Electromagnetic waves are produced by the accelerated charges not by the mechanical vibrations of particles. It travels with speed equals to the speed of light in vacuum.

(i) Ampere- Maxwell law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
\(\oint_{i} \vec{B} \cdot \overrightarrow{d l}=\mu_{0}i =\mu_0(i_c+i_d)\)
(ii) \(\oint_{i} \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i_{C}+\mu_{0} \varepsilon_{0} \frac{dΦ_E}{d t}\)  where \(\vec{B}\) is the magnetic field.
(iii) This equation shows that both conduction current and displacement current produce magnetic field.

i) The surface integral of magnetic field over a closed surface is zero. Mathematically, \(\oint \vec{B} \cdot d\vec{A}=0\) (Gauss's law for magnetism) where \(\vec{B}\) is the magnetic field.
ii) This equation implies that the magnetic lines of force form a continuous closed path. In other words, it means that no isolated magnetic monopole exists.

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

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

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.

Dielectric constant (relative permittivity of the medium) is ε_r = 2.25
Magnetic permeability is μ_r = 2.5
Refractive index of the medium,
n = \(\sqrt { { \varepsilon }_{ r }{ \mu }_{ r } } =\sqrt { 2.25\times 2.5 } \) = 2.37