Question of a problem on Poynting vector

In summary: It is so obvious that I just miss it all together. "j" translate to e to the power j(pi/2) which change the cos() to -sin().
  • #1
yungman
5,723
242
I have been working on this problem and I cannot get the solution as shown in the book. I scan the question and solution of the book and my work. It is too difficult for me to type it in, please open the attach scanned pdf file. THe top part is the copy of the book, the lower part is my work. You can see the solution on my part does not agree with the book and I check over my work many times already. Please tell me what did I do wrong
Thanks
 

Attachments

  • Question8_8.pdf
    363.4 KB · Views: 367
Physics news on Phys.org
  • #2
We can't see your attachment until it is approved by admin. It would be quicker if you uploaded your pdf to imageshack.us and then posted a link to it.
 
  • #3
gabbagabbahey said:
We can't see your attachment until it is approved by admin. It would be quicker if you uploaded your pdf to imageshack.us and then posted a link to it.

It can be opened now.
Thanks
 
  • #4
yungman said:
It can be opened now.
Thanks

Not by me it can't. It still says that it is pending approval.
 
  • #5
gabbagabbahey said:
Not by me it can't. It still says that it is pending approval.

I just opened already! I wonder why!
 
  • #6
You can open it before it has been approved since you are the one that attached it. Everyone else has to wait 'til its approved.

Again, you can save a lot of time by uploading it to a free imagehosting site and then just posting a link to it.
 
  • #7
I don't think your expression for the real part of [tex]\mathbf{E}(R,\theta,t)[/tex] and [tex]\mathbf{H}(R,\theta,t)[/tex] are correct. Remember that the original phasor expression for E and H has 'j' inside, meaning that you have to multiply throughout by j, so that makes the formerly imaginary part real and the formerly real part imaginary.
 
  • #8
Defennder said:
I don't think your expression for the real part of [tex]\mathbf{E}(R,\theta,t)[/tex] and [tex]\mathbf{H}(R,\theta,t)[/tex] are correct. Remember that the original phasor expression for E and H has 'j' inside, meaning that you have to multiply throughout by j, so that makes the formerly imaginary part real and the formerly real part imaginary.

Thanks again. It is so obvious that I just miss it all together. "j" translate to e to the power j(pi/2) which change the cos() to -sin().

Thanks again.
 

Related to Question of a problem on Poynting vector

1. What is the Poynting vector?

The Poynting vector is a mathematical concept used in electromagnetism to describe the directional flow of energy in an electromagnetic field. It is represented by the symbol S and is a vector quantity, meaning it has both magnitude and direction.

2. How is the Poynting vector calculated?

The Poynting vector is calculated using the formula S = E x H, where E is the electric field and H is the magnetic field. This formula takes into account both the electric and magnetic components of an electromagnetic wave to determine the direction and magnitude of energy flow.

3. What does the Poynting vector represent?

The Poynting vector represents the flow of energy in an electromagnetic field. It shows the direction and rate at which energy is being transferred from one point to another. It is an important concept in understanding the behavior of electromagnetic waves.

4. How is the Poynting vector related to the intensity of an electromagnetic wave?

The magnitude of the Poynting vector is directly proportional to the intensity of an electromagnetic wave. This means that as the Poynting vector increases, the intensity of the electromagnetic wave also increases. This relationship allows scientists to measure the intensity of electromagnetic radiation using the Poynting vector.

5. What are some real-world applications of the Poynting vector?

The Poynting vector has many practical applications, including in the fields of telecommunications, solar energy, and medical imaging. It is used to understand and optimize the transmission of electromagnetic waves in various technologies, such as cell phones and radio antennas. It is also used in solar panels to determine the efficiency of energy conversion. In medical imaging, the Poynting vector is used to measure the amount of energy absorbed by different tissues in the body during an MRI scan.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
969
Replies
8
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
977
  • Other Physics Topics
Replies
2
Views
4K
  • Advanced Physics Homework Help
4
Replies
118
Views
13K
  • Advanced Physics Homework Help
Replies
1
Views
940
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
10
Views
5K
Replies
2
Views
2K
Back
Top