Electromagnetics Test 1

As per Maxwell’s equation,

\(\Rightarrow \bar{\nabla }\times \bar{H}=\bar{J}+\frac{\partial \bar{D}}{\partial t}\)

\(\Rightarrow \mathop{\iint }_{s}\left( \bar{\nabla }\times \bar{H} \right).d\bar{s}=\mathop{\iint }_{s}\left( \bar{J}+\frac{\partial \bar{D}}{\partial t} \right).d\bar{s}\)

Using Stokes Theorem, we can write:

\(\mathop{\oint }_{C}\bar{H}.d\bar{l}=\mathop{\iint }_{s}\left( \bar{J}+\frac{\partial \bar{D}}{\partial t} \right).d\bar{s}\)

Concept:

The distance between two adjacent maxima or minima points for standing waves is given by:

\(d=\frac{λ}{2}\)

λ = wavelength of the operating wave

Also, frequency and wavelength are related by the relation:

c = ν λ 

c = Speed of light = 3 × 10¹⁰ cm/s

Calculation:

d = 37.5 - 12.5 = 25 cm

\(25=\frac{λ}{2}\)

λ = 50 cm

\(ν =\frac{3\times10^{10}}{50}~Hz\)

∴ The operating frequency = 600 MHz

Concept:

The boundary condition for electrostatic fields are defined as:

E1t = E2t (Tangential components are equal across the boundary surface)

Also, the normal component satisfies the following relation:

D1n – D2n = ρs

For a charge-free boundary, ρs = 0, the above expression becomes:

D1n – D2n = 0

D1n = D2n

Calculation:

As the dielectric slab occupies the region 0 < z < d and the field intensity in the free space is in +a_z direction, the field will be normal to the boundary of the plane dielectric slab.

So from the boundary condition, the field normal to the surface are related as:

ε E_in = ε₀ E

With ε = 4ε₀, we can write:

\({E_{in}} = \frac{{{{\rm{\varepsilon }}_0}}}{{4{{\rm{\varepsilon }}_0}}}{E_0}{a_z} = \frac{{{E_0}}}{4}{a_z}\)

∴ The electric field intensity (Ein) inside the dielectric slab will be \(\frac{E}{4}{a_z}\) 

\({D_{in}} = {\rm{\varepsilon \;}}{{\rm{E}}_{{\rm{in}}}} = 4{{\rm{\varepsilon }}_0}\frac{{{E_0}}}{4}{a_z}\)

\({D_{in}} = {{\rm{\varepsilon }}_0}{E_0}{a_z}\)

And the electric flux density (Din) inside the dielectric slab will remain unchanged.

Concept:

The general equation of magnetic field intensity is given by:

H = H0 cos (ωt - βz)ax A / m

The phase velocity is defined as the rate at which the phase of the wave propagates in space.

Mathematically this is calculated as:

\({v_p} = \frac{ω }{β }\)

Calculation:

Given magnetic field intensity in the non-magnetic medium is:

H = 1.5 cos (10⁹ t – 5z) a_x A / m

Comparing both the expressions, we get:

ω = 10⁹ rad / sec

And β = 5

So, the phase velocity of the wave in the medium will be:

\({v_p} = \frac{ω }{β } = \frac{{{{10}^9}}}{5} = 2 \times {10^8}m/s\)

Concept:

Guided wavelength:

\({{\lambda }_{g}}=\frac{\lambda }{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}\)

Where λ is the free space wavelength, f is the operating frequency and fc is the cut-off frequency.

∵ \(\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}\) is always less than 1

∴ λg > λ always

∴ Statement 1 is incorrect.

Phase velocity:

\({{v}_{p}}=\frac{c}{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}\)

Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.

We can readily see that the phase velocity is a non-linear function of the operating frequency.

∴ Statement 2 is incorrect

The intrinsic/waveguide impedance 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

∴ Statement 3 is incorrect

For TE mode:

Wave impedance

\({{\eta }_{TE}}=\frac{\eta }{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}\)

For TM mode:

Wave impedance

\({{\eta }_{TM}}=\eta \sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}\)

Where η is the free space impedance

∴ Wave impedance can be greater or less than the intrinsic impedance. It depends on the mode of propagation.

