How is current induced in superconductors?

[Kevin J]

Assume resistor R/heated segment is 5 Ohms, while the path in the right hand side remains 0, thus, total resistance is 0 Ohms, my question is how could current be induced if potential difference is 0(due to zero total resistance) Or is it probably because of the batteries internal resistance?

 

[darth boozer]

A fairly comprehensive explanation of the whole process is available below.
https://en.wikipedia.org/wiki/Superconducting_magnet

 

[Dale]

You are correct that the superconducting side does have 0 resistance, however resistance is not the only means of creating a voltage. The superconducting side also has a non zero inductance which means: $v=L \frac{d}{dt}i$

Typically the inductance L is quite large and v is small so the current increases slowly, often taking hours to ramp up to the desired current. Once the desired current is obtained then the voltage can be reduced to 0 and the current will not change. The resistive section can be cooled and the current source removed.

 

[Kevin J]

You are correct that the superconducting side does have 0 resistance, however resistance is not the only means of creating a voltage. The superconducting side also has a non zero inductance which means: $v=L \frac{d}{dt}i$

Typically the inductance L is quite large and v is small so the current increases slowly, often taking hours to ramp up to the desired current. Once the desired current is obtained then the voltage can be reduced to 0 and the current will not change. The resistive section can be cooled and the current source removed.

Really? They don't use resistance for creating a voltage, then why is the heater even placed there?

 

[Dale]

Really? They don't use resistance for creating a voltage, then why is the heater even placed there?

There needs to be a return path for the current. At the beginning all of the current goes through the resistor. As the inductor current ramps up less and less of the current goes through the resistor.

 

[Kevin J]

There needs to be a return path for the current. At the beginning all of the current goes through the resistor. As the inductor current ramps up less and less of the current goes through the resistor.

If I somehow manage creating an ideal current source, will there still be an induced current?

 

[Dale]

If I somehow manage creating an ideal current source, will there still be an induced current?

Yes. In this circuit the difference between ideal and nearly ideal is small.

For a current source the internal resistance is a large resistance in parallel with the source. So the large internal resistance is in parallel with the small external resistance (heated superconductor material). The parallel equivalent of the two is almost the same as the small resistance alone.

 

[Kevin J]

Why should there be a heater there, and what's the point of parallel circuit, why don't use a series circuit and use the source's internal resistance to produce potential difference? (I'm still new to this area)

 

[Dale]

what's the point of parallel circuit, why don't use a series circuit

Physically that just wouldn’t be possible. This is a superconducting magnet, so like all electromagnets it is wound in a big loop, usually with multiple turns. Further, when you disconnect the power you want the entire loop to be superconducting so the magnetic field can be permanent, so the loop is completely closed. Anywhere you connect an external circuit would therefore have two paths through the loop, which makes it a parallel circuit. There simply is no way around that.

Now, given that we must have a parallel circuit then the goal is to get a large current into the high inductance side. To do that requires a voltage, which we cannot get across the low inductance side without making it resistive.

 

[Kevin J]

Physically that just wouldn’t be possible. This is a superconducting magnet, so like all electromagnets it is wound in a big loop, usually with multiple turns. Further, when you disconnect the power you want the entire loop to be superconducting so the magnetic field can be permanent, so the loop is completely closed. Anywhere you connect an external circuit would therefore have two paths through the loop, which makes it a parallel circuit. There simply is no way around that.

Now, given that we must have a parallel circuit then the goal is to get a large current into the high inductance side. To do that requires a voltage, which we cannot get across the low inductance side without making it resistive.

But doesn't it matter how much resistance is in the segment heated by the heater, like I asked on my short circuit problem, it doesn't matter how much resistance you have in this path(5 Ohms), because the actual current would be the voltage divided by the internal resistance (10V/2Ohms), and what's the difference between the superconducting induction and the circuit I draw, doesn't it work the same way?(or am I confusing voltage source and current source?)

