Faraday's law -- circular loop with a triangle

[timetraveller123]

ok so in general to find out the electrostatic potential difference between two points i have to (ir) - induced emf in the branch is that right? where the i can be calculated using kirchhoff rules(faraday's law)

then the reading by voltmeter can be calculated using kirchhoff adding the magnetic flux if needed

 

[cnh1995]

Is there a disagreement here? Let's make sure we are considering the same setup. We have a circular ring with B field confined to the interior of the ring.

I think the voltmeter would read iR as rude man says, where R is the resistance of the green section. This would be true whether or not the ring has a uniform resistance per unit length.

Yes, that is correct. But when rude man said the voltmeter loop does not enclose any flux and when I repeated it in bold pink letters, I had this situation in mind, where the voltmeter probes are parallel to the B-field. Here, the voltmeter will read zero if the resistance of the ring is uniform.

I was going to edit it after reading #63, but somehow, I forgot it.
Thanks!

 

[cnh1995]

have to (ir) - induced emf in the branch is that right? where the i can be calculated using kirchhoff rules(faraday's law)

then the reading by voltmeter can be calculated using kirchhoff adding the magnetic flux if needed

Yes. But while considering the electrostatic voltage, you should interpret its sign correctly, since it aids the induced emf in some parts (higher resistance) and opposes the induced emf in other parts (lower resistance).

 

[TSny]

Yes, that is correct. But when rude man said the voltmeter loop does not enclose any flux and when I repeated it in bold pink letters, I had this situation in mind, where the voltmeter probes are parallel to the B-field. Here, the voltmeter will read zero if the resistance of the ring is uniform.
View attachment 213396

I was going to edit it after reading #63, but somehow, I forgot it.
Thanks!

OK. That's why I wanted to clarify if you and rude man were considering the same situation. It appears that there is a difference in interpretation of what is meant by the "voltmeter loop that encloses no flux." Rude man chooses the voltmeter loop to be a closed loop consisting of the meter, its leads, and part of the circular ring. Whereas, I believe your voltmeter loop doesn't need to include part of the ring. Is that right?

Can we consider one more example? Suppose we pick two points $a$ and $b$ on the ring which are separated by one quarter of the circumference. The ring is uniform in resistance.

I think Rude man's voltmeter loop is shown in Fig. 1. Do we all agree that the meter here would read $\varepsilon/4$ ,where $\varepsilon$ is the total induced emf in the ring?

I think your voltmeter loop would be as in Fig. 2, where the meter and leads lie in a plane perpendicular to the plane of the ring. What does the meter read for this configuration?

 

[cnh1995]

Rude man chooses the voltmeter loop to be a closed loop consisting of the meter, its leads, and part of the circular ring. Whereas, I believe your voltmeter loop doesn't need to include part of the ring. Is that right?

That's right.

I think Rude man's voltmeter loop is shown in Fig. 1. Do we all agree that the meter here would read ε/4ε/4\varepsilon/4 ,where εε\varepsilon is the total induced emf in the ring?

Yes.

I think your voltmeter loop would be as in Fig. 2, where the meter and leads lie in a plane perpendicular to the plane of the ring. What does the meter read for this configuration?

It should read the line integral along the staright line joining a and b. As per my calculations, it is (r²/2)dB/dt= ε/2π . Is that correct?

Anyway, it is NOT zero although the resistance of the ring is uniform and the voltmeter probes are in a perpendicular plane.

 

[cnh1995]

Am looking forward to the answer to Fig.2!

As per my calculations, it is (r2/2)dB/dt= ε/2π

Don't we get a finite amount of flux thru this loop area? The loop surface is some kind of weirdo 3-D thing isn't it? So B dot A ≠ 0?

I chose the loop which doesn't enclose any flux i.e. along the straight line joining a and b. So the voltmeter should read the line integral along this straight line path (since Vm+line integral along line segment ab=0, no flux enclosed).

 

[cnh1995]

Agreed. So you get zero volts.

No, for TSny's fig 2 in #74, the voltmeter reads ε/2π and not zero.

I chose the loop which doesn't enclose any flux i.e. along the straight line joining a and b. So the voltmeter should read the line integral along this straight line path (since Vm+line integral along line segment ab=0, no flux enclosed).

So Vm= line integral along the line segment joining a and b=ε/2π.

 

[cnh1995]

@TSny, suppose the resistance of this ring is uniform and the total emf induced in the ring is 10V, anticlockwise. Let's call the top and bottom terminals of the voltmeter V₂ and V₁ respectively.

So the voltmeter should read 5V, with V₂ -ve (top) and V₁ +ve (bottom).

Now, the "electrostatic voltage" between A and V₂ should not be zero even though the resistance of the wire connecting them is zero. Ditto for B and V₁.

