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Find intensity of radiating electric dipole

  1. Dec 7, 2014 #1

    QuantumCurt

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

    Suppose a radiating electric dipole lies along the z axis. Let ##I_1## be the intensity of the radiation at a distance of 10 m and an angle of 90 degrees. Find the intensity (in terms of ##I_1##) at (a) a distance of 30 m and an angle of 90 degrees, (b) a distance of 10 m and an angle of 45 degrees, and (c) a distance of 20 m and an angle of 30 degrees.

    2. Relevant equations

    $$Intensity=I=\frac{P_{source}}{Area}$$

    3. The attempt at a solution

    I've been at a complete loss as to how to solve this. We never covered anything at all similar to this in class, and the textbook has no examples that are similar to this. The solution to this problem states that

    $$I(r, \theta)~\alpha ~\frac{sin^2\theta}{r^2}$$

    The 'initial conditions' given are 10 m and 90 degrees which is ##I_1##, and the solution then states that

    $$I(r,\theta)=I_1~sin^2\theta~(\frac{10m}{r})^2$$

    Where are they getting the factor of ##sin^2\theta## from? How does that proportionality lead them to this equation? I was trying to puzzle this out for about an hour last night, and I just spent another half hour staring at it. I feel like I'm missing something really simple right now. Any hints would be very appreciated.
     
  2. jcsd
  3. Dec 7, 2014 #2

    QuantumCurt

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    Oh! I'm apparently getting myself into too much of a rush with finals starting tomorrow. There's a note above this section in the problems.

    Note: All of the problems in this section are based on the following information. Refer to figure 30-11. It can be shown that the intensity of radiation from a radiating electric dipole at a field point far from the antenna is proportional to ##\frac{sin^2\theta}{r^2}##, where ##\theta## is the angle between the electric dipole moment vector ##\vec p## and the position vector ##\vec r## of the field point relative to the enter of the antenna. The pattern of radiation from this type of antenna is independent of the azimuthal angle, that is, you can rotate the pattern about the antenna axis and it does not change shape.

    Apparently I should slow down and read things first...lol
     
  4. Dec 7, 2014 #3

    rude man

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    The derivation of the electric field from a dipole radiator is not trivial. One way is to start with the vector potential A for a short radiator:

    E = (-1/ε) ∇ {∇* ∫A dt} - μ ∂A/∂t

    where " * " signifies the dot in dot-product).
    One then assumes a point at least several wavelengths away from the radiator, which simplifies the expansion you get from the above equation (I notice the problem didn't give the wavelength but assume λ << 10m or at least less than about 2 meters):

    Eθ = η il sinθ/2rλ

    where η = √(μ/ε)
    i = radiator current = i0sin(wt)
    l = length of radiator
    θ = angle away from z axis towards x-y plane
    r = distance away from radiator to observation point, assumed >> λ (actually, several wavelengths is already OK).

    So obviously I write all this just to show you the derivation is not trivial & am not trying to help you derive it in any detail.
    But now you can say that Eθ is proportional to sinθ and inversely proportional to r. So how are E and intensity I related?
     
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