Potential difference defined in non-conservative electric field

Silver2007
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TL;DR Summary: There is a loop of wire placed in a changing magnetic field, there will be an induced electric field here. I am wondering if there can be a potential difference here?

Here is the problem:
A conducting square loop symmetrically encloses a solenoid of radius ##R##. The magnetic field inside the solenoid is directed into the page and increases linearly with time as: ##B = \alpha t##
Neglecting the magnetic field outside the solenoid, find the potential difference between points 1 and 2, which are equidistant from the corners of the square (see figure).

1750760725551.webp

$$\vec{E} = -\nabla V -\frac{\partial \vec{A}}{\partial t}$$
There will be a non-conservative electric field here, and I wonder if it is possible to have a conservative electric field and therefore a potential difference here?
 
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Why not use Faraday's law to find the electric field ##\vec E(\vec r)## and then do the line integral from 1 to 2?
 
Silver2007 said:
There will be a non-conservative electric field here, and I wonder if it is possible to have a conservative electric field and therefore a potential difference here?
The wording used in the question is potentially (pun intended) misleading and may be the source of confusion.

There will be an induced emf, ##\mathscr E##, in the wire loop. It seems you are being asked to find the emf which is generated in the section of wire between points 1 and 2.

The term ‘potential difference’ is generally used in the context of conservative fields so is probably not the best choice of terminology here.
 
kuruman said:
Why not use Faraday's law to find the electric field ##\vec E(\vec r)## and then do the line integral from 1 to 2?
Because the electric field is not conservative, the question is to find the potential difference between 1 and 2. So I wonder how to find the potential difference when there is no conservative electric field here
 
Steve4Physics said:
The wording used in the question is potentially (pun intended) misleading and may be the source of confusion.

There will be an induced emf, ##\mathscr E##, in the wire loop. It seems you are being asked to find the emf which is generated in the section of wire between points 1 and 2.

The term ‘potential difference’ is generally used in the context of conservative fields so is probably not the best choice of terminology here.
But the question wants to find the potential difference between 1 and 2, I just wonder if there can be a conserved electric field here?
1750821935287.webp

Here is the solution, personally I think it is impossible to define electric potential in this case, because the electric field is not conservative.
 
Silver2007 said:
Here is the solution, personally I think it is impossible to define electric potential in this case, because the electric field is not conservative.
I agree that the field is not conservative, but so what? Is there not an electric field at all points outside the solenoid that can be determined using Faraday's law? Is the potential difference (or the work per unit charge) not given by $$\Delta V=-\int_1^2\mathbf E \cdot d\mathbf l~?$$The analog in mechanics would be to calculate the work per unit mass done by friction when a box is dragged across a floor. It can be done even though the force of friction cannot be derived from a potential.
 
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Silver2007 said:
But the question wants to find the potential difference between 1 and 2, I just wonder if there can be a conserved electric field here?
EDIT/CORRECTION

The electric field here is non-conservative. But it doesn't matter.

If you connect a voltmeter between points 1 and 2 you will get a reading, ##\Delta V_{12}##. That's what you are being asked to find.


The above (struck) paragraph is wrong. Connecting a voltmeter will not give a meaningful reading. See @hutchphd's Post #8

Just use the term 'voltage' instead of 'potential difference'. Instead of '##\phi_1 - \phi_2##' use ##\Delta V_{12}##.
 
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Steve4Physics said:
If you connect a voltmeter between points 1 and 2 you will get a reading, ΔV12. That's what you are being asked to find.
But this number will depend upon the exact routing of the wires to the voltmeter (the line integral is path dependent). This is Prof. Walter Lewin's point in his controversial and badly received "lecture 16"
 
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hutchphd said:
But this number will depend upon the exact routing of the wires to the voltmeter (the line integral is path dependent). This is Prof. Walter Lewin's point in his controversial and badly received "lecture 16"
The video is 50+ minutes long and I don't have time to watch it yet - but will check it out later.

But I see your point about the problem with the (unavoidable) induced emf in the voltmeter wires.

Immediate thoughts are that the question (and official solution) are could be at fault. The setup seems analogous to a short-circuited battery: the emf equals the 'lost volts' so the 'terminal pd' is zero.

Edited.
 
  • #10
There will be some static charge distribution on the square loop when it is carrying current. The field of this charge “distorts” the induced electric field produced by the solenoid alone. Thus, some people like to think of the net electric field as a superposition of the field of the static charge and the induced field of the solenoid. $$\mathbf{E}_{\rm net}(\mathbf{r}) = \mathbf{E}_{\rm st \, ch}(\mathbf{r}) + \mathbf{E}_{\rm sol}(\mathbf{r}).$$

The field associated with the static charge is conservative. So you can define a potential function ##\phi## associated with this field and write $$\mathbf{E}_{\rm net}(\mathbf{r}) = -\nabla \phi(\mathbf{r}) + \mathbf{E}_{\rm sol}(\mathbf{r})$$

Integrating this along the wire from point 1 to point 2: $$\int_1^2 \mathbf{E}_{\rm net} \cdot \mathbf{dl} = \phi_1 - \phi_2 + \int_1^2 \mathbf{E}_{\rm sol} \cdot \mathbf{dl}.$$ The left side equals ##i r_{12}## where ##i## is the current and ##r_{12}## is the resistance of the wire segment 1-2. This is the equation in the solution written as ##ir_{12} = \phi_1 - \phi_2 + E_{12}##

Apparently, the “potential difference” asked for in this problem is the potential difference ##\phi_1 - \phi_2## associated with the static charge field.

I don’t think the potential difference ##\phi_1 - \phi_2## has much physical significance. What’s important is the net electric field, which is nonconservative. For example, if a quantity of charge ## q## flows in the wire from point 1 to point 2, the heat produced in this part of the circuit is ##q \int_1^2 \mathbf{E}_{\rm net}\cdot \mathbf{dl}##, where the integration is along the wire from 1 to 2. The heat does not equal the change in “electrostatic potential energy” ##q \Delta \phi##.
 
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Steve4Physics said:
The video is 50+ minutes long and I don't have time to watch it yet - but will check it out later.
You can skip to the last 11 minutes if you want. It's worth it.
 
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  • #12
hutchphd said:
This is Prof. Walter Lewin's point in his controversial and badly received "lecture 16"
Your talking about me, right?

FactChecker said:
You can skip to the last 11 minutes if you want. It's worth it.
Yes, it's a good lecture, but it's about 5x too long and parts are annoyingly dramatic.

The phrase "straw man" comes to mind. The dude is stuck on Kirchhoff in a context where it obviously doesn't apply. He could just say that clearly once and move on.

Kirchhoff's laws are a book keeping method for circuit representations. It's not the real world of EM. If he's going to pretend that what he drew was a "circuit" then he needs to put in a bunch of tiny coupled inductors. Then he could spend a whole bunch of time talking about circuit modeling of various EM scenarios, I guess.

Nearly always, you are either using some form of Maxwell's equations, or you are using circuit theorems like Kirchhoff, Miller, Thevenin, etc. They seldom mix well.

--- end of rant ---

PS: Sorry, let's not veer off onto an EE sort of circuit/Kirchhoff tangent. Neither that or Prof. Lewin are the point of this thread.
 
  • #13
DaveE said:
Your talking about me, right?
I don't think so. The comment was to "just put a Voltmeter on it to measure V12" This is exactly what Lewin was warning against. It is my understanding that folks at MIT accused him of faking the demo.....nontrivial turf wars.......so not just a failure of communication. It was a failure of comprehension IMHO. Needs to be reiterated whenever possible.
 
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