Option 4 : Impedance

The **intrinsic impedance of the wave is defined as the ratio of the electric field and the magnetic field **phasor (complex amplitude), i.e.

\(η =\frac{E}{H}=\frac{j\omega \mu }{\gamma }\)

It has a unit of Ohms and is defined as the opposition offered by the medium to the propagating electromagnetic field.

For free-space, the intrinsic impedance is a real quantity, i.e.

η = η0 = 120π

η ≈ 377 Ω

**Since the unit of impedance is also Ohms, they are of the same dimensions.**

__Notes__:

A uniform plane wave has the following properties:

- The electric and magnetic fields are perpendicular to each other which are also perpendicular to the direction of propagation.
- They are also called as (Transverse electromagnetic wave) TEM waves.
- The ratio \(\frac{{\left| E \right|}}{{\left| H \right|}}\) for the wave is equal to the intrinsic impedance of the medium, η.

Option 3 : equal to the particle velocity

__Concept:__

The group velocity of a wave is the velocity with which the overall envelope shape of the wave's amplitudes known as the modulation or envelope of the wave propagates through space.

The group velocity is defined by the equation:

\({v_g} = \frac{{d\omega }}{{dk}}\)

Where ω = wave’s angular frequency

k = angular wave number = 2π/λ

Wave theory tells us that a wave carries its energy with the group velocity. For matter waves, this **group velocity** is the **velocity u of the particle**.

Energy of a photon is given by the planck as:

E = hν

With ω = 2πν

ω = 2πE/h ----- (1)

Wave number is given by:

k = 2π/λ = 2πp/h ----(2)

where λ = h/p (de broglie)

Now from equations 1 and 2, we get:

\(d\omega = \frac{{2\pi }}{h}dE;\)

\(dk = \frac{{2\pi }}{h}dp;\)

\(\frac{{d\omega }}{{dk}} = \frac{{dE}}{{dp}}\)

By definition: \({v_g} = \frac{{d\omega }}{{dk}}\)

v_{g} = dE/dp ---- (3)

If a particle of mass m is moving with a velocity v, then

\(E = \frac{1}{2}m{v^2} = \frac{{{p^2}}}{{2m}}\)

\(\frac{{dE}}{{dp}} = \frac{p}{m} = {v_p}\) ---(4)

Now from equations 3 and 4:

**v _{g} = v_{p}**

Option 3 : \(\lambda = \frac{300}{f}\)

Wavelength (λ) is equal to the distance traveled by the wave during the time in which any one particle of the medium completes one vibration about its mean position. It is the length of one wave.

Frequency (f) of vibration of a particle is defined as the number of vibrations completed by the particle in one second. It is the number of complete wavelengths traversed by the wave in one second.

The relation between velocity, frequency, and wavelength:

c = f × λ

Where,

c is the speed of light in vacuum = 3 x 10^{8} m/s

\(\lambda = \frac{3\times10^8}{f}\)

But since it is given that the frequency is in MHz, we can write:

\(\lambda = \frac{300\times10^6}{f(MHz)}\)

Since 1 MHz = 10^{6} Hz, the above can be written as:

\(\lambda = \frac{300~MHz}{f(MHz)}\)

\(\lambda = \frac{300}{f}\)

Option 4 : transverse in nature

**Uniform plane wave**is a wave In which the Electric field, E̅ and Magnetic field H̅ lie in a plane and have the same value everywhere in that plane at any fixed instant.- Uniform plane wave traveling in the z-direction, the spatial variations of E̅ and H̅ are zero over z = constant plane.
- A plane wave is transverse, i.e. E̅ and H̅ are perpendicular to the direction of propagation.
- So they are called Transverse Electromagnetic waves (TEM).

The plane wave propagating through the dielectric has the magnetic field component as H = 20 e^{-ax} cos (ωt – 0.25x) a_{y} A/m (a_{x}, a_{y}, a_{z} are unit vectors along x, y, and z-axis respectively)

Option 2 : -a_{z}

__Concept__**:**

1) The direction of the electric field is considered as the polarization of the electromagnetic wave.

2) The direction of propagation of EM wave is given by the cross-product of the vector product of direction of electric field and magnetic field, i.e.

\({\hat a_p} = {\hat a_E} \times {\hat a_H}\)

This is an application of the Poynting theorem.

