(a) Ultra violet radiation
i) It is produced by Sun, arc and ionized gases. Its frequency range is 8 x 10¹⁴ HZ   to 10¹⁷ HZ. It has less penetrating power.
ii) It can be absorbed by atmospheric ozone and is harmful to human body. It is used to destroy bacteria in sterilizing the surgical instruments, burglar alarm, to detect the  invisible writing, finger prints and also in the study of atomic structure.
(b) Infrared radiation
i) It is produced by hot bodies (also known as heat waves) and also by when the molecules undergoing rotational and vibrational transitions.
ii) The frequency range is 10¹¹Hz to 4 X 10¹⁴ Hz. It provides electrical energy  to satellites by means of solar cells.
iii) 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.
(c) Gamma rays
i) It is produced by transitions of radioactive nuclei and decay of certain elementary particles. They produce chemical reactions on photographic plates, fluorescence, ionisation, diffraction. The frequency range is 10¹⁸ Hz and above.
ii) Gamma rays have higher penetrating power than X-rays and ultraviolet radiations, it has no charge but harmful to human body.
iii) Gamma rays provide information about the structure of atomic nuclei. It is used in radio therapy for the, treatment of cancer and tumour, in food industry to kill pathogenic micro-organism.

(i) In order to understand how the changing electric field induces magnetic field, let us consider a situation of charging a parallel plate capacitor which contains non-conducting medium, between the plates.
(ii) Let a time-dependent current i_c called conduction current be passed through the wire to charge the capacitor.

(iii) Ampere's circuital law can be used to find the magnetic field produced around the current  carrying wire.
(iv) To calculate the magnetic field at a point P near the wire and outside the  capacitor, let us draw a circular Amperian loop which encloses the circular surface S_{1 .} Using Ampere's circuital law for this loop, we get
\(\oint_{\text {enclosing } S_1} \vec{B} \cdot \mathrm{d} \vec{l}=\mu_0 i_c\)   ... (1)
Where, \(\mu_0\) is the permeability of free space.

Given : Frequency of ૪ - rays = 3 x 10²⁰ Hz
Formula:
Energy of gamma rays,
E = h૪
= 6.63 x 10^-34 x 3 x 10²⁰
E = 19.8 x 10^-14 J
∵ [ 1 eV = 1.6 x 10 - 19J; 1 T = \(\frac{1}{1.6\times 10^{-19}}\)  eV]
Solution:
\(E=\frac { 19.8\times { 10 }^{ -14 } }{ 1.6\times { 10 }^{ -19 } } \)
E = 1.24 x 10⁶ eV

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

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.

Given: Velocity of electromagnetic wave is
c = 3 x 10⁸ m/s.
Relative electric permittivity ε_r = 2
Relative magnetic permeability μ_r = 1
To find:
Velocity of Electromagnetic in a medium is
\(v=\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \varepsilon }_{ r }.{ \mu }_{ 0 }{ \mu }_{ r } } } =\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } \times \sqrt { { \varepsilon }_{ r }{ \varepsilon }_{ r } } } \)
Solution:
\(\therefore v=\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } } ,v=\frac { 1 }{ \sqrt { { \varepsilon }_{ r }{ \mu }_{ R } } } \)
\(\therefore v=\frac { 3\times { 10 }^{ 8 } }{ \sqrt { 2\times 1 } } =\frac { 3 }{ \sqrt { 2 } } \times 10^{ 8 }m/s\)

Given: Displacement current,
\({ I }_{ d }={ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \)
\(={ \varepsilon }_{ 0 }\frac { d(EA) }{ dt } \quad (\therefore { \phi }_{ E }=EA)\)
\({ I }_{ d }={ \varepsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) \left( \because E=\left( \frac { V }{ d } \right) \right) \)
\({ I }_{ d }=\frac { { \varepsilon }_{ 0 }A }{ d } \left( \frac { dV }{ dt } \right) (\because C=\frac { { \varepsilon }_{ 0 }A }{ d } )\)
Formula:
\({ I }_{ d }=C.\frac { dV }{ dt } \)
Solution: 
\(\frac { dV }{ dt } =\frac { { I }_{ d } }{ C } =\frac { 1.0 }{ 1\times { 10 }^{ -6 } } ={ 10 }^{ 6 }\) 

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 = \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ E }^{ 2 }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ (CB) }^{ 2 }\)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }.\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }=\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }={ U }_{ B }\)
\(\therefore { U }_{ E }={ U }_{ B }\)

(i) \(\lambda =\frac { 2\pi }{ 300\pi } =\frac { 1 }{ 150 } m\)
\([{ B }_{ y }={ B }_{ o }sin2\pi \left( \frac { x }{ \lambda } +\frac { t }{ r } \right) ]\)
(ii) \(\\ { E }_{ o }=c,{ B }_{ o }=3\times { 10 }^{ 8 }\times 8\times { 10 }^{ -6 }=2400{ Vm }^{ -1 }\)

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.

Electric field between the capacitor plates
\(E=\frac { \sigma }{ \varepsilon _{ 0 } } =\frac { q }{ \varepsilon _{ 0 }A } \)
Where q is the charge accumulated on the positive plate.
The electric flux through this plate
\({ \phi }_{ E }=EA=\frac { q }{ \varepsilon _{ 0 }A } .A=\frac { q }{ \varepsilon _{ 0 } } \)
∴ Displacement current
\(\\ { I }_{ d }=\varepsilon _{ 0 }.\frac { d\phi }{ dt } =\varepsilon _{ 0 }\frac { d }{ dt } \left[ \frac { q }{ \varepsilon _{ 0 } } \right] =\frac { dq }{ dt } \)
\(\frac { dq }{ dt } \) is the rate at which charge flows to a positive plate through the conducting wire.
I_d = I_c
i.e. displacement current between the capacitor plates = conduction current through the wire.

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
\(\oint _{ s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\left[ { I }_{ c }+{ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \right] \)