# Instantaneous power at infinite distance from current element

1. Jul 11, 2011

### 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

2. Jul 12, 2011

### 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" [Broken]

Last edited by a moderator: May 5, 2017
3. Jul 12, 2011

### avjt

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

Last edited by a moderator: May 5, 2017