What Defines Induced EMF in Circuit Analysis?

[Amplitude]

Homework Statement

Homework Equations

The Attempt at a Solution

First part is easy . Area of the triangular loop is A = (1/2)(2R)(R) = R²

Induced EMF in loop ABC = Rate of change of flux = dΦ/dt = AdB/dt

= R²(dB/dt) which is correct .

Now the second part has stumped me .

I would really appreciate if somebody could help me with this .

Thank you .

 

[TSny]

My interpretation of part (ii) is that it is asking for the magnitude of the line integral $\int \mathbf E \cdot \mathbf {dl}$ along the straight line AB.

If so, then use the result from (i) along with symmetry arguments.

 

[Amplitude]

Thanks for replying .

The induced electric field lines will be concentric circles . So I can't see how line integral can be calculated .

How would we use symmetry in this ?

 

[TSny]

What's the value of the line integral for the entire triangular circuit?

What's the value of the line integral for the horizontal segment AC?

 

[Amplitude]

You made it look so easy :) . I have been thinking about it for last couple of days .

What's the value of the line integral for the entire triangular circuit?

R²(dB/dt)

What's the value of the line integral for the horizontal segment AC?

As we move along from C to A along CA , direction of induced electric field would be perpendicular at every point of the path CA as CA is a a diameter and induced electric field lines would be concentric circles .The line integral would be zero along CA .

Since field lines are circles , the paths AB and AC would look similar and their line integral would be equal .

$\int \mathbf E \cdot \mathbf {dl}$ along the closed loop ABC =
$\int \mathbf E \cdot \mathbf {dl}$ along the straight line CA + $\int \mathbf E \cdot \mathbf {dl}$ along the straight line AB +$\int \mathbf E \cdot \mathbf {dl}$ along the straight line BC .

$\int \mathbf E \cdot \mathbf {dl}$ along the straight line AB =
R²(dB/2dt) .

Is that alright ?

 

[TSny]

Yes, that looks right to me.

 

[Amplitude]

Thank you . Please clarify some points .

1) Is KVL applicable in a closed circuit in the presence of induced EMF ? It does seem to apply in closed circuit ABC in this problem but one of the books I have suggests otherwise .It states that when electric field is non conservative Kirchoff's Voltage Law no longer holds true .

2) In the branch AB what does the product iR_AB signify ? It surely isn't the potential difference between points A and B . i²R_AB is the power dissipated in wire AB .But what does iR_AB tell us ?

3) The line integral $\int \mathbf E \cdot \mathbf {dl}$ along the straight line AC is zero whereas the line integral along AB and BC is non zero . Is it something like having a battery in each of the branches AB and BC , but no battery and only resistance in branch AC ? Current is determined by the net EMF of the loop ?

4) Is there an induced electric field line also on the circumference of radius R , i.e on the solenoid ?

 

[cnh1995]

Is KVL applicable in a closed circuit in the presence of induced EMF ? It does seem to apply in closed circuit ABC in this problem but one of the books I have suggests otherwise .It states that when electric field is non conservative Kirchoff's Voltage Law no longer holds true .

You'll have to modify the equation. In absence of a varying B-field, the closed loop integral of E.dl (in any loop) is zero. If there's a varying B-field, closed loop integral of E.dl is equal dΦ/dt, where Φ is the magnetic flux through that loop.

In the branch AB what does the product iRAB signify ?

It is the resultant emf in branch AB i.e.sum of electrostatic emf(conservative) and induced emf (non-conservative).

The line integral ∫E⋅dl∫E⋅dl\int \mathbf E \cdot \mathbf {dl} along the straight line AC is zero whereas the line integral along AB and BC is non zero . Is it something like having a battery in each of the branches AB and BC , but no battery and only resistance in branch AC ? Current is determined by the net EMF of the loop ?

Yes.

Is there an induced electric field line also on the circumference of radius R , i.e on the solenoid ?

Yes.

 

[Amplitude]

It is the resultant emf in branch AB i.e.sum of electrostatic emf(conservative) and induced emf (non-conservative).

Is there an electrostatic EMF in branch AB ? I think there is only non conservative EMF along the entire loop ABC .

