# Instantaneous power at infinite distance from current element

• avjt
In summary: The problem is that you are neglecting the radiation resistance of the sphere. The sphere's surface will stop the power from propagating away quickly.
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

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"

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chrisbaird said:
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

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## 1. What is "instantaneous power at infinite distance from current element"?

"Instantaneous power at infinite distance from current element" refers to the amount of power that is being transmitted or radiated from a current element (such as a wire or antenna) at an infinite distance away. It is a theoretical concept used in electromagnetics and wireless communications to calculate the power received by a distant receiver.

## 2. How is the instantaneous power at infinite distance calculated?

The instantaneous power at infinite distance from a current element is calculated using the equation P = I^2*R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms at the distance of interest. This equation is based on the relationship between current, voltage, and resistance known as Ohm's Law.

## 3. What factors affect the instantaneous power at infinite distance?

The instantaneous power at infinite distance is affected by several factors including the current flowing through the element, the resistance of the element, and the distance between the element and the receiver. Other factors such as the frequency of the signal, the environment, and any obstacles between the element and the receiver can also impact the power received.

## 4. Why is the concept of instantaneous power at infinite distance important?

The concept of instantaneous power at infinite distance is important in the field of electromagnetics and wireless communications as it allows engineers to calculate the power received by a distant receiver. This information is crucial in designing and optimizing wireless systems, such as antennas and satellite communications, to ensure efficient and reliable transmission of signals over long distances.

## 5. How is the instantaneous power at infinite distance used in practical applications?

The instantaneous power at infinite distance is used in practical applications such as wireless communication systems, radio astronomy, and satellite communications. It is also used in the design and testing of antennas and other wireless devices to ensure they can transmit and receive signals effectively over long distances.

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