Induced current in a circular loop

[cnh1995]

Hi friends..I'm struggling with an interesting puzzle...Suppose there is a circular conducting loop of radius 1/π cm, one half of which is made of a material with resistance 1Ω and the other half from a material with resistance 5Ω. If somehow, one managed to set up a changing magnetic field through that loop such that 5V emf is induced in the loop, what will be the current through the loop? What will the voltmeter read if connected across the two diametrically opposite junction points? Is it really as straightforward as it looks?

 

[Simon Bridge]

Suppose there is a circular conducting loop of radius 1/π cm, one half of which is made of a material with resistance 1Ω and the other half from a material with resistance 5Ω.

You mean the circumference is 2cm ... so you take two 1cm lengths of wire and join them (and a voltmeter) to for a loop?

If somehow, one managed to set up a changing magnetic field through that loop such that 5V emf is induced in the loop, what will be the current through the loop? What will the voltmeter read if connected across the two diametrically opposite junction points? Is it really as straightforward as it looks?

I don't know - how does it look to you?
i.e. please show us your best attempt with reasoning.

 

[cnh1995]

You mean the circumference is 2cm ... so you take two 1cm lengths of wire and join them (and a voltmeter) to for a loop?

I don't know - how does it look to you?
i.e. please show us your best attempt with reasoning.

If the voltmeter part weren't there in the problem, I would say the current will be simply E/R i.e 5/(1+5)=5/6 A. But the real problem is voltage drop IR... Which R should be taken? 1Ω or 5Ω?? Tricky..isn't it?

 

[Simon Bridge]

If the voltmeter part weren't there in the problem...

Yes it is - in post #1 you wrote:

What will the voltmeter read if connected across the two diametrically opposite junction points?

... my emphasis.

Anyway - in any problem that asks about voltage it is useful to insert an ideal voltmeter in the appropriate place ;)

I would say the current will be simply E/R i.e 5/(1+5)=5/6 A. But the real problem is voltage drop IR... Which R should be taken? 1Ω or 5Ω?? Tricky..isn't it?

How so? How would you normally get the volt-drop for two resistors in series?
When you say "volt drop" which voltage is being dropped?
Where are the probes of the multimeter?

(Note: you use neither and both.)

 

[cnh1995]

If the voltmeter part weren't there in the problem, I would say the current will be simply E/R i.e 5/(1+5)=5/6 A. But the real problem is voltage drop IR... Which R should be taken? 1Ω or 5Ω??

Yes it is - in post #1 you wrote:
... my emphasis.

Anyway - in any problem that asks about voltage it is useful to insert an ideal voltmeter in the appropriate place ;)

How so? How would you normally get the volt-drop for two resistors in series?
When you say "volt drop" which voltage is being dropped?
Where are the probes of the multimeter?

(Note: you use neither and both.)

Probes are on the two diametrically opposite junction points.Resistors are in series..But I can't apply KVL to that loop just like we do for the circuits containing sources. I am stuck at this point..And what about the 1/π cm radius and 2cm circumference? Do they really matter?

 

[Jony130]

You meant something like this ?


https://www.physicsforums.com/attachments/kvl-question-b-pdf.30979/
https://www.physicsforums.com/threads/is-mit-prof-lewin-wrong-about-kirchhoffs-law.453575/page-13

 

[cnh1995]

You meant something like this ?


https://www.physicsforums.com/attachments/kvl-question-b-pdf.30979/
https://www.physicsforums.com/threads/is-mit-prof-lewin-wrong-about-kirchhoffs-law.453575/page-13

Yeah...I have watched these...The problem is similar to the one discussed by Prof Lewin. But in my problem, geometry of the circuit is given in detail..And I couldn't apply Prof Lewin's logic there..Frankly, I didn't understand that logic too...I strongly feel it has something to do with the geometry...What do you think?

 

[Simon Bridge]

Consider - take the 6Ohm bit of wire by itself.
If you put the voltmeter probes at each end of the wire, the resistance between the terminals is 6Ohms - If you put one probe at one end and the other probe at the center of the wire, what is the resistance between the probes?

