What Defines Induced EMF in Circuit Analysis?

[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?]

 

[Amplitude]

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.

Ok .

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?]

Wouldn't the induced electric field lines be circular in the above case as well where B field extends to infinity ?

I think they would be circular , concentric with the loop . Please correct me .

In this case also the induced field lines would be concentric circles with common center at the center of the larger grey circle .

There would be an induced current in the loop, but induced electric field lines would be non uniform at different points of the loop . ∫Edl around the loop would not be equal to E(2πR) . Is that so ?

 

[TSny]

For the B field that extends to infinity imagine that there is no wire loop. If the E field forms concentric circles, what determines the location of the center of these circles?

 

[Amplitude]

For the B field that extends to infinity imagine that there is no wire loop. If the E field forms concentric circles, what determines the location of the center of these circles?

Induced Electric field lines form closed loops around the magnetic field lines .

But then concentric circular loops do not make sense .

Sorry , I don't have an answer .

 

[TSny]

Induced Electric field lines form closed loops around the magnetic field lines .

But then concentric circular loops do not make sense .

Sorry , I don't have an answer .

Right. I don't think there is an answer. A similar problem is to propose an infinite space with a uniform electric charge density. What would the electric field lines look like? Gauss' law requires that there is a net electric flux through any closed surface. But, the symmetry of the situation seems to imply that at any point, there is no reason why E should point in a particular direction.

 

[Amplitude]

But in the first picture in post#52 , there would be a uniform electric field along the periphery of blue loop ?

Have I stated correctly about the second picture in post#52 ?

 

[TSny]

But in the first picture in post#52 , there would be a uniform electric field along the periphery of blue loop ?

Yes, that sounds right.

Have I stated correctly about the second picture in post#52 ?

Here's how I see this, but I could be mistaken. Without any conducting loop, the E-field lines would of course be circular and centered at the midpoint of the B-field region.

With the wire loop placed off-center in the B field, the E field will get distorted by static charge that builds up on the loop. The field inside the wire loop will be essentially uniform and circumferential around the loop, even thought the loop is not centered in the B field. The reason for this is as follows. The current must be uniform around the loop if the resistivity $\rho$ and cross-sectional area of the loop is uniform. The current density $j$ must therefore be essentially uniform around the loop. Since $j = E / \rho$, the E field must be uniform around the loop. Outside the loop, the E field will be complicated and will depend on where the loop is placed in the B field.

 

[TSny]

The E field inside the circular ring is uniform around the ring as shown by solving ∇ x E = -∂B/∂t in cylindrical coordinates, giving E_θ = (r/2) (-∂B/∂t) i.e. not a function of θ and therefore uniform around the ring. So the emf is equally developed about the two dissimilar halves.

Assuming uniform resistivity ρ the ring's cross-sectional area of course has to have different values (2:1) to get r and 2r. But the current is constant around the ring, so current density j varies 2:1 also. But that is inconsistent with $j = E / \rho$.since that says E is not uniform around the entire ring.

How to resolve the paradox? One (or more) assumption above would seem to be incorrect.

The equation ∇ x E = -∂B/∂t alone does not determine the solution E_θ = (r/2) (-∂B/∂t) . This solution corresponds to also imposing the condition ∇ ⋅ E = 0 everywhere and the condition that the solution be rotationally invariant about the center of the circular B field region. Note that you could add a constant E field to your solution and still satisfy ∇ x E = -∂B/∂t and ∇ ⋅ E = 0 everywhere.

What if you added a fixed point charge q external to the B field region?

The net E field would still satisfy ∇ x E = -∂B/∂t , but the E field would not be given by E_θ = (r/2) (-∂B/∂t).

Likewise, if you put the charge inside the region of B, ∇ x E = -∂B/∂t would still hold, but the field would not be given by
E_θ = (r/2) (-∂B/∂t).

If you put your nonuniform ring into the B field, there will be charge induced on the ring such that the E field produced by this charge combines with the induced field to create just the right nonuniform E field inside the ring to produce the necessary nonuniform current density.

