Faraday's law -- circular loop with a triangle

[cnh1995]

I suspect the problem with the above is that the wire is also generating em

Exactly!

My thinking here was as follows: without the triangle we all agree the voltage between any 2 points along the circle = 0. So running a wire between those two points can't change voltages anywhere.

I agree with this part (voltage="electrostatic" voltage).

However, *instead of putting a triangle inside, if you just joined A and B, everything will change. ^#But if there is an equilateral triangle inscribed in the circle with all the sides having "equal" resistances, then again the electrostatic voltage will be zero.

Basically, KCL has to be satisfied at every node.
In the former case (*), in order to satisfy KCL at A and B, currents through the two parts of circumference would be unequal, because some of the current would flow through the wire joining A and B (Shape of this wire matters too.)
In the latter case(#), currents through all three sides of the equilateral triangle will be equal and currents through all three arcs of the circumference would also be equal and KCL will be satisfied at every node without developing any electrostatic voltage.

In OP's problem, the sides of the triangle have unequal resistances, hence, there is an electrostatic voltage present.

 

[cnh1995]

wait why is that so

See if this helps.
https://www.physicsforums.com/posts/5859178/
The page I've linked is from an excellent book. See if you can buy it if you want to learn circuits at a deeper, intuitive level.

 

[timetraveller123]

@cnh1995
sir thanks for the recomendations but right now it would be helpful if you could clear up just some doubts
so induced emf is like a non conservative battery
and
the potential difference between two points like in this question is the current in the branch * resistance of the branch
so in this question v_ab = I₁R₂
is these right
and you said when the resistances are the same in triangle there is potential diference between A and B is it because due to symetry
thanks for the help

 

[cnh1995]

the potential difference between two points like in this question is the current in the branch * resistance of the branch
so in this question vab = I1R2

No. I₁R₂ gives the "net" emf in branch AB.
This net emf has two components: one is electromagnetically induced emf and the other is electrostatic voltage developed because of surface charges. This electrostatic voltage is the potential difference between A and B. It is same for all the paths from A to B. Concept of potential can be applied only in electrostatic case.

 

[timetraveller123]

ok now it makes sense
just last doubt
in principle i could have gone through all the possible loops to calculate the net emf as well right?
and when you say electromagnetically induced emf it means kA_o/3 right?

 

[cnh1995]

in principle i could have gone through all the possible loops to calculate the net emf as well right?

Yes.

when you say electromagnetically induced emf it means kAo/3 right?

Yes, Ao is the area of the triangle.

 

[timetraveller123]

ok sir if i have this following circuit the

say the magnetic field is increasing into the page and the right resistor in r1 and left resistor is r2 there will be current in the clockwise direction induced after that how to solve the problem

 

[cnh1995]

ok sir if i have this following circuit the

I am 22 and I just graduated. No need to address me as 'sir'.

say the magnetic field is increasing into the page and the right resistor in r1

The current will be anticlockwise.
Let's take only one voltmeter and connect it first on the left and then on the right.
(Two voltmeters at the same time might make the problem complicated. I'll have to think about it.)
You can write KVL equation for the rectangular loop as follows: I(r1+r2)=AdB/dt.
This will give you the current I.

About the voltmeter reading:
If the voltmeter is placed on the left,
IR1+V_left=AdB/dt
So,
V_left=AdB/dt-IR₁

Similarly, if the voltmeter is placed on the right,
IR₂+V_right=AdB/dt

V_right=AdB/dt-IR₂.

Note that the polarities of the two voltmeter readings will be opposite. Can you say why?

 

[cnh1995]

Check out this video.
The exact same problem is discussed here by Prof Lewin.
.

 

[timetraveller123]

I am 22 and I just graduated. No need to address me as 'sir'.

no formalism but i am a indian i just address those who teach me as sir
actually i am coming from that video to here

Note that the polarities of the two voltmeter readings will be opposite. Can you say why?

wait why is it because of the different paths i have no clue
is this concept supposed to be hard or am i not getting it?

my question why can't the same reasoning be applied to a circuit with the magnetic field replaced with a battery i still can go in loops like this

 

[timetraveller123]

@rude man
what i am having trouble with is understanding all these terms and how they relate to each other
so if have the triangle without the circle all with same resistances then the emf induced in each of the legs will kA/3 the current through the triangle is kA/3r hence with this information how can i calculate the potential difference between two vertex

what does net emf mean ?
do i really have to know vector calculus to undestand all of this
thanks sir

 

[cnh1995]

do i really have to know vector calculus to undestand all of this

No, but having some intution about the line integral of E.dl will certainly help.

what does net emf mean ?

The line integral of E_net.dl, where E_net is the resultant of the induced electric field (non-conservative) and electrostatic field (conservative).

what i am having trouble with is understanding all these terms and how they relate to each other

Ok.
Maybe you should try a similar but simpler problem first.
The rate of change of flux through the following circular loop is 3V. Resistances of the green half and red half are 1 ohm and 2 ohm respectively.
What is the "potential difference" between A and B?

 

[cnh1995]

Assuming the voltmeter loop doesn't enclose any flux..

I told you the voltmeter reading between any two points along the circle without the triangle is zero. That was incorrect. If you hook up the voltmeter so that the leads and meter form a loop which does not encircle any B field, then the voltmeter reading would be iR where i = πa⋅dB/dt and R the resistance between the two points.

