Is the electric field inside a circuit non-conservative?

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SUMMARY

The discussion clarifies that the electric field inside a circuit is not conservative due to the presence of electromotive force (emf) which increases the energy of charge carriers as they complete a loop. While the electric field in conductive wires can be considered conservative (with a line integral of zero), the emf source introduces a non-conservative element. The conversation emphasizes that in circuit theory, the electric field is not directly dealt with, as it requires knowledge of physical dimensions and voltage changes across circuit elements. Energy conservation in circuits is maintained through the work done by external forces, rather than the electric field itself.

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  • Understanding of circuit theory and its limitations
  • Familiarity with electromotive force (emf) and its role in circuits
  • Knowledge of Faraday's Law of electromagnetic induction
  • Basic concepts of conservative and non-conservative forces
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Nikitin
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As charge moves around a circuit, their energy increases by each loop.

However, I just heard somebody say the total electric field inside a circuit is conservative.

How is that possible? Is the field conservative or not? Why?

Thanks for all help! :)
 
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Wires are conductors, so within them yes- the line integral of the electric field is 0. But within the emf, this is not true, so the field is not conservative there.
 
In circuit theory you don't deal directly with electric fields, so I don't know how this question arises. In circuit theory the length of a wire or the physical size of a lumped element is unknown. The E-field is the spatial gradient of the voltage, so to know it you would have to know the physical size of the circuit element, as well as the voltage change acorss it. So the electric field is simply something that you do not need and cannot determine within the constraints of circuit theory.

However, energy is conserved in circuit theory. The work done on a circuit is always equal to the work done by the circuit plus the energy stored in the circuit.
 
You can answer this very generally: Whenever and wherever there is a time-dependent magnetic field the electric field cannot be a potential field, because then and there according to Faraday's Law one has
\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.
For a field to have a scalar potential, i.e. to have
\vec{E}=-\vec{\nabla} \vec \Phi,
its curl necessarily must vanish, which is not the case when a time-dependent magnetic field is present.
 
So let's say you have a circuit without any magnetic field inside to speak of. Then the E-field inside the circuit is defined as conservative??

However, how do you then explain the emf increasing the energy of every charge going a loop through the circuit?
 
Nikitin said:
However, how do you then explain the emf increasing the energy of every charge going a loop through the circuit?
What needs to be explained about it?
 
isn't the emf-source part of the circuit and its electric field?
 
Yes. Although, as I mentioned earlier the fields are outside the scope of circuit theory.
 
I'm a complete noob (1st year uni) so don't overestimate me: How can a field be conservative if it does work on a particle going one loop around?
 
  • #10
Note that inside the source charge carriers are pushed "uphill" against the electric field, so there must be some other force (chemical, mechanical etc) acting on them. This force does all the work.
 
  • #11
ohhh, so it's not an electrical field which does the work?

Allright, then I get it. thanks for all the help, guys!
 
  • #12
Nikitin said:
I'm a complete noob (1st year uni) so don't overestimate me: How can a field be conservative if it does work on a particle going one loop around?
Think of another conservative force, like gravity. If water goes downhill (e.g. a river) then gravity does work on the water. The water can be used to do work on other things (e.g. a turbine). Then, if you want to continue the process you have to pump the water back up the hill.
 

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