BTW in your post #50 whom were you addressing?

That was meant for @Amplitude. I was questioning his/her method for solving such a circuit using a "superposition of currents" approach.

 

[rude man]

I was taught that the E field is uniquely solved by ∇x E = - ∂B/∂t AND ∇⋅E = 0 AND
the field has to go to zero at infinity and not be infinite anywhere.

OK I think I see the problem.
In cylindrical coordinates ∇⋅B = ∂E_r/∂r + E_r/r + (1/r)∂E_θ/∂θ + ∂E_z/∂z
and I ignored the fact that the third term is not necessarily zero due to the asymmetrically distributed resistance.

The other two conditions I learned are that (3) the field must go to zero at infinity and (4) it must not be infinite in magnitude anywhere.

I have not run into your other criterion of "the condition that the solution be rotationally invariant about the center of the circular B field region" and in fact I'm not sure what it says. I'm thinking it's subsumed in one or more of the other stated conditions, I guess the 3rd and/or 4th.

 

[TSny]

I was taught that the E field is uniquely solved by ∇x E = - ∂B/∂t AND ∇⋅E = 0 AND
the field has to go to zero at infinity and not be infinite anywhere.

OK, I think that's right.

I have not run into your other criterion of "the condition that the solution be rotationally invariant about the center of the circular B field region" and in fact I'm not sure what it says. I'm thinking it's subsumed in one or more of the other stated conditions, I guess the 3rd and/or 4th.

The curl in cylindrical coordinates is

$\nabla \times \mathbf E = \left( \frac{1}{r} \frac{\partial E_z}{\partial \theta} - \frac{\partial E_{\theta}}{\partial z} \right) \hat {\mathbf r} + \left( \frac{\partial E_r}{\partial z} - \frac{\partial E_z}{\partial r}\right) \hat {\mathbf \theta} + \frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) - \frac{\partial E_r}{\partial \theta}\right) \hat {\mathbf z}$

The rotational invariance about the z axis, along with the fact that $E_z = 0$ and none of the components of E depend on z, simplifies the curl to

$\nabla \times \mathbf E = \frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) \right) \hat {\mathbf z}$

So, $\frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) \right) = B_0$, where $B_0$ is the constant rate of change of the B field. The solution of this is your solution. So, the rotational invariance helps to simplify the differential equation. If the region of the magnetic field were square rather than circular, you would get a more complicated differential equation to solve.

 

[rude man]

Ξ

OK, I think that's right.
The curl in cylindrical coordinates is

$\nabla \times \mathbf E = \left( \frac{1}{r} \frac{\partial E_z}{\partial \theta} - \frac{\partial E_{\theta}}{\partial z} \right) \hat {\mathbf r} + \left( \frac{\partial E_r}{\partial z} - \frac{\partial E_z}{\partial r}\right) \hat {\mathbf \theta} + \frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) - \frac{\partial E_r}{\partial \theta}\right) \hat {\mathbf z}$

The rotational invariance about the z axis, along with the fact that $E_z = 0$ and none of the components of E depend on z, simplifies the curl to

$\nabla \times \mathbf E = \frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) \right) \hat {\mathbf z}$

So, $\frac{1}{r}\left( \frac{\partial}{\partial r}(r E_{\theta}) \right) = B_0$, where $B_0$ is the constant rate of change of the B field. The solution of this is your solution. So, the rotational invariance helps to simplify the differential equation. If the region of the magnetic field were square rather than circular, you would get a more complicated differential equation to solve.

Yes, I looked at that for your square loop of a few posts ago & concluded you wind up with a PDE in x and y of Ex and Ey ( 2 dependent, 2 independent variables!).
BTW I re-did the computation for the voltage between the two r/2r junctions of my non-uniform-resistance ring and came out with zero again. So I guess the idea is that, since emf and resistance in a closed dB/dt loop always go together such that E = ρj the voltage difference between any 2 points is always zero regardless of resistance variations.

 

[Amplitude]

Thanks TSny .

