Lenz's law and conservation of energy

[annie122]

I don't understand how lenz's law is a form of energy.
An explanation said if the induced current did not oppose change, we would be obtaining energy without doing any work.

I think my problem here is that I can't follow the

no change -> get free energy

logic.

 

[tiny-tim]

hi annie122!

An explanation said if the induced current did not oppose change, we would be obtaining energy without doing any work.
…
no change -> get free energy

no, it means that if the induced current "supports" the change in magnetic flux, we'd get free energy … both the current and the flux would be increasing in energy!

(and if there's no induced current, there's no problem)

eg if you have two circular wires, with a current in one inducing a current in the other, then the second current will tend to reduce the first current rather than increase it …

the energy in the second circuit is being drained from the first circuit! ​

 

[annie122]

Thanks. :)

the energy in the second circuit is being drained from the first circuit!

I got this part.
But I realized I've been missing something really important.

both the current and the flux would be increasing in energy!

I'm going to ask a really fundamental question;so current and flux are forms of energy?

In currents, electrons are moving, so more current -> more KE in energy?

But what about flux?
How is flux related to energy?

 

[rude man]

I don't understand how lenz's law is a form of energy.
An explanation said if the induced current did not oppose change, we would be obtaining energy without doing any work.

I think my problem here is that I can't follow the

no change -> get free energy

logic.

Take a permanent magnet bar and move it towards a coil, inducing a current in the coil. Say you move the N pole towards the coil.

If the current induced went the other way you would have made a magnet out of the coil with its S pole facing the bar's N pole, attracting the bar to the coil. You would have generated heat in your coil - free energy! That violates the conservation of energy. Plus, the coil's B field (and flux) was built up, again with no energy input.

In reality, your coil would actually have generated a N pole facing the bar magnet, repelling the bar from the coil. If you still want to move the bar towards the coil you would have to push the bar towards the coil. This time you generated the energy necessary to heat up the coil and build up its magnetic field.

Building up a magnetic field takes energy. It can be shown that the energy density (joules per unit volume) of a B field = B²/2μ.

 

[annie122]

Take a permanent magnet bar and move it towards a coil, inducing a current in the coil. Say you move the N pole towards the coil.

If the current induced went the other way you would have made a magnet out of the coil with its S pole facing the bar's N pole, attracting the bar to the coil. You would have generated heat in your coil - free energy! That violates the conservation of energy. Plus, the coil's B field (and flux) was built up, again with no energy input.

In reality, your coil would actually have generated a N pole facing the bar magnet, repelling the bar from the coil. If you still want to move the bar towards the coil you would have to push the bar towards the coil. This time you generated the energy necessary to heat up the coil and build up its magnetic field.

Building up a magnetic field takes energy. It can be shown that the energy density (joules per unit volume) of a B field = B²/2μ.

It can be shown that the energy density (joules per unit volume) of a B field = B²/2μ.

Okay, that got me cleared up on that, thanks.

You would have generated heat in your coil - free energy! That violates the conservation of energy.

Is this because since there's always resistance in the coil, inducing current produces heat by the I^2R law?

 

[rude man]

Okay, that got me cleared up on that, thanks.



Is this because since there's always resistance in the coil, inducing current produces heat by the I^2R law?

Right.

If the coil were a true superconductor with zero resistance then you could not move the bar towards the coil at all! I believe somebody is hoping to develp superconductor rails for levitating trains which would work along the same idea.

 

[annie122]

Right.

If the coil were a true superconductor with zero resistance then you could not move the bar towards the coil at all! I believe somebody is hoping to develp superconductor rails for levitating trains which would work along the same idea.

Can you explain this part?

If the coil were a true superconductor with zero resistance then you could not move the bar towards the coil at all!

So to summarize:
If current went in the same way,
1) the B-field, which requires energy to make stronger, would be strengthened, and
2) the coil would give off more heat without me doing any work (or rather would it be negative work, since the field is in the direction of motion?)
Therefore the current must be in the opposite direction.

 

[rude man]

Can you explain this part?


