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
\(E=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}\)
\(\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}\)

Here, R = 10 cm = 0.1 cm;
C = 200 pF = 200 x 10^-12 F = 2 x 10^-10 F;
E_rms = 200 V; \(\omega\) = 200 rad s^-1 ;
r = 2.0 x 10^-2 m.
(a) \({ I }_{ rms }=\frac { { E }_{ rms } }{ 1/\omega C } =\omega C{ E }_{ rms }\)
= 200 x (2 x 10^-10) x 200
(b) Yes, because I_D = 1
(c) \({ I }_{ 0 }=\sqrt { 2 } { I }_{ rms }=\sqrt { 2 } \times 8\times { 10 }^{ -6 }\)
= 11.312 x 10^-6 A
(d) Consider a loop of radius r between two circular plates of parallel plate capacitor placed coaxially with them. The area of this loop \({ A }^{ \prime }=\pi { r }^{ 2 }\)
By symmetry, the magnetic field \(\overrightarrow { B } \) is equal in magnitude and is tangentially to the circle at every point. In this case, only a part of displacement current I_D will cross the loop of area \({ A }^{ \prime }\) . Therefore, the current passing through the area \({ A }^{ \prime }\)
\({ I }^{ \prime }=\frac { { I }_{ D } }{ \pi { R }^{ 2 } } \times \pi { r }^{ 2 }=\frac { { I }_{ D } }{ { R }^{ 2 } } { r }^{ 2 }\)
Using Ampere's Maxwell law we have, \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\times \)(total current through the area \({ A }^{ \prime }\)) or \(B=\frac { { \mu }_{ 0 }{ I }_{ 0 }r }{ 2\pi { R }^{ 2 } } =\frac { 4\pi \times { 10 }^{ -7 }\times 11.312\times { 10 }^{ -6 }\times 2\times { 10 }^{ -2 } }{ 2\pi \times { \left( 0.1 \right) }^{ 2 } } \)
or \(2\pi rB={ \mu }_{ 0 }\frac { { I }_{ 0 } }{ { R }^{ 2 } } { r }^{ 2 }\)
= 4.525 x 10^-12 T

Here, \({ I }_{ D }=2.0A, \ C=1\mu F={ 10 }^{ -6 }F.\)
We know, \({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { d }{ dt } \left( EA \right) \)
\(={ \epsilon }_{ 0 }A\frac { dE }{ dt } ={ \epsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) ={ \epsilon }_{ 0 }\frac { A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(\left( \because E=\frac { V }{ d } \right) and \ \left( C=\frac { { \epsilon }_{ 0 }A }{ d } \right) \)
\(or \ \frac { dV }{ dt } =\frac { { I }_{ D } }{ D } =\frac { 2.0 }{ { 10 }^{ -6 } } =2\times { 10 }^{ -6 }V{ s }^{ -1 }\)
Thus a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 x 10⁶ Vs^-1 .

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)

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.

(a)  \(\lambda =\frac { c }{ v } =\frac { 3\times 10^{ 8 } }{ 2.0\times 10^{ 10 } } \)
\( =1.5\times 10^{ -2 }m\)
\(E=48Vm^{ -1 }\)
(b) \({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 48 }{ 3\times { 10 }^{ 8 } }\)
or  \({ B }_{ 0 }=1.6\times 10^{ -7 } \ T\)
(c) Energy density in E field, 
\({ U }_{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }\)
Energy density in B field, 
\({ U }_{ B }=\frac { 1 }{ 2\mu _{ 0 } } B^{ 2 }\)
Using \(E=cB\) and \(c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } ,\)
We find \({ U }_{ E }={ U }_{ B }\).

Eye is most sensitive to the light of wavelength \(\lambda=5600 \dot{A}\)
\( T=\frac{1}{v}=\frac{\lambda}{c}=\frac{5600 \times 10^{-10}}{3 \times 10^8} \\ =1.87 \times 10^{-15} \mathrm{~s} \)

Radio waves,  microwaves, UV rays, X-rays,

Ratio = 1; because of the velocity of both the wave. length in a vacuum is same \(\left(=3 \times 10^8 \mathrm{~ms}^{-1}\right)\)

Pressure exerted by reflected wave on the surface is 
\(P=\frac{2 l}{c}=\frac{2 \times 1}{3 \times 10^8}=6.67 \times 10^{-9} \mathrm{~N} / \mathrm{m}^2\)

Time taken by laser beam to move through a distance 90 cm is
\(t=\frac{90}{c}=\frac{90 \mathrm{~cm}}{3 \times 10^{10} \mathrm{~cm} / \mathrm{s}}=3 \times 10^9 \mathrm{~s}\)
The energy contained in 90 cm length of laser beam is
\( U=p t p(10 \mathrm{~mW}) \times\left(3 \times 10^{-9} \mathrm{~s}\right) \\ =\left(10 \times 10^{-3} \mathrm{Js}^{-1}\right) \times\left(3 \times 10^{-9} \mathrm{~s}\right) \\ =30 \times 10^{-12} \mathrm{~J} \)

Maxwell, Hertz, Bose and Marconi

Not necessarily, A displacement current such as that between the plates of a charging capacitor) can also produce loops of \(\vec B\)

Electric flux through plates of capacitor, \(\phi_E=\frac{q}{\varepsilon_0}\).
where, charge, q= constant (as the capacitor is fully charged)
Displacement current, \(I_d=\varepsilon_0 \frac{d \phi_E}{d t}=\varepsilon_0 \frac{d\left(\frac{q}{\varepsilon_0}\right)}{d t}=0\)
Conduction current, \(I_c=C \frac{d V}{d t}=0\) (as voltage becomes constant when the capacitor becomes fully charged).

The quantity \(\epsilon_0 \frac{d \phi_E}{d t}\) has the dimension of current and this current exists in a region between the two plates of the capacitor when a displacement of charges occurs there. i.e., during charging and discharging of the capacitor.

No, the displacement current also produces magnetic field between two plates of capacitor during charging and discharging of capacitor

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.

(i) (c) : \(\frac{1}{2} \varepsilon_{0} E^{2}=\text { energy density }=\frac{\text { Energy }}{\text { Volume }}\)
\(\therefore\left[\frac{1}{2} \varepsilon_{0} E^{2}\right]=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{3}}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
(ii) (b): \(\text { As } \varepsilon_{0}=\frac{q_{1} q_{2}}{4 \pi F R^{2}}(\text { from Coulomb's law) }\)
\(\varepsilon_{0}=\frac{C^{2}}{N m^{2}} \frac{[A T]^{2}}{M L T^{-2} L^{2}}=M^{-1} L^{-3} T^{4} A^{2}\)
(iii) (d): The frequency of the electromagnetic wave remains same when it passes from one medium to another. Refractive index of the medium \(n=\sqrt{\frac{\varepsilon}{\varepsilon_{0}}}=\sqrt{\frac{4}{1}}=2\)
Wavelength of the electromagnetic wave in the medium, \(\lambda_{\mathrm{med}}-\frac{\lambda}{n}-\frac{\lambda}{2}\)
(iv) (b): \(\beta\)-rays consists of electrons which are not electromagnetic in nature.
(v) (b): The velocity of electromagnetic waves in free space (vacuum) is equal to velocity of light in vacuum (i.e., 3 x 10⁸m S^-1).
 

(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.