Now try again with the loop - put one probe at the join between the 6 and 1 Ohms sections, and the other probe 1/4 the way around on the 6Ohm side - what is the resistance between the probes? Now same setup except the second probe is 1/4 way around on the 1Ohm side: what is the resistance?

 

[cnh1995]

Consider - take the 6Ohm bit of wire by itself.
If you put the voltmeter probes at each end of the wire, the resistance between the terminals is 6Ohms - If you put one probe at one end and the other probe at the center of the wire, what is the resistance between the probes?

Now try again with the loop - put one probe at the join between the 6 and 1 Ohms sections, and the other probe 1/4 the way around on the 6Ohm side - what is the resistance between the probes? Now same setup except the second probe is 1/4 way around on the 1Ohm side: what is the resistance?

For the stretched wire with one probe at the center, resistance will be 6/2=3Ω..
When loop is formed,
First case- the resistance will be 6/4Ω (total resistance*part length covered by the voltmeter)
Second case- resistance will be 1/4Ω...
Just the principle of uniform resistance gradient i.e. fixed resistance per unit length...

 

[Simon Bridge]

Careful - each wire forms half the loop ... so when you put one probe 1/4 the way around the loop, half the wire is between the probes.
You should draw a picture to check.

So - if you move the red probe 1/4 way around the loop on the 6Ohm side and the black probe 1/4 around the loop on the 1Ohm side - then the probes are diametrically opposite each other.

What is the resistance from the red probe to the join?
What is the resistance from the join to the black probe?
What is the net resistance between the probes?
What is the current through that resistance?
What is the voltage between the probes?

 

[cnh1995]

Careful - each wire forms half the loop ... so when you put one probe 1/4 the way around the loop, half the wire is between the probes.
You should draw a picture to check.

So - if you move the red probe 1/4 way around the loop on the 6Ohm side and the black probe 1/4 around the loop on the 1Ohm side - then the probes are diametrically opposite each other.

What is the resistance from the red probe to the join?
What is the resistance from the join to the black probe?
What is the net resistance between the probes?
What is the current through that resistance?
What is the voltage between the probes?

Oh...you mean both the probes are in the middle of the two half loops?? Oh god how do I put this...You mean line joining the probes is perpendicular to the line joining the junction points??

 

[Simon Bridge]

Well done. Like I said - draw a picture.
If the loop is around the center of a compass rose, and we put the join at due north, then the open ends will be at S while the probes are E and W.
That's how I read the problem.

The other way to read it is that the wire really forms a complete loop ... no break anywhere. This means there are two joins - which would be the "junction points".
But that would not fit the full description because, in that case, where do you measure the first voltage?

 

[cnh1995]

Oh...you mean both the probes are in the middle of the two half loops?? Oh god how do I put this...You mean line joining the probes is perpendicular to the line joining the junction points??

Well done. Like I said - draw a picture.
If the loop is around the center of a compass rose, and we put the join at due north, then the open ends will be at S while the probes are E and W.
That's how I read the problem.

The other way to read it is that the wire really forms a complete loop ... no break anywhere. This means there are two joins - which would be the "junction points".
But that would not fit the full description because, in that case, where do you measure the first voltage?

Well I get what you're saying...But what do you mean by 'open ends at S?' If it is open at S, how will the current flow? For that there must not be a break anywhere...And the problem specifies the two joins...
But from your point of view (keeping S end open), answers to your questions above would be-
1. 3Ω
2. 0.5Ω
3. 3.5Ω
4.current will be 0
5.Voltage will be 5V(induced) because there is a break at S.

If there is no break at S,
1. 3Ω||(3Ω+1Ω)=1.71Ω
2. 0.5Ω||(0.5Ω+6Ω)=0.464Ω
3. 3.5Ω||3.5Ω=1.75Ω
4. don't know
5.don't know..that's where I'm stuck..

 

[Simon Bridge]

So sketch the situation for the first setup - where is the voltmeter and the ammeter?
Did you answer that the current was zero for the first question?