Assuming the voltmeter doesn't draw any current, the "electrostatic potential" along the wire from B to A through the meter should look something like this.

It won't be all straight lines as the induced emf varies along the length of the wire, but this gives an idea about the electrostatic voltage gradient along the voltmeter probes.

Do you agree with this?

 

[cnh1995]

I see no other way to reconcile cnh's post 72 with what we (I hope) believe would be emf/2 with fig. 1 assuming a and b are 180 deg. apart.

They are 90 degrees apart as TSny has mentioned in #74.

 

[cnh1995]

So flux is non-zero in the surface of fig. 2 and there is an emf developed which will give a different reading in fig. 2 than in fig. 1.

Yes, in fig1, the voltmeter reads emf/4 while in fig2, it reads emf/2π.

 

[TSny]

That's right.

Yes.It should read the line integral along the staright line joining a and b. As per my calculations, it is (r²/2)dB/dt= ε/2π . Is that correct?

Anyway, it is NOT zero although the resistance of the ring is uniform and the voltmeter probes are in a perpendicular plane.

Yes. ε/(2π). Good.

 

[TSny]

Meanwhile it'll be bugging me too ... I wonder about the "loop" in fig. 2. Don't we get a finite amount of flux thru this loop area? The loop surface is some kind of weirdo 3-D thing isn't it?

Yes, if you close the voltmeter loop by going from b to a along the ring. And then
B dot A ≠ 0?

 

[TSny]

@TSny, suppose the resistance of this ring is uniform and the total emf induced in the ring is 10V, anticlockwise. Let's call the top and bottom terminals of the voltmeter V₂ and V₁ respectively.
View attachment 213406
So the voltmeter should read 5V, with V₂ -ve (top) and V₁ +ve (bottom).

Now, the "electrostatic voltage" between A and V₂ should not be zero even though the resistance of the wire connecting them is zero. Ditto for B and V₁.

Assuming the voltmeter doesn't draw any current, the "electrostatic potential" along the wire from B to A through the meter should look something like this.View attachment 213405
It won't be all straight lines as the induced emf varies along the length of the wire, but this gives an idea about the electrostatic voltage gradient along the voltmeter probes.

Do you agree with this?

Yes, I agree with all of this.

 

[TSny]

So VM reading is $\varepsilon$/4 - (dB/dt)(a²/4)(π-2)
= (dB/dt){(πa²/4 - (πa²/4)(π-2)}
= (dB/dt)(a²/2) = (dB/dt)πa²/2π = $\varepsilon$/2π
with a = radius.

Whew!

Yes, that looks good to me.

 

[timetraveller123]

sorry for the very very late reply my o levels were going on @cnh1995

Yes, that is correct. But when rude man said the voltmeter loop does not enclose any flux and when I repeated it in bold pink letters, I had this situation in mind, where the voltmeter probes are parallel to the B-field. Here, the voltmeter will read zero if the resistance of the ring is uniform.
View attachment 213396

I was going to edit it after reading #63, but somehow, I forgot it.
Thanks!

ok in this case there is no electrostatic voltage due to uniform resistance then to calculate voltmeter reading it is
there is (somewhat of not formally speaking) rise of $\frac{\epsilon}{2}$ and a drop of ir and ir also equals $\frac{\epsilon}{2}$ thus voltmeter reads am i right?

ok then as for problem by @TSny i understand figure 1 it is emf/4 then for figure 2 looking from on top
the area enclosed like what @rude man said is $\frac{a^2}{4}(\pi - 2)$ then the equations become
as what rude man i got the same result also
are my reasonings correct?

 

[cnh1995]

sorry for the very very late reply my o levels were going on @cnh1995

ok in this case there is no electrostatic voltage due to uniform resistance then to calculate voltmeter reading it is
there is (somewhat of not formally speaking) rise of $\frac{\epsilon}{2}$ and a drop of ir and ir also equals $\frac{\epsilon}{2}$ thus voltmeter reads am i right?

ok then as for problem by @TSny i understand figure 1 it is emf/4 then for figure 2 looking from on top
the area enclosed like what @rude man said is $\frac{a^2}{4}(\pi - 2)$ then the equations become
as what rude man i got the same result also
are my reasonings correct?

Yes, if you've got the same results as we did.

 

[timetraveller123]

@chn1995 thanks now returning to walter lewin problem

say the magnetic field is causing emf e in the clockwise direction

by symetry the induced emf along ab and cd will be the same

but to make the current same in the circuit electrostatic voltage will develop across a and b and c and d to even out the current

current induced in entire circuit is $\frac{e}{r_1 + r_2}$
assuming $r_2 > r_1$ then $v_B > v_A$ then the equation is
$\frac{v_B - v_A}{r_2} + \frac{e}{2 r_2} = \frac{e}{r_1 + r_2}$
where the voltmeter measures $v_B - v_A$
is this correct?

