(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Designers of electrical circuits often take the maximum amplitude of the radiated electric field (at a large distance r) produced by an alternating current I_{0}flowing in a loop of area A cm^{2}to be given by

E=[itex]\frac{2.6AI_{0}f^{2}}{r}[/itex][itex]\mu[/itex]Vm^{-1}

where f is the frequency in MHz. Calculate the ratio energy radiated when the loop carries a current of 1A at 10MHz to that for 0.01A at 500MHz. Comment on your result. Remember that for a wave the energy carried is proportional to the square of the amplitude.

2. Relevant equations

E=[itex]\frac{2.6AI_{0}f^{2}}{r}[/itex][itex]\mu[/itex]Vm^{-1}

3. The attempt at a solution

For state 1 (where I_{0}=1A and f=10MHz):

E=[itex]\frac{2.6A(1)(10*10^{6})^{2}}{r}[/itex]

=[itex]\frac{2.6*10^{14}A}{r}[/itex]

For state 2 (where I_{0}=0.01A and f=500MHz):

E=[itex]\frac{2.6A(0.01)(5*10^{8})^{2}}{r}[/itex]

=[itex]\frac{6.5*10^{15}A}{r}[/itex]

As the energy carried is proportional to the square of the amplitude this gives:

=[itex]\frac{\sqrt{6.5*10^{15}}}{\sqrt{2.6*10^{14}}}[/itex]

=5

Showing that state 2 has 5 times more energy radiated than state 1. Also the higher the frequency and current, the higher the amount of energy radiated. Is there anything else that I've missed out since this seems like state of the obvious stuff to point out as a result?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Calculating ratio energy radiated in a loop and comment on result

**Physics Forums | Science Articles, Homework Help, Discussion**