What Defines Induced EMF in Circuit Analysis?

[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 ?

 

[TSny]

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

I'm not following this. Suppose you let $\varepsilon$ equal the rate of change of magnetic flux through the circular loop, which is assumed to be given. Then Faraday's law for the circular loop is $\oint Edl = \varepsilon$.

This may be written as $\int_{ALB} Edl + \int_{BRA} Edl = \varepsilon$. Then substitute $\int_{ALB} Edl = i_{_L} r$ and $\int_{BRA} Edl =i_{_R} r$, where $r$ is the resistance of half the circular loop. This will give you one equation involving the unknowns $i_{_L}$ and $i_{_R}$.

Similarly, get a second equation using loop BVAB.

Finally, use the junction rule for currents to get a third equation.

 

[Amplitude]

i_Lr+i_Rr=ε

i_vr_v-i_Rr=0

i_L=i_R+i_v

From these three equations , there are non zero currents and non zero EMF in the three branches .

And voltmeter reading is non zero ?

If this is correct , then I think I misunderstood you in post#31

the voltmeter will not read zero. The reading will depend on the shaded area (assuming the region of magnetic field encompasses this area).

I interpreted the statement in the parenthesis as - "if the magnetic field is not present in the coloured region then voltmeter will read 0 ."

 

[rude man]

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

Do you agree to my correcting the last line in your post #71 as follows:

∴V=2ε/3-ε/2=-ε/3+ε/2=ε/6.

Thanks again for your help.

 

[TSny]

i_Lr+i_Rr=ε

i_vr_v-i_Rr=0

i_L=i_R+i_v

From these three equations , there are non zero currents and non zero EMF in the three branches .

And voltmeter reading is non zero ?

Yes, that's right.

If this is correct , then I think I misunderstood you in post#31
I interpreted the statement in the parenthesis as - "if the magnetic field is not present in the coloured region then voltmeter will read 0 ."

Sorry, I can see how my statement was misleading. I was only commenting on the case where the field was present everywhere. I wasn't trying to imply anything about the case where there is no field in the shaded region.

The case where the field is only inside the loop is interesting. If you start with the voltmeter on the right, the voltmeter will read nonzero (as you showed). Keeping the voltmeter attached to A and B, you can imagine moving the voltmeter to the left side of the loop. The meter on the left side will read the opposite voltage from what it read on the right side; that is, opposite sign. During the movement of the meter, the reading would pass through zero.

 

[Amplitude]

This was my first thread and I must acknowledge I learned very good concepts in this discussion .Thanks for your patience :)

Just a quick confirmation to make sure I have understood things properly .

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.

A circular ring is placed in this type of B field (circular boundary ) concentric with the center of the region of the B field .The ring is of uniform cross section but resistance of the two halves is different .

The induced field will be uniform along the circumference of the ring . But from E_net = ρj , net electric field in the two halves will be different .

From ∫Edl = -dΦ/dt , the total induced EMF in the loop does not depend on the resistance , but induced EMF in the two halves is different because resistance as well as net electric field in the two halves is different .

Is this valid ?

Edit : In an earlier post you said in ∫Edl = -dΦ/dt , E is the net field .But closed loop integral of conservative field is 0 so basically in ∫Edl = -dΦ/dt , E should be the induced field ?

 

[Amplitude]

Possibly one last question :)

The four equations I wrote in post#74 are equivalent to the three equations I wrote in post#76 giving the same result .

In post#74 , I treated ∫Edl in a branch just like a battery in a DC circuit . Is this incorrect ? It yielded the same equations as in post#76 .

You wrote ∫Edl = iR for each branch .

Something similar is Ohm's law V=IR .

So is ∫Edl = iR nothing but Ohm's Law ?

 

[TSny]

This was my first thread and I must acknowledge I learned very good concepts in this discussion .Thanks for your patience :)

I have enjoyed thinking about these ideas.

A circular ring is placed in this type of B field (circular boundary ) concentric with the center of the region of the B field .The ring is of uniform cross section but resistance of the two halves is different .

The induced field will be uniform along the circumference of the ring . But from E_net = ρj , net electric field in the two halves will be different .

OK. Personally, I don't like to mentally break up the E field as partly "induced" and partly "electrostatic". So, in this situation, my brain tends to suppress the statement, "the induced field will be uniform along the circumference of the ring". To me, there is only one (net) electric field which is not uniform around the ring. But, that's just my preference. There could be times where it's nice to think in terms of the superposition of an "induced field" and a "static field".