With the wire loop placed off-center in the B field, the E field will get distorted by static charge that builds up on the loop. The field inside the wire loop will be essentially uniform and circumferential around the loop, even thought the loop is not centered in the B field.

Now if a voltmeter is placed across two diametrically opposite points just like in post#43 , the reading might not be 0 .Since E field lines inside the circular area of the wire loop is unknown , ∫Edl along the diameter is not necessarily 0 .

Is that so ?

 

[TSny]

Now if a voltmeter is placed across two diametrically opposite points just like in post#43 , the reading might not be 0 .Since E field lines inside the circular area of the wire loop is unknown , ∫Edl along the diameter is not necessarily 0 .

Is that so ?

The reading of the voltmeter will still be zero no matter where the loop is placed in the field as long as the loop is entirely within the field region. This is because the rate of change of flux through the left-half area of the loop will be equal to the rate of change of flux through the right-half area of the loop. ∫Edl along the diameter would be 0.

If you replace the uniform loop with @rude man 's nonuniform loop where the resistances of the left and right halves of the loop are different, then you would get a nonzero reading on the voltmeter.

Or, if you have a uniform loop but you bend one of the voltmeter leads as shown, then the meter will read a nonzero value. The rate of change of flux is no longer the same through the two shaded regions.

 

[cnh1995]

emf1 = 2emf2

Induced emf does not depend on the resistance of the path. If you are considering two "halves" of the ring, the induced emfs E1 and E2 will be equal, say E. The total induced emf is therefore 2E.

The net electric field in 2r should be more than that in r. Hence, electrostatic voltage V will have a polarity such that it adds to the emf in 2r and subtracts from the emf in r.

Using KCL,
(E+V)/2r=(E-V)/r
Simplify this and you'll get V in terms of E i.e. V=E/3 (which is not zero).

If the resistances of both the halves were equal, say R, you can see from the equation above that the electrostatic voltage would be 0.

 

[Amplitude]

If you replace the uniform loop with @rude man 's nonuniform loop where the resistances of the left and right halves of the loop are different, then you would get a nonzero reading on the voltmeter.

Call leftmost point on the circumference L and rightmost point R . Resistance of the left semicircle is R₁ and that of right semicircle is R₂ .

From E = ρj , electric field E₁ in left semicircle will be different from electric field E₂ in right semicircle . Call E₃ electric field in the middle branch .

Since change in flux in the two halves is same total EMF induced in the two halves will be same .

In the left half moving anticlockwise along A →L→B→A , total EMF = ∫E₁dl + ∫E₃dl

In the right half moving anticlockwise along B →R→A→B , total EMF = ∫E₂dl -∫E₃dl

∫E₁dl + ∫E₃dl = ∫E₂dl -∫E₃dl

∫E₃dl = (1/2)( ∫E₂dl - ∫E₁dl )

∫E₃dl ≠ 0

Net EMF in the middle branch is non zero .

@TSny , is that valid ?

 

[TSny]

View attachment 212938

Since change in flux in the two halves is same total EMF induced in the two halves will be same .

In the left half moving anticlockwise along A →L→B→A , total EMF = ∫E₁dl + ∫E₃dl

In the right half moving anticlockwise along B →R→A→B , total EMF = ∫E₂dl -∫E₃dl

∫E₁dl + ∫E₃dl = ∫E₂dl -∫E₃dl

∫E₃dl = (1/2)( ∫E₂dl - ∫E₁dl )

∫E₃dl ≠ 0

Net EMF in the middle branch is non zero .

@TSny , is that valid ?

Looks very good to me.

 

[cnh1995]

Why? emf is the line integral of the electric field and the two E fields are different while the line distance is the same.

I was talking about the "induced" electric field because of the change in magnetic flux. That emf has to be same in both the halves
The electrostatic voltage developed will aid the induced electric field in 2r and the same electrostatic voltage will oppose the induced electric field in r. Hence the "net" electric fields in both the halves will be not be same.
In 2r, the line integral of E.dl will be E+E/3 i.e. 4E/3 and in r it will be E-E/3 i.e. 2E/3.

