Poynting vector in static electromagnetic field

AI Thread Summary
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
Messages
5
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
Back
Top