 

[TSny]

Is there an electrostatic EMF in branch AB ? I think there is only non conservative EMF along the entire loop ABC .

If the branches BC and CA are removed so that only AB is sitting there, then charge will be induced on the surface of AB to create a static E field that cancels the induced E field everywhere inside AB. This will happen almost instantaneously. So, E_net = 0 inside AB. Thus, no work would be required to move a test charge from A to B inside the branch AB. In that sense, there would be zero "net emf" between AB. (@cnh1995, please correct me if this is wrong.)

That's one of the reasons I wasn't sure of the interpretation of part (ii). I interpreted it to mean: find the line integral of the nonconservative field between A and B for the straight path from A to B. I interpreted it that way because I thought it made an interesting question.

 

[Amplitude]

I interpreted it to mean: find the line integral of the nonconservative field between A and B for the straight path from A to B.

Does the answer obtained in post 5 depend on whether BC and AC are present or not ?

Why did the question asked to remove BC and AC ?

 

[TSny]

If you interpret (ii) as asking for the line integral of the nonconservative field along path AB then the answer does not depend on whether or not BC and AC are removed. But, maybe my interpretation of what (ii) is asking for is incorrect.

 

[Amplitude]

Your interpretation must be correct as the answer obtained matches with the answer given in the book

Do you think KVL can be applied in a closed loop where induced EMF is present irrespective of whether DC battery is present or not ?

 

[TSny]

Your interpretation must be correct as the answer obtained matches with the answer given in the book

OK, good.

Do you think KVL can be applied in a closed loop where induced EMF is present irrespective of whether DC battery is present or not ?

Yes. The induced EMF in the loop just gets added in with the DC battery emf. That's esssentially what you are doing when you use emf = -LdI/dt for an inductor in the loop.

This is worth watching:

 

[cnh1995]

So, Enet = 0 inside AB

Wouldn't that depend on the resistances of AB and the circumference?
I have worked it out using some numerical values for dB/dt, radius and resistances of the segments and I am not getting zero net field inside AB.
If you have time, I will post it here. (It's a bit lengthy to follow).

 

[TSny]

@cnh1995, Here's how I'm thinking about it. With sections BC and CA removed, you just have an isolated piece of wire AB that is sitting in the induced E field of the solenoid. This E field will induce a complicated surface charge density on AB such that the net field is zero everywhere inside AB. It's like putting a chunk of metal between the charged plates of a capacitor. Induced surface charge on the chunk will create an E field such that the total field is zero at every point inside the chunk.

 

[cnh1995]

Induced surface charge on the chunk will create an E field such that the total field is zero at every point inside the chunk.

Here's how I thought about it: The induced charges will create an electric field in all the three parts (two arcs and segment AB), such that the sum of currents through smaller arc and AB is equal to the current through the larger arc. The field due to induced charges will depend upon the resistances of the segments.
Does that sound correct?

 

[TSny]

I interpret the circular arcs as representing the current in the windings of the solenoid. The circuit that we are interested in is the triangle ABCA. In (ii) this is dismantled with BC and CA removed so that just AB is left.

 

[cnh1995]

Here's my working with some actual numbers.

Let's assume radius=1. So, the area of the circle= 3.1415 unit, area of the smaller loop=0.2854 unit (π/4-1/2) and area of the bigger loop=2.8561 unit.

Assume dB/dt=5.
So, emf induced in the entire circle=AdB/dt=3.1415×5=15.70V.

Emf induced in the bigger loop=(2.8561/3.1415)*15.70=14.2736V

Emf induced in AB =R²/2*dB/dt=5/2= 2.5V (from A to B).

∴Emf induced along the bigger arc= 14.2736-2.5= 11.7732V. (B to A)

Emf induced in the smaller loop=15.70- 14.2736=1.4264V

∴Emf induced along smaller arc= 2.5+1.426= 3.9268V (A to B).

Now for each loop, these values satisfy closed loop ∫E.dl= dΦ/dt.

Is this correct so far?

 

[TSny]

I agree with your numbers. I'm interested in where you are going with this .

Edit: Added figure for reference

 

[cnh1995]

I agree with your numbers. I'm interested in where you are going with this .

