Electromagnetics Test 3

Concept:

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

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

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

For a given Electric field E0 incident at the interface, the reflected wave is given by:

\({E_r} = {\rm{\Gamma }}{E_0}\)

Also, the intrinsic impedance (η) of a medium is calculated as:

\(\eta = \sqrt {\frac{\mu }{\varepsilon }} \)

Calculation:

The intrinsic impedance of the dielectric medium (μ₀, 4ϵ₀) will be:

\(\eta = \sqrt {\frac{\mu }{\varepsilon }} = \sqrt {\frac{{{\mu _0}}}{{4{\varepsilon _0}}}} \)

\(\eta = \frac{{{\eta _0}}}{2}\)

∴ The reflection coefficient will be:

\({\rm{\Gamma }} = \frac{{\frac{{{\eta _0}}}{2} - {\eta _0}}}{{\frac{{{\eta _0}}}{2} + {\eta _0}}}\)

\({\rm{\Gamma }} = \frac{{\frac{1}{2} - 1}}{{\frac{1}{2} + 1}} = - \frac{1}{3}\)

Now, for the given Electric Field magnitude E₀, the magnitude of the electric field of the reflected wave will be:

\({E_r} = {\rm{\Gamma }}{E_0} = - \frac{{{E_0}}}{3}\)

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.

For a good conductor σ ≫ωϵ

Intrinsic impedance:

\(\eta = \frac{E}{H} = \sqrt {\frac{{j\omega \mu }}{{σ + j\omega \varepsilon }}}\)

For a good conductor, σ >> 1. The above expression then becomes:

\(\eta = \frac{E}{H} = \sqrt {\frac{{j\omega \mu }}{{σ }}}\)

Writing the above in phasor domain, we get:

\(\eta = \sqrt {\frac{{\omega \mu }}{σ }} \angle 45^\circ\)

Concept:

For a propagating electromagnetic wave, the field satisfies the following Maxwell’s equation in charge free regions:

\(\nabla.\vec E=0\) (∵ ρ = 0, for charge free region)

\(\nabla \times E = - \frac{{\partial B}}{{\partial t}} \ne 0\)

Analysis:

Field P:

P = 60 sin (ωt + 10x) a_z

So, ∇ ⋅ P = 0

And ∇ × P = -600 cos (ωt + 10x) a_y ≠ 0

i.e. P is a possible EM field.

Field Q:

\(Q = \frac{{10}}{ρ }\cos \left( {\omega t - 2ρ } \right){a_\phi }\)

∇ ⋅ Q = 0

\(\nabla \times Q = \frac{1}{ρ }\frac{{\partial }}{{\partial ρ }}\left[ {10\cos \left( {\omega t - 2ρ } \right){a_z} \ne 0} \right]\)

i.e. Q is a possible EM field

Field R:

\(R = 3{ρ ^2}\cot \phi {a_ρ } + \frac{1}{ρ }\cos \phi {a_\phi }\)

\(\nabla \cdot R = \frac{1}{ρ }\frac{\partial }{{\partial ρ }}(3{ρ ^2}\cot \phi )\frac{{\sin \phi }}{ρ } \ne 0\)

i.e. R is not a possible EM field.

Field S:

\(S = \frac{1}{r}\sin \theta \sin \left( {\omega t - 6r} \right) {a_\theta }\)

\(\nabla \cdot S = \frac{1}{{{r^2}\sin \theta }}\sin \left( {\omega t - 6r} \right)\frac{{\partial ({{\sin }^2}\theta )}}{{\partial \theta}} \ne 0\)

i.e. S is not an EM field.

Thus, the possible EM fields are P and Q. 

Concept:

For a high loss medium, the attenuation constant is given as:

\(\alpha =\sqrt{\pi f{{\mu }_{0}}\sigma }.\)

Calculation:

First, we must check if seawater should be treated as a conductor or as a lossy dielectric for the given frequency.

To see if seawater is a conductor or not at 1 MHz,

We check for \(\frac{\sigma }{\omega \epsilon }~,~i.e.\)

\(\frac{\sigma }{\omega \epsilon }=\frac{4}{2\times \pi \times {{10}^{6}}\times 81\times 8.854\times {{10}^{-12}}}=888\gg 1\)

This is clearly a high-loss medium, so the attenuation constant is

\(\propto =\sqrt{\pi f{{\mu }_{0}}\sigma }=\sqrt{\pi \times {{10}^{6}}\times 4\pi \times {{10}^{-7}}\times 4}\)

= 3.974 (Np/m)

The lowest magnitude of magnetic field intensity received by the submarine = 1μA/m = 10^-6 A/m.

The maximum range is calculated from the following relation:

\( ~H~=~{{H}_{0}}{{e}^{-\alpha d}}\)

\( ~{{10}^{-6}}=~10,000~{{e}^{-\alpha d}}\)

\( ~{{10}^{-10}}=~{{e}^{-\alpha d}}\)

ln (10¹⁰) = αd