Finding and using the Poynting vector

In summary, the conversation discusses a physics question regarding a parallel-plate capacitor formed by a gap in a fat wire. Parts a) and b) involve finding the electric and magnetic fields, as well as the energy density and Poynting vector in the gap. For part c), the total energy in the gap can be calculated by integrating the energy density, and the power input can be found by taking the derivative of the total energy with respect to time.
  • #1
Vuldoraq
272
1

Homework Statement



Hi, this is a repeat post, I fear I put it in the wrong forum to start with (I put it in advance physics and I think it may be more of an introductory physics question). I'm really sorry if this contravenes any of the rules.

A fat wire, radius a, carries a constant current I, uniformly distributed over it's cross section. A narrow gap in the wire, of width w<<a, forms a parllel-plate capacitor (see attachment, this question refers to question 3, sorry I couldn't cut and paste the picture).

a)Find the electrical and magnetic fields inside the gap, as functions of the distance s from the axis and the time t. Assume the charge is zero at time t=0.

b)Find the energy density uem in the gap and the Poynting vector, S in the gap. Note especially the direction of S

c)Determione the total energy in the gap, as a function of time. Calculate the total power into the gap, by integrating the Poynting vector of the appropriate surface. Check that the power inout is equal to the rate of increase of energy in the gap.

Homework Equations



In the following:
s=distance from the axis of the wire
a=radius of the wire
t=time
I=current
and I'm working in cylinrical polars.

For a parallel plate capacitor,

[tex]\underline{E}=\frac{\sigma}{\epsilon_{0}}[/tex]

For an amperian loop (with Maxwells fix),

[tex]\oint\underline{B}\cdot d\underline{l}=\mu_{0}I_{enc}+\mu_{0}\epsilon_{0}\int\frac{\partial\underline{E}}{\partial t} \cdot d\underline{a}[/tex]

Energy density,

[tex]U_{em}=\frac{1}{2}(\epsilon_{0}E^{2}+\frac{1}{\mu_{0}}B^{2})[/tex]

Poynting Vector,

[tex] \frac{1}{\mu_{0}}\underline{E}\times\underline{B}[/tex]

The Attempt at a Solution



For part a) I found the electric to be,

[tex]\underline{E}=\frac{It}{\epsilon_{0}}\widehat{z}[/tex]

and the magnetic field to be,

[tex] \underline{B}=\frac{\mu_{0}}{2\pi}\frac{Is}{a^2}\widehat{\phi}[/tex]

For part b) I found the energy density to be,

[tex]\frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}})[/tex]

and the Poynting vector to be,

[tex]-\frac{\mu_{0}}{2\pi \epsilon_{0}}\frac{I^{2}t s}{a^{2}}\widehat{s}[/tex]

For part c) I have no idea even where to begin.

Please could someone tell me if parts a) and b) are correct and give me a hand for part c)? If you want me show more working just ask (I left it out to keep the post a bit smaller).

Thanks!
 
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  • #2



Hi there, thank you for reaching out to the forum for help with your question. Your solutions for parts a) and b) look correct to me. For part c), you can use the energy density equation to calculate the total energy in the gap as a function of time. This will involve integrating the energy density over the volume of the gap. Once you have the total energy, you can take the derivative with respect to time to find the rate of change of energy in the gap, which should be equal to the power input calculated in part b). Let me know if you have any further questions or need more clarification. Good luck with your calculations!
 
  • #3


As a scientist, it appears that you have correctly applied the relevant equations and have found the electric and magnetic fields, as well as the energy density and Poynting vector for this system. Your calculations for parts a) and b) seem to be correct.

For part c), you can determine the total energy in the gap by integrating the energy density over the volume of the gap. This will give you an expression for the total energy as a function of time. To calculate the total power into the gap, you can integrate the Poynting vector over the appropriate surface (in this case, the surface of the gap). This will give you an expression for the power as a function of time. Then, you can check if the power input is equal to the rate of increase of energy in the gap, as stated in the question. This would confirm that your calculations are correct.

I hope this helps. Keep up the good work!
 

1. What is the Poynting vector and why is it important in science?

The Poynting vector is a mathematical concept that describes the direction and magnitude of electromagnetic energy flow. It is important in science because it helps us understand the behavior of electromagnetic fields and how energy is transferred through space.

2. How is the Poynting vector calculated?

The Poynting vector is calculated by taking the cross product of the electric field and magnetic field at a specific point in space.

3. What are some real-life applications of the Poynting vector?

The Poynting vector is used in many real-life applications, such as designing antennas, understanding the behavior of light in optical fibers, and studying the effects of electromagnetic radiation on living organisms.

4. Can the Poynting vector be negative?

Yes, the Poynting vector can be negative in cases where the direction of energy flow is opposite to the direction of the vector. This can happen, for example, when an electromagnetic wave is reflected off a surface.

5. How can the Poynting vector be used to calculate the power of an electromagnetic wave?

The magnitude of the Poynting vector at a specific point in space is equal to the power per unit area of an electromagnetic wave at that point. Therefore, by calculating the magnitude of the Poynting vector, we can determine the power of an electromagnetic wave at a given location.

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