Ok.
Let's assume the resistance of the entire circumference=1 ohm and resistance of AB=1 ohm as well.
Therefore, resistance of smaller arc=0.25 ohm and resistance of bigger arc=0.75 ohm.
Current in the circuit is counterclockwise and there is current division at points A and B, such that
Current through AB+current through smaller arc=current through bigger arc...(1)

Now, assume point B at an electrostatic potential V w.r.t. A.
Using KCL at point B (refer 1), we get
(2.5-V)/1 + (3.9268-V)/0.25= (11.7732+V)/0.75.

Solving this, we get V=0.3970V, from B to A i.e. B is at 0.3970V w.r.t. A.

As you can see, the net field inside AB is not zero and current flows from A to B.

Did I miss something here?

 

[TSny]

We are interpreting the problem differently. I was taking the large outer circle as representing the solenoid and I was not thinking of it as part of the circuit in which there is an induced current. I was taking only the triangular circuit as the circuit with an induced current. So, in part (ii), the only part of the circuit left was the straight piece of wire AB. This piece of wire is an open circuit and can't carry current. So, total E is zero inside the piece of wire.

But you are taking the outer circle to be part of the circuit. After removing BC and CA, you have a closed circuit consisting of the outer circle and the segment AB. So, you have a circuit with a "larger loop", "smaller loop", etc.

That makes an interesting circuit to think about.

Let's assume the resistance of the entire circumference=1 ohm and resistance of AB=1 ohm as well.
Therefore, resistance of smaller arc=0.25 ohm and resistance of bigger arc=0.75 ohm.
Current in the circuit is counterclockwise and there is current division at points A and B, such that
Current through AB+current through smaller arc=current through bigger arc...(1)

OK. I think I'm with you up to here.

Now, assume point B at an electrostatic potential V w.r.t. A.

As you know, this is a dangerous assumption when we have a nonconservative E field.

Using KCL at point B (refer 1), we get
(2.5-V)/1 + (3.9268-V)/0.25= (11.7732+V)/0.75.

I don't think it's correct to say I_ABR_AB = 2.5 V for the wire AB. (See your first term in the equation quoted above.) From part (i), we know that the integral of the induced E field of the solenoid along the straight path A to B is 2.5 V. But I_ABR_AB will be the integral of the net E field along the straight path from A to B. The net field inside AB will include contributions from induced charges at various parts of the circuit.

Solving this, we get V=0.3970V, from B to A i.e. B is at 0.3970V w.r.t. A.

Does this contradict your use of 2.5 V in the first term of your KCL equation?

If my calculations are correct, the value of I_ABR_AB would be about 2.1 V instead of 2.5 V. To get this I used two loop equations (for voltages) and one junction equation (for currents).

 

[cnh1995]

If my calculations are correct, the value of IABRAB would be about 2.1 V instead of 2.5 V.

It is 2.1V in my calculations as well.
See the green term in my equation.

(2.5-V)/1 + (3.9268-V)/0.25= (11.7732+V)/0.75.

Induced emf along AB is 2.5V and V is 0.397V. That means, the net emf or i_ABR_AB=2.5-0.397=2.1V...

 

[TSny]

OK. I was misreading your expressions. I was going too quickly and interpreted 2.5-V as 2.5 Volts rather than a subtraction of 2.5 and V. Sorry about that. Glad we actually agree.

 

[Amplitude]

TSny ,

I came across a much simpler question than this which has left me a bit puzzled .

Induced EMF ε across circular loop = Rate of change of flux = πR²B₀

Current i flowing in the loop = ε/r . r is the resistance of the loop .

Resistance of semicircular wire AB = r/2 .

∫Edl between A and B along the circumference = ε/2

Now how do I calculate potential difference between A and B ?

Is Potential difference between A and B = (Induced EMF between A and B) - (iR_AB) ?

The two terms are equal resulting in 0 potential difference .

I am quite perplexed by this result .

 

[cnh1995]

Now how do I calculate potential difference between

It is zero. There is no electrostatic voltage present because of the uniform resistance of the ring. The electrostatic potential difference appears because of non-uniform resistance and the fact that KCL has to be satisfied at every point. Here, the resistance is uniform, so there is no electrostatic voltage.

