• B
A former colleague asked me a simple question about Faraday's law. As shown below, the configuration is just a circular conductor and a changing B field, and then tried to measure the voltage across the diameter. But I am not sure what the answer is?

Any help would be greatly appreciated.

Last edited:
Delta2

jtbell
Mentor
I am not sure what the answer is?
Nevertheless, what are your thoughts, uncertainties, etc., about it? Then people here can address them directly, and try to clear up any errors in your thinking, instead of simply giving you the answer (which is frowned upon here).

hutchphd
Homework Helper
You nowhere say exactly how the B field is changing in time (uniformly?).
When you have a changing field like this it is no longer a "circuits' problem and you need to consider the fields explicitly. The reading on the voltmeter will depend upon the exact path of the wires, not just their topology. Are you assuming perfect symmetry? So make an analysis.
This is of course Walter Lewin's famous headscratcher.

Nevertheless, what are your thoughts, uncertainties, etc., about it? Then people here can address them directly, and try to clear up any errors in your thinking, instead of simply giving you the answer (which is frowned upon here).

My idea is that from the front view, there are two loops connected to the voltmeter, the EMF induced on the right side seems to be +0.5V, and the EMF induced on the left side seems to be -0.5V. There seems to be a contradiction, maybe there is no contradiction at all. I think it is possible to add two values. So the answer should be that the maximum chance is 0V.

Is my inference correct?

You nowhere say exactly how the B field is changing in time (uniformly?).
When you have a changing field like this it is no longer a "circuits' problem and you need to consider the fields explicitly. The reading on the voltmeter will depend upon the exact path of the wires, not just their topology. Are you assuming perfect symmetry? So make an analysis.
This is of course Walter Lewin's famous headscratcher.
You make a lot of sense. Let's assume that the B field changes linearly with time, so the induced E field is a constant value. The path of the voltmeter connection is shown in the diagram, which is vertically to the plane of the circular conductor and located behind it. In this way relative to the plane of two parallel connecting lines of the voltmeter, the left and right are completely symmetrical. Therefore, the connection line of the voltmeter should be perpendicular to the induced electric field, which means that no EMF will be generated on them.

In addition, of course the resistivity of each point of the circular conductor is also the same, and the internal resistance of the voltmeter is infinite, etc.

Last edited:
vanhees71
Gold Member
Of course it depends on the geometry of the wires connected to the volt meter. You don't have a voltage here but an EMF, because the electric field is not conservative since ##\vec{\nabla} \times \vec{E}=-\partial_t \vec{B} \neq 0##. The volt meter in fact measures a (usually very small) current going through it. So you measure the EMF as calculated along the wires connecting the volt meter.

Delta2, hutchphd and alan123hk
hutchphd
Homework Helper
So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.

Delta2 and alan123hk
Of course it depends on the geometry of the wires connected to the volt meter. You don't have a voltage here but an EMF, because the electric field is not conservative since ∇→×E→=−∂tB→≠0. The volt meter in fact measures a (usually very small) current going through it. So you measure the EMF as calculated along the wires connecting the volt meter

So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.
Your explanation is very logical and reasonable. I think you may not intend to explain some details for the time being, but your explanation has no flaws anyway.

Delta2
Homework Helper
Gold Member
So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.
I, on the other hand, believe that even in the case of asymmetry (that is even if we put the voltmeter leads anywhere along the circumference of the conductor) we will have again a zero reading, because there is no charge separation happening along the circumference of the conductor. A voltmeter measures indeed a current according to my opinion too, but this current must be a result of a scalar potential which is result of charge separation.

alan123hk
hutchphd
Homework Helper
But the definition of a scalar potential assumes ##\nabla \times \vec E =0## so I don't know what scalar potential you mean. The internal workings of the Voltmeter in fact work by comparing magnetic force of an unknown current to a that of a spring

vanhees71
Gold Member
alan123hk and hutchphd
I personally tried to think more deeply about this problem, and believe that now I have found a very simple way to describe and predict the behavior of such circuits

I, on the other hand, believe that even in the case of asymmetry (that is even if we put the voltmeter leads anywhere along the circumference of the conductor) we will have again a zero reading, because there is no charge separation happening along the circumference of the conductor. A voltmeter measures indeed a current according to my opinion too, but this current must be a result of a scalar potential which is result of charge separation.
I think you made the very crucial point, that is, in fact, there is no charge separation happening along the circumference of the conductor, so from the perspective of the external circuit, there is no potential difference along the circumference of the circular conductor. This means that as long as the connecting wire of the voltmeter does not picked up the induced electric field generated by the changing magnetic field, regardless of the position of the connecting wire, the voltage measured by the voltmeter is still zero.

Delta2
hutchphd
Homework Helper
I would like to comment but I have no idea what you are saying:
This means that as long as the connecting wire of the voltmeter does not picked up the induced electric field generated by the changing magnetic field, regardless of the position of the connecting wire,
The Princeton citation from @vanhees71 is above is pretty comprehensive. Have at it.

vanhees71 and alan123hk
Delta2
Homework Helper
Gold Member
Well the paper states that
"The fact that the two meters give different readings when a = c and b = d (in sec. 2.1) indicates that the meter readings are not simply proportional to the electric scalar potential V at those points, as would be the case for a DC circuit"
I might be wrong in what i said in post #9 afterall.

Delta2
Homework Helper
Gold Member
However the experiment setup is a bit different from what is discussed here so i am not completely sure.

I would like to comment but I have no idea what you are saying:
Sorry, my expression may not be clear enough, or I did not express my thoughts correctly.

Referring to the example I put forward on #1, the induced electric field generated by the changing B field will not only be concentrated on the circular conductor, but will also extend to all surrounding spaces, including the two connecting wires of the voltmeter.