__Analysis__**:**

Given:

\(\vec H = 20\;{e^{ - \alpha x}}\cos \left( {\omega t - 0.25x} \right){\hat a_y}\)

\({\hat a_H} = {\hat a_y}\)

\({\hat a_p} = {\hat a_x}\)

\({\hat a_p} = {\hat a_e} \times {\hat a_H} \Rightarrow {\hat a_x} = {\hat a_E} \times {\hat a_y}\)

\({\hat a_E} = - {\hat a_z}\)

Now, the polarization of the wave = direction of the electric field, i.e.

\(= - {\hat a_z}\)Option 3 : the wave impedance of free-space divided by the refractive index of the medium

__Absolute refractive index:-__

The refractive index of a material describes how fast light travels through that material.

More specifically, it is the ratio of the speed of light in a vacuum to the speed of light in that medium.

It is always greater than one.

\(\text{Refractive Index} = {\text{Speed of light in Vacuum} \over \text{Speed of light in the medium}} \)

\(n = {c \over v}\)

The velocity of the EM wave in any medium is given by:

\(v_p = \frac{1}{{√ {{μ _r}{ε _r}{μ _0}{ε _0}} }}\) ----(1)

The velocity of the EM wave in free space is given by:

\(c = \frac{1}{{√ {{}{}{μ _0}{ε _0}} }}=3×10^8~m/s\)-----(2)

\(v_p = \frac{c}{n}\)

n = √ ϵ_{r} -- (3)

__Wave impedance:__

It is defined as:

\(η = η_0 \sqrt {\left(\frac{μ_r}{\epsilon_r}\right)}\)

Where η_{0 }is the characteristic impedance

The value of μ_{r }= 1 (if not given)

\(η = η_0 \sqrt {\left(\frac{1}{\epsilon_r}\right)}\)

From, equation 3 we can write:

\(η = \frac{\eta_0}{n}\)

∴ The wave impedance of a medium is equal to the wave impedance of free-space divided by the refractive index of the medium

Option 2 : The polarisation is the direction of the electric field in an electromagnetic wave

The direction of the electric field is considered as the polarization of the electromagnetic wave.

The direction of propagation of EM wave is given by the cross-product of the vector product of direction of electric field and magnetic field, i.e.

\({\hat a_p} = {\hat a_E} \times {\hat a_H}\)

This is an application of the Poynting theorem.

The electric field and magnetic field are perpendicular to the direction of propagation of EM waves.

The electric field in an electro-magnetic wave (in vacuum) is described by E = E_{max} sin(Kx - ωt)

Where

E_{max} = 100 N/C and K = 1 × 10^{7} m^{-1}, speed of light is 3 × 10^{8} m/s.

What is the amplitude of the corresponding magnetic wave?

Option 3 : 3.33 × 10^{-7} T

__Concept__:

The intrinsic impedance of the wave is defined as the ratio of the electric field and the magnetic field phasor (complex amplitude), i.e.

\(η =\frac{E}{H}\)

For free-space, the intrinsic impedance is a real quantity, i.e.

η = η0 = 120π

__Analysis__:

E = Emax sin(Kx - ωt)

With E_{max} = 100 N/C, the magnetic field will be:

With \(η =\frac{E}{H}\):

\(H =\frac{E}{\eta}\)

\(H =\frac{100}{120\pi}A/m\)

B = μ_{0} H

\(B=4\pi × 10^{-7}× \frac{100}{120\pi }\)

**B = 3.33 × 10 ^{-7} T**

Option 3 : \(\frac{\pi }{2}rad/m\)

**Concept:**

A uniform plane wave is propagating in the material which is lossless i.e. there is no loss.

Hence R = G = 0, σ = 0 only L and C of the material is considered.

L is represented by μ_{0}μ_{r}

And C represented by ε_{0} ε_{r} in the wave.

