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

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: 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: 
\(\frac { dV }{ dt } =\frac { { I }_{ d } }{ C } =\frac { 1.0 }{ 1\times { 10 }^{ -6 } } ={ 10 }^{ 6 }\) 

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