Flow of energy in a cylindrical conductor

[kent davidge]

Homework Statement

A cylindrical conductor with a circular cross section has a radius a and a resistivity ρ and carries a constant current I. What is the flow of energy into the volume occupied by a length l of the conductor? Discuss why the energy dissipated in a current carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

Homework Equations

no

The Attempt at a Solution

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(first of all, sorry my bad english)

I found it to be ρ I² l / π a²

Now my answer for why can we think the flow as entering on the conductor is that the field energy is the cause of the increase of the thermal energy, and thus it must be equal each other. Is it ok?

 

[haruspex]

I found it to be ρ I² l / π a²

Do you mean $\rho I^2l/(\pi a^2)$? If so, that does not have the right dimension for power.
Even if you don't mean that, it does not look right to me.
The second part of the question strikes me as strange. Is it a translation? The only thing that comes to mind is how the power is distributed along the length.

 

[DrClaude]

Homework Equations

no

Really?

 

[kent davidge]

DrClaude

haruspex, is this the second part you've mentioned? "Discuss why the energy dissipated in a current..." If so, it's how the problem was written by the author.
I'll show you how I found my answer for the flow.

The magnitude of the electric and magnetic fields are ρ I / (π a²) and μ I / (2 π a), respectivelly. The magnitude of the Poynting vector is the ratio of power (rate of flow energy) and area. Then, considering the length as L = 2 π a, I multiplied it by both the area and the intensity of energy (magnitude of the Poynting vector) to find the power.

 

[haruspex]

The magnitude of the electric and magnetic fields is ρ I / (π a²) and μ I / (2 π a), respectivelly.

What is $\rho$ in that expression? I don't see how resistivity will affect the electric field. I don't see the relevance of the magnetic field either.

. The magnitude of the Poynting vector is the ratio of power (rate of flow energy) and area. Then, considering the length as L = 2 π a, I multiplied it by both the area and the intensity of the energy (magnitude of the Poynting vector) to find the power.

You are asked for the flow of energy, not the flow per unit area.
L = 2 π a makes no sense. a is an area, and the length given is not a circumference.

Seems to me you are making the question far more complicated than it is. What is the resistance of the conductor?

 

[kent davidge]

No, a is the radius of the cylindrical conductor. But It's okay. Now I would like to know if my explanation about why the thermal and field energies were equal is correct.

 

[TSny]

Kent, I think it would have helped if in your first post you had made it clear that you were dealing with the Poynting vector. (Therefore, DrClaude's comment)

You did not show any of your work in the first post, so it was not evident how you got your expression for the rate at which field energy is flowing into the section of the conductor.

Does your answer agree with what you would calculate to be the rate at which energy is being dissipated inside the material due to the current flowing through a resistance?

 

[haruspex]

No, a is the radius of the cylindrical conductor.

Sorry, my mistake. But the rest of my comments stand.