So,before you place the ring in the B field, the (net) E field will be uniform around the circle where you will eventually place the loop. But, after the loop is in place, the (net) E field will be nonuniform around the loop. Maxwell's equations only deal with the net field. So, E in any situation is the solution of Maxwell's equations with appropriate boundary conditions. So, there is no need to worry about the induced part versus the static part of the field. However, if someone asks for the physical reason why the E field changed from uniform circular to nonuniform circular when the loop is placed, then I think I would have to mention the static charge built up on the ring that creates a static field that adds to the "induced" field. Oh well.

From ∫Edl = -dΦ/dt , the total induced EMF in the loop does not depend on the resistance , but induced EMF in the two halves is different because resistance as well as net electric field in the two halves is different .

Is this valid ?

Yes, I think that's right.

Edit :

In an earlier post you said in ∫Edl = -dΦ/dt , E is the net field .But closed loop integral of conservative field is 0 so basically in ∫Edl = -dΦ/dt , E should be the induced field ?

In the Maxwell equation ∫Edl = -dΦ/dt, I always think of E as the net field. That is, E is the force per unit test charge.

 

[Amplitude]

Thanks again for sharing your knowledge .These things are not discussed in so much detail in textbooks .

Please let me know what do you think about post#80 .

 

[TSny]

The four equations I wrote in post#74 are equivalent to the three equations I wrote in post#76 giving the same result .

In post#74 , I treated ∫Edl in a branch just like a battery in a DC circuit . Is this incorrect ? It yielded the same equations as in post#76 .

You wrote ∫Edl = iR for each branch .

I was a little confused by your equations in #74. (Probably my misinterpretation.)

You wrote

-i_L r + ∫E_Ldl -i_R r + ∫E_Rdl = 0

To me this is an identity that doesn't convey information. If you substitute E_Ldl = i_L r and ∫E_Rdl = i_R r, then the equation just says

-i_L r + i_L r -i_R r + i_R r = 0, which is clearly true but doesn't say much.

Similarly for your second equation.

But I'm probably misinterpreting something here.

So is ∫Edl = iR nothing but Ohm's Law ?

Yes, that's how I see it.

 

[Amplitude]

Thanks a lot for this amazing discussion . Hope I get a chance to interact with you again :)

 

[TSny]

Thanks a lot for this amazing discussion . Hope I get a chance to interact with you again :)

Thank you. I enjoyed it.

 

[cnh1995]

So, electrostatic voltage= net emf+ electromagnetically induced emf

∴V=2ε/3-ε/2=ε/3+ε/2=5ε/6.

V=2ε/3-ε/2=-ε/3+ε/2=ε/6.

Yes, I made a couple of stupid algebraic mistakes. You are right, it should be ε/6.

 

[cnh1995]

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/ρj=E/ρj = E / \rho.since that says E is not uniform around the entire ring.

How to resolve the paradox?

I see you've worked it out using some higher math, but here is how I see it:
j=E/ρ.
Assuming r has twice the cross sectional area of 2r,
j_2r=2j_r

j_r/j_2r= E_r/E_2r
∴E_r/E_2r=1/2
Or, E_2r=2E_r.

Since the induced electric field is same in both the halves, an electrostatic voltage is developed in the loop so as to satisfy the equation above. Hence, the electrostatic field adds to the induced electric field in 2r and subtracts from the induced electric field in r.

 

[Amplitude]

@TSny

When the magnetic field is confined in the circular region and decreasing and there is no magnetic field in the colored region , this situation is similar to placing of a voltmeter across a branch in the usual DC circuits where there is no magnetic field anywhere .

R is the rightmost point of the circle .

The total induced EMF in the loop ARBVA is zero which means ∫Edl along BRA is equal to ∫Edl along BVA .

So now we have ∫Edl along BRA = ∫Edl along BVA = i_Rr = i_vr_v

The voltmeter is essentially in parallel to the right circular branch .

Is is correct till now ?

Now if instead of the two diametrically opposite points we place the voltmeter across any two points C and D of the right semicircle , the voltmeter would read i_vr_v which would be equal to ir_CD . So voltmeter would always measure the induced EMF ∫Edl along CD ( = ir_CD )

Is that so ?

 

[TSny]

So voltmeter would always measure the induced EMF ∫Edl along CD ( = ir[SUB]CD[/SUB] )

Is that so ?

Yes, I think everything you said is correct. As you noted, the only restriction is that there is no changing magnetic flux through the shaded region.

 

[Amplitude]

Thanks .

Is EMF whether induced or not always represented by the line integral ∫Edl ?

