Question

A microwave transmitter emits electromag-

netic waves of a single wavelength. The max-
imum electric field 0.963 km from the trans-
mitter is 8.04 V/m.
The speed of light is 2.99792 x 108 m/s
and the permeability of free space is 47 x
10-7 N/A².

a)Assuming that the transmitter is a point
source and neglecting waves reflected from
the Earth, calculate the maximum magnetic
field at this distance.
Answer in units of T.

b)Calculate the total power emitted by the
transmitter.
Answer in units of W.

Answer 1

a) To calculate the maximum magnetic field at this distance, we can use the relationship E = cB, where E is the electric field, B is the magnetic field, and c is the speed of light.

E = cB
B = E/c

In this case, E = 8.04 V/m, c = 2.99792 x 108 m/s

B = 8.04 V/m / 2.99792 x 108 m/s = 2.68 x 10^-9 T

The maximum magnetic field at 0.963 km from the transmitter is 2.68 x 10^-9 T.

b) To calculate the total power emitted by the transmitter, we can use the formula P = E^2/ Z, where P is the power, E is the electric field, and Z is the impedance of free space which is (377 ohm).

P = E^2 / Z

In this case, E = 8.04 V/m, Z = 377 ohm

P = 8.04 V/m² / 377 ohm = 2.13 x 10^-4 W

The total power emitted by the transmitter is 2.13 x 10^-4 W.