 

[Amplitude]

It is zero. There is no electrostatic voltage present because of the uniform resistance of the ring. The electrostatic potential difference appears because of non-uniform resistance and the fact that KCL has to be satisfied at every point. Here, the resistance is uniform, so there is no electrostatic voltage.

Sorry , I find this line of reasoning quite unconvincing .

What has uniformity/non-uniformity of resistance got to do with the presence of voltage ?

Does @TSny agree with the above post ?

 

[TSny]

@Amplitude. cnh1995 posted a reply while I was constructing a reply. I’ll go ahead and post my thoughts with the warning that the way I define “emf for a path” in the remarks below is not the same as cnh1995’s “electrostatic voltage” that he refers to.

Let me start with some basic remarks just to clear the way for addressing your question. I apologize if you’re already familiar with this.

When you have a changing B field, it doesn’t really make sense to ask for the potential difference between the points A and B.

In electrostatics with no changing B field, you have a potential function V such that the work needed to move a unit charge from A to B is independent of path and equal to V_B – V_A. The concept of potential difference between A and B is meaningful.

In the situation you have posed, the work needed to move a unit charge from A to B cannot be written as the difference in value of a potential function V. Such a function does not exist in this case because the work needed depends on the path from A to B. (I guess you could try to separate out the part of the E field that is due to static charge and define a potential function associated with that field, but I don't see any use in doing that.)

Now, even when there is a changing B field, there is a value of the net E field at each point of space. So, for a particular path from A to B, there is a well-defined value of the line integral of E_net. I don’t see anything wrong with calling the negative of this integral “the emf for this path”. The emf will be path dependent. I guess you could call this emf, “the potential difference for the path”. But that’s kind of misleading since it does not represent the difference in a potential function.

OK, with that out of the way, how do we interpret the problem you have stated in post #25 where it asks for the “potential difference between two diametrically opposite points A and B”? Once you pick a path between A and B, you can ask for the emf of this path (defined as the negative of the line integral of E_net along the path). But the answer will depend on the path. If you take the path to be the diameter of the circle from A to B, then you will get zero for the emf of this path.

 

[Amplitude]

Thanks a lot !

I agree with you . It doesn't make sense to ask for potential difference but that is the way questions are asked .

What will a voltmeter read when put across A and B ?

 

[cnh1995]

What has uniformity/non-uniformity of resistance got to do with the presence of voltage ?

Suppose half of the ring has a resistance R and othe other half has a resistance 2R.
"Induced" elevtric field in the ring will be same at every point, regardless of the resistance. So, for both the halves, the emf induced will be E/2.
Now, compute the currents through both the halves assuming just the "induced" field as the driving quantity. The currents are not same in both the halves. To make the currents same in both the halves, there is an electrostatic voltage developed in the circuits such that it increases the net electric field in the part of high resistance (2R) and decreases the net electric field in the part of lower resistance (R), so that the currents through both the parts become equal.

Check out your original problem worked out with some actual numbers in #21.

 

[TSny]

What will a voltmeter read when put across A and B ?

The reading of the voltmeter will depend on how the meter is placed between A and B. If it is placed in the center of the loop with the leads of the meter going straight to points A and B, the meter will read zero. See left figure below.

If the meter is placed as shown on the right, the voltmeter will not read zero. The reading will depend on the shaded area (assuming the region of magnetic field encompasses this area).

 

[cnh1995]

What will a voltmeter read when put across A and B ?

Zero, assuming the leads are not inside the magnetic field.
Edit: Posted a moment after TSny..

 

[Amplitude]

The reading of the voltmeter will depend on how the meter is placed between A and B. If it is placed in the center of the loop with the leads of the meter going straight to points A and B, the meter will read zero. See left figure below.

If the meter is placed as shown on the right, the voltmeter will not read zero. The reading will depend on the shaded area (assuming the region of magnetic field encompasses this area).
View attachment 212763

Does a voltmeter read ∫Edl along the path from A to B through it ?

If so , how does it read the line integral ?

In a DC circuit a voltmeter measures the potential difference between two points which in turn is directly proportional to the current flowing through the voltmeter ( a Galvanometer converted into a voltmeter ) .

 

[cnh1995]

Does a voltmeter read ∫Edl along the path from A to B through it ?