∴ Statement 4 is Correct.

The power received P_r = P_avg A_e

P_avg = Time average power

\({P_r} = \frac{{E_0^2}}{{2\;\eta }}{A_e}\)

Concept

Reluctance is the opposition offered by a magnetic circuit to the establishment of magnetic flux.

where, S = Reluctance

l = Length

μ = Permeability

A = Cross-Sectional Area

Explanation

For two magnetic circuits connected in parallel, the total reluctance is found exactly like resistances in parallel (but with reluctance instead of resistance).

For two magnetic circuits connected in series, the reluctance of the circuit is:

S_T = S1 + S2

Relative Permeability of Non-Magnetic Materials

Definition: Relative permeability (μ_r) is a measure of a material's ability to conduct magnetic flux relative to the magnetic permeability of free space (μ₀). It is a dimensionless quantity defined as the ratio of the material's absolute permeability (μ) to the permeability of free space (μ₀).

For non-magnetic materials, the relative permeability is usually very close to 1, indicating that they do not significantly enhance or hinder the magnetic field. These materials do not exhibit strong magnetic properties and are not easily magnetized. Examples of non-magnetic materials include air, water, wood, and most non-metals.

Correct Option Analysis:

The correct option is:

Option 3: Unity

This is the correct answer because the relative permeability of non-magnetic materials is approximately equal to unity (1). This means that the material does not significantly alter the magnetic field passing through it. In mathematical terms:

μ_r = μ / μ₀

For non-magnetic materials, μ ≈ μ₀, which results in:

μ_r ≈ 1

Examples of such materials include air, vacuum, and most non-metallic substances. These materials do not have magnetic domains and hence do not exhibit strong magnetic properties.

Magnetomotive Force (MMF) in Magnetic Circuits

Definition: Magnetomotive force (MMF) is a measure of the magnetic potential in a magnetic circuit. It represents the ability to drive magnetic flux through the circuit. MMF is analogous to electromotive force (EMF) in electrical circuits and is measured in ampere-turns (AT).

Concept:

The general equation of electric field intensity of an EM wave propagating in a_x direction in a medium is given as:

E = E₀ cos (ωt - βx) a_y A/m

ω = Angular velocity defined as:

\(\omega =\frac{2\pi }{T}\)

T = Time period

Calculation:

Given:

E = 9 cos (4 × 10⁸ t - βx) az V / m

Comparing it with the given expression of electric field intensity, we get:

ω = 4 × 10⁸ rad / s

So, the time period of the wave in the air will be:

\(T = \frac{{2\pi }}{ω } = \frac{{2\pi }}{{4 \times {{10}^8}}}\)

= 15.71 ns

Since in one time period, the wave covers one wavelength (λ), ∴ The time taken by the wave to travel λ / 4 distance will be:

Concept:

The velocity of propagation is given by:

\(v = \frac{1}{{\sqrt {LC} }}\)

Characteristic Impedance:

\({Z_o} = \sqrt {\frac{L}{C}} \)

Calculations:

\(v = \frac{1}{{\sqrt {LC} }}\)

\(L = \frac{1}{{C{v^2}}} \)

\(L= \frac{1}{{\left( {30 \times {{10}^{ - 12}}} \right){{\left( {2.5 \times {{10}^8}} \right)}^2}}} \)

\(L= 533\;nH/m\)

Characteristic Impedance will be:

\({Z_o} = \sqrt {\frac{L}{C}} = \sqrt {\frac{{533 \times {{10}^{ - 9}}}}{{30 \times {{10}^{ - 12}}}}} \)

\(Z_0= 133.29\;{\rm{\Omega }}\)

Calculation:

Guided wavelength is given by:

\({\lambda _g} = \;\frac{\lambda }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} \;}}\) ; where

f_c = Cut-off frequency

And f = Operating frequency.

