Displacement of does not arise due to motion of charge carries but it arises due to time variation of electric flux.

Microwaves are used. They are considered suitable for radar systems for aircraft navigation due to their short wavelength or high frequency. Its wavelength range 1 x 10^-3 m to 3 x 10^-1 m and frequency range is 3 x 10¹¹Hz to 1 x 10⁹ Hz.

Infrared radiation waves are produced by hot bodies and molecules so it is referred as heat waves. (eg. Sun)
(i) Electromagnetic waves having shorter λ than Infrared radiation are visible, U - v, X-rays, and ૪ - rays.
(ii) Electromagnetic waves having longer λ than Infrared radiation are microwaves, radiowaves.

(i) Speed of Electromagnetic wave \(=\frac{Peak\ value\ of\ Electric\ field}{Peak\ value\ of\ magnetic\ field}\)
\(c=\frac { { E }_{ 0 } }{ { B }_{ 0 } } \)
(ii) Yes, As Electromagnetic waves contain both electric and magnetic fields, there is a non-zero energy density associated with it.
\(E=\frac { hc }{ \lambda } \)
Momentum p \(=\frac{Total\ energy\ transferred\ to\ the\ surface}{Velocity\ of\ light\ in\ vacuum}\)
i.e.p = \(\frac{U}{c}=mc\)
U - total energy transferred to the surface.
EM waves carry not only energy and momentum but also angular momentum.

It is used in radar systems for aircraft navigation, speed of the vehicle, microwave oven for cooking, and very long-distance wireless communication through satellites.

Inductance  L = 1μH = 1\(\times\)10^-6 H
Capacitance C = 1μF = 1 \(\times\)10^-6 F
∴ Frequency \(f =\frac{1}{2 \pi \sqrt{L C}} \)
\(f =\frac{1}{2 \pi \sqrt{1 \times 10^{-6} \times 1 \times 10^{-6}}} \)
Frequency of electromagnetic wave, f \(=\frac{1}{2 \pi \times 10^{-6}} Hz\)
∴ Wavelength of electromagnetic wave (⋋) = \(\frac{C}{f}\)
\(⋋ = 3 \times 10^8 \times 2\pi \times10^{-6}\)
\(=6.28 \times 10^{-6} \times 3 \times 10^{8} \)
Wavelength, ⋋ = 18.84 x 10² m

When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

\(\oint _l\vec{B} \cdot \overrightarrow{d l}=\mu_{o} i_{\text {c }}+\mu_{o} \varepsilon_{o} \frac{d}{d t} \oint _s \vec{E} \cdot \overrightarrow{d A}\)

An electromagnetic waves are the waves that are radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.

The displacement current can be defined as the current which comes into play in the region in which the electric field or the electric flux is changing with time.

The amplitude of the electric field, E₀ = 3 x 10⁴ NC^-1
The amplitude of the magnetic field, B₀ = 2 x 10^-4 T. Therefore, speed of the electromagnetic wave in a medium is
v = \(\frac { 3\times { 10 }^{ 4 } }{ 2\times { 10 }^{ -4 } } \)  = 1.5 x 10⁸ ms^-1.

(a) 10⁹ Hz - Radiowaves
Application: For radio & television communication.
(b) 10¹¹ Hz - microwaves
Application: used in Radar for aircraft navigation, (microwave oven for cooking).

(i) u - v rays
(ii) X - rays
(iii) ૪ - rays.

(a) X - rays are used as diagnostic tool in medicine. X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(b) Radio are produced by oscillators in electric circuits. It obeys reflection and diffraction. It is used in radio and communication system and cellphones.

During charging, the electric flux between the plates capacitor keeps on changing. This results in the production of displacement current between the plates.
\({ I }_{ d }=\varepsilon _{ 0 }\left( \frac { d{ \phi }_{ E } }{ dt } \right) \)

Conduction current Ie = Displacement current ID
\({ I }_{ C }={ I }_{ s }=\frac { dq }{ dt } =\frac { d }{ dt } ({ q }_{ 0 }cos2\pi \gamma t)\)
\(=-2\pi { q }_{ 0 }\gamma sin2\pi \gamma t\)

(i) It is produced by transitions of atomic nuclei and decay of certain elementary particles. They produce chemical reactions on photographic plates, fluorescence, ionization, diffraction.
(ii) Gamma rays provide information about the structure of atomic nuclei. It is used in radiotherapy for the treatment of cancer and tumor, in the food industry to kill pathogenic microorganisms.

