx - \omega t} \right)} dt = {1 \over 2} ]
{c^2}{\varepsilon _o}{E_o}\left( {{{{E_o}} \over c}} \right)\left( {{1 \over 2}} \right) = {1 \over 2}c{\varepsilon _o}E_o^2 = {1 \over {2c{\mu _o}}}E_o^2

PROPERTIES OF EM WAVES

(i)   The direction of variation of electric field E and magnetic field B are perpendicular to each other as well as to the direction of propagation.
(ii)  Speed of EM wave is 3 x 108 m/s in space, in any other medium it depends on the electric and magnetic properties of the medium and is independent of the amplitude of field variation.
(iii) The electric and magnetic field variation are in phase.
(iv) EM waves are produced by accelerating charged particle. An oscillating charge in an LC circuit also produces EM waves. These waves required no medium for its propagation.

EM WAVE SPECTRUM

The whole spectrum of electromagnetic spectrum and the modern terminology used for various sections of the spectrum are shown in table. Various regions do not have sharply defined boundaries.

Various layers of our atmosphere

Various layers in which the atmosphere has been divided are shown in figure. The actual boundaries are not sharp and the distances are approximate.
Ozon layer: The upper region of stratosphere is rich in ozone and is called ozone layer. The ozone layer absorbs ultraviolet radiation from the sun and prevents them from reaching the earth's surface, otherwise these can cause damage to life.

Green house effect

The phenomenon of heating of earth's atmosphere due to trapping of infra-red rays, which are radiated from earth's surface, by carbon dioxide layer in the atmosphere is called green house effect.

Modulation and demodulation

The process by which an audio signal is superimposed over the carrier waves is known as modulation. The modulated signal is sent through transmitting antenna.

On the receiving end the audio signal is filtered from the carrier waves and this process is known as demodulation.

 



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