Poynting theorem and electromagnetic density

[LCSphysicist]

Homework Statement

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Relevant Equations

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I am a little confused with the Poynting theorem https://en.wikipedia.org/wiki/Poynting's_theorem .
When we use this equation, the energy density that enters in $$\partial u / \partial t$$ is the one due only to the fields generated by charges/source itself? That is, if we have a magnetic field generated by a current varying in time, it will produce an induced electric field. So, $U$ will be $U = B^2 / 2 \mu$ or $U = B^2 / 2 \mu + \epsilon E^2 /2$?

I am asking because i was doing this exercise: "A time-dependent current, $I = I(t) = I_{0} t$, flows through the coils of an infinitely long, cylindrical solenoid. The solenoid has radius a and n turns per unit length." And i have noticed that the flux of the poyting vector will account only for the variation of the magnetic energy density.

 

[Orodruin]

Both the electric and magnetic fields are generated by the source. Maxwell’s equations are linear.

 

[LCSphysicist]

Both the electric and magnetic fields are generated by the source. Maxwell’s equations are linear.

With generated by the source i mean generated by charge and current density.

 

[Orodruin]

With generated by the source i mean generated by charge and current density.

Yes? As opposed to?

 

[LCSphysicist]

Yes? As opposed to?

I think i understand what you are talking about. But if so, why the electric energy density does not enter on the calculation $$\partial u / \partial t + \nabla . \vec S = 0 $$ inside the solenoid?

 

[Orodruin]

I think i understand what you are talking about. But if so, why the electric energy density does not enter on the calculation $$\partial u / \partial t + \nabla . \vec S = 0 $$ inside the solenoid?

Who says it doesn’t?

 

[LCSphysicist]

Who says it doesn’t?

$$B = \mu n i$$
$$2 \pi r E = \pi r^2 \partial B / \partial t$$
$$E = \mu r n \partial i/ \partial t / 2$$

$$U_b = \pi r^2 h (\mu n i)^2 /2 \mu$$
$$\partial U_b / \partial t = \frac{\pi r^2 h (\mu n)^2 i \partial i / \partial t}{\mu}$$

$$|S . da| = \frac{2 \pi r^2 h (\mu n )^2 i \partial i/ \partial t}{2 \mu} = \frac{\pi r^2 h (\mu n )^2 i \partial i/ \partial t}{\mu} $$

 

[LCSphysicist]

Never mind, just realized that it is because the current is linear on time... Oh god damn it