Is the electric field inside a circuit non-conservative?

In summary, the electric field is not always conservative in a circuit, because the curl of the field must vanish for it to be conservative.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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
[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.[/tex]
For a field to have a scalar potential, i.e. to have
[tex]\vec{E}=-\vec{\nabla} \vec \Phi,[/tex]
its curl necessarily must vanish, which is not the case when a time-dependent magnetic field is present.
 
  • #5
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?
 
  • #6
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?
 
  • #7
isn't the emf-source part of the circuit and its electric field?
 
  • #8
Yes. Although, as I mentioned earlier the fields are outside the scope of circuit theory.
 
  • #9
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.
 

1. What is a non-conservative electric field?

A non-conservative electric field is one in which the electric potential energy of a charge changes when it moves from one point to another within the field. This means that the work done by the electric field on a charge is path dependent, and the amount of work done will differ for different paths taken by the charge.

2. How can I tell if the electric field inside a circuit is non-conservative?

To determine if the electric field inside a circuit is non-conservative, you can calculate the work done by the electric field on a charge as it moves around a closed loop within the circuit. If the amount of work done is not equal to zero, then the electric field is non-conservative.

3. What causes the electric field inside a circuit to be non-conservative?

The presence of non-conservative forces, such as friction or resistance, can cause the electric field inside a circuit to be non-conservative. These forces dissipate energy and contribute to the change in electric potential energy of a charge as it moves within the circuit.

4. Does a non-conservative electric field violate the law of conservation of energy?

No, a non-conservative electric field does not violate the law of conservation of energy. While the work done by the electric field on a charge may not be equal to zero, the overall energy in the system remains constant. The energy dissipated by non-conservative forces is converted into other forms, such as heat or light.

5. Can a circuit with a non-conservative electric field still function properly?

Yes, a circuit with a non-conservative electric field can still function properly. Non-conservative forces may cause some energy to be lost, but as long as the circuit is designed to account for this energy loss, it can still function as intended. In fact, many electronic devices, such as computers and smartphones, rely on non-conservative electric fields to function.

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