Read the following passage and answer the questions that follow :

Antennas have become increasingly importance to the society and at present, they are indispensable. They are being used every places. They are available in vast varieties. They are operating at various frequencies which are depending on different application. They operate on the principle of Maxwell's equation. They have different types of radiation patterns. There are several atmospheric losses in the way of propagation of waves. Due to which signal fades down, when it travels from transmitter to receiver antennas.

Based on the above para, answer the following questions :

 This is the curl-of-curl identity, and it is the single most used vector identity in electromagnetics:

\(\nabla\times\nabla\times\vec{A}=\nabla\left(\nabla\cdot\vec{A}\right)-\nabla^{2}\vec{A}\)

Since \(\nabla^{2}\vec{A}=\nabla\cdot\nabla\vec{A}\), that is option 4.

A dimensional check on the operators disposes of the others. The left-hand side is a vector, being the curl of a vector, so the right-hand side must be a vector too.

Note the distinction that options 1 and 4 turn on: \(\nabla\cdot\vec{A}\) is the divergence, a scalar, but \(\nabla\left(\nabla\cdot\vec{A}\right)\) is the gradient of that scalar, which is a vector. The extra \(\nabla\) is what makes option 4 legitimate.

Why this identity matters, and why it appears under a passage about antennas. It is the step that turns Maxwell's equations into the wave equation. Taking the curl of Faraday's law and substituting Ampère's law gives

\(\nabla\times\nabla\times\vec{E}=-\mu\varepsilon\dfrac{\partial^{2}\vec{E}}{\partial t^{2}}\)

and applying the identity, with \(\nabla\cdot\vec{E}=0\) in a source-free region, collapses the left side to \(-\nabla^{2}\vec{E}\), leaving

\(\nabla^{2}\vec{E}=\mu\varepsilon\dfrac{\partial^{2}\vec{E}}{\partial t^{2}}\)

— the wave equation, from which the velocity \(1/\sqrt{\mu\varepsilon}\) follows. Without this identity there is no wave, and no radiation for an antenna to launch.

Hence, ∇×∇×A = ∇(∇·A) − ∇·∇A.

A small loop's radiation resistance depends on the square of its area measured in wavelengths:

\(R_{r}=31171\left(\dfrac{A}{\lambda^{2}}\right)^{2}\ \Omega\)

Step 1 — the area in wavelengths. For a square of side \(l=\lambda/10\):

\(A=l^{2}=\dfrac{\lambda^{2}}{100}\quad\Rightarrow\quad \dfrac{A}{\lambda^{2}}=0.01\)

Step 2 — substitute.

\(R_{r}=31171\times\left(0.01\right)^{2}=31171\times10^{-4}=3.12\ \Omega\)

— option 2.

Option 3 is the deliberate trap. 73 Ω is the radiation resistance of a half-wave dipole, a figure so familiar that it is easy to reach for. But it belongs to a resonant, half-wavelength structure, whereas this loop is a tenth of a wavelength on a side — electrically tiny. Option 4's 273 Ω is another standard figure, that of a folded dipole with its impedance stepped up, and is equally irrelevant here.

Why such a small resistance is a serious problem. The loop's conductor also has an ohmic loss resistance \(R_{L}\), and the radiation efficiency is

\(\eta=\dfrac{R_{r}}{R_{r}+R_{L}}\)

With \(R_{r}\) only a few ohms, a loss resistance of the same order halves the efficiency — and for a loop of \(\lambda/100\) the radiation resistance falls to milliohms and almost all the input power is dissipated as heat. This is the fundamental difficulty of every electrically small antenna.

Note the fourth-power dependence on size. Since \(R_{r}\propto A^{2}\propto l^{4}\), halving the side reduces the radiation resistance by sixteen times. Two remedies follow directly: use N turns, which multiplies \(R_{r}\) by \(N^{2}\), or fill the loop with ferrite to raise the effective permeability — which is exactly what the ferrite rod aerial in a portable radio does.

Hence, the radiation resistance is approximately 3.12 Ω.

This is the Friis transmission formula written in terms of effective apertures — option 3.

\(P_{r}=P_{t}\dfrac{A_{et}A_{er}}{r^{2}\lambda^{2}}\)

Two features identify it at once. It involves an effective aperture for each of two antennas, so it must describe a one-way link between a transmitter and a receiver; and it falls as \(1/r^{2}\), which is the inverse-square spreading of a wave over a single path.

Why option 4 is the discriminating distractor. The radar equation looks similar but falls as \(1/r^{4}\):

\(P_{r}=\dfrac{P_{t}G^{2}\lambda^{2}\sigma}{\left(4\pi\right)^{3}r^{4}}\)

because the wave travels out to the target and back, suffering inverse-square spreading twice. The expression given has \(r^{2}\), so it cannot be a radar equation. It also contains a target cross-section \(\sigma\) in the radar case and none here.