Look at the usual description for induced currents in your textbook - look at the diagram - how do they handle voltages and currents in the loop?
It's all down to how you draw the diagram.

 

[cnh1995]

This is the original question diagram...Bold half is 5Ω and the other is 1Ω...

 

[Simon Bridge]

OK I see the problem.
Try this and see if it looks familiar.

Draw a dotted line horizontally through the center of the loop - label the left and right intersections C and D respectively.
For the sake of argument, set V_A=5V and V_B=0 ...
Then, surely, V_C=V_D=2.5V right?
... that should take care of the second part.
That is, if I am reading the question right.

For the first part - try tracing the energy that a charge gains and loses as it traverses the loop expressed in terms of current.

Meantime I'm going to have more of a think.

 

[DrZoidberg]

I believe professor Lewin explained this topic very well. If you place the voltmeter on the left side (as in your drawing) you will measure 5V/6Ω * 5Ω = +4.17V. If instead you place the voltmeter on the right side you measure -5V/6Ω * 1Ω = -0.83V. Despite the fact that the leads are connected to the exact same points in both cases you measure very different voltages. In other words V_AB is NOT equal to -V_BA. If you move the voltmeter slowly from left to right, keeping it connected the whole time, you will see the voltage slowly drop from +4.17V to -0.83V.
The reason for this behavior is that the voltmeter and it's leads are also inside the magnetic field. They form one winding around an (air) core so to speak.

 

[cnh1995]

I believe professor Lewin explained this topic very well. If you place the voltmeter on the left side (as in your drawing) you will measure 5V/6Ω * 5Ω = +4.17V. If instead you place the voltmeter on the right side you measure -5V/6Ω * 1Ω = -0.83V. Despite the fact that the leads are connected to the exact same points in both cases you measure very different voltages. In other words V_AB is NOT equal to -V_BA. If you move the voltmeter slowly from left to right, keeping it connected the whole time, you will see the voltage slowly drop from +4.17V to -0.83V.
The reason for this behavior is that the voltmeter and it's leads are also inside the magnetic field. They form one winding around an (air) core so to speak.

Okay..here the voltmeter loop is in the plane of the given circular loop, so it is inside the magnetic field.But what if the voltmeter loop is placed perpendicular to the given loop i.e.if I move the voltmeter such that the A-V-B-A loop is perpendicular to the conducting loop?? Now the voltmeter leads will be touching A and B from above.. Also in the original diagram, the voltmeter is at the left. If I move it to further left, it will change the voltmeter loop area and voltmeter will read different. Does that mean the voltmeter reading is affected by the distance at which it is placed from the loop? Sounds odd...

 

[cnh1995]

OK I see the problem.
Try this and see if it looks familiar.

Draw a dotted line horizontally through the center of the loop - label the left and right intersections C and D respectively.
For the sake of argument, set V_A=5V and V_B=0 ...
Then, surely, V_C=V_D=2.5V right?
... that should take care of the second part.
That is, if I am reading the question right.

For the first part - try tracing the energy that a charge gains and loses as it traverses the loop expressed in terms of current.

Meantime I'm going to have more of a think.

Okay...But we can't talk in terms of potentials of individual points like V_A or V_B here...Non conservative fields have nothing to do with the potential...

 

[DrZoidberg]

It will read different only if the flux through the loop changes. So if instead of an air core you have e.g. a ferromagnetic ring then the flux will stay the same if you move the meter. But of course that would also mean you can't move the meter to the other side without separating the connection.
I mean, of course you could move it but it wouldn't change the flux since the lead wires can't cut through the ring.

 

[cnh1995]

It will read different only if the flux through the loop changes. So if instead of an air core you have e.g. a ferromagnetic ring then the flux will stay the same if you move the meter. But of course that would also mean you can't move the meter to the other side without separating the connection.
I mean, of course you could move it but it wouldn't change the flux since the lead wires can't cut through the ring.