Propagation velocity, \({V_p} = \frac{\omega }{\beta }\)

And \({V_p} = \frac{C}{{\sqrt {{\mu _r}{\varepsilon _r}} }}\;\;\;\;\;\;\;\left( {C = 3 \times {{10}^8}\;m/s} \right)\)

**Calculation:**

Now, given that: freq. f = 50 × 10^{6} Hz

Relative permeability μ_{r} = 2.25

Relative permittivity ε_{r} = 1

Since the material is lossless, σ = 0

To find β (propagation constant):

\({V_p} = \frac{{3\; \times \;{{10}^8}}}{{\sqrt {2.25\; \times \;1} }} = \frac{{2\pi \left( {50\; \times \;{{10}^6}} \right)}}{\beta }\)

\(\Rightarrow \beta = \frac{\pi }{2}rad/s\)

Option 4 : 2.50 mm

__CONCEPT__:

- The skin effect is the tendency of an alternating electric current to become distributed within a conductor such that the current density is largest near the surface of the conductor and decreases with greater depths in the conductor.
- The electric current flows mainly at the skin of the conductor, between the outer surface and a level called the skin depth.
- Skin depth is defined as the reciprocal of attenuation constant i.e.,

\(\delta=\frac{1}{\alpha}\)

- The attenuation constant is given by:

\(\Rightarrow\alpha=\omega\sqrt{\frac{\mu\epsilon}{2}\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]}\)

- For an ideal conductor with σ ≈ ∞,

\(\Rightarrow \delta=\frac{1}{\alpha}=\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]≈\frac{σ}{\omega\epsilon}\)

__CALCULATION__:

Given α = 400 /m

Skin depth is defined as the reciprocal of attenuation constant i.e.,

\(\Rightarrow \delta=\frac{1}{\alpha}\)

\(\Rightarrow \delta=\frac{1}{400}=2.5~mm\)

α is the attenuation constant given by:

\(α=\omega\sqrt{\frac{\mu\epsilon}{2}\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]}\)

ω = Operating frequency

μ = Permeability of the material

σ = Conductivity of the material

For a good conductor σ >> 1. The above expression for the attenuation constant can be approximated as:

\(α=\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]≈\frac{σ}{\omega\epsilon}\)

∴ The attenuation constant becomes:

\(α=\omega \sqrt{\left(\frac{\muσ}{2\omega}\right)}\)

\(α=\sqrt{\left(\frac{\omega\muσ} {2}\right)}\)

\(\therefore~α =\sqrt{\pi f\muσ}\)

Thus, the skin depth becomes:

\(\delta=\frac{1}{\sqrt{\pi f \muσ}}\)

__Note__: The skin depth is inversely proportional to the square root of frequency.

Option 2 : 0.225 mm

Relative permeability (μ_{r}) = 200

Conductivity (σ) = 5 × 10^{6} mho/m

Outer diameter of conductor = 8 mm

Length of conductor = 2 mm

Total current carried by conductor, i(t) = 2 cos (π × 10^{4}t) Amp

⇒ ω = π × 10^{4} rad/sec

⇒ f = 0.5 × 10^{4} Hz

\(\delta = \frac{1}{{\sqrt {\pi f\;\sigma \;{\mu _0}{\mu _r}} }}\)

\(\delta = \frac{1}{{\sqrt {\pi\; \times \;0.5\; \times \;{{10}^4}\; \times \;4\pi\; \times \;{{10}^{ - 7}}\; \times \;200\; \times \;5 \;\times\; {{10}^6}} }}\)

\(\Rightarrow \delta = \frac{1}{{\sqrt {2{\pi ^2}\; \times \;{{10}^6}} }} = 0.225\;mm\)

Option 4 : \(\frac{1}{{\left( {30\pi } \right)}}A/m\)

__Concept__:

The intrinsic impedance of the wave is defined as the ratio of the electric field and the magnetic field phasor (complex amplitude), i.e.

\(η =\frac{E}{H}=\frac{j\omega \mu }{\gamma }\)

For free-space, the intrinsic impedance is a real quantity, i.e.

η = η0 = 120π

__Calculation__:

Given \(|\vec E|=4~V/m\)

\(η_0=\frac{|\vec E|}{|\vec H|}\)

\(|\vec H|=\frac{|\vec E|}{\eta_0}\)

\(|\vec H|=\frac{4}{120\pi}=\frac{1}{30\pi}\)

Option 4 : Visible light

- Under the electromagnetic spectrum,
**visible light**has the lowest frequency among the given option. - The spectrum ranges from radio waves having low frequency till gamma rays, having the highest one.

Option 3 : -1.0

when a transmission line is loaded with impedance it is represented as follows:

V = incident voltage

V’= reflected voltage

V’’= refracted or transmitted voltage.

