# Homework Help: Poynting vector over a cylinder

1. Feb 11, 2010

### fluidistic

1. The problem statement, all variables and given/known data
There's a cylindrical resistor of length L, radius a and resistivity $$\rho$$ in which a current i flows.
1)Calculate the Poynting vector over the surface of the resistor.
2)In which direction does the energy that is transported by the electromagnetic field point.
3)Show that the rate P at which flows this energy over the cylindrical surface (given by the Poynting vector) is equal to the rate of which the thermal energy is generated, namely i²R.
2. Relevant equations
None given.

3. The attempt at a solution

By Ohm's law, $$R=\frac{\rho L}{\pi a^2}$$.
Poynting vector is defined as $$\vec S =\frac{\vec E}{ \mu _0} \times \vec B$$.
So I must calculate the E and B field over the surface of the cylinder.
I start with the E field: $$\oint \vec E d\vec A =4 \pi k Q _{\text{enclosed}} \Rightarrow E \cdot \pi a^2 L=4 \pi k Q _{\text{enclosed}}$$. But the problem is that I don't have the value of $$Q _{\text{enclosed}}$$. I'm stuck on this. I know that $$i=\frac{dq}{dt}$$, so $$q(t)=\int _{t_0}^{t_1} idt$$. i is constant, so q is a matter of time.
Hmm, I made another attempt: $$\oint \vec J \cdot d\vec A=i \Rightarrow J=\frac{i}{\pi a^2}$$. J is the current density. So... is $$Q _{\text{enclosed}}=JL$$? I don't think so... I'm really stuck on this.

I calculated the B field to be $$\frac{\mu _0 I}{2\pi a}$$.
Dealing with vectors, if $$\hat E =\hat z$$, $$\hat B=\hat j$$, then $$\hat S=\hat i$$, which answers question 2).

For question 3), I have to show that $$EB=\frac{i^2\rho L}{\pi a^2}$$. As I already have B, I can get E by this method, but it's cheating.
So E would be worth $$\frac{2\rho L i}{\mu _0 a}$$ and Q would be worth $$\frac{L^2 i a \rho}{2k}$$, assuming I didn't make an error while calculating B.
I've absolutely no clue about how to reach the answer. Any help is greatly appreciated.

2. Feb 11, 2010

### vela

Staff Emeritus
Actually, the total flux through the cylinder would be 0. When a current flows, there's no net charge in the resistor. The current is just the result of the already present charges moving. It's like water in a pipe. The amount of water in the pipe doesn't change when the water is flowing. The flow of water is just a result of the water moving.

The electric field in the resistor is external, due to the rest of the circuit. What you know is that there is a potential difference V going from one end of the resistor to the other, a distance of L. If you assume the E field is uniform, you have E=V/L.

3. Feb 11, 2010

### fluidistic

I still have the sensation that there are free charges within the resistor (the one moving that creates the current) and that this number is fixed since i is constant through the resistor. I understand that there is the same amount of free charges entering the resistor and leaving it, and so this number of free charge is constant, but it's not worth 0.

Ah... I think I'm getting you. Despite the number of free charges, there's still no net charge. So Q=0... And hence, as you said, the E field inside the conductor exists because of the external E field. I wouldn't have figure this myself. Thanks a lot.
Following your tip, V=-iR. I already wrote down what R is so I should be able to go further.
I thank you very much for your help. I'm extremely stressed, tomorrow is the final exam which is worth 100% of my grade in this course...
Any further problem I encounter, I'll let you know.

4. Feb 11, 2010

### fluidistic

I still don't reach the answer, but I'm not that far.
$$E=-\frac{i\rho}{\pi a^2}$$.

$$S=-\frac{i^2 \rho}{2\pi ^2 a^3}$$. But I think I should find $$i^2R$$, thus $$\frac{i^2 \rho L}{\pi a^2}$$ according to question 3).
Do you see any error?

5. Feb 11, 2010

### vela

Staff Emeritus
The Poynting vector is the rate of energy transfer per unit area. You need to calculate the total power over the surface of the cylinder.

6. Feb 11, 2010

### fluidistic

Ah thanks! So I just have to multiply my answer of S by $$2\pi a L$$, which indeed reach the desired result.
Thank you very much! The problem is solved.