So to summarize:
If current went in the same way,
1) the B-field, which requires energy to make stronger, would be strengthened, and
2) the coil would give off more heat without me doing any work (or rather would it be negative work, since the field is in the direction of motion?)
Therefore the current must be in the opposite direction.

If the were no resistance then if you could move your magnet bar you would be generating a finite emf and infinite current = emf/0, creating an infinitely strong magnet from your coil, so you couldn't move the bar!

I'd say yes for 1) and 2). Don't know about 'negative work' but I suppose yes, the work would be negative if the bar were sucked into the coil instead of havig to push it in.

To make your life more complex: a permanent magnet will attract iron pieces, which means work is done by the magnet since the pieces had to be accelerated, which means force, and force times distance = energy (work). So where didthat energy come from? What happened to energy conservation?

The answer is that the "amperian currents" surrounding a permanent magnet, making a sort of electromagnet like a solenoid, see no resistance and so do not dissipate energy. This gets into quantum mechanics with which I have as little to do as possible!

 

[anhnha]

If the were no resistance then if you could move your magnet bar you would be generating a finite emf and infinite current = emf/0, creating an infinitely strong magnet from your coil, so you couldn't move the bar!

Impressive!

I'd say yes for 1) and 2). Don't know about 'negative work' but I suppose yes, the work would be negative if the bar were sucked into the coil instead of having to push it in.

Can you explain why the work is negative?

To make your life more complex: a permanent magnet will attract iron pieces, which means work is done by the magnet since the pieces had to be accelerated, which means force, and force times distance = energy (work). So where didthat energy come from? What happened to energy conservation?

The answer is that the "amperian currents" surrounding a permanent magnet, making a sort of electromagnet like a solenoid, see no resistance and so do not dissipate energy. This gets into quantum mechanics with which I have as little to do as possible!

I am still not clear with this answer. Can you give it more detail or a link explaining about that?

 

[rude man]

Impressive!

Can you explain why the work is negative?

I am still not clear with this answer. Can you give it more detail or a link explaining about that?

If the coil had sucked the bar into the coil the work would be negative because F would be opposite in direction to displacement - I mean the force you would have to "exert" to move the bar. Instead of you exerting force on the bar, the bar wouild be exerting force on you!

Or how about this: you take a relxed spring and pull it a distance x. You have exerted work = integral F dx = kx^2/2 and that energy is stored in the spring.

You then allow the spring to get back to its relaxed position, but still holding on. Now you have "exerted" negative work on the spring and the spring stored energy = 0. The spring's stored energy went to pulling your hand thru the distance x. You have performed negative work on the spring.

 

[anhnha]

Thanks, I have problem in understanding negative work.

Or how about this: you take a relxed spring and pull it a distance x. You have exerted work = integral F dx = kx^2/2 and that energy is stored in the spring.

Yes, in this case I understand. I have to exert a force onto the spring.

You then allow the spring to get back to its relaxed position, but still holding on. Now you have "exerted" negative work on the spring and the spring stored energy = 0. The spring's stored energy went to pulling your hand thru the distance x. You have performed negative work on the spring.

This part is confusing me. In this situation, I do nothing. There is a force of spring exert on my hand and the work of spring is positive work.
But I do nothing and I don't exert a force onto spring then how can the negative work possible?

 

[milesyoung]

To make your life more complex: a permanent magnet will attract iron pieces, which means work is done by the magnet since the pieces had to be accelerated, which means force, and force times distance = energy (work). So where didthat energy come from? What happened to energy conservation?

The answer is that the "amperian currents" surrounding a permanent magnet, making a sort of electromagnet like a solenoid, see no resistance and so do not dissipate energy. This gets into quantum mechanics with which I have as little to do as possible!

I think that's stretching it a bit. It's akin to asking where the energy comes from for gravity to accelerate mass.

The system of the permanent magnet and iron pieces will have potential energy by virtue of its configuration of magnetic moments. The principle of energy conservation still holds.

 

[rude man]

I think that's stretching it a bit. It's akin to asking where the energy comes from for gravity to accelerate mass.