 

[Dale]

like I asked on my short circuit problem

The short circuit problem is a different circuit. It does not behave the same. In the short circuit there is no resistance and no inductance. In the superconductor there is no resistance but a very large inductance. That makes a huge difference in the circuit behavior. Again, resistance is not the only way to get a voltage.

 

[Kevin J]

The short circuit problem is a different circuit. It does not behave the same. In the short circuit there is no resistance and no inductance. In the superconductor there is no resistance but a very large inductance. That makes a huge difference in the circuit behavior. Again, resistance is not the only way to get a voltage.

So it behaves totally different? And If you don't mind, can you explain briefly how the superconducting loop works? Just briefly. Thankyou

 

[Dale]

can you explain briefly how the superconducting loop works? Just briefly.

The loop has a very high inductance, L, which means $v=L\frac{d}{dt}i$. So a voltage does not cause an infinite current, it causes a slowly increasing current.

 

[Kevin J]

The loop has a very high inductance, L, which means $v=L\frac{d}{dt}i$. So a voltage does not cause an infinite current, it causes a slowly increasing current.

What do you mean by the term 'very high inductance'? (I'm kind of new) and I can you explain me in details how a permanent current is formed in the loop?

 

[Dale]

What do you mean by the term 'very high inductance'? (I'm kind of new)

You should start with the hyperphysics page and maybe the Wikipedia page on inductance

can you explain me in details how a permanent current is formed in the loop?

First we heat up the small segment and then we turn on the current source. At the beginning all of the current goes through the resistive side, but over time the current slowly increases through the loop until all of the current is going through the loop. Then the heater is turned off, the resistive segment becomes superconducting, the current becomes permanent, and the current source is disconnected.

 

[Kevin J]

How could the 'current slowly goes to the loop, until no current passes the resistance'? Is there a method doing that?

 

[Dale]

How could the 'current slowly goes to the loop, until no current passes the resistance'? Is there a method doing that?

The inductance. Please read the hyperphysics and wiki pages. You may also want to search for a page on “RL circuits”

 

[Kevin J]

The inductance. Please read the hyperphysics and wiki pages. You may also want to search for a page on “RL circuits”

*and why does the current moves through the resistive side 'first'?

 

[Dale]

*and why does the current moves through the resistive side 'first'?

Again, search for an introductory page on “RL circuits”

 

[Kevin J]

Ok thanks

 

[Mister T]

What do you mean by the term 'very high inductance'?

When you study circuits that have a steady current (that is, a current that doesn't change with time) then we can say the resistance $R$ is the only thing that impedes the current. That is, the more resistance you have along a conducting path the smaller the current through that path.

But when you study circuits where the current does change with time, you find that the resistance $R$ alone is not enough to describe how the current is impeded. This is because, for example, as you try to increase the current through a conducting path, the magnetic field (whose source is that current) in the vicinity of that wire increases. The increasing magnetic field generates an electrical voltage across that conducting path that also impedes the current. We call that the induced voltage, and refer to the phenomenon as inductance. For a superconductor $R$ makes no contribution to the impedance, but the inductance does.

 

[Kevin J]

When you study circuits that have a steady current (that is, a current that doesn't change with time) then we can say the resistance $R$ is the only thing that impedes the current. That is, the more resistance you have along a conducting path the smaller the current through that path.

But when you study circuits where the current does change with time, you find that the resistance $R$ alone is not enough to describe how the current is impeded. This is because, for example, as you try to increase the current through a conducting path, the magnetic field (whose source is that current) in the vicinity of that wire increases. The increasing magnetic field generates an electrical voltage across that conducting path that also impedes the current. We call that the induced voltage, and refer to the phenomenon as inductance. For a superconductor $R$ makes no contribution to the impedance, but the inductance does.

Thnx I understand now, but shouldn't it happen in any parallel circuit then? What I mean is that if we simply change the batteries to a higher voltage rating, while keeping the same resistance, doesn't that mean the magnetic field also increases, and could possibly induce a current across it?

