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

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.

Amplitude of magnetic field B = 10^-5 T
Frequency, f = 10¹⁰ HZ
(i) Wavelength of the wave,
\({\lambda}=\frac{c}{f} =\frac{3 \times 10^8}{ 10^{10}} \)
= 3 x 10^{8 - 10}
Wavelength = 3 x 10^-2 m
(ii) Electric field E(x,t)\(\hat i\)
Angular frequency \(\omega =2 \pi f \)
\(\omega=2 \times 3.14 \times 10^{10}=6.28 \times 10^{10} rads^{-1}\)
\(k=\frac{2 \pi}{\lambda} =\frac{2 \times 3.14}{3 \times 10^{-2}} \)
\(=\frac{6.28}{3 \times 10^{-2}}=\frac{628}{3}=2.09 \times 10^{2} \)
k = 2.09 x 10²
(iii) E_o = B_oC
E_o = 10^-5 x 3 x 10⁸
E_o = 3 x 10³ V m^-1
The required expression for electric field is
\(\vec{E}(x, t)=E_o \sin \left(\frac{2 \pi}{\lambda} x-2 \pi f t\right) \hat{i} N C^{-1} \)
\(\vec{E}(x , t)=3 \times 10^{3} \sin \left(2.09 \times 10^{2} \mathrm{x}-6.28 \times 10^{10} \mathrm{t}\right) \hat(-{k}) N C^{-1} \)

Frequency of the electromagnetic waves given, f = 2450 MHz
The corresponding angular frequency is
ω = 2πf = 2 x 3.14 x 2450 x 10⁶
= 15,386 x 10⁶ Hz
= 1.54 x 10¹⁰ s^-1
The required magnetic field, B = \(\frac { { m }_{ e }\omega }{ |q| } \)
Mass of the electron, m_e = 9.11 x 10^-31 kg
Charge of the electron,
q = -1.60 x 10^-19C
⇒ |q| = 1.60 x 10^-19 C
B = \(\frac { (9.11\times { 10 }^{ -31 })(1.54\times 10^{ 10 }) }{ (1.60\times 10^{ -19 }) } \)  = 8.7683 x 10^-2T
B = 0.08768 T
This magnetic field can be easily produced with a permanent magnet. So, electromagnetic waves of frequency 2450 MHz can be used for heating and cooking food because they are strongly absorbed by water molecules.