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Poynting vector power flow question

  1. Nov 29, 2008 #1
    Im aware that the complex electromagnetic powerflow from a solid is described by the poynting vector (see 1.1), however Im not sure how the 1/2 term arises in the formula

    [tex]1/2\oint\vec{E} \times \vec{H}.ds[/tex] (1.1)

    where [tex]\times[/tex] = vector cross product and ds denotes a variance in the surface of the object

    From the textbook I read, it says that [tex]\vec{E}[/tex] and [tex]\vec{H}[/tex] are peak phasors, not RMS which is meant to explain the 1/2 term, but from my understanding, the Vrms of a sinusoidal is always Vpeak/sqrt(2) where Vpeak = peak voltage of sinusoid. So Im really confused

    Can anyone help me out here ?

  2. jcsd
  3. Nov 30, 2008 #2


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    E and H each get divided by sqrt{2}, so their product is divided by 2.
  4. Dec 2, 2008 #3
    Ok I get the maths, but why is the E and H vectors divided by sqrt(2) ? In calculating electromagnetic power flow, are we only interested in the effective power being emitted from a solid ?
  5. Dec 2, 2008 #4

    Ben Niehoff

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    The equation is wrong. The Poynting vector should be

    [tex]\frac12 \Re \{ \vec E \times \vec H^* \}[/tex]

    That is, with the complex conjugate of H. This expression gives the real power emitted (hence the real part being taken). The imaginary part gives the reactive power, which is not part of the net energy transfer.

    The factor of 1/2 comes because we are using complex phasors to represent real fields by means of

    [tex]\vec E_{\textrm{real}} = \Re \{ \vec E (\cos \omega t + i \sin \omega t) \}[/tex]

    Therefore, every phasor field actually represents two oscillatory fields superimposed over each other, 90 degrees out of phase. Thus when we calculate the power or intensity, we will overshoot by a factor of 2.
  6. Dec 3, 2008 #5
    Ok how do you obtain the 2 mathematically ?

    Is it from realising that

    cos(theta) = [exp(j*theta) + exp(-j*theta)] /2

    sin(theta) = [exp(j*theta) - exp(-j*theta)] / (2*j)

    Or that when you integrate a cos(theta)^2 term you end up with a 1/2 constant in your result ?

    Note j = i
  7. Dec 3, 2008 #6
    Cos^2(theta) + Sin^(theta) = 1

    So the time average of Cos^2(theta) must be 1/2.

    Thats all there is to the 1/2.

    Note: Cos^2(theta) is the same as Cos(Theta)^2
  8. Dec 6, 2008 #7
    All right, to show my understanding of this concept, I just want to run some things through with everyone

    1) We take the real part of E x H divided by a half since practically, we are interested in the real power, not the reactive power

    2) The reason for dividing by 2 arises from the fact that the E or H field consists of 2 sinusoidal components and the definition of power being the average energy over a period of time

    3) the cos^2(theta) is a mathematical product of the E x H cross product operation

    Please feel free to correct me
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