## Monday, March 28, 2011

### Electrodynamics solutions

We left off in Basic Electrodynamics with three problems, as follows:

1) A radio wave transmits 25 W/m^2 of power per uit area. A plane surface of area 2.4 m x 0.7 m is perpendicular to the direction of propagation of the wave. Calculate the radiation pressure P_R on the surface if it is assumed to be a perfect absorber.

2) An AM radio station broadcast isotropically with an average power of 4 kW. A dipole receiving antenna 65 cm long is located 4 miles from the transmitter. Find the Emf induced by this signal between the ends of the receiving antenna.

3) A community plans to build a solar power conversion station, i.e. to convert solar radiation into electrical power. They require 1 MW (megawatt) of power, and the final system is assumed to have an efficiency of 30% (30% of the solar energy incident on the surface is converted to electrical energy). What must be the effective area, A, of an assumed perfectly absorbing surface to be used in such an installation? Assume a constant solar energy flux incident of 1000 W/m^2.

Now, let's look at the solutions of each in turn:

1) The intensity of an EM wave is :

I = S (av) = c U(av) = P_R(A)

where the area A = 2.4 m x 0.7 m = 1.68 m^2 and P_R denotes the radiation pressure, or P_R = S/c (for complete absorption)

Now, S = Power/ area = (25 W/m^2)/ (1.68 m^2) = 14.88 W

And:

P_R = S(av) / c = 14.88 W/ (3 x 10^8 m/s) = 3.3 x 10^-9 N/m^2

(2) We have: P(av) = 4 kw = 4 x 10^3 W

Assume spherical symmetry for area affected so area of the applicable spherical surface is:

A = 4 π r^2 r = 4 mi. = 6.44 x 10^3 m

Area = 4 π (6.44 x 10^3 m) ^2 = 8.09 x 10^4 m^2

Therefore the average Poynting vector associated with the transmission is:

S(av) = P(av)/ A = (4 x 10^3 kw)/ (8.1 x 10^4 m^2) = 7.7 x 10^-6 W/m^2

But recall:

S(av) = E(max)^2/ 2 u_o c Therefore, solving for E(max):

E(max) = [2S(av) u_o c]^½

Then: E(max)= [2(7.7 x 10^-6 W/m^2) (4 π x 10^-7 H/m)(3 x 10^8 m/s)]^½

E(max) = 7.6 x 10^-2 V/m

Then the emf induced in a 65 cm long (L =0.65m) antenna is:

Emf = E(max) L = (0.076 V/m) x (0.65m) = 0.049 Volts

(3) Assume: P(solar) = 10^3 W/m^2

But because efficiency is relevant we need P(in). Thus,

eff = P(out)/ P(in) = 0.3 = 1 MW/ P(in)

Where 1 MW = 10^6 watts is the desired energy to come out, or be produced.

To get this, the power we need to put in, is:

P(in) = P(out)/ eff = (10^6 w)/ 0.3 = 3.33 x 10^6 W

Then:

Area A = P(in)/ S = (3.33 x 10^6 W)/ (10^3 W/m^2)
= 3.33 x 10^3 m^2