Solving Wave in a Conductor Exercise 24-12

[stunner5000pt]

Homework Statement

Wangsness exercise 24-12
A plane wave travels in the postiive z direction in a conductor with real conductivity.

a) Find the instantaeous and time average power loss per unit volume due to resisitive heating for any z

b) Find the total power loss per unit area between z= 0- and z approaching infinity

c) Find the time average Poynting vector at any z

Homework Equations

Poynting vector is given by
$$\vec{S} = \frac{1}{2} \Re (\vec{E} \times \vec{H*})$$
For a conducting medium

$$E = E_{0a} e^{-\beta z} e^{i(az-\omega t+ \upsilon}$$
$$B= \frac{|k|}{\omega} \hat{z} \times E_{0a} e^{-\beta z} \cos (\alpha z-\omega t + \upsilon + \Omega)$$

The Attempt at a Solution

Im a little stumped here

per unit area it would be the $$<S> \bullet \vec{n}$$

but per unit volume??

The instantaneous would simply the derivative of the prveious quantity with respect to time.. right/

FOr total power loss we need to integrate the power oss per unit volume over hte whole volume. What is the shape of the condutor though??

 

[Jhero]

Hi there,

For part a), you are correct in thinking that the instantaneous power loss per unit volume would be the derivative of the Poynting vector with respect to time. However, since the Poynting vector is a complex quantity, you would need to take the real part of the derivative. Additionally, since we are dealing with a conductor, we need to take into account the conductivity in our calculations. The equation for power loss per unit volume due to resistive heating is given by:

P = \sigma |\vec{E}|^2

where \sigma is the conductivity and |\vec{E}|^2 is the magnitude of the electric field.

For part b), we can use the equation for power loss per unit volume to find the total power loss per unit area by integrating over the entire volume. Since we are dealing with a plane wave traveling in the positive z direction, the conductor would have a uniform cross-sectional area in the xy plane. Therefore, the shape of the conductor does not matter in this case.

For part c), the time average Poynting vector at any z can be found by taking the time average of the Poynting vector equation you provided. This would involve integrating over one period of the wave, which would result in the time average of the cosine term being zero. The remaining terms would give you the time average Poynting vector.

I hope this helps! Let me know if you have any further questions.