on line. 10 km from the transmitter, what will be the values of Erms and Hrms at a distance of 100 km? Example 2.1.1 Given: Data of Example 2.1.1 applied to a receiver at The electric field strength at a receiving station is mea- 10 km from transmitter. sured and found to have an r.m.s. value of 10 microvolts/ Required: (a) Erms at 100 km, (b) Hrms at 100 km. m. Calculate (a) the magnetic field strength; (b) the amount of power incident on a receiving aerial with an effective area of 5 m². Solution Given: Electric field strength =10 microvolts/m. Using Equation 2.1.9 at a distance of 100 km, the power Required: (a) Magnetic field strength, (b) incident density will be reduced by a factor (10/100)2 = 0.01, SO power on a receiving aerial with effective area of 5 m ². power density = 2.65 X 10-15 W/m2. Now, power density = Erms X Hrms and since Hrms = Erms/377 Solution (Equation 2.1.7) Using equation 2.1.8 (a) Hrms = 10 uV/m/377 O = 2.65 X 10- A/m (b) Power density is given by Hence Erms X Hrms = 10 x 10-6 2.65 Erms = 10-15 = 1 uV/m 10-8W/m2 = 2.65 X 10-13 W/m2 This is the amount of power incident on a surface of Hrms = 1uV/m/377 d =2.65 x 10-9 A/m area 1 m². For an aerial with area 5 m², the total incident power will be Summary of propagation principles P = 2.65 X 10-13 W/m2 = 1.33 pW Several important points have been established in Sec- tion 2.1.4. Power density R.F. energy is radiated by way of travelling electric The product Erms X Hrms has the dimensions of 'volts and magnetic fields which together constitute an per metre' times "amps per metre', giving watts per electromagnetic wave propagating in free space with square metre. This is equivalent to the amount of r.f. the velocity of light. power flowing through one square metre of area per- In a plane wave, the electric and magnetic fields vary pendicular to the direction of propagation and is known in phase and act at right-angles to each other. Both as the power density of the wave. The power density fields are at right-angles to the direction of measures t