If so , how does it read the line integral ?

It doesn't. It reads the eletrostatic potential difference which is caused by surface charges. In circuits with no varying B-field, this potential difference is equal to the line integral of E.dl along that path (or along any path joining the two points, since the field is conservative).
In this circuit, there is a non-conservative field present and hence, the line integral of E.dl is not same as the electrostatic potential difference here.

 

[Amplitude]

It doesn't. It reads the eletrostatic potential difference which is caused by surface charges. In circuits with no varying B-field, this potential difference is equal to the line integral of E.dl along that path (or along any path joining the two points, since the field is conservative).
In this circuit, there is a non-conservative field present and hence, the line integral of E.dl is not same as the electrostatic potential difference here.

If the voltmeter doesn't read the ∫Edl then how does it read 0 in the left figure ?

Isn't ∫Edl = 0 along the diameter ?

 

[TSny]

Does a voltmeter read ∫Edl along the path from A to B through it?

Yes, I believe so.

If so , how does it read the line integral ?

In a DC circuit a voltmeter measures the potential difference between two points which in turn is directly proportional to the current flowing through the voltmeter ( a Galvanometer converted into a voltmeter ) .

In both cases a very small current flows through the meter which is proportional to the line integral of E_net through the meter.

 

[cnh1995]

Isn't ∫Edl = 0 along the diameter

It is zero because E=0 "along" the diameter.

It doesn't. It reads the eletrostatic potential difference which is caused by surface charges.

This is assuming the voltmeter loop is not inside the magnetic field. If it is in the magnetic field, the current (or net E-field) through the voltmeter will be due to the sum of electrostatic and induced fields, hence, it will read the integral E.dl. through the meter.

 

[TSny]

I found the following paper by Robert Romer to be very helpful
http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf

 

[Amplitude]

Yes, I believe so

Thanks .

In both cases a very small current flows through the meter which is proportional to the line integral of E_net through the meter.

Is E_net purely induced electric field ?

In the question I posted in post#25 (no voltmeter present ) I hope there is no conservative electric field present .

But in the two pictures you have posted where there is a voltmeter , is conservative electric field also present along with induced electric field ?

If so , what is the source of this field ? Which charges are responsible for this conservative electric field ?

 

[TSny]

Is E_net purely induced electric field ?

No, not in general. $E_{net}$ at a point of space is the total field due to all sources. As I understand it, there is never any need to worry about how much of this field is "induced" and how much is "electrostatic" (e.g., due to charge build up on elements of the circuit).

In the question I posted in post#25 (no voltmeter present ) I hope there is no conservative electric field present .

But in the two pictures you have posted where there is a voltmeter , is conservative electric field also present along with induced electric field ?

If so , what is the source of this field ? Which charges are responsible for this conservative electric field ?

If the region of changing magnetic field is circular, but the loop is placed "off center" as shown, then I think there would be a complicated, nonuniform electrostatic charge on the loop.

But there is no need to worry about this. Only the net E field matters. In the law $\oint \mathbf E \cdot \mathbf {dl} = -\frac{d\Phi}{dt}$ , $E$ is the net field.

 

[cnh1995]

Is Enet purely induced electric field ?

No, it is the sum of induced electric field (non-coulomb) and electrostatic (coulomb) field.

In the question I posted in post#25 (no voltmeter present ) I hope there is no conservative electric field present .

Right.

But in the two pictures you have posted where there is a voltmeter , is conservative electric field also present along with induced electric field ?

Yes, because the voltmeter loop has different resistances in all the parts.

If so , what is the source of this field ? Which charges are responsible for this conservative electric field ?

The surface charges that result from different resistances of the parts.

Suppose half of the ring has a resistance R and othe other half has a resistance 2R.
"Induced" elevtric field in the ring will be same at every point, regardless of the resistance. So, for both the halves, the emf induced will be E/2.
Now, compute the currents through both the halves assuming just the "induced" field as the driving quantity. The currents are not same in both the halves. To make the currents same in both the halves, there is an electrostatic voltage developed in the circuits such that it increases the net electric field in the part of high resistance (2R) and decreases the net electric field in the part of lower resistance (R), so that the currents through both the parts become equal.

