Calculating Radio Source Intensity and Momentum

Click For 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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
2
Views
2K
Replies
8
Views
3K
Replies
2
Views
4K