Transmission line impedance is surge impedance Z_{s}.

Load impedance is Z_{L}.

Short circuited is line is considered when Z_{L} = 0

Coefficient of reflection is given by the expression \({V_{reflection}} = \frac{{V'}}{V}\)

\({V_{reflection}} = \frac{{{Z_L} - {Z_s}}}{{{Z_L} + {Z_S}}}\)

__Calculations: __

\({{\rm{V}}_{{\rm{reflection}}}} = \frac{{0 - {Z_S}}}{{0 + {Z_S}}}\)

V _{reflection }= -1

Therefore coefficient of reflection of voltage for a short circuited line is -1.

In spherical coordinates, let â_{θ}, â_{ϕ} denote unit vectors along the θ, ϕ directions.

\(E = \frac{{100}}{r}\sin \theta \cos \left( {\omega t - \beta r} \right){\hat a_\theta }\;V/m\)

\(H = \frac{{0.265}}{r}\sin \theta \cos \left( {\omega t - \beta r} \right){\hat a_\phi }\frac{A}{m}\)

Represents the electric and magnetic field components of the EM wave at large distance r from a dipole antenna, in free space, the average power (W) crossing the hemispherical shell located at r = 1 km, 0 ≤ θ ≤ π/2 is _________Concept__**:**

Average Power of an EM wave is given by:

∫∫ P_{avg}⋅ds

\({P_{avg}} = \frac{1}{2}{E_0}{H_0}\)

E_{0} = Maximum value of Electric field Intensity

H_{0} = Maximum value of Magnetic field Intensity

P_{avg} = Average Power Density

__Calculation__**:**

Given:

\({E_0} = \frac{{100}}{r}\sin \theta \)

\({H_0} = \frac{{0.265}}{r}\sin \theta \)

The average power will be:

\({P_{avg}} = \frac{1}{2}\left( {\frac{{100}}{r} \times \frac{{0.265}}{r}} \right){\sin ^2}\theta \)

\({P_{avg}} = \frac{{26.5}}{{2\;{r^2}}}{\sin ^2}\theta \)

The differential area in spherical coordinate is defined as:

ds = r^{2} sin θ dθ dϕ

The net power of the EM wave will be:

\(P = \int\!\!\!\int \left( {\frac{{26.5}}{{2\;{r^2}}}{{\sin }^2}\theta } \right){r^2}\sin \theta \;d\theta \;d\phi \)

Limits:

For hemisphere θ: 0 to π/2

ϕ : 0 to 2π

∴ The power of the EM wave will be:

\( = \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\pi /2} \frac{{26.5}}{2}{\sin ^3}d\theta \;d\phi \)

\( = \frac{{26.5}}{2}\;\mathop \smallint \limits_0^{2\pi } \left[ {\mathop \smallint \limits_0^{\frac{\pi }{2}} {{\sin }^3}\theta \;d\theta } \right]d\phi \)

Using Trigonometric Identity:

sin 3θ = 3 sin θ – 4 sin^{3} θ

\({\sin ^3}\theta = \frac{{3\sin \theta - \sin 3\theta }}{4}\;\)

Power of EM wave can be written as:

\( = \frac{{26.5}}{2}\;\mathop \smallint \limits_0^{2\pi } \left[ {\mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{3\sin \theta - \sin 3\theta }}{4}d\theta } \right]d\phi \)

\( = \frac{{26.5}}{2}\mathop \smallint \limits_0^{2\pi } \left[ {\frac{3}{4}\left( { - \cos \theta } \right)_0^{\frac{\pi }{2}} - \left( {\frac{1}{3}\frac{{\left( {\cos 3\theta } \right)}}{4}} \right)_0^{\frac{\pi }{2}}} \right]d\phi \)

\( = \frac{{26.5}}{2}\mathop \smallint \limits_0^{2\pi } \left[ {\frac{3}{4}\left( {0 - \left( { - 1} \right)} \right) - \left( {\frac{1}{2}\left( {0 - \left( { - 1} \right)} \right)} \right)} \right]d\phi \)

\( = \frac{{26.5}}{2}\mathop \smallint \limits_0^{2\pi } \left( {\frac{3}{4} - \frac{1}{{12}}} \right)d\phi = \frac{{26.5}}{2}\left( {\frac{2}{3}} \right)\mathop \smallint \limits_0^{2\pi } d\phi \)

\( = \frac{{26.5}}{2} \times \frac{2}{3} \times 2\pi \)

\( = \frac{{26.5 \times 2\pi }}{3}\)

P = 55.5 W

__Extra Information__:

In the given question:

Also, the direction of the power flow of the EM wave is given by the direction of E × H.

