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:
\(\therefore v=\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } } ,v=\frac { 1 }{ \sqrt { { \varepsilon }_{ r }{ \mu }_{ R } } } \)
\(\therefore v=\frac { 3\times { 10 }^{ 8 } }{ \sqrt { 2\times 1 } } =\frac { 3 }{ \sqrt { 2 } } \times 10^{ 8 }m/s\)

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

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