For waveguide R­₁ : f₁ = 2GHz and f_c1 = 1 GHz

For waveguide R₂ : f₂ = 4 GHz_. We are to find fc­₂.

Given λ_g1 = λ_g2

Equating the wavelength for Both waveguides ;

\( \Rightarrow \frac{{{\lambda _1}}}{{\sqrt {1 - {{\left( {\frac{{{f_{c1}}}}{{{f_1}}}} \right)}^2}} }} = \;\frac{{{\lambda _2}}}{{\sqrt {2 - {{\left( {\frac{{f{c_2}}}{{{f_2}}}} \right)}^2}} }}\)

\(\frac{{c/{f_1}}}{{\sqrt {1 - {{\left( {\frac{{f{c_1}}}{{{f_1}}}} \right)}^2}} }} = \frac{{c/{f_2}}}{{\sqrt {1 - {{\left( {\frac{{{f_{c2}}}}{{{f_2}}}} \right)}^2}\;} }}\)

\(\sqrt {f_1^2 - f_{c1}^2} = \;\sqrt {f_2^2 - f_{c2}^2} \;\)

\( \Rightarrow \sqrt {4 - 1} = \sqrt {16 - f_{c2}^2} \)

⇒ 3 = 16 – fc₂²

fc₂² = 13

\({f_{c2}} = \sqrt {13} \)

Concept:

The electric field at a distance is given by:

E = E_oe^-αd

Where ‘α’ is the attenuation constant

For conductors, the attenuation constant is the inverse of the skin depth, i.e. the attenuation constant is given by:

\({\rm{\alpha }} = \sqrt {{\rm{\pi f}}{{\rm{\mu }}_0}{\rm{\sigma }}}\)

Calculations:

If the Amplitude of election filed intensity is E₀ on one side of the sheet, then the magnitude of the Electric field intensity on the other side is 10^-6 E_o

\({\rm{\alpha }} = \sqrt {{\rm{\pi f}}{{\rm{\mu }}_0}{\rm{\sigma }}}\)

\(= \sqrt {{\rm{\pi }} \times {{10}^6} \times 4{\rm{\pi }} \times {{10}^{ - 7}} \times 3.7 \times {{10}^7}} \)

= 1.21 × 10⁴ Np/m

The electric field is related as:

10^-6 E_o = E_o e^-αd

Taking log both sides

-6 ln 10 = -αd

\({\rm{d}} = \frac{{6{\rm{ln}}\left( {10} \right)}}{{\rm{\alpha }}} = 0.00142{\rm{\;m}}\)

Concept:

The reflection coefficient for a wave propagating from medium 1 to medium 2 is defined as:

\({\rm{Γ }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)   ---(1)

Where η1 and η2 are the intrinsic impedance of the two mediums respectively.

For a given Electric field Ei incident at the interface, the reflected wave is given by:

\({E_r} = {\rm{Γ }}{E_i}\)

Analysis:

Since 20% of the energy in the incident wave is reflected at the boundary, we can write:

\((\frac{E_r}{E_i})^2 = Γ^2 =0.2\)

\(\left| {\rm{Γ }} \right| = \sqrt {0.2} = \pm 0.447\), i.e. for both Γ = +0.447 and Γ = - 0.447, the reflected power will be 20% of the incident power.

Using equation (1), we can write:

\(\frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}} = \pm 0.447\)

\(\frac{{{\eta _0}\sqrt {\frac{{{\mu _{r2}}}}{{{\varepsilon _{r2}}}}} - {\eta _0}\sqrt {\frac{{{\mu _{r1}}}}{{{\varepsilon _{r1}}}}} }}{{{\eta _0}\sqrt {\frac{{{\mu _{r2}}}}{{{\varepsilon _{r2}}}}} + {\eta _0}\sqrt {\frac{{{\mu _{r1}}}}{{{\varepsilon _{r1}}}}} }} = \pm 0.447\)

With εr1 = μ3r1, and εr2 = μ3r2, the above expression becomes:

