Calculating Radio Source Intensity and Momentum

AI Thread Summary
The discussion revolves around calculating the intensity and momentum of a radio source. The intensity at 25m from a 100 MHz isotropic source is given as 120 W/m², leading to a calculated rms electric field of 213 V/m. For a surface reflecting 3/4 of the energy over two hours, the momentum is determined to be 8.06 x 10^-6 kg·m/s. The total power transmitted by the source is calculated to be 9.42 x 10^5 W. Participants are seeking clarification on the calculations, particularly for part C.
huntingrdr
Messages
24
Reaction score
0

Homework Statement



At a distance of 25m from an isotropic 100 MHz radio source the intensity is 120W/m2.

A) What is the rms electric field at 25m?

The answer should be 213 V/m.

B) Over a 2 hour period, what is the momentum given to a 4cm x 4cm surface (at the 25m distance) perpendicular to the source that reflects 3/4 of the energy?

The answer should be 8.06*10^-6.

C) What is the total power transmitted by the source?

The answer should be 9.42*10^5.

Homework Equations



I=P/a = E^2max/2(u)(c)

The Attempt at a Solution



Emax = 274587.4
Erms=Emax/sqrt(2)

I can not get any of these parts right. I tried them but can't get the same results. Can someone show me how to get these? Thanks for the help!
 
Physics news on Phys.org
Hi huntingrdr! :smile:

(have a mu: µ and an epsilon: ε and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
huntingrdr said:
At a distance of 25m from an isotropic 100 MHz radio source the intensity is 120W/m2.

A) What is the rms electric field at 25m?

The answer should be 213 V/m.

I=P/a = E^2max/2(u)(c)

You mean I=P/a = E2max/2(µ)(c)?

No, you're confusing electricity and magnetism :redface:

it's I = P/A = ε0E2max/2c …

see http://en.wikipedia.org/wiki/Energy_density#Energy_density_of_electric_and_magnetic_fields" :wink:
 
Last edited by a moderator:
I figured out parts a &b. Can anyone help me with c?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top