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Amplitude of emf

  1. Oct 14, 2005 #1
    Sry, i think i put this in the wrong forum last time.


    Here is the problem i have been working on.

    An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.20 kW. A dipole receiving antenna 70.0 cm long is at a location 4.00 miles from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.

    I got P= 3200 W
    L antenna = .40m

    I did A= 4pir^2 in which i get 6437.376*4pi =520748007.7

    Next i know that S= energy/(area*time) but power is equal to energy/time, so I did 3200w/520748007.7 and i get 6.145006706e-6

    From here i Know s= Emax/2mu_oC so i did Emax = S* 2mu_oC in which i get 6.145006706e-68 *(3*10^8)(4pi*10^-7)*2 and i get .0046332259. Now i know that Vmax= Emax(L) so i multiply that answer by .70 and i get . .00324 v or 3.24mv, but appearantly this is wrong... Where did i mess up?
     
  2. jcsd
  3. Oct 15, 2005 #2

    vanesch

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    Seems correct (didn't check the numbers): the power is distributed over a sphere with radius 4 miles (to be converted in meter of course).
    Ok, so this should be the power flux per square meter at the level of the antenna (in W/m^2), which you can identify with the power flux of a plane wave by now.
    Check the formula which links the Emax and the power: I'd suspect a quadratic relationship ! But the idea is right:

    From the flux of power of an EM plane wave, you can indeed obtain the maximum E-field, which would be indeed the EMF over the distance of 70 cm.
    But depending on the sophistication of the problem (the level of the course), one is maybe not interested by what would be the voltage difference between two points 70 cm apart, but by the actual voltage seen by the dipole antenna. That's quite more difficult, and I don't know enough about it myself to help you there.
    If you didn't see things like equivalent cross sections of antennae and things like that, then I think your approach is, in principle, correct. But check units and numbers...
    cheers,
    Patrick.
     
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