This would be true only if the circle does NOT have a uniform resistance, which is not the case with OP's circuit. The resistance of the circle in OP's diagram is uniform (3r₁) and hence, the voltmeter will read zero between any two points (without the triangle inscribed in the circle).

Edit: @rude man, I believe I have misinterpreted your setup. What you wrote is actually correct. See #72.

 

[cnh1995]

it is a triangle then wouldn't the magnitude of electric fields change along the triangle

Yes, but the integral of E.dl along the three sides will be the same, thanks to the symmetry of equilateral triangle. It is the line integral that you should be interested in, and not the actual electric field.

This is interesting! We have agreed that the line integral of E.dl along the three sides is same. It turns out that although the electric field at every point on the triangle is different in magnitude, its component "along" the side of the triangle is same at every point. You can see it by drawing concentric field lines. The electric field at some points is larger in magnitude, but its angle with the side is also bigger, so the components of the fields along the sides are equal at every point.

 

[timetraveller123]

This is interesting! We have agreed that the line integral of E.dl along the three sides is same. It turns out that although the electric field at every point on the triangle is different in magnitude, its component "along" the side of the triangle is same at every point. You can see it by drawing concentric field lines. The electric field at some points is larger in magnitude, but its angle with the side is also bigger, so the components of the fields along the sides are equal at every point.

yeah that part i got it when you related it to symetry

i was recently reading the other thread and you (@cnh1995) said given a circle with two halves of different resistances then the emf induced will be the same in both halves but due to different resistance the current due to induced emf will be different in the halves hence a electrostatic voltage is developed to even out the current in both the halves this made a whole lot of sense
is this so as to make the current in both halves the same?

now the question you gave me

emf induced along green loop = emf along red loop = 1.5 v
current through the loop = 1A
then how is the voltmeter connected ? doesn't the answer depend on that

 

[cnh1995]

is this so as to make the current in both halves the same?

Yes.

then how is the voltmeter connected ? doesn't the answer depend on that

There is no voltmeter. I asked for "potential difference", which means the electrostatic voltage between A and B. It is path independent.

Voltmeter reading will depend on how the meter is connected and you can work it out using the KVL equation for the voltmeter loop.

 

[timetraveller123]

ok then net emf along green is iR = 1v
induced emf = 1.5 v
so electrostatic voltage is -0.5v for green

for red it is 2-1.5 = 0.5v is this correct sir ?

 

[cnh1995]

so electrostatic voltage is -0.5v for green

for red it is 2-1.5 = 0.5v is this correct sir ?

It should be same for both the paths (or for all the paths joining A and B).

Assume the induced emf to be anticlockwise. What is the potential difference between A and B?

 

[timetraveller123]

okay is it

for red:
V_a -(i r₁ - ε/2) = v_b
for green :
V_a +(i r₂ - ε/2) = V_b
both gets -0.5v is that correct?

 

[cnh1995]

okay is it

for red:
V_a -(i r₁ - ε/2) = v_b
for green :
V_a +(i r₂ - ε/2) = V_b
both gets -0.5v is that correct?

Vb-Va=0.5V, means Vb is at 0.5V higher potential than Va. You can see the electrostatic voltage adds to the induced emf in 2 ohm and subtracts from the induced emf in 1 ohm.

 

[cnh1995]

If the voltmeter loop is to the left of the circle such that there is no B field anywhere inside the loop then the reading will be VB - VA = 1A x 1Ω = +1V, clockwise current assumed.

If the voltmeter loop is to the right of the circle such that there is no B field anywhere inside the loop then the reading will be VA - VB = 1A x 2Ω = +2V, clockwise current assumed.

Note the difference in polarity of the two voltages. B-A in one, A-B in the other.
@cnh1995 did I get that right?

Yes.

 

[timetraveller123]

wait so the potential between a and b is 0.5 v

@rude man
i thought we were using a anti clockwise current
so

how can i write the kvl for this voltmeter loop

 

[TSny]

@TSny
i am getting these are these correct
$I_1 = \frac{k(6A_1 - 1.5A_0 - 2A)}{12r_2}\\ I_3 = \frac{k(6A_1 + 1.5 A_o + 2A)}{-18r_2}$

Yes

 

[TSny]

If you hook up the voltmeter so that the leads and meter form a loop which does not encircle any B field, then the voltmeter reading would be iR where i = πa⋅dB/dt and R the resistance between the two points. Remember: the loop cannot encircle any B field. If the loop did include any B field you should expect the readings to be different and not just iR.

Assuming the voltmeter loop doesn't enclose any flux..

This would be true only if the circle does NOT have a uniform resistance, which is not the case with OP's circuit. The resistance of the circle in OP's diagram is uniform (3r₁) and hence, the voltmeter will read zero between any two points (without the triangle inscribed in the circle).

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.

 

[timetraveller123]

The problem as stated in post 1 would generate a clockwise current. But in your post 49 you changed the sign of dB/dt so now with that change yes the current is counterclockwise.

oh @rude man i think we are talking about different questions i think i have my initial problem figured out now i am talking about the sample problem cnh 1995 gave me in post 55

 

[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.