Note that for the entire circular loop, the induced emf (or line integral of E.dl) is still the same i.e. 2E. This is because the closed loop integral of electrostatic field is zero.

P.S. : A similar thread once again, posted a few minutes ago , check out the provided solution.
https://www.physicsforums.com/threa...ld-in-circuit-with-varying-resistance.928336/

 

[cnh1995]

No recourse to any "electrostatic voltage" necessary. It's all emf.

Yes. You can work it out wihout even mentioning electrostatic voltage.

For example, in you circular loop with r-2r resistances, assume the current to be i and net induced emf=AdB/dt=ε.
So, i=ε/(r+2r)=ε/3r.
Line integral of E.dl along r =net emf along r= i*r.
∴Net emf along r=ε/3

Similarly, net emf along 2r is 2ε/3.
Their sum is the total emf ε.
All emf and no mention of any electrostatic voltage.

Which is fine except I still don't see the explanation for induced emf being the same in both halves.

By "induced" emf, I meant electromagnetically induced emf. That is same in both the halves due to symmetry. In this case, it is ε/2. But since the "net" emfs are different, you can see that there is an electrostatic voltage present in the circuit.
So, electrostatic voltage= net emf+ electromagnetically induced emf

∴V=2ε/3-ε/2=ε/3+ε/2=5ε/6.

Anyway, thank you for your patience, I will probably not address this problem again.

No probs at all.

 

[Amplitude]

Looks very good to me.

Thank you .

Call leftmost point on the circumference L and rightmost point R .Resistance is uniform across the circular loop . Resistance of voltmeter is r_v .

Suppose there is no magnetic field in the coloured region . But there is magnetic field in the circular loop .

∫Edl around circular loop ALBA is not zero . Current in the branch ARB (right semicircle ) should be non zero . ∫Edl in branch ARB ≠ 0 .

There would be no change in flux in loop AVBA (coloured loop) . ∫Edl around closed loop AVBA will be zero . Currents in branch AVB (through voltmeter) and branch ARB (right semicircle ) should also be zero .

@TSny , I feel perplexed .Why is there an apparent contradiction in terms of current in the branch ARB ?

Is there a current through the voltmeter ?

How is ∫Edl in the branch AVB zero ?

 

[TSny]

Call leftmost point on the circumference L and rightmost point R .Resistance is uniform across the circular loop . Resistance of voltmeter is r_v .

Suppose there is no magnetic field in the coloured region . But there is magnetic field in the circular loop .

OK

∫Edl around circular loop ALBA is not zero . Current in the branch ARB (right semicircle ) should be non zero . ∫Edl in branch ARB ≠ 0 .

OK

There would be no change in flux in loop AVBA (coloured loop) . ∫Edl around closed loop AVBA will be zero .

OK

Currents in branch AVB (through voltmeter) and branch ARB (right semicircle ) should also be zero.

To find out if this is true, set up three equations for the three unknowns i_L, i_R, and i_V, were the subscripts L, R, V refer to the left semicircle, the right semicircle, and the voltmeter, respectively.

 

[Amplitude]

To find out if this is true, set up three equations for the three unknowns i_L, i_R, and i_V, were the subscripts L, R, V refer to the left semicircle, the right semicircle, and the voltmeter, respectively.

Assume i_L flows from A to B , i_R flows from B to A , i_v flows from B to A via voltmeter .

Moving anticlockwise in the circular loop and applying KVL ,

-i_L r + ∫E_Ldl -i_R r + ∫E_Rdl = 0

Moving anticlockwise in the loop BVAB and applying KVL ,

-i_v r_v+ ∫E_vdl +i_R r - ∫E_Rdl = 0

Since total induced EMF in loop BVAB is zero ,
∫E_vdl - ∫E_Rdl = 0

Edit : Using KCL , i_L=i_R+i_v

Are these equations written correctly ?

If so , there are non zero currents and non zero EMF in the three branches .

And voltmeter reading will be non zero ?