\(\frac{{\sqrt {\frac{{{\mu _{r2}}}}{{\mu _{r2}^3}}} - \sqrt {\frac{{{\mu _{r1}}}}{{\mu _{r1}^3}}} }}{{\sqrt {\frac{{{\mu _{r2}}}}{{\mu _{r2}^3}}} + \sqrt {\frac{{{\mu _{r1}}}}{{\mu _{r1}^3}}} }} = \pm 0.447\)     

\(\frac{{{\mu _{r1}} - {\mu _{r2}}}}{{{\mu _{r1}} + {\mu _{r2}}}} = \pm 0.447\)

\(\frac{{{\mu _{r1}}}}{{{\mu _{r2}}}} = \frac{{1 \pm 0.447}}{{1 \mp 0.447}}\)

\(\frac{{{\mu _{r1}}}}{{{\mu _{r2}}}} = 2.62\;or\;0.38\)

So, \(\frac{{{\varepsilon _{r2}}}}{{{\varepsilon _{r1}}}} = {\left( {\frac{{{\mu _{r2}}}}{{{\mu _{r1}}}}} \right)^3}\)

Concept:

Electric Flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity, because it's the dot product of two vector quantities, electric field and the perpendicular differential area.
   ϕ = E.A = EA cosθ 
• The SI unit of the electric flux is N-m2/C.

Electric flux density (D) is a vector quantity because it is simply the product of the vector quantity electric field and the scalar quantity permittivity of the medium, i.e.

Its unit is Coulomb per square meter.

EXPLANATION:

1. Forefinger (Index finger): Represents the direction of the magnetic field (magnetic flux). Therefore option 3 is correct.
2. Middle finger: Represents the direction of motion of charge (current).
3. The thumb : Represents the direction of force or motion experienced by positively charged particles.

\({P_r} = \frac{{{P_t}\left( {{G_r}} \right)\left( {{G_t}} \right)}}{{{d ^2}{4 \pi}}}\)

Pr= Power received 

Pt = Power of transmitting antenna 

Gr= Gain of receiving antenna

Gt= Gain of transmitting antenna

d= distance between the two antenna

\(Directivity = \frac{{4\pi {A_e}}}{{{\lambda ^2}}}\)

Directivity = 1.63 for half wave dipole

\({P_r} = \frac{{{P_t}\left( {\frac{{4\pi {A_{eT}}}}{{{\lambda ^2}}}} \right)\left( {\frac{{4\pi {A_{eR}}}}{{{\lambda ^2}}}} \right)}}{{4\pi {d^2}}}\)

Where Ae is the effective area of the antenna'

\(=\frac{{80\left( {A_e^2} \right)4\pi }}{{{d^2}{\lambda ^4}}}\)

\(\lambda = \frac{C}{f} = \frac{{3 \times {{10}^8}}}{{100 \times {{10}^6}}} = 3m.\)

substituting the values

For the optical power

\(\alpha = \frac{{10}}{L}\log \left( {\frac{{{P_{in}}}}{{{P_{out}}}}} \right)\) 

\(L = \left( {\frac{{10}}{{1.41}}} \right)\log \left( {\frac{1}{{0.5}}} \right)\) 

\(L = \left( {\frac{{10}}{{1.41}}} \right)\log \left( 2 \right)\) 

L = 2.134 km

Concept:

The voltage at load is superposition of 2 waves

1) Forward travelling wave

2) Reflected wave

If V⁺ is the amplitude of forward travelling wave

Then voltage at load

V_L = V⁺ + V_R

V_L = V⁺ + Γ V⁺

V_L = V⁺ (1 + Γ)

Where V_R = Reflected wave voltage

Also:

V_max = |V⁺| (1 + (Γ(z)))

\(=\left| {{V}^{+}} \right|\left( 1+\frac{SWR-1}{SWR+1} \right)\)

\(=\left| {{V}^{+}} \right|\left( \frac{2~S\omega R}{S\omega R+1} \right)\)

Similarly,

\({{V}_{min}}=\left| {{V}^{+}} \right|\left( 1-\text{ }\!\!\Gamma\!\!\text{ }\left( z \right) \right)\)