Option 4 : \(\frac{{3\pi }}{{10}}\;m\)

**Concept:**

The wavelength of the plane wave is defined as the ratio of the speed of light and the frequency

i.e. \(\lambda = \frac{C}{f}\)

**Calculation:**

Electric field intensity, H = 20 A/m in air medium wave propagation is a_{z} direction

ω = 2 × 10^{9} rad/sec

\(\lambda = \frac{C}{f}\)

\(f = \frac{\omega }{{2\pi }} = \frac{{2 \times {{10}^9}}}{{2\pi }} = \frac{{{{10}^9}}}{\pi }\)

\(\Rightarrow \lambda = \frac{{3 \times {{10}^8}}}{{\frac{{{{10}^9}}}{\pi }}} = \frac{{3\pi }}{{10}}\;m\)Option 2 : Power/Area

__Poynting vector__:

It states that the cross product of electric field vector (E) and magnetic field vector (H) at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point that is

\( \vec S = \vec E \times \vec H\)

Where P = Poynting vector,

E = Electric field and

H = Magnetic field

Or, it can be written as,

\(S = \frac{Power}{Area}\)

The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves.

The unit of the Poynting vector is watt/m2.

__Additional Information__

1) The Poynting vector (i.e. energy flow per unit area per unit time) for a plane electromagnetic wave is given by

\( \vec S = \frac{1}{\mu_o}(\vec E \times \vec H)\)

\( S = \frac{1}{\mu_o}(E H\, sin\theta )\)

2) From the above, it is clear that E and H are mutually perpendicular and also they are perpendicular to the direction of propagation of the wave.

Thus, the direction of the Poynting vector is along the direction of the propagation of the electromagnetic wave.

The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves per unit volume.

∴ The dimensions of both the Poynting vector and the electromagnetic power density are the same, i.e. M1 L-1 T-2

where M = Mass, L = Length, T = Time

Option 1 : v_{p} v_{g} = c^{2}

**Phase velocity:**

It is the rate at which the phase of the wave propagates in space. Mathematically:

\({V_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}\)

ω_{c} = Cut off frequency

ω = Frequency of operation

**Group velocity:**

It is the velocity at which the overall envelope of the wave travels.

\({V_g} = c\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} \)

Multiplying both velocities, we get:

\({V_p}{V_g} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }} \times c\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} \)

\({V_p}{V_g} = {c^2}\)

Option 3 : lower than in free space

__Concept:__

The phase velocity (Vp) of a plane electromagnetic wave is given as-

\({V_p} = \frac{1}{{\sqrt {\mu \varepsilon } }}\)

Here, μ = μoμr

ε = εo εr

\({v_p} = \frac{1}{{\sqrt {{\mu _o}{\mu _r}{\varepsilon _o}{\varepsilon _r}} }}\)

\({v_p} = \frac{c}{{\sqrt {{\mu _r}{\varepsilon _r}} }}\)

\(c = \frac{1}{{\sqrt {{\mu _o}{\varepsilon _o}} }} = 3 \times {10^8}\;m/sec\)

μ0 = permeability in free space 4π × 10-7 H/m

εo = permittivity in free space 8.854 × 10-12 C2/Nm2

__Application:__

The velocity of an electromagnetic wave is inversely proportional to the relative permittivity.

Therefore, the velocity of the electromagnetic wave through dielectric is lower than in free space.

Option 4 : UV and infrared

The **electromagnetic spectrum** is the range of frequencies (the spectrum) of electromagnetic radiation and their respective wavelengths and photon energies.

The electromagnetic spectrum can be shown in the figure below:

The wavelength and frequency of different colors are shown in the following table:

Sl no. | Colour | Wavelength | Frequency |
---|---|---|---|

1 | Violet | 400 to 440 | 668 THz to 789 THz |

2 | Blue | 460 to 500 | 606 THz to 668 THz |

3 | Green | 500 to 570 | 526 THz to 606 THz |

4 | Red | 620 to 720 | 400 THz to 484 THz |