View Single Post
EmilyRuck
#1
Jan31-13, 03:37 AM
P: 38
Good morning,
in circuit theory I know that reacting power arise from phasors and represents a power which can't be used, because not delivered to any load, but continuously flows back and forth between the load and the generator with a zero mean during one period.
I can't understand very well, anyway, the meaning of this power in a field context, with electromagnetic sources in free space.
Let's consider a hertzian dipole, which has several fields component with several dipendence from the distance [itex]r[/itex] (in spherical coordinates).

[itex]H_{\varphi} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{jk}{r} + \frac{1}{r^2} \right) \sin \theta
\\
E_r = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{2 \eta}{r^2} + \frac{2}{j \omega \epsilon r^3} \right) \cos \theta
\\
E_{\theta} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{j \omega \mu}{r} + \frac{\eta}{r^2} + \frac{1}{j \omega \epsilon r^3} \right) \sin \theta[/itex]

Components proportional to [itex]1/r^2[/itex] and [itex]1/r^3[/itex] are called near field components; components proportional to 1/r are called far-field components. [itex]I_0[/itex] is the phasor of the current in the dipole and [itex]h[/itex] its length.

I calculate the power as flux of the Poynting vector through a surface [itex]S[/itex]:

[itex] P = \displaystyle \oint_S \mathbf{E} \times \mathbf{H}^* \cdot d\mathbf{S}
[/itex]

where the dipole is in the origin and [itex]S[/itex] is a sphere with radius [itex]r[/itex] centered in the origin too.

I find for the power [itex]P[/itex] a real part, which is the power that goes away from the dipole, and an imaginary part, which is the reactive power and is only stored near the dipole. In the waveguide theory, reactive power is carried by modes that cannot propagate, that is, modes that have a purely imaginary propagation constant: they attenuate exponentially along the waveguide. But now the reactive power is carried by a field that propagate, because it has the [itex]e^{-jkr}[/itex] term like the "far field" components: how can it is possible?
Moreover, what about the meaning of this reactive power? Books say that it is due to the [itex]1/r^2[/itex] and [itex]1/r^3[/itex] components, which are related to *static* fields. So, should I state that reactive power is that carried from static fields? A static field stores an energy and I can *use* this energy if I place a charge in the field, because the field will move the charge, making a work: but this is a sort of active power, a suitable power, isn't it?
Thank you for having read and sorry for my confusion.
Bye!

Emily
Phys.Org News Partner Physics news on Phys.org
Scientists uncover clues to role of magnetism in iron-based superconductors
Researchers find first direct evidence of 'spin symmetry' in atoms
X-ray laser probes tiny quantum tornadoes in superfluid droplets