\(=\left| {{V}^{+}} \right|\left( \frac{2}{SWR+1} \right)\)

Calculations:

Reflection coefficient (Γ_L)

\({{\text{ }\!\!\Gamma\!\!\text{ }}_{L}}=\frac{{{Z}_{L}}-{{Z}_{0}}}{{{Z}_{L}}+{{Z}_{0}}}\)

\(=\frac{50-100}{50+100}\)

\(=\frac{-1}{3}\)

\(=\frac{1}{3}{{e}^{j\pi }}\)

\(\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|=\frac{1}{3}~,~{{\theta }_{L}}=-\pi \)

\(SWR=\frac{1+\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|}{1-\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|}=\frac{3+1}{3-1}=2\)

To calculate minimum and maximum voltage first calculate forward voltage amplitude.

\({{V}_{L}}={{V}^{+}}\left( 1+{{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right)\)

\({{V}^{+}}=\frac{{{V}_{L}}}{\left( 1+{{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right)}=\frac{50}{\left( 1-\frac{1}{3} \right)}=\frac{50}{\frac{2}{3}}\)

= 75 V

\({{V}_{max}}=\left| {{V}^{+}} \right|\left( \frac{2~SWR}{SWR+1} \right)\)

\(=75\left( \frac{2\times 2}{2+1} \right)\)

CONCEPT:

• Magnetic field strength or magnetic field induction or flux density of the magnetic field is equal to the force experienced by a unit positive charge moving with unit velocity in a direction perpendicular to the magnetic field.
  • The SI unit of the magnetic field (B) is weber/meter2 (Wbm-2) or tesla.
• The CGS unit of B is gauss.

​1 gauss = 10-4 tesla.

EXPLANATION:

• From the above explanation, we can see that the relation between tesla and Weber/m2 is given by:

1 tesla = 1 Weber/m2

Concept:

Coulomb's law:

It states that the magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance r between them.

• It is represented mathematically by the equation:

Concept:

Magnetic Field Strength (H): the amount of magnetizing force required to create a certain field density in certain magnetic material per unit length.

The intensity of Magnetization (I): It is induced pole strength developed per unit area inside the magnetic material.

The net Magnetic Field Density (Bnet) inside the magnetic material is due to:

• Internal factor (I)
• External factor (H)

∴ Bnet ∝  (H + I)

Bnet = μ0(H + I) …. (1)

Where μ0 is absolute permeability.

Note: More external factor (H) causes more internal factor (I).

∴ I ∝  H

I = KH …. (2)

And K is the susceptibility of magnetic material.

From equation (1) and equation (2):

Bnet = μ0(H + KH)

Bnet = μ0H(1 + K) …. (3)

Dividing equation (3) by H on both side

In a waveguide the EM energy (signals) travel at the speed equal to group velocity:

\({V_g} = {\frac{c}{\sqrt\epsilon_r}}\sqrt {{{1-\left( {\frac{{{f_c}}}{f}} \right)}^2}}\)

\({f_c} = \frac{c}{{\sqrt 4 \times 2 \times {{10}^{ - 2}}}}\sqrt {{{\left( {\frac{2}{{2.5}}} \right)}^2} + {{\left( {\frac{1}{1}} \right)}^2}}\)

= 9.6 GHz

For 13 GHz pulse

\({v_g} = {\frac{3}{2}} \times {10^8}\sqrt {1 - {{\left( {\frac{{9.6}}{{13}}} \right)}^2}}\)

= 1.01 × 10⁸m/s

For 18 GHz pulse

\({v_g} = {\frac{3}{2}} \times {10^8}\sqrt {1 - {{\left( {\frac{{9.6}}{{18}}} \right)}^2}}\)

= 1.27 × 10⁸m/s

The time delay between two pulses at the end of 200 m tunnel is

\({\rm{\Delta }}t = \frac{{200}}{{1.01 \times {{10}^8}}} - \frac{{200}}{{1.27\times {{10}^8}}}\)

Concept:

The magnetic field intensity (H) of a circular coil is given by