Question

An airplane flying at a distance of 8.3 km from a radio transmitter receives a signal of intensity 15 μW/m2. What is the amplitude of the (a) electric and (b) magnetic component of the signal at the airplane? (c) If the transmitter radiates uniformly over a hemisphere, what is the transmission power?

Answer 1

a) Amplitude of the electric field, E = 0.1063 V/m

b) Magnetic field, B = = 3.545 x 10⁻¹⁰ Tesla

c) Transmission power, P = 6492.73 W

Step-by-step explanation:

Intensity, I = 15 μW/m²

Distance, R = 8.3 km = 8300 m

a) The electric field amplitude

Relationship between the amplitude and intensity of a signal is given by:

`I = (E^(2) )/(2c \mu)`..........(1)

c = 3 * 10⁸ m/s

μ = 4 π * 10⁻⁷

Substitute these values into equation (1)

`15 * 10^(-6) = (E^(2) )/(2* 3 * 10^(8) * 4\pi * 10^(-7) )\\15 * 10^(-6) * 240\pi = E^(2)\\ E^(2) = 0.0113097\\E = 0.1063 V/m`

b)Amplitude of the magnetic field component at the airplane

B = E / c

B = = 0.1063 / (3 x 10⁸ )

B = = 3.545 x 10⁻¹⁰ Tesla

3) The transmission power:

Intensity,
`I = (Power)/(Area)`

P = I A

Area of the hemisphere, A = 2πR²

A = 2π * 8300²

A = 432848635.8116 m²

Power, P = 15 * 10⁻⁶ * 432848635.8116

P = 6492.73 W

Answer 2

a) = 0.106 V/m

b) = 354.52 pT

c) = 6.5 W

Step-by-step explanation:

Given,

Intensity, I = 15 μW/m²

Distance, d = 8.3 km

E = √(2Iμc)

where

c = speed of light = 2.99*10^8

μ = permeability of free space = 4π*10^-7

E = √(2 * 15*10^-6 * 4*3.142*10^-7 * 2.99*10^8)

E = √(30*10^-6 * 1.257*10^-6 * 2.99*10^8)

E = √1.13*10^-2

E = 0.106 V/m

B = E/c

B = 0.106 / 2.99*10^8

B = 354.52 pT

P = IA

Where A = 1/2 * 4πr²

A = 0.5 * 4 * 3.142 * (8.3*10^3)²

A = 4.33*10^8

P = I A

P = 15*10^-6 * 4.33*10^-8

P = 6495 W

P = 6.5 kW

Therefore

a) = 0.106 V/m

b) = 354.52 pT

c) = 6.5 W