(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.

Expression for Ampere circuital law
\(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }{ I }_{ c }+{ \mu }_{ 0 }\varepsilon _{ 0 }\frac { d{ \phi }_{ E } }{ dt } \)
\(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }{ (I }_{ c }+{ I }_{ d })\)
(i) In case of steady current in a conducting wire, the electric field does not change with time, condition current I_c exists but I_d may be zero
so \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }.{ I }_{ c }.\)
(ii) In the large region of space, no condition current (I_c) but there is only displacement current (I_d) to time-varying electric field (i.e. flux).
so \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\varepsilon _{ 0 }\frac { d{ \phi }_{ E } }{ dt } \)

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

i) Maxwell's prediction was experimentally confirmed by Heinrich Rudolf Hertz in 1888. The experimental set up used is shown in Figure.
ii) It consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns. This is to produce very high electromotive force (emf).
iii) Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced.
iv) The gap between electrode (ring type - not completely closed and has a small gap in between) kept at a distance also gets spark. This implies that the energy is transmitted from electrode to the receiver (ring electrode) as a wave, known as electromagnetic waves.

v) If the receiver is rotated by 90° - then no spark is observed by the receiver. This confirms that electromagnetic waves are transverse waves as predicted by Maxwell.
vi) Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light (3 x 10⁸m S^-1).

(a) Microwaves:
It is produced by special vacuum tubes such as klystron, magnetron and gunn diode. The frequency range of microwaves is 10⁹ Hz to 10¹¹ Hz. These waves undergo reflection and can be polarised.
Uses:
It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.
(b) X-rays:
lt is produced when there is sudden stopping of high speed electrons at high-atomic number target, and also by electronic transitions among the innermost orbits of atoms. The frequency range of X-rays is from 10¹⁷ Hz to 10¹⁹ Hz. X-rays have more penetrating power than ultraviolet radiation.
Uses:
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.
(c) Radio waves:
It is produced by accelerated motion of charges in conducting wires. The frequency range is from few Hz to 10⁹ Hz. It obeys reflection and diffraction.
Uses:
It is uses in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.
(d) Visible light:
Visible light is produced by incandescent bodies and also it is radiated by excited atoms in gases. The frequency range is from 4 x 10¹⁴ Hz to 8 x 10¹⁴ Hz. It obeys the laws of interference, diffraction and can be polarised. It exhibits photo-electric effect also.
Uses:
It can be used to study the structure of molecules, arrangement of electrons in external shells of atoms and it causes sensation of vision.

Potential difference between the plates of the capacitor,
\(V=V_{\max } \sin 2 \pi f t\) 
\(=230 \sqrt{2} \sin (2 \pi \times 50 t)\) 
\(\therefore V=325 \sin 100 \pi t\) 
d = 1 mm = 1 x 10^–3 m
A = 20 cm² = 20 x 10^–4 m²
Displacement current, \(i_{d}=\epsilon_{0} \frac{d \Phi_{E}}{d t}=\epsilon_{\circ} \frac{d(\mathrm{EA})}{d t}\) 
\(\therefore i_{d}=\frac{\epsilon_{0} A}{d}\left[\frac{d V}{d t}\right] \quad\left[\because E=\frac{V}{d}\right]\) 
\(=\frac{\epsilon_{0} A}{d}(325)(100 \pi) \cos 100 \pi t\)
\(=\left(\begin{array}{l} 8.85 \times 10^{-12} \times 20 \times 10^{-4} \times 325 \\ \times 100 \times 3.14 \times \cos (100 \pi \times 1) \end{array}\right) /\left(1 \times 10^{-3}\right)\) 
\(\begin{aligned}=1.81 \times 10^{-6} \mathrm{~A}=1.81 \mu \mathrm{A}[\because \cos (100 \pi \times 1)=1] \end{aligned}\)

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

UV light is absorbed by the ozone layer
(i) Source: Sun, arc and ionized gases.
(ii) Uses: To destroy bacteria, sterilizing the surgical instruments, burglar alarm etc

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.

૪ - rays, microwaves, IR, UV rays.
(i) Microwaves < IR < UV < ૪ - rays
(ii) Microwaves < IR > UV > ૪ - rays.

(i) During charging or discharging there is a displacement current, but no conduction current between the plates of the capacitor.
(ii) The displacement current is equal to conduction current I_d = I_c, so I_d = 0.25 A.

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

Area of the capacitor = A
Radius = R
Current in the wire connected to the plates I = 5 A
The electric field, between the plates of a parallel plate capacitor,
\(E=\frac{\sigma}{\varepsilon_0}  \)
\(E=\frac{Q}{A \varepsilon_0}\)
Q is the charge accumulated at the positive plate.
The flux of this field, \(\phi_E=\frac{Q}{A \varepsilon_0} \times A=\frac{Q}{\varepsilon_0}\)
Displacement current \(i_d=\varepsilon_0 \frac{d \phi_E}{d t}\)
\(=\varepsilon_0 \frac{d}{d t}\left(\frac{Q}{\varepsilon_0}\right)=i_c\)
\(\therefore \mathrm{i}_{\mathrm{d}}=5 \mathrm{~A} \quad\left(\because\right.\) The current through the capacitor i_c = 5 A)
Displacement current = 5 A

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.

The energy crossing per unit area per unit time and perpendicular to the direction of propagation of the electromagnetic wave is called the intensity.
Intensity, I = [u] c or \(I=\frac { total\ electromagnetic\ energy(U) }{ Surface\ area(A)\times time(t) } \)
\(=\frac { Power(P) }{ Surfacearea(A) } \) 

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

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.

E and B vectors must be in x and y directions.
Formula: We know \(\lambda =\frac { v }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 30\times { 10 }^{ 6 } } \)
λ = 10m.

Since displacement current = conduction current
\({ I }_{ d }=\frac { V }{ { X }_{ C } } =\frac { 220 }{ 100 } =2.2A\)

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 \(\oint_l \vec {B}. \vec {d l}=\mu_{o} i_c+\mu_{o} \varepsilon_{o} \frac{d}{d t} \oint_s \vec{E} \cdot {d} \vec A\)