It provides electrical energy to satellites by means of solar cells. It is used to produce dehydrated fruits, in green houses to keep the plants warm, heat therapy for muscular pain or sprain, TV remote as a signal carrier, to look through haze fog or mist and used in night vision or infrared photography.

μ_r = 1.0,  ε_r = 2.25
Refractive index of the medium \(n= {\sqrt{\varepsilon_{r} \mu_{r}}} =\sqrt{1 \times2.25}=1.5\)
Speed of electromagnetic waves in this medium,
\(V_{m}=\frac{c}{n}=\frac{3 \times 10^{8}}{1.5}\)
V_m = 2 x 10⁸ m/s

Formula:
Intensity, I = \(\frac{Power\ of\ visible\ light}{Area}\)
\(I=\frac { \frac { 5 }{ 100 } \times 100 }{ 4\pi { (1) }^{ 2 } } =0.4{ W/m }^{ 2 }\)

Given: The amplitude of magnetic field of an Electromagnetic wave B = 1.6 x 10^-7 T
To find:
The amplitude of electric field of an Electromagnetic wave E = ?
frequency ૪ = 30 Mhz = 30 x 10⁶ Hz.
To find: Angle velocity ω =?
Wavelength of Electromagnetic wave λ = ?
(i) Ampere of electric field E = ?
\(\frac { E }{ B } =C\Rightarrow E=C.B\Rightarrow 3\times { 10 }^{ 8 }\times 1.6\times { 10 }^{ -7 }\)
E = 48Vm^-1.
(ii) Angle velocity, ω = 2π૪
ω = 2 x 3.14 x 30 x 10⁶
ω = 1.885 x 10⁸ rad /s.
(iii) Wavelength of Electromagnetic wave, λ = \(\frac{C}{\gamma}\)
\(\gamma=\frac{3\times 10^8}{30\times 10^6}\) = 10m
λ = 10m

(i) Wavelength \(\lambda =\frac { c }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 1.5\times { 10 }^{ 10 } } =2\times { 10 }^{ -2 }m\)
(ii) \(B=\frac { E }{ c } =\frac { 36 }{ 3\times { 1 }0^{ 8 } } =12\times { 10 }^{ -8 }T\)
Formula: (or) 1.2 x 10^-7T
Average energy of magnetic field \(\overrightarrow { E } \) \({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ E }^{ 2 }\)
The average energy density of electric field \(\overrightarrow { B } \) \({ U }_{ E }=\frac { 1 }{ 2{ \mu }_{ 0 } } .{ B }^{ 2 }\)
But E = CB & C² = \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ E }^{ 2 }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ (CB) }^{ 2 }\)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }.\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }=\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }={ U }_{ B }\)
\(\therefore { U }_{ E }={ U }_{ B }\)

Given:
The frequency of Electromagnetic wave
૪ = 5 x 10⁸Hz
λ = 0.6m
\(\lambda =\frac { c }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 5\times { 10 }^{ 8 } } =0.6\)
(ii) The amplitude of magnetic field B_o = 10⁷ T
To find:
Tile amplitude of electric field E_o = ?
E_o = c B_o = 3 x 10⁸ x 10^-7 = 30 V m^-1
The expression for oscillating electric field
E_z =?
\(E={ E }_{ 0 }sin2\pi (vt+\frac { 1 }{ \lambda } .x)\)
E = 30 sin 2π (3 x 10⁸ t + 1.66 x) Vm^-1.