Speaking of the air core part you mentioned, how do the leads form a winding? You mean A to V, V to B and again B to A via the 1Ω curve??

 

[DrZoidberg]

Yes, it forms a circle.
And a changing magnetic field induces a circular electric field around itself. Every circular path around the core will therefore have a voltage induced in it according to the integral of the electric field around that path which of course will be equal to the magnetic flux change per time through the area enclosed by said path.

 

[cnh1995]

Yes, it forms a circle.
And a changing magnetic field induces a circular electric field around itself. Every circular path around the core will therefore have a voltage induced in it according to the integral of the electric field around that path which of course will be equal to the magnetic flux change per time through the area enclosed by said path.

Yeah that's true for the circular loop. But what about the voltmeter loop? Are you saying that the current through the voltmeter is also due to this magnetic field?

 

[DrZoidberg]

Yes. Every loop around the core will have the same 5V induced in it as long as the entire magnetic flux is inside the loop. That is the case e.g. for a transformer since there the flux is concentrated in the core material. You have two loops here: A-V-B-1Ω-A and A-5Ω-B-1Ω-A (if the voltmeter is on the left side). Both loops have 5V induced in them.
So you can say:
V_V + V_1Ω = 5V
V_5Ω + V_1Ω = 5V
I_V + I_5Ω = I_1Ω
Now you can simply solve that.
One thing you can see instantly is that V_V and V_5Ω must be equal.

 

[vanhees71]

I think, you really have to do the calculation carefully, and I don't think that the question is put concise enough. One needs an assumption about the magnetic field. The answer in #17 is correct, if the magnetic field is negligible outside the shaded circular area.

It is also very misleading to say that there's a voltage between the points A and B, and this leads to the confusion. There's no electric potential in this case but to the contrary an electromotive force, i.e., a curl of the electromagnetic field!

Then you can use Faraday's Law in integral form for the path ABV (i.e., along the left half of the wire making the circuit and then through the loop formed by the connection to the volt-meter). According to the above assumption, then there's no magnetic flux through this loop, and thus the volt meter shows
$$U=5 \; \mathrm{V} \cdot 5 \Omega/6 \Omega \simeq 4.17 \;\mathrm{V}.$$
If you put the voltmeter in the right half by the same argument you get
$$U=5 \; \mathrm{V} \cdot 1 \Omega/6 \Omega \simeq 0.83 \;\mathrm{V}.$$
I've assumed I've plugged the voltmeter such that it always shows a positive voltage, which is given by the direction of electric field (i.e. the direction of current-density vector, because of Ohm's Law, $\vec{E}=\sigma \vec{j}$) in the wire of the loop. I also assumed the voltmeter to be ideal, i.e., of very large resistance.

If there's a time-changing magnetic field outside of the circular region, you cannot say, what the volt-meter measures except you know the precise time change of the magnetic flux there.

 

[cnh1995]

Yes. Every loop around the core will have the same 5V induced in it as long as the entire magnetic flux is inside the loop. That is the case e.g. for a transformer since there the flux is concentrated in the core material. You have two loops here: A-V-B-1Ω-A and A-5Ω-B-1Ω-A (if the voltmeter is on the left side). Both loops have 5V induced in them.
So you can say:
V_V + V_1Ω = 5V
V_5Ω + V_1Ω = 5V
I_V + I_5Ω = I_1Ω
Now you can simply solve that.
One thing you can see instantly is that V_V and V_5Ω must be equal.

I think, you really have to do the calculation carefully, and I don't think that the question is put concise enough. One needs an assumption about the magnetic field. The answer in #17 is correct, if the magnetic field is negligible outside the shaded circular area.

It is also very misleading to say that there's a voltage between the points A and B, and this leads to the confusion. There's no electric potential in this case but to the contrary an electromotive force, i.e., a curl of the electromagnetic field!

