91Ó°ÊÓ

Write the expression for the magnitude of the intensity of radiation.

$\mathit{I}\mathbf{=}<S>$ …… (1)

Here, is the Poynting vector which is given as:

Here,${\mathrm{μ}}_0$ is the permeability of magnetic field in free space,$m_0$ is the maximum value of the magnetic dipole moment,$\mathrm{Ó\neg }$ is the angular frequency, c is the speed of light, r is the shortest distance from the source towards the observer, and$\mathrm{θ}$ is the angle made by the displacement vector r with the vertical.

(a)

Draw the given situation.

From the above figure, the data is observed as,

$\begin{array}{rcl}{r}^2& =& R^2+h^2\\ {\sin }^2\mathrm{θ}& =& \frac{R^2}{r^2}\end{array}$

Write the expression for the total radiated power.

…… (2)

Substitute in equation (1).

$$\begin{gathered} I=\left(\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\right)\frac{{\sin }^2\mathrm{θ}}{r^2} \\ I=\left(\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\right)\frac{\left({\displaystyle \frac{R^2}{r^2}}\right)}{r^2} \\ I=\left(\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\right)\left(\frac{R^2}{r^2}×\frac{1}{r^2}\right) \\ I=\left(\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\right)\left(\frac{R^2}{{(h^2+R^2)}^2}\right)........\left(3\right) \end{gathered}$$

From equations (2) and (3),

(b)

The measurement are taken by the engineer for the maximum intensity. The first derivative the magnitude of the intensity of flux will corresponds to zero. Hence, from equation (4),

On further solving, the above equation becomes,

$$\begin{gathered} \frac{2R^2}{R^2+h^2}=1 \\ 2R^2=R^2+h^2 \\ {R}^2=h^2 \\ h=R \end{gathered}$$

Substitute the value of R in equation (4).

…… (5)

Therefore, the observation made at the location is$h=R$ and it corresponds to .

(c)

Re-write the equation (5) in terms of h.

Substitute$P=35\mathrm{kW}$ and$h=200\mathrm{m}$ in the above expression.

$$\begin{gathered} I=\frac{3\left(35\mathrm{kW}×{\displaystyle \frac{{10}^3\mathrm{W}}{1\mathrm{kW}}}\right)}{32\mathrm{Ï€}{\left(200\mathrm{m}\right)}^2} \\ I=0.02611\mathrm{W}/{\mathrm{m}}^2×\left(\frac{{10}^{-6}\mathrm{μ°Â}}{1\mathrm{W}}\right)\left(\frac{1{\mathrm{m}}^2}{{10}^4{\mathrm{cm}}^2}\right) \\ I=2.611\mathrm{μ°Â}/{\mathrm{cm}}^2 \end{gathered}$$

The value of city’s radio emission limit is $200\frac{\mathrm{μ°Â}}{{\mathrm{cm}}^2}$,the KRUD is in compliance with the city’s radio emission limits as the value is $2.611\frac{\mathrm{μ°Â}}{{\mathrm{cm}}^2}$.

Therefore, Yes, the KRUD is in compliance.