Question

During a test, a NATO surveillance radar system, operating at $\mathbf{12}\mathbf{\text{GHz}}$at$\mathbf{180}\mathbf{\text{kW}}$ of power, attempts to detect an incoming stealth aircraft at$\mathbf{\text{90km}}$.Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft’s location? The aircraft reflects radar waves as though it has a cross-sectional area ofonly${\mathbf{\text{0.22m}}}^{\mathbf{2}}$.(b) What is the power of the aircraft’s reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value of the magnetic field of the reflected radar beam?

Short answer

(a) The intensity of the beam when it reaches the aircraft’s locationis .$3.5\text{μW}/{\text{m}}^{\text{2}}$

(b) The power of the aircraft’s reflectionis .$0.78\text{μW}$

(c) The intensity of the reflected radar beamis .$1.5\times {10}^{-17}{\text{W/m}}^{\text{2}}$

(d) The maximum value of the electric field vector of the reflected radar beam is $1.1\times {10}^{-7}\text{V/m}$.

(e) The rms value of the magnetic field of the reflected radar beam is .$0.25\text{fT}$

Step-by-step solution

The given data

• The power of the incident wave, $P=180\text{kW}=180\times {10}^3\text{W}$
• The distance of aircraft from the radar, $r=\text{90km}=90\times {10}^3\text{m}$
• The radar beam and reflected beam from the aircraft is spread uniformly over a hemisphere.
• The waves are reflected fromtheaircraft through an effective area, $A={\text{0.22m}}^{\text{2}}$

Understanding the concept of radiated wave: 

An electromagnetic wave is an incident on an object; it may get absorbed or reflected. The intensity and hence the electric and magnetic field components of the wave change when it is reflected from the surface of an object.

Formulae:

The intensity of the radiation is,

$I=\frac{P}{A}$….. (i)

Here, $I$ is the intensity, $P$ is the power, and$A$ is the area.

The intensity of a radiation wave due to reflection,

$I=\frac{1}{c{\mu }_0}{E}_{rms}^2$….. (ii)

Here, $c$is the speed of light having a value of data-custom-editor="chemistry" $3\times {10}^8\text{m}/\text{s}$and ${\mu }_0$ is thepermeability of free space having a value $4\pi \times {10}^{-7}\text{H}/\text{m}$.

The RMS value of electric field,

$E_{rms}=\frac{E_m}{\sqrt{2}}$….. (iii)

The RMS value of the magnetic field,

$B_{rms}=\frac{E_{rms}}{c}$…..(iv)

(a) Calculation of the intensity of the radar beam: 

The beam spreads over a hemisphere having a radius that is equal to the distance of aircraft from the radar. Thus, the intensity of the radar beam as it reaches the aircraft is calculated using equation (i) as follows:

$I=\frac{P}{2\pi r^2}$

Here, the area of the hemisphere is$2\pi r^2$ .

Substitute known values in the above equation.

$\begin{array}{c}I=\frac{180\times {10}^3\text{W}}{2\times 3.14\times {(90\times {10}^3\text{m})}^2}\\ =3.5\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}\\ =3.5\text{μW}/{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the intensity is.$3.5\text{μW}/{\text{m}}^{\text{2}}$

(b) Calculation of the power of aircraft’s reflection:

The beam is reflected from the aircraft having effective area of.$0.22{\text{m}}^2$ Thus, the power of the reflected beam is calculated using equation (i) as follows:

$P_{ref}=I\times A$

Substitute ${\text{0.22m}}^{\text{2}}$for $A$ and $3.5\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}$for $I$ in the above equation.

$\begin{array}{c}{P}_{ref}=(3.5\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}})\times \left({\text{0.22m}}^{\text{2}}\right)\\ =0.78\times {10}^{-6}\text{W}\\ =0.78\text{μW}\end{array}$

Hence, the value of the power is.$0.78\text{μW}$

(c) Calculation of the reflected radar beam: 

The beam is reflected from the aircraft towards the radar. The intensity of this reflected beam is determined using equation (i) and the above value of power of the reflected beam as follows.

Since, the beam spreads over a hemisphere having a radius is equal to the distance of aircraft from the radar. Therefore,

$\begin{array}{c}{I}_{ref}=\frac{P_{ref}}{A}\\ =\frac{P_{ref}}{2\pi r^2}(\because areaofhemisphere=2\pi r^2)\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}{I}_{ref}=\frac{0.78\times {10}^{-6}}{2\times 3.14\times {(90\times {10}^3)}^2}\\ =1.5\times {10}^{-17}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the intensity is.$1.5\times {10}^{-17}{\text{W/m}}^{\text{2}}$

(d) Calculation of the maximum value of the electric field: 

The maximum value of the electric field is calculated by substituting the rms value of equation (iii) in equation (i) as follows:

$\begin{array}{l}{I}_{ref}=\frac{1}{c{\mu }_o}\frac{{E_m}^2}{2}\\ E_m^2=2c{\mu }_0{I}_{ref}\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}{E}_m^2=2\times (3\times {10}^8\text{m}/\text{s})\times (4\pi \times {10}^{-7}\text{H}/\text{m})\times (1.5\times {10}^{-17}{\text{W/m}}^{\text{2}})\\ =1.13\times {10}^{-14}{\text{V}}^2/{\text{m}}^2\end{array}$

$E_m=1.1\times {10}^{-7}\text{V}/\text{m}$

Hence, the value of maximum electric field is.$1.1\times {10}^{-7}\text{V}/\text{m}$

(e) Calculation of the rms value of magnetic field:

The rms value of the magnetic field is calculated by substituting equation (iii) in equation (iv) as follows:

$B_{rms}=\frac{E_m}{c\sqrt{2}}$

Putting known values in the above equation, you have

$\begin{array}{c}{B}_{rms}=\frac{1.1\times {10}^{-7}\text{V}/\text{m}}{3\times {10}^8\text{m}/\text{s}\times \sqrt{2}}\\ =2.5\times {10}^{-16}\text{T}\\ =0.25\times {10}^{-15}\text{T}\\ =0.25\text{fT}\end{array}$

Hence, the value of the magnetic field is.$0.25\text{fT}$$0.25\text{fT}$