Question

Problem 1. The emitting antenna of a 100 kW radio station radiates equally in all directions. What are the magnitudes Emax and Bmax (a) 100 m from the antenna and (b) 50 km from the antenna? (c) For these two distances from the antenna, calculate the maximum potential difference caused between the ends of a receiving antenna that is 1.0 m long, assuming the antenna is aligned perfectly with the electric field of the radio wave

Answer 1

Given,

P = 100 kW ;

(a)r = 100 m

We know that, I = P/A

I = 100 x 10^3/(4 x 3.14 x 100^2) = 0.796 W/m^2

Erms = sqrt (I/ce0)

Erms = sqrt(0.796/(3 x 10^8 x 8.85 x 10^-12)) = 17.32

Emax = sqrt(2)17.32 = 24.5 V/m

We know that, c = E/B => B = E/c

B = 24.5/3 x 10^8 = 8.17 x 10^-8 T

Hence, Emax = 24.5 N/C ; Bmax = 8.17 x 10^-8 T

b)r = 50 km = 50 x 10^3 m

I = P/A

I = 100 x 10^3/(4 x 3.14 x (50 x 10^3)^2) = 3.18 x 10^-6 W/m^2

Erms = sqrt (I/ce0)

Erms = sqrt(3.18 x 10^-6/(3 x 10^8 x 8.85 x 10^-12)) = 0.035 V/m

Emax = sqrt(2)0.035 = 0.05 V/m

We know that, c = E/B => B = E/c

B = 0.05/3 x 10^8 = 1.67 x 10^-10 T

Hence, Emax = 0.05 N/C ; Bmax = 1.67 x 10^-10 T

c)E = V/d => V = Ed

Vmax1 = 24.5 x 1 = 24.5 V

Vmax2 = 0.05 x 1 = 0.05 V

Hence, Vmax1 = 24.5 V ; Vmax2 = 0.05 V