 

[Dale]

What I mean is that if we simply change the batteries to a higher voltage rating, while keeping the same resistance, doesn't that mean the magnetic field also increases, and could possibly induce a current across it?

Yes, but in most circuits the inductance of the wires and the resistors is very low, so we can ignore it. In a superconducting coil the inductance is very large, by design, so it cannot be ignored.

 

[Kevin J]

Yes, but in most circuits the inductance of the wires and the resistors is very low, so we can ignore it. In a superconducting coil the inductance is very large, by design, so it cannot be ignored.

Ok, thankyou very much

 

[Kevin J]

You should start with the hyperphysics page and maybe the Wikipedia page on inductance

First we heat up the small segment and then we turn on the current source. At the beginning all of the current goes through the resistive side, but over time the current slowly increases through the loop until all of the current is going through the loop. Then the heater is turned off, the resistive segment becomes superconducting, the current becomes permanent, and the current source is disconnected.

Sorry for the late response, but I'm just wondering the sentence you stated that current goes through the resistor's path, but doesn't electricity/electrons choose the least path of resistance, which means in this case the loop?

 

[Dale]

doesn't electricity/electrons choose the least path of resistance,

No. That is a statement you should actively forget. Remove it from your mind.

It is not an accurate description of the physics. It is more like one of those charicatures that you can buy at amusement parks. If you already know the person then you can see the vague resemblance, but it is not enough of a resemblance for a stranger to recognize the person based on the charicature.

 

[Kevin J]

No. That is a statement you should actively forget. Remove it from your mind.

It is not an accurate description of the physics. It is more like one of those charicatures that you can buy at amusement parks. If you already know the person then you can see the vague resemblance, but it is not enough of a resemblance for a stranger to recognize the person based on the charicature.

That was not what I meant, I know that the term 'path of least resistance' is totally inaccurate, but I was just wondering why current moves throught the segment/path on which it is heated(creating resistance), shouldn't current flow through the zero ohm path (the loop that has the coil winding on it), I'm referring this to the case of the 'short circuit' as they are configurated the same way, or am I wrong?

 

[Dale]

I know that the term 'path of least resistance' is totally inaccurate,

So then reasoning based on it is also totally inaccurate.

I'm referring this to the case of the 'short circuit' as they are configurated the same way, or am I wrong?

It is not a short circuit. It has a very high inductance. Did you read about inductance? Do you have any questions about what you read?

 

[gneill]

That was not what I meant, I know that the term 'path of least resistance' is totally inaccurate, but I was just wondering why current moves throught the segment/path on which it is heated(creating resistance), shouldn't current flow through the zero ohm path (the loop that has the coil winding on it), I'm referring this to the case of the 'short circuit' as they are configurated the same way, or am I wrong?

Inductance, inductance, inductance. It's been mentioned several times that you need to look into inductance before you can understand what is happening in this circuit. The very large inductance of the loop simply will not allow its current to change instantaneously, so it appears as a very high impedance to the suddenly applied voltage (or current). The heated portion will have very low inductance and a small resistance, so it'll initially take essentially all of the current. As time goes by the current in the loop will ramp up until it's eventually taking (for all intents and purposes) all of the current.

 

[Mister T]

shouldn't current flow through the zero ohm path

There is no zero ohm path while the current is changing. The superconductor path has an impedance $Z$.

 

[Kenneth L Bridges]

I understand that the source of the heater is the resistance R, that being said, what governs the level of resistance and when it is changed what affect does
this have on the superconductor. Can Kevin explain. And if the R can be modulated through or by another means making it a variable, what does this do
to the equasion?

 

[Dale]

I understand that the source of the heater is the resistance R

The source of the heat is not the resistance R. The heat is supplied by a completely separate circuit that does not interact electrically with the superconductor

 

[SanderStols]

I enjoyed reading the discussion and I am glad to say that I now understand the whole process of setting up a large current in a superconducting coil.