 

[cnh1995]

@chn1995 thanks now returning to walter lewin problem

View attachment 215129
say the magnetic field is causing emf e in the clockwise direction

by symetry the induced emf along ab and cd will be the same

but to make the current same in the circuit electrostatic voltage will develop across a and b and c and d to even out the current

current induced in entire circuit is $\frac{e}{r_1 + r_2}$
assuming $r_2 > r_1$ then $v_B > v_A$ then the equation is
$\frac{v_B - v_A}{r_2} + \frac{e}{2 r_2} = \frac{e}{r_1 + r_2}$
where the voltmeter measures $v_B - v_A$
is thius correct?

The voltmeter reads either ir₁ or ir₂ depending on where it is placed. It is not necessarily V_A-V_B.

 

[timetraveller123]

oh wait sorry i got mixed with voltmeter readings and electrostatic voltage
the first term is electrostatic voltage
the second term is em induced emf
the right hand side is net emf
is that correct

 

[mabilde]

Find the right answer on:
Greetings from Belgium, Cyriel Mabilde

 

[rude man]

I need to study the video at least 1-2 more times but it reinforces the interpretation I had before engaging in this lengthy thread. Thank you so much. We owe you a great debt of gratitude.

Perhaps you care to comment on another area of disagreement I have with Dr Lewin's lecture, where he presents a battery in series with a resistor. He claims there is an E field in the battery such that the integral of the E field around the loop is zero. I disagree and claim the net E field inside the battery, other than that accounted for by any internal resistance, is zero. Do you think it plausible that there are two equal and opposite E fields in the battery, one conservative and one non-conservative, canceling each other? (again ignoring internal resistance).

EDIT: by now I'm convinced this is the case. Proof: 2 fields, E_s and E_m. Contour integral of E_s is zero. Therefore a field E_s must exist in the battery, pointing from + to -. Also there must exist a field E_m since the battery is a source of emf, pointing - to + and of equal magnitude to E_s. That's QED.

(I don't pretend to have come to this conclusion by myself. Prof. H.H. Skilling of Stanford seems to state the same thing (indirectly) in his Fundamentals of Electric Waves (eq.185, p. 75, 2nd edition, 10th printing, John Wiley & Sons). Dr Skilling presents this equation as a prelude to electromagnetism where the E integral around a loop is of course non-zero. His obvious implication is that no source of emf by itself (e.g. without any internal resistance) has a net E field such that its own line integral is non-zero.

I find it inconceivable that Dr Skilling's textbook could have survived 9 printings and 14 years of wide use without someone having pointed out the error.)

 

[rude man]

Also Lewin's statement that "Kirchhoff is Wrong" has always puzzled me. Kirchhoff never talked about E fields, he talked only about voltage drops around any circuit = 0.

 

[Charles Link]

You may also want to see this thread: https://www.physicsforums.com/threads/walter-lewin-paradox.948122/#post-6003552 Instead of putting a voltmeter between the points A and B, you can replace the voltmeter by a large resistor, and compute the product of the current flow and the resistance. $\\$ Perhaps there is a shortcut, but it looks like a lengthy algebraic exercise. If you compute the current in the resistor between A and B, I think you could just take that current and multiply it by the resistor value, and that's what you get for a large resistor in parallel to it since the flux (and thereby the rate of flux change) between the large voltmeter resistor and this resistor $r_2$ is zero, if you put them directly parallel to each other. That greatly simplifies the algebra. You don't need to include the extra voltmeter resistor. $\\$ In any case, you will get a different answer if you put the voltmeter resistor parallel to $r_2$, (directly adjacent to it), or across the arc with resistor $r_1$. In general the answer will be path dependent.

 

[timetraveller123]

hi all this thread has cleared up a lot of doubts for me however this still some doubts i would like to clarify

how exactly does current flow?
like when it flows through the conductor there is still some non neglible resistance there which needs electric field to overcome i know the electric field in conductor is only zero when it is electrostatic but i don't understand what is different here
like when pd is applied across a conductor electron rush to negate the electric field but once they reach the end of conductor they sort of leak out of the conductor and another set of electrons comes to fill the space essentially setting up a current but at every instant there is electrons there to negate the pd so why would there be non zero electric field even when current is flowing
i am sorry if the question doesn't make sense i am quite confused and not even sure of what i don't understand thanks

 

[rude man]

It may be remarked that the Mabilde setup does not measure potentials directly. Rather, by virtue of putting the meter leads inside the B flux, a secondary emf equal to the measured potential is generated, the attendant potential drop around the meter loop being what the voltmeter measures. The data is correct since the area inscribed by the coil section plus the meter leads is the same fraction of the total coil flux area as the fraction of the potential around the coil. The setup is therefore not empirical proof of the existence of a potential drop over the measured coil segment. Also, his description of the coupling between the coil and the meter circuit is misleading, as can be readily seen by rearranging the meter leads in one of his diagrams, which would alter the degree of coupling. Cf. my recent Insight tutorial on Dr. Lewin's Conundrum.