In the present problem the induced EMF in any branch of the circuit is given by the line integral ∫Edl from one end point to another .

In a battery the EMF is defined as the work done per unit charge against the electric field from one terminal to the other .

But is the EMF of a battery in DC circuits also defined as the line integral ∫Edl ?

 

[TSny]

Thanks .

Is EMF whether induced or not always represented by the line integral ∫Edl ?

In the present problem the induced EMF in any branch of the circuit is given by the line integral ∫Edl from one end point to another .

In a battery the EMF is defined as the work done per unit charge against the electric field from one terminal to the other .

But is the EMF of a battery in DC circuits also defined as the line integral ∫Edl ?

Uh-oh, now we're entering the slippery realm of definition of terms. I doubt if you could ever get everyone to agree on one definition of EMF.

I generally think of the "EMF along a path between two points" as ∫Edl along the path. Here, E is the net electric field. Since E is the force per unit charge, this is the same as saying the EMF is the work per unit charge in moving charge along the path. So, in particular, for the battery you can think of the EMF of the battery either in terms of ∫Edl through the battery or in terms of work per unit charge.

I see very little need to break up E into an "induced part" and an "electrostatic part". All that matters is the net field.

What I think is truly important is Faraday's law. The line integral of the net electric field E around any closed loop equals the negative of the rate of change of magnetic flux through the circuit. That's the basis for handling circuits with (or without) time dependent B fields. You also need the fact that iR for a resistor is ∫Edl through the resistor, $\varepsilon_{\rm battery}$ is ∫Edl through the battery, etc. But the basic law is Faraday's. You can give names such as "EMF" to the line integral, but the names don't matter too much.

I don't know if that helps. I expect others will have different opinions on this.

 

[rude man]

@TSny

View attachment 213891When the magnetic field is confined in the circular region and decreasing and there is no magnetic field in the colored region , this situation is similar to placing of a voltmeter across a branch in the usual DC circuits where there is no magnetic field anywhere .

R is the rightmost point of the circle .

The total induced EMF in the loop ARBVA is zero which means ∫Edl along BRA is equal to ∫Edl along BVA .

So now we have ∫Edl along BRA = ∫Edl along BVA = i_Rr = i_vr_v

The voltmeter is essentially in parallel to the right circular branch .

Is is correct till now ?

Now if instead of the two diametrically opposite points we place the voltmeter across any two points C and D of the right semicircle , the voltmeter would read i_vr_v which would be equal to ir_CD . So voltmeter would always measure the induced EMF ∫Edl along CD ( = ir_CD )

Is that so ?

Agree also. Those loops obey Kirchhoff's laws.

 

[rude man]

Thanks .
Is EMF whether induced or not always represented by the line integral ∫Edl ?
In the present problem the induced EMF in any branch of the circuit is given by the line integral ∫Edl from one end point to another .

In a battery the EMF is defined as the work done per unit charge against the electric field from one terminal to the other .

But is the EMF of a battery in DC circuits also defined as the line integral ∫Edl ?

I think this is controversial. Consider a circuit with ideal battery with emf zero internal resistance, and resistor R connected together. The voltage across the resistor is obvioulsly emf/R. The question is, what is the E field INSIDE the battery? MIT professor Lewin says it points from + to - and we have ∫E⋅dl = -emf for just the battery so the total integral = 0, i.e. obeying Kirchhoff.

My own text however says ∫E⋅dl around the entire loop = emf, directly implying that E inside the battery = 0 since obviously ∫E⋅dl through inside (or outside) the resistor = emf. So this way Kirchhoff is NOT obeyed.

I think this boils down to nothing more than semantics. After all, what would happen to a test charge placed inside the battery? Probably get sticky or wet from the electrolyte but what woud be the force on it which would define E? So I agree with TSny and Dr. Lewin: the important thing is to hold onto Faraday at all costs and don't get too hung up on what the various E fields really are.

BTW if you close the loop AROUND the battery instead of THROUGH it then I think we all agree that that loop IS Kirchhoff-obeying. There is then no emf around the loop; everything is electrostatic voltages.

Additional thought: One could postulate another circuit like your figure in post 84 except now put the votmeter with leads (the entire voltmeter loop) INSIDE the circle. Connect the leads to your points C and D as you did. The the voltmeter reading would be i⋅R_CD + or - the emf generated about the loop (depending on voltmeter polarity connection) where emf = A⋅dB/dt with A the area of the voltmeter loop. If C and D coincide then you obviously would just read that emf.

Comments welcomed.