The system of the permanent magnet and iron pieces will have potential energy by virtue of its configuration of magnetic moments. The principle of energy conservation still holds.

Good point. What I wanted to do was to distinguish between the magnetic field generated by the coil, which requires work to set it up, and the existence of the permanent magnet's field which does not.

Thanks.

 

[rude man]

Thanks, I have problem in understanding negative work.

Yes, in this case I understand. I have to exert a force onto the spring.

This part is confusing me. In this situation, I do nothing. There is a force of spring exert on my hand and the work of spring is positive work.
But I do nothing and I don't exert a force onto spring then how can the negative work possible?

You don't 'do nothing'. You resist the force pulling the spring back to its rest state. So the force you apply is opposite to the displacement and work = force (vector) doted into displacement (vector) which here is clearly negative.

 

[anhnha]

Thanks for your patience.

You don't 'do nothing'. You resist the force pulling the spring back to its rest state. So the force you apply is opposite to the displacement and work = force (vector) doted into displacement (vector) which here is clearly negative.

The force that I exert to the spring and the force of spring pulling back are equal?
If so how can the spring move back to the rest state because it starts at rest when spring are pulled to the maximum position?

 

[rude man]

I don't understand the question.

1. You pull the spring from rest position to position x. You perform positive work in doing so. You hold the spring at position x.

2. You allow the spring to go back to its rest position while still holding onto the spring. You have to resist the spring pulling itself back. The force you apply in going from x to zero is negative and so the work you perform is negative.
.

 

[anhnha]

2. You allow the spring to go back to its rest position while still holding onto the spring. You have to resist the spring pulling itself back. The force you apply in going from x to zero is negative and so the work you perform is negative.

Let's me express my quesion more detail.
In this case the force that I exert to spring is in the opposite direction of spring's moving.
There are two force exert to spring.
The pulling force from x to 0 and the force that I exert with the opposite direction.
Do the two forces have the same magnitude but oppsite directions?
If so(two forces equals in manginute) then how can the spring move to the rest state?

 

[rude man]

Let's me express my quesion more detail.
In this case the force that I exert to spring is in the opposite direction of spring's moving.
There are two force exert to spring.
The pulling force from x to 0 and the force that I exert with the opposite direction.
Do the two forces have the same magnitude but oppsite directions?
If so(two forces equals in manginute) then how can the spring move to the rest state?

good question.

As you allow the spring to return to its rest position you are gradually reducing your restraining force. In fact, the restraining force at any point x is kx as you go along from x to zero.

If you do this infinitely slowly then the two forces are always equal and opposite. If you do it faster then you'll have to ask an ME what goes on ... there will be accelerations and decelerations which implies unequal forces but I'm not sure what the equivalent mass would be - your hand or ?.
.

 

[anhnha]

Thanks, got it

If you do it faster then you'll have to ask an ME what goes on ...

What is "ME"?

 

[milesyoung]

Do the two forces have the same magnitude but oppsite directions?
If so(two forces equals in manginute) then how can the spring move to the rest state?

For simplicity's sake, let's assume the spring itself is massless and it's the displacement of a mass attached to its movable endpoint we're considering. If the force exerted by the spring and your hand on the mass are equal in magnitude and opposite in direction, and there are no other forces applied to it, there is no net force on the mass and it won't accelerate.

I think your confusion comes from the fact that you're applying Newton's third law incorrectly. Consider the mass alone:
- The spring exerts a force on the mass and the mass thus exerts a force equal in magnitude and opposite in direction on the spring.
- Your hand exerts a force on the mass and the mass thus exerts a force equal in magnitude and opposite in direction on your hand.

This is true whether or not the force exerted by the spring and your hand on the mass are equal in magnitude.

 

[anhnha]

Thanks, milesyoung.
But that is not my problem. My problem was that I didn't know that the process spring going back to its rest position has to do infinitely slowly. And we only have to accelerate and decelerate it a bit at the start and the end of process.

 

[milesyoung]

My problem was that I didn't know that the process spring going back to its rest position has to do infinitely slowly. And we only have to accelerate and decelerate it a bit at the start and the end of process.

