(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

There's a cylindrical resistor of length L, radius a and resistivity [tex]\rho[/tex] 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, [tex]R=\frac{\rho L}{\pi a^2}[/tex].

Poynting vector is defined as [tex]\vec S =\frac{\vec E}{ \mu _0} \times \vec B[/tex].

So I must calculate the E and B field over the surface of the cylinder.

I start with the E field: [tex]\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}}[/tex]. But the problem is that I don't have the value of [tex]Q _{\text{enclosed}}[/tex]. I'm stuck on this. I know that [tex]i=\frac{dq}{dt}[/tex], so [tex]q(t)=\int _{t_0}^{t_1} idt[/tex]. i is constant, so q is a matter of time.

Hmm, I made another attempt: [tex]\oint \vec J \cdot d\vec A=i \Rightarrow J=\frac{i}{\pi a^2}[/tex]. J is the current density. So... is [tex]Q _{\text{enclosed}}=JL[/tex]? I don't think so... I'm really stuck on this.

I calculated the B field to be [tex]\frac{\mu _0 I}{2\pi a}[/tex].

Dealing with vectors, if [tex]\hat E =\hat z[/tex], [tex]\hat B=\hat j[/tex], then [tex]\hat S=\hat i[/tex], which answers question 2).

For question 3), I have to show that [tex]EB=\frac{i^2\rho L}{\pi a^2}[/tex]. As I already have B, I can get E by this method, but it's cheating.

So E would be worth [tex]\frac{2\rho L i}{\mu _0 a}[/tex] and Q would be worth [tex]\frac{L^2 i a \rho}{2k}[/tex], 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.

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# Homework Help: Poynting vector over a cylinder

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