Instantaneous power at infinite distance from current element

[avjt]

Hi all,

Consider a 'current element' of length 'dl' carrying a current of $I_{0} cos ({\omega}t)$. I understand the computation of the electric and magnetic fields due to this current element at any point $(r, \theta, \phi)$, the computation of the Poynting vector at that point, and how the radiation resistance $R_{r}$ is derived from the time-averaged Poynting vector integrated over the surface of an enclosing sphere.

Now, I am trying to derive the instantaneous power passing through the surface of a sphere or radius 'r' centered at the current element (by integrating the Poynting vector directly over the surface of that sphere) and take the limit, at $r\rightarrow\infty$, so as to eliminate the effect of the near field entirely.

Given that the current is $I_{0} cos ({\omega}t)$, I was expecting to get a result of $I_{0}^2R_{r} cos^2({\omega}t)$. But I keep getting $I_{0}^2R_{r} sin^2({\omega}t)$, or $I_{0}^2R_{r} cos^2({\omega}t+{\pi}/2)$ if you like.

Am I making a mistake (if so, can anyone point me to the correct derivation)? Or is there really a 90-degree phase shift between the current and the instantaneous power (if so, can anyone suggest some kind of physical explanation)?

Can anyone help me out? Thanks...

Avijit

 

[chrisbaird]

I don't know what you are doing wrong. In the far-field, radiation is a transverse spherical traveling wave, not much different than the transverse plane waves we learn about in Freshman physics. The derivation of the instantaneous power of a radiated wave is nearly identical to that of plane wave. http://faculty.uml.edu/cbaird/95.658%282011%29/Lecture1.pdf"

 

[avjt]

I don't know what you are doing wrong. In the far-field, radiation is a transverse spherical traveling wave, not much different than the transverse plane waves we learn about in Freshman physics. The derivation of the instantaneous power of a radiated wave is nearly identical to that of plane wave. http://faculty.uml.edu/cbaird/95.658%282011%29/Lecture1.pdf"

Thanks Chris.

The problem I seem to be having is that, for the current element ${\delta}{\ell}$ carrying a current $I_{0} cos({\omega}t)$, I get far-field values of $\vec{B}=-{\mu}_0 \frac{I_{0}\: {\delta}{\ell}\: sin({\theta})}{4{\pi}}(\frac {\omega} {cr})sin({\omega}t')\hat{\phi}$ and $\vec{E}=-\frac {1} {{\epsilon}_0} \frac{I_{0}\: {\delta}{\ell}\: sin({\theta})}{4{\pi}}(\frac {\omega} {c^2 r})sin({\omega}t')\hat{\theta}$, or, in other words, the electric and magnetic fields in the far field, though in phase with each other, seem to be 90-degree out of phase with the current (though the $\frac {1} {r^2}$ terms in the near field are nicely in phase with the current).

Computing the Poynting vector for these values of the far field gives me $\frac {1} {c\:{\epsilon}_0}(\frac{I_{0}\: {\delta}{\ell}}{4{\pi}})^2(\frac {\omega}{c \: r})^2 sin^2 {\theta} \: sin^2 ({\omega}t') \hat{r}$, and the surface integral of this gets me to ${I_{0}}^2 R_r\:sin^2 {\omega}t'$

Do I have a mistake there?

Avijit