Non conservative electric field

[aim1732]

If an incomplete loop is placed in a time varying magnetic field that is radially symmetrical with a defined axis are charges accumulated at it's ends? I can positively say yes, but I am having difficulty calculating it.
If we invoke the generalised Ohm's Law
εnet = ∫E.dl + ∫ (v×B).dl + F_b.dl =i*R

For this case the current is zero hence the net emf is zero. The loop is fixed hence (v×B).dl is zero and there is no battery in the path so F_b is zero.. Hence ∫E.dl should be zero too. Since E_{non-conservative} is present hence the only possible way for the total term to be zero is for conservative field to be present---thus proving the presence of static charges.

But when i think of ways to evaluate this field,supposing the loop to be actually a straight wire I get stuck. I need to calculate E_nc.dl along a straight wire with the field lines curling all around.
Can anybody help?

 

[Andrew Mason]

If an incomplete loop is placed in a time varying magnetic field that is radially symmetrical with a defined axis are charges accumulated at it's ends? I can positively say yes, but I am having difficulty calculating it.
If we invoke the generalised Ohm's Law
εnet = ∫E.dl + ∫ (v×B).dl + F_b.dl =i*R

For this case the current is zero hence the net emf is zero. The loop is fixed hence (v×B).dl is zero and there is no battery in the path so F_b is zero.. Hence ∫E.dl should be zero too. Since E_{non-conservative} is present hence the only possible way for the total term to be zero is for conservative field to be present---thus proving the presence of static charges.

But when i think of ways to evaluate this field,supposing the loop to be actually a straight wire I get stuck. I need to calculate E_nc.dl along a straight wire with the field lines curling all around.
Can anybody help?

I am not familiar with this equation as a generalized Ohm's law. Can you provide a reference? But it must be equivalent to Faraday's law, which is:

$$\oint E\cdot dL = - \frac{d}{dt} \oint B\cdot dA$$

I am not clear on the direction of B relative to the loop. Can you clarify that?

Be careful. The emf does not depend upon the presence of a current. The fact that there is no current does not mean that there is no electric potential between the ends of the conducting loop.

AM

 

[aim1732]

I really have no reference. I have tried finding it before but no luck.My teacher calls it the generalized Ohm's Law. It seems to me that it is a restatement of Faraday's Law but is applicable on all kinds of conducting paths and not just loops.


There is no specified direction of B intended it is just a question that popped out of my mind. Any direction,not parallel to the plane of course, and maybe perpendicular to the plane for simplicity.

The last thing about emf. You say there can be a potential difference b/w the ends of an open loop --- agreed. I said that too. But I just checked the definition for it. It is work per unit charge to produce electric P.D.---read conservative electric field. So the E in the expression for Ohm's Law should actually be the nonconservative field.

 

[aim1732]

http://en.wikipedia.org/wiki/Electromotive_force"
Please check "Formal Definition of EMF".

 

[Andrew Mason]

I really have no reference. I have tried finding it before but no luck.My teacher calls it the generalized Ohm's Law. It seems to me that it is a restatement of Faraday's Law but is applicable on all kinds of conducting paths and not just loops. There is no specified direction of B intended it is just a question that popped out of my mind. Any direction,not parallel to the plane of course, and maybe perpendicular to the plane for simplicity.

The last thing about emf. You say there can be a potential difference b/w the ends of an open loop --- agreed. I said that too. But I just checked the definition for it. It is work per unit charge to produce electric P.D.---read conservative electric field. So the E in the expression for Ohm's Law should actually be the nonconservative field.

Are you referring to the field produced by induction? The integral $\int E\cdot dl$ around a closed path is non-zero if the path encloses a time dependent magnetic field. That field is often referred to as nonconservative but that is a bit of a misnomer, I think. It is an induced field. The only difference is that, unlike in a non-inductive circuit where the sum of all potentials = 0 ($\int E\cdot dl = 0$), in an inductive circuit it is $-d\phi/dt$.

AM

 

[aim1732]

I think the law I stated is not popular enough to be understood well. Actually I have never seen it in any of the books or sites I use. When I said net emf was zero I meant ∫E_nc.dl + ∫E_c.dl was zero across the length of the open loop(nc-induced field if you like)..That is like a battery because in steady state the battery force must exactly cancel the conservative E field such that their work/charge also cancels. But the battery still has emf--it is the work per charge of the external battery force or the negative of work/charge of the conservative E field.

Same token here. The induced E field is like the battery force.