What I mean is that since the circular conductor does not exert a potential difference on the two connecting wires, as long as the force of the induced electric field (generated by the changing B) acting on the charges on the two connecting wires does not generate a voltage across the voltmeter, the measured voltage on the voltmeter must be zero.

Last edited:
Delta2
After rethinking this issue, I think that there is no paradox.

We should not treat a and b are exactly same, even if there is no voltage drop due to Ohm's law, there does exist an induced electric field between them, so they are indeed two different points.

As long as we define the Electromotive force (EMF) and PD (potential difference) as follows,

Electromotive force, ##~EMF_{ab} =−\int_a^b E \cdot dl ##
Potential difference, ##~PD_{ab} = \int_a^b E \cdot dl ##

and put the EMF (created by the induced electric field) into the equations, the correct calculation result can be obtained.

For example, we can easily prove that if the resistance of each point of the circular conductor is the same, there is indeed no potential difference along the circumference of the conductor.

Total length ## ~ L=\sum L_i ##
Total EMF ## ~ \mathcal E=\sum \mathcal E_i=\sum C_1L_i##
Total Resistance ## ~R= \sum R_i= \sum C_2L_i ##

where ## ~C_1## and ## C_2 ~## are constants

For any arc length ##~ L_i ~## on the circumference, the actual potential difference should be : -

$$\mathcal E_i - \frac {\sum \mathcal E_i} {\sum R_i} R_i$$
$$C_1 Li - \frac {\sum C_1L_i} { \sum C_2L_i } ~ C_2L_i ~~ = ~ C_1 Li - \frac {C_1} { C_2 } ~ C_2L_i =~0$$

vanhees71
Gold Member
Well the paper states that
"The fact that the two meters give different readings when a = c and b = d (in sec. 2.1) indicates that the meter readings are not simply proportional to the electric scalar potential V at those points, as would be the case for a DC circuit"
I might be wrong in what i said in post #9 afterall.
Sigh, well, I forgot that this imprecision is in this text. Please bear in mind that there is NO POTENTIAL in this case, because (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B} \neq 0$$
and that's the reason why the meters in Lewin's example show different readings. What the (idealized) volt meter measures is the EMF (NOT VOLTAGE!) along the wires. Note that it's a CLOSED line integral:
$$\mathcal{E}=\int_{\text{wires}} \mathrm{d} \vec{r} \cdot \vec{E}=-\dot{\Phi}_{\vec{B}},$$
where ##\Phi_{\vec{B}}## is the magnetic flux through an arbitrary surface with the wires making its boundary (assuming that the wires are at rest).

Delta2
Homework Helper
Gold Member
@vanhees71 the scalar potential is defined even in the case of ##\nabla\times\vec{E}\neq\vec{0}##, simply by the solution to the wave equation (in the lorentz gauge)$$\nabla^2 V-\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\epsilon_0}$$ but of course as you notice the voltmeter doesn't measure that (it usually does but not in this weird setup).

Last edited:
hutchphd
Homework Helper
What the (idealized) volt meter measures is the EMF (NOT VOLTAGE!) along the wires. Note that it's a CLOSED line integral:
I believe that part of the confusion in the OP is from the geometry of the Voltmeter as the nexus of two such closed curves of wire. For me it is easier to understand the measurement result as the sum of two opposite induced currents (but I have never been on really friendly terms with the EMF).

vanhees71
vanhees71
Gold Member
@vanhees71 the scalar potential is defined even in the case of ##\nabla\times\vec{E}\neq\vec{0}##, simply by the solution to the wave equation (in the lorentz gauge)$$\nabla^2 V-\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\epsilon_0}$$ but of course as you notice the voltmeter doesn't measure that (it usually does but not in this weird setup).
That's of course true. I misunderstood the statement to mean a static potential such that ##\vec{E}=-\vec{\nabla} V##, which is of course wrong in the general case. Here the relations of the em. field to the potentials is (in SI units)
$$\vec{E}=-\vec{\nabla} V-\partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
In the Lorenz gauge
$$\Box V=\frac{1}{\epsilon_0} \rho, \quad \Box \vec{A}=\mu_0 \vec{j},$$
where the d'Alembert operator is defined as
$$\Box=\frac{1}{c^2} \partial_t^2 -\Delta.$$

etotheipi and Delta2
Homework Helper
Gold Member
2020 Award
The voltmeter is perhaps best modeled as a large resistor, with the voltage reading given by the product of the current and the voltage, as the resistance approaches infinity. The solution is to write out the loop equations, how Professor Lewin does, making sure to include any EMF from the changing magnetic flux, if it occurs inside the loop.
The location of the resistor of the voltmeter, along with the connecting leads, will affect what the voltmeter reads, rather than being able to assign a reading by just considering the two points in the circuit where the leads are attached.

vanhees71, etotheipi and Delta2
I would like to express my views on the issue of non-conservative electric fields and potential differences.

Although the induced electric field is non-conservative, but for the fixed path and direction in the electric field, such as R1 and R2 shown in the figure below, there can indeed be a potential difference, because their voltage will respectively vary with their resistance, and they consume energy. But, of course, the placement and direction of the two voltmeters and their leads may affect the measured value.

Consider again the transformer shown in the figure, the secondary output voltage is also produced by the non-conservative electric field induced by changing magnetic field, therefore, its corresponding line integral of the electric field should only be along the path of the secondary coiled wire. If the line integral follows other paths, of course there may be different results.

In addition, We can see that the placement and direction of the secondary external output circuit loop has no (or extremely little) impact on the output voltage. The reason is obvious, because for the same magnetic flux , it only has one turn, but the number of turns inside the transformer is usually much larger than one turn.

Delta2