Then you can use Faraday's Law in integral form for the path ABV (i.e., along the left half of the wire making the circuit and then through the loop formed by the connection to the volt-meter). According to the above assumption, then there's no magnetic flux through this loop, and thus the volt meter shows
$$U=5 \; \mathrm{V} \cdot 5 \Omega/6 \Omega \simeq 4.17 \;\mathrm{V}.$$
If you put the voltmeter in the right half by the same argument you get
$$U=5 \; \mathrm{V} \cdot 1 \Omega/6 \Omega \simeq 0.83 \;\mathrm{V}.$$
I've assumed I've plugged the voltmeter such that it always shows a positive voltage, which is given by the direction of electric field (i.e. the direction of current-density vector, because of Ohm's Law, $\vec{E}=\sigma \vec{j}$) in the wire of the loop. I also assumed the voltmeter to be ideal, i.e., of very large resistance.

If there's a time-changing magnetic field outside of the circular region, you cannot say, what the volt-meter measures except you know the precise time change of the magnetic flux there.

Ok..There is very negligible field outside the loop...I got the concept...Thank you..Just out of curiosity, I want to know what the voltmeter will show if the voltmeter loop is perpendicular to the circular loop i.e. say the probes touch A and B from above..In this case, there is no flux through the voltmeter loop (safely assumed to be 0). What will it show?? Zero??

 

[cnh1995]

Yes. Every loop around the core will have the same 5V induced in it as long as the entire magnetic flux is inside the loop. That is the case e.g. for a transformer since there the flux is concentrated in the core material. You have two loops here: A-V-B-1Ω-A and A-5Ω-B-1Ω-A (if the voltmeter is on the left side). Both loops have 5V induced in them.
So you can say:
V_V + V_1Ω = 5V
V_5Ω + V_1Ω = 5V
I_V + I_5Ω = I_1Ω
Now you can simply solve that.
One thing you can see instantly is that V_V and V_5Ω must be equal.

That was very helpful.. I now got what prof Lewin was saying in his lecture...Thank you.. I'm curious about that perpendicular voltmeter loop case I mentioned in an earlier post in which the probes touch A and B from above.. What will the voltmeter read??

 

[DrZoidberg]

If there is no solid core you can move the voltmeter above the coil and get a voltage that's somewhere in between +4.17 and -0.83 or -4.17 and + 0.83 depending on which way around you connected the meter. At some point it will read 0 but that point is probably not exactly above the coil because the setup is not symmetric since the two resistances are different.

 

[cabraham]

Is the frequency given? If we know that and the thickness of the wire, we can compute the loop inductance and reactance. If the loop inductive reactance is substantial, it must be added to the 1 ohm and 5 ohm in quadrature. If the loop inductive reactance is negligible, i.e. well under 1 ohm, then we can just compute the current as 5 volts divided by (5+1) ohms or 0.833 amp. The voltage drop across the 5.0 ohm semi-loop is 4.167 volt. The voltage drop around the 1.0 ohm semi-loop is 0.833 volt. Did I help?

Claude

 

[cnh1995]

Is the frequency given? If we know that and the thickness of the wire, we can compute the loop inductance and reactance. If the loop inductive reactance is substantial, it must be added to the 1 ohm and 5 ohm in quadrature. If the loop inductive reactance is negligible, i.e. well under 1 ohm, then we can just compute the current as 5 volts divided by (5+1) ohms or 0.833 amp. The voltage drop across the 5.0 ohm semi-loop is 4.167 volt. The voltage drop around the 1.0 ohm semi-loop is 0.833 volt. Did I help?

Claude

Yes Claude..Thanks

 

[cnh1995]

If there is no solid core you can move the voltmeter above the coil and get a voltage that's somewhere in between +4.17 and -0.83 or -4.17 and + 0.83 depending on which way around you connected the meter. At some point it will read 0 but that point is probably not exactly above the coil because the setup is not symmetric since the two resistances are different.

Well I have studied the circuit using surface charge feedback mechanism too..My calculations for voltmeter reading match for both the cases (left and right side) with those done using your suggested method.. Exactly above the coil, the voltmeter will read 1.67 V, not 0 ,as you correctly predicted using the symmetry argument..Thanks again for your help..