Why the other two are wrong in kind. The Poynting vector is a power density in watts per square metre at a point, not a total power received; and a "power gain factor" would be a dimensionless ratio, whereas this expression is explicitly labelled in watts.

Deriving it in two steps shows why the aperture form is natural. The transmitter spreads its power over a sphere, and its aperture concentrates it:

\(S=\dfrac{P_{t}G_{t}}{4\pi r^{2}}\ \text{W/m}^{2}\)

and the receiving antenna collects \(P_{r}=SA_{er}\). Substituting the relation between gain and aperture,

\(G=\dfrac{4\pi A_{e}}{\lambda^{2}}\)

converts the familiar gain form into the aperture form given.

What the formula says about design. The \(\lambda^{2}\) in the denominator means that for fixed physical apertures, a shorter wavelength delivers more received power — which is why microwave links use dishes. That runs opposite to the more familiar gain form, where \(\lambda^{2}\) sits in the numerator for fixed gains; the two are consistent, and which one applies depends on whether the antennas are specified by size or by gain.

Hence, the expression is the Friis transmission formula.

 The two orthogonal components carry power independently, so their contributions simply add:

\(P_{avg}=\dfrac{E_{x}^{2}+E_{y}^{2}}{2\eta_{0}}\)

Substituting, with the free-space intrinsic impedance \(\eta_{0}=377\ \Omega\):

\(E_{x}^{2}+E_{y}^{2}=1.5^{2}+3^{2}=2.25+9=11.25\)

\(P_{avg}=\dfrac{11.25}{2\times377}=\dfrac{11.25}{754}=0.0149\ \text{W/m}^{2}\approx15\ \text{mW/m}^{2}\)

— option 2.

Why the 75° phase makes no difference to the power. This is the point the question is testing. The phase angle determines the polarisation — here elliptical, since the components are unequal and out of phase — and it decides the shape and tilt of the ellipse the field vector traces. But power is the sum of the two components' squares, and since the x and y directions are orthogonal there is no cross term to depend on the relative phase. The same two amplitudes at 0° would give a linearly polarised wave, and at 90° with equal amplitudes a circular one, yet the power would be identical.

Where the factor of 2 comes from. The amplitudes given are peak values, so the RMS value of each is \(E/\sqrt{2}\) and the time average of \(E^{2}\) is \(E^{2}/2\). Omitting it would give 30 mW/m² — which is precisely what option 3 offers, and the most likely error.

A note on the printed exponent. The wave is stated to travel in the +z direction, so the phase term should read \(\omega t-\beta z\) rather than \(\omega t-\beta x\). The slip does not affect the answer, since the power depends only on the amplitudes.

Hence, the average power per unit area is 15 mW/m².

 A propagation constant is always the square root of a product, never of a ratio — which identifies option 2 immediately.

\(\gamma=\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}=\alpha+j\beta\)

Product against ratio — the distinction that settles every option. The two fundamental quantities of any propagating medium are formed from the same two parameters in different ways:

Option 1 is the ratio form, so it is \(Z_{0}\), not \(\gamma\). Option 3 is also a ratio — \(\sqrt{j\omega\mu/(\sigma+j\omega\varepsilon)}\) is the intrinsic impedance of a medium, the field-theory counterpart of \(Z_{0}\). The corresponding propagation constant would be the product \(\sqrt{j\omega\mu\left(\sigma+j\omega\varepsilon\right)}\), which is not among the options. Option 4 is the reflection coefficient at a boundary between two media, and is dimensionless.

What the two parts of γ mean. Separating real and imaginary parts,

\(e^{-\gamma z}=e^{-\alpha z}e^{-j\beta z}\)

The real part \(\alpha\) is the attenuation constant in nepers per metre, describing how the amplitude decays; the imaginary part \(\beta\) is the phase constant in radians per metre, from which the wavelength and phase velocity follow:

\(\lambda=\dfrac{2\pi}{\beta},\qquad v_{p}=\dfrac{\omega}{\beta}\)

The lossless case shows the structure plainly: with R = G = 0,

\(\gamma=j\omega\sqrt{LC}\)

so \(\alpha=0\) — no attenuation — and \(\beta=\omega\sqrt{LC}\), giving a velocity independent of frequency and therefore no dispersion. That is the ideal a real line only approximates.

A memory aid worth keeping : \(\gamma\) and \(Z_{0}\) are built from the same two ingredients, one by multiplying and one by dividing — and their product recovers the series impedance while their quotient recovers the shunt admittance.

Hence, the propagation constant is √[(R + jωL)(G + jωC)].