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Calculating ratio energy radiated in a loop and comment on result

  1. Sep 27, 2011 #1
    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 I0 flowing in a loop of area A cm2 to be given by

    E=[itex]\frac{2.6AI0f2}{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{2.6AI0f2}{r}[/itex][itex]\mu[/itex]Vm-1

    3. The attempt at a solution
    For state 1 (where I0=1A and f=10MHz):
    E=[itex]\frac{2.6A(1)(10*106)2}{r}[/itex]
    =[itex]\frac{2.6*1014A}{r}[/itex]

    For state 2 (where I0=0.01A and f=500MHz):
    E=[itex]\frac{2.6A(0.01)(5*108)2}{r}[/itex]
    =[itex]\frac{6.5*1015A}{r}[/itex]

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

    =[itex]\frac{\sqrt{6.5*1015}}{\sqrt{2.6*1014}}[/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
     
  2. jcsd
  3. Sep 28, 2011 #2
    Doesn't matter, worked out what I did wrong.
     
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