According to Ampere's circuital law,
\(\oint _{ s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }I\quad ...(1)\)
As the current flows across the area bounded by loop S₁, so
\(\oint _{ { s }_{ 1 },s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }I\quad ...(2)\)
But the area bounded by S₂ lies in the region between the plates capacitor where no current flows across it.
\(\therefore \oint _{ { s }_{ 1 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } =0\)
Consider that loops enclosing S₁ & S₂ are infinitesimally close to each other. Then
\(\oint _{ { s }_{ 1 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } =\oint _{ { s }_{ 2 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } \)
This equation is inconsistent with equations (2) & (3). To remove this maxwell said that a changing electric field (during charging) between the capacitor plates must induce a magnetic field which in turn must be associated with current I_d.
\({ I }_{ d }={ \varepsilon }_{ 0 }\left( \frac { d{ \phi }_{ E } }{ dt } \right) \) [\(\frac { d{ \phi }_{ E } }{ dt } \) change in electric flux]
The total current must be
I = I_conduction + I_displacement
\({ I }_{ c }={ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \)
Hence the generalized from of Ampere's circuital law is
\(\oint _{ s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\left[ { I }_{ c }+{ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \right] \)

Emission spectra:
When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:
(i) Continuous emission spectra (or continuous spectra) :
(a) If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours.
(b) Thus, it consists of wavelengths containing all the visible colours ranging from violet to red (in the figure). Examples: spectrum obtained from carbon arc, incandescent solids.
(ii) Line emission spectrum (or line spectrum) :
(a) Suppose light from hot gas is allowed to pass through a prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra consists of  sharp lines of definite wavelengths or frequencies.
(b) Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element  and are different for different elements. Examples: spectra of atomic hydrogen, helium, etc.
(iii) Band emission spectrum (or band spectrum) :
(a) Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces.
(b) This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited.
(c) Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube, etc.

(i) We have stated Ampere's law as \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_oi\)
(ii) Where, i is the electric current crossing a surface bounded by a closed curve and the line integral of \(\vec{B}\) is calculated along that closed curve. This equation is valid only when the electric field at the surface does not change with time.
(iii) Maxwell strongly believed that when the time varying magnetic field produces an electric field, the time varying electric field must produce a magnetic field.
(iv) To understand how a varying electric field produces magnetic field, let us consider a situation of charging a parallel plate capacitor.

(v) Let i_c be the conduction current. To calculate the magnetic field at P (fig. 1 ) an amperian loop. S₁ is drawn. Applying Ampere circuital law for the surface S₁, we get
\(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 i_c\) Where, \(\mu_0\) is permeability of free space.
(vi) Applying the same for the surface S₂, we get \(\oint \vec{B} \cdot \overrightarrow{d l}=0.\)
Because the surface S₂ nowhere touches the wire carrying conduction current. Therefore for the point P at one surface (S₁) it has some value and at another surface (S₂) it has zero value.
(vii) So, Maxwell believed that there must be a current associated with the changing electric field in between the capacitor and he called that current as displacement current.
(viii) Applying Gauss law to the electric flux between the plates of the capacitor \(\phi_E=\oint \vec{E} \cdot \overrightarrow{\mathrm{dA}}=E A=\frac{q}{\varepsilon_0}\) where, A is the area of the plate.
The change in electric flux is \(\frac{d \phi_F}{d t}=\frac{1}{\varepsilon_0} \frac{d q}{d t} (or) \frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{i}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\), where i_d is the displacement current.
(ix) The displacement current can be defined as the current which comes into play in the region in which the electric field and electric flux are changing with time.
(x) So, Maxwell modified Ampere's law \(\oint_{l} \vec{B} \cdot d \vec{l}=\mu_{0} i_c+\mu_{0}-i_d\) which means the total current enclosed by the surface is sum of conduction current and displacement current.

MaxWell's equations in integral form
i) Gauss law in electricity, \(\oint _s\vec{E} \vec{d} A=\frac{Q_{\text {enclosed }}}{\varepsilon_{o}}\)
ii) Gauss law in magnetism \(\oint _s \vec{B} \cdot \vec{d} A=0\)
iii) Faraday's law \(\oint_l \vec E. \vec {d l}=-\frac{d \phi _B}{d t}\)
iv) Ampere-Maxwell's law \(\oint_l \vec {B}. \vec {d l}=\mu_{o} i_c+\mu_{o} \varepsilon_{o} \frac{d}{d t} \oint_s \vec{E} \cdot {d} \vec A\)