Poynting vector in static electromagnetic field

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In a static electromagnetic field, both electric and magnetic fields can exist simultaneously, leading to a non-zero Poynting vector, S=ExH, indicating energy flow despite the energy density remaining constant (dU/dt=0). This situation arises when a static magnetic field is generated by a constant electric current, which implies energy is being transferred into matter due to resistance. The conservation of energy is expressed through the equation ∂u/∂t + ∇·S = -J·E, clarifying the relationship between energy density and flux. The concept of "hidden momentum" is mentioned but not elaborated upon, with a recommendation to consult Jackson's book on Electrodynamics for further understanding. Static fields indeed exhibit energy flux, confirming the initial confusion.
zql
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There is a situation, we have an electric field and a magnetic field, both are static. And we know the density of energy is u=E·D/2+B·H/2, so dU/dt=0, but Poynting vector S=ExH is not zero, which means energy is flowing. This confused me. Static field also has energy flux?
 
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Yes, correct. Look also for "hidden momentum".
 
can you give me more details?thanks.
 
zql said:
There is a situation, we have an electric field and a magnetic field, both are static. And we know the density of energy is u=E·D/2+B·H/2, so dU/dt=0, but Poynting vector S=ExH is not zero, which means energy is flowing. This confused me. Static field also has energy flux?
The static magnetic field is produced by a constant electric current. That means there is resistance, and energy is flowing into matter. "Hidden" momentum is not involved.
 
zql said:
There is a situation, we have an electric field and a magnetic field, both are static. And we know the density of energy is u=E·D/2+B·H/2, so dU/dt=0, but Poynting vector S=ExH is not zero, which means energy is flowing. This confused me. Static field also has energy flux?

The relevant expression, from the conservation of energy, is:

\frac{\partial u}{\partial t} + \nabla \bullet S = -J \bullet E.

Does this help?
 
Andy Resnick said:
The relevant expression, from the conservation of energy, is:

\frac{\partial u}{\partial t} + \nabla \bullet S = -J \bullet E.

Does this help?

I think I got it, thanks.
 
And how about "hidden momentum"? I didn't know about this.
 
Probably the best source on these matters is still Jackson's book on Electrodynamics.
 

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