Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

It is given that a plane electromagnetic wave travels in vacuum along z-direction and the frequency of the electromagnetic wave is 30MHz.
We can say that electric field and magnetic field will be in x-plane because the electromagnetic wave travels along the z-direction and both fields are mutually perpendicular to each other.
The formula of the wavelength of a wave is,
λ=c/ν
Substitute the values in the above expression,
λ=(3×10⁸)/(30×10⁶)=10m.
Thus, the value of wavelength is 10 m and the direction of electric and magnetic fields will be in x-y plane.

It is given that a charged particle oscillates about its mean equilibrium position and the frequency of charged particle is 10⁹  Hz.
The frequency of an electromagnetic wave produced by the oscillator will be same to the value of frequency of a charged particle that is oscillating about its mean position.
Thus, the value of the frequency of an electromagnetic wave produced by oscillator is 10⁹ Hz.

(a) Comparing the given equation with
\(B_{y}=B_{0} \sin \left[2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)\right]\)
We get, \(\lambda=\frac{2 \pi}{0.5 \times 10^{3}} \mathrm{~m}=1.26 \mathrm{~cm}\)
and \(\frac{1}{T}=v=\left(1.5 \times 10^{11}\right) / 2 \pi=23.9 \mathrm{GHz}\)
(b) E₀ = B₀c = 2 x 10^–7 T x 3 x 10₈ m/s = 6 x 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as
E_z = 60 sin (0.5 x 10³x + 1.5 x 10¹¹ t) V/m.

The speed of light (3 × 10⁸ m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

conduction current is due to flow of electrons in the circuit. It exists even if the flow of electrons is at uniform rate. Displacement current is due to the time-varying electric field. It does not exist under steady condition.

Frequency of the electromagnetic wave, ν = 2.0 x 10¹⁰ Hz
Electric field amplitude, E₀ = 48 V m⁻¹
Speed of light, c = 3 x 10⁸ m/s
(a) Wavelength of a wave is given as:
\(\lambda=\frac{c}{v}\)
\(=\frac{3 \times 10^{8}}{2 \times 10^{10}}=0.015 \mathrm{~m}\)
(b) Magnetic field strength is given as:
\(B_{0}=\frac{E_{0}}{c}\)
\(=\frac{48}{3 \times 10^{8}}=1.6 \times 10^{-7} T\)
(c) Energy density of the electric field is given as:
\(U_{E}=\frac{1}{2} \in_{0} E^{2}\)
And, energy density of the magnetic field is given as:
\(U_{B}=\frac{1}{2 \mu_{0}} B^{2}\)
Where,
∈₀ = Permittivity of free space
μ₀ = Permeability of free space
We have the relation connecting E and B as:
E = cB … (1)
Where,
\(c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} \ldots \text { (2) }\)
Putting equation (2) in equation (1), we get
\(E=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}\)
Squaring both sides, we get
\(E=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}\)
\(\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}\)

(i) (d): All electromagnetic waves travel in vacuum with the same speed.
(ii) (c): Cathode rays (beam of electrons) get deflected in an electric field.
(iii) (c) : \(\Upsilon\)-rays are detected by ionization chamber.
(iv) (b): Size of particle \(=\lambda=\frac{c}{v}\)
\(v=\frac{c}{\lambda}=\frac{3 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}}{3 \times 10^{-4} \mathrm{~cm}}=3 \times 10^{14} \mathrm{~Hz}\)
(v) (a): Every body at a temperature T > 0 K emits radiation in the infrared region.

(i) (b): Electromagnetic waves propagate in the direction of \((b) \vec{E} \times \vec{B}\)
(ii) (a): Photon is the fundamental particle in an electromagnetic wave.
(iii) (a): Polarisation establishes the wave nature of electromagnetic waves.
(iv) (c): Frequency D remains unchanged when a wave propagates from one medium to another. Both wavelength and velocity get changed.
(v) (c): The electric and magnetic fields of an electromagnetic wave are in phase and perpendicular to each other.