Can you please clearly state the difference between induced and non-conservative field? I thought they were the same.

 

[Andrew Mason]

I think the law I stated is not popular enough to be understood well. Actually I have never seen it in any of the books or sites I use. When I said net emf was zero I meant ∫E_nc.dl + ∫E_c.dl was zero across the length of the open loop(nc-induced field if you like)..That is like a battery because in steady state the battery force must exactly cancel the conservative E field such that their work/charge also cancels. But the battery still has emf--it is the work per charge of the external battery force or the negative of work/charge of the conservative E field.

I am having trouble following you there. Where is the battery? Is it in the open loop?

Lets suppose first that there is no induced voltage in the conductor - there is just a battery - say 12 volts. That means there is a potential difference of 12 V from one end of the open loop to the other. If there is no circuit there is no current flowing. So what would cause the potential of a 12 volt battery connected to that open conductor to go from 12V to 0?

Now suppose that, in addition to the battery, there is a time dependent magnetic field through the plane of the open loop. That creates an induced electric field along the path of the open loop. Why would the two fields not simply add together? Why would they add to 0?

Now connect the ends of the loop and turn off the induced field. Now you have a circuit and current flowing and work being done. The potential of the battery pulls or pushes charges through the conductor causing the molecules in the metal conductor to become agitated by the interactions of the electric fields of the charges and atoms. So the electric potential of the battery + all the IR drops around the circuit sum to 0. $\oint E\cdot dl = 0$.

Finally, add a time dependent magnetic field through the plane of the open loop. That creates an induced electric field along the path of the open loop which adds to the previous fields. So now $\oint E\cdot dl = 0 - d\phi/dt \ne 0$.

Same token here. The induced E field is like the battery force.

Not quite. The difference is that $\oint E\cdot dl = 0$ for a circuit containing a battery and no induced field ie. $d/dt (\int B\cdot dA)= 0$ over the area enclosed by the circuit. That is the fundamental difference.

Can you please clearly state the difference between induced and non-conservative field? I thought they were the same.

I don't like the term "non-conservative" because it suggests that energy is being lost. It is not being lost. It is just that the work done on charges in an induced field is stored as magnetic potential energy rather than electrical potential energy. So, technically, it is not like a normal electric field (in which potential depends only on position, not path, so $\oint E\cdot dl = 0$). Since $\oint E\cdot dl \ne 0$ around that complete closed path, a charge going around that path and returning to its original position does not return to its original potential - so position does not determine potential. But that does not make it non-conservative. It just makes it different than a static electric field. The energy is conserved - always. Just in a different form if the field is an induced field.

AM

 

[aim1732]

So what would cause the potential of a 12 volt battery connected to that open conductor to go from 12V to 0?

That would be because there is an electric field inside the battery and outside it--the ∫E.dl around a loop of which will be zero.

Why would the two fields not simply add together? Why would they add to 0?

Yes they would. But I used the battery as an example. The original question did not have it. I used the example of an isolated battery to bring out a certain point.

The electric field produced by the battery and the induced field do not add to zero but they do produce a static electric field by causing charges to be accumulated at the ends of the ring.That is like a capacitor. That field stops further accumulation of charges and gives us a finite potential.

Not quite. The difference is that for a circuit containing a battery and no induced field ie. over the area enclosed by the circuit. That is the fundamental difference.

The ∫E.dl is -dφ/dt but that happens over a cycle. The loop is incomplete and hence no area or flux can be associated with the integral.

I am sorry but I have trouble putting my points across.

 

[Andrew Mason]

The electric field produced by the battery and the induced field do not add to zero but they do produce a static electric field by causing charges to be accumulated at the ends of the ring.That is like a capacitor. That field stops further accumulation of charges and gives us a finite potential.

The charges in the conductor, being free to move, distribute themselves so as to make the electric field inside the conductor 0. They do that whether an electric potential is produced by a battery or by inductance. So I would say that the electric field must be between the ends of the conductor, not in the conductor or between the battery terminals and the ends of the conductor.

The ∫E.dl is -dφ/dt but that happens over a cycle. The loop is incomplete and hence no area or flux can be associated with the integral.

But the closed path is still there. And there is time-dependent flux through the area enclosed by that path. It is just that part of the path includes a conductor. So I would say this means that there is no electric field where the conductor is placed so there must be between the ends of the conductor. Integrating around that path completely (ie the path includes conductor, battery, conductor and space between conductor ends), you would get $\oint E\cdot dl = V_{batt} - V_{batt} -d\phi/dt$

AM