 

[rude man]

Just to wrap up the OP’S original post::

Let
d1 = length of one arc (=2πa/3),
d2 = length of one side of the triangle,
S = area A-B,

The strategy is to irradiate (apply the B field to) the triangle area A-B-C and area A-B separately, then vectorially add the various E_s and E_m (static and emf) fields by superposition. The other segments, not abutting the A-B loop, can be ignored altogether; whether or not they are irradiated does not affect the E fields in the two A-B segments.

We first note that when just the triangle area is irradiated. there is no E_s field. This means that the triangle area can also be ignored since potential differences are line-integrated E_s fields only.

So we irradiate S only:
If E_m = emf field around the A-B loop and i the corresponding current,
emf = E_m(d1 + d2) = -S dB/dt = i(r1 + r2) so we know E_m and i;

and if
E_s1 = static field along arc AB,
E_s2 = static field along the triangle side AB,

from the basic relation j = σE, j = current density, E = E_m + E_s, we get

i = (E_m - E_s1)d1/r1 or
E_s1 = (1/d1)(E_m d1 - i r1) = E_m - i r1/d1

and similarly
E_s2 = r2 i/d2 - E_m.
Then ΔV = E_s1 d1 = E_s2 d2.
The rest is geometry and substitution.

Note that this computation is not that of a voltmeter (by all we have discussed so far) measuring voltage between A and B. The voltmeter reading is in fact VB - VA = -r1 i.

EDIT: I forgot that the bottom side of the triangle has R=2r2, not r2. That unfortunately invalidates the computation. May look into it further but it looks bad.

 

[timetraveller123]

Just to wrap up the OP’S original post::
View attachment 231017
Let
d1 = length of one arc (=2πa/3),
d2 = length of one side of the triangle,
S = area A-B,

The strategy is to irradiate (apply the B field to) the triangle area A-B-C and area A-B separately, then vectorially add the various E_s and E_m (static and emf) fields by superposition. The other segments, not abutting the A-B loop, can be ignored altogether; whether or not they are irradiated does not affect the E fields in the two A-B segments.

We first note that when just the triangle area is irradiated. there is no E_s field. This means that the triangle area can also be ignored since potential differences are line-integrated E_s fields only.

So we irradiate S only:
If E_m = emf field around the A-B loop and i the corresponding current,
emf = E_m(d1 + d2) = -S dB/dt = i(r1 + r2) so we know E_m and i;

and if
E_s1 = static field along arc AB,
E_s2 = static field along the triangle side AB,

from the basic relation j = σE, j = current density, E = E_m + E_s, we get

i = (E_m - E_s1)d1/r1 or
E_s1 = (1/d1)(E_m d1 - i r1) = E_m - i r1/d1

and similarly
E_s2 = r2 i/d2 - E_m.
Then ΔV = E_s1 d1 = E_s2 d2.
The rest is geometry and substitution.

Note that this computation is not that of a voltmeter (by all we have discussed so far) measuring voltage between A and B. The voltmeter reading is in fact VB - VA = -r1 i.

thanks for reply
i guess i was highly coufused on how actually the placement of voltmeter could affect the reading and the buildup of charge in the circuit now i get it thanks to everyone involved in this very lengthy thread
sorry for the late reply i hadn't logged in for a long time

 

[rude man]

You started quite a rolling ball didn't you! I'm glad you did because it got me to revisit and re-learn old matrial myself. Lots of fun!
PS did you ever get an answer from your instructor?

 

[timetraveller123]

You started quite a rolling ball

apparently so another similar post to this also was very several pages long
https://www.physicsforums.com/threads/induced-emf-between-two-points.926298/

but so much more of this makes more sense now that i have learned a bit of vector calculus

did you ever get an answer from your instructor?

i got it from a regional physics Olympiad paper and apparently according to my current teacher no one solved it completely during that competition

thank you for you continued interest in physics forum(i find this far superior to stack exchange ;) ) many of those in this thread have helped in many of my previous threads including you @rudeman

 

[rude man]

Good for you for persevering! That put you around 80% above the others who ask for homework help, IMO.

The key thing I'd like to impart here is the importance of understanding that there are two kinds of E fields, namely emf-generated and electrostatic. It's very confusing to analyze this kind of problem without that understanding.

If you want to pursue this further you may want to read my recent Insight article:
https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/

Keep it up!