 

[cnh1995]

@Amplitude, you might want to read this book. It explains circuit theory at electronic level.
https://books.google.co.in/books?id...ved=0ahUKEwjdsbeoh-nWAhWIPo8KHcCJCGcQ6AEIJDAA
(Wait a few seconds if the page doesn't load.)

 

[Amplitude]

The current through the voltmeter is 0 .Can I prove it by Kirchoff's Voltage Law ?

Call the leftmost point on the circle L and rightmost point on the circle R i.e ALB and ARB are semicircles.

In the left loop ALB , assume a current i₁ flowing anticlockwise . In the right loop ARB , assume a current i₂ flowing anticlockwise .

∫Edl in the closed loop ALB = ∫Edl along semicircle ALB = EπR = ε/2 .

ε is total induced EMF in the circular loop .r_v is resistance of middle branch having voltmeter .r is the total resistance of the circular loop .

In the left semicircular loop ALB , applying KVL , +ε/2 -i₁r/2 -i₁r_v =0

i₁= ε/(r+2r_v)

Similarly in the right semicircular loop ARB , applying KVL , +ε/2 -i₂r/2 -i₂r_v =0

i₂= ε/(r+2r_v)

Current in the middle branch = i₁-i₂=0

@TSny , Is that valid ?

 

[TSny]

In the left semicircular loop ALB , applying KVL , +ε/2 -i₁r/2 -i₁r_v =0

i₁= ε/(r+2r_v)

Similarly in the right semicircular loop ARB , applying KVL , +ε/2 -i₂r/2 -i₂r_v =0

i₂= ε/(r+2r_v)

Current in the middle branch = i₁-i₂=0

@TSny , Is that valid ?

I agree with the result that there is no current through the voltmeter (middle branch). But I don't agree with your result for the current in the loop. What information would you get by applying KVL to the entire loop?

I've never particularly liked the method of solving circuits where you superimpose currents for different loops. I like to assign a current to each branch. Then solve for the currents using the junction rule and the loop rule. Thus, assign a current i_L for the left side of the loop, i_R for the right side of the loop, and i_V for the current through the voltmeter.

 

[Amplitude]

But I don't agree with your result for the current in the loop. What information would you get by applying KVL to the entire loop?

Sorry . I do not understand what you are objecting to .

Is it an improper application of KVL or is it that KVL itself is not applicable ? But I thought you yourself cleared in an earlier post that KVL is valid in circuits where induced EMF's are present .

Or you didn't find superposition of currents appropriate . But I think it is valid while solving for currents in the circuit .

 

[TSny]

Since there is no current through the middle branch, you just have one current i_o going around the loop. So, for the loop as a whole you have i_o r = ε. Or, i_o = ε/r. Does this agree with your result for the current in the loop?

 

[Amplitude]

Does this agree with your result for the current in the loop

I see .

If I consider individual currents in the branches then i_L = i_R = i₀

This is quite interesting that superposition of currents works quite well in DC circuits but is failing in this case .

 

[Amplitude]

The induced electric field lines are concentric circles around the magnetic field lines .But what determines the center of the induced electric field lines ?

Please consider the above two figures . In the left figure it seems the induced field lines are concentric with the circular loop .

But what about the right figure ? The field lines would still be circular . But would they be concentric with the circular loop (complete circle minus the voltmeter ) ?

 

[TSny]

This is quite interesting that superposition of currents works quite well in DC circuits but is failing in this case .

Does your method of superposition work for the DC circuit shown below (no B field)?

 

[TSny]

The induced electric field lines are concentric circles around the magnetic field lines .But what determines the center of the induced electric field lines ?

You get circular E field lines only if the boundary of the region of the B field is also a circular. In that case, the circular E field lines are concentric with the center of the region of the B field.

If the boundary of the region of the B field is a square, say, then the induced E field would not be circular. I'm not sure how you would figure out the shapes of the closed E field lines in this case. $\mathbf E$ as a function of space would be the solution of $\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$ and $\nabla \cdot \mathbf E = 0$ with appropriate boundary conditions. But it looks difficult to work out.

[An old puzzler is to imagine that the region of the uniform B field extends to infinity. So it doesn't have a boundary. If the field changes at a uniform rate, what would the induced E field look like?]