I'm not sure what you mean by this, but you're free to accelerate the mass however much you'd like.

 

[rude man]

Thanks, got it

What is "ME"?

"Mechanical Engineer".

 

[anhnha]

I'm not sure what you mean by this, but you're free to accelerate the mass however much you'd like.

I think the work done is calculated under the constraint of being quasi-static, meaning that it is the force needed to move the spring without accelerating it. The "quasi" part let's you accelerate a bit at the beginning and declerate it the same amount at the end.

This is what I understand from rude man. Is my understanding correct?

 

[milesyoung]

I think the work done is calculated under the constraint of being quasi-static, meaning that it is the force needed to move the spring without accelerating it. The "quasi" part let's you accelerate a bit at the beginning and declerate it the same amount at the end.

This is what I understand from rude man. Is my understanding correct?

There's no need for any constraint on the acceleration of the mass. However you choose to go about it, if your hand brings the mass-spring-system to rest at its equilibrium position, then the mass-spring-system has transferred its initial potential energy to the hand-system. That means the hand-system has done negative net work on the mass-spring-system (if it was positive, the energy of the mass-spring-system should have increased).

 

[rude man]

I should point out that when I came up with the spring analogy I had no mass attached to it besides the experimenter's hand.

The idea was to point out that there is a continuous equality between the potential energy of the spring and the work (force times displacement) applied by the hand, going in both directions.

 

[milesyoung]

I should point out that when I came up with the spring analogy I had no mass attached to it besides the experimenter's hand.

I used a lumped element model so it would be (arguably) easier to identify what's accelerating, what has kinetic energy etc.

The conclusion remains the same, though, if the hand-system brings the spring-system to rest in its equilibrium configuration.

Edit:

The idea was to point out that there is a continuous equality between the potential energy of the spring and the work (force times displacement) applied by the hand, going in both directions.

I wasn't trying to argue against your point, my conclusion even supported it. I wanted to get rid of that quasistatic business, which I don't think you necessarily meant as a constraint. The OP, though, might have considered it as such.

 

[anhnha]

Now I will post what I understand here and hope you can correct it.

For simplicity's sake, let's assume the spring itself is massless and it's the displacement of a mass attached to its movable endpoint we're considering. If the force exerted by the spring and your hand on the mass are equal in magnitude and opposite in direction, and there are no other forces applied to it, there is no net force on the mass and it won't accelerate.

If the spring is massless then there is no net force on the mass and the force exerted by the spring and my hand on the mass are equal in magnitude and opposite in direction no matter what the acceleration is.
F = ma = 0 always no matter what value of a is.
It means that I can choose to accelerate it or not accelerate it the result of work done by my hand still the same.
F = 0 => Fspring + F my hand = 0 => F of my hand is opposite direction of F spring.
And I can imagine how the work done by my hand calculated by using F*s with F in the opposite direction of spring's moving. Therefore, the work done by my hand is negative.
But how about if we take the mass of spring into consideration.
F = ma ≠ 0
How can I know the force of my hand is in the opposite direction of spring force?

 

[milesyoung]

In my description, there are 3 distinct objects: The massless spring S, the mass attached to its movable endpoint M and your hand H.

S is considered massless as a mathematical abstraction and it's only there to remind you that there's a force exerted on M that's proportional to its displacement from some reference point. You could have left out the spring completely and just considered M to be within a field that exerts an equivalent force on M. Applying Newton's second law to S won't make much sense.

Anyway, to review what I originally responded to, from post #17:

In this case the force that I exert to spring is in the opposite direction of spring's moving.
There are two force exert to spring.
The pulling force from x to 0 and the force that I exert with the opposite direction.

Here I consider the spring to be a deformable solid. It has two external forces applied to it. One is from whatever keeps one of its endpoints fixed and the other is from your hand. Whatever else is internal to the spring.

If you were instead to consider a small mass element of its movable endpoint then that object would also have two forces applied to it. One is the force from its connection with the spring and the other is from your hand. In that case, this:

Do the two forces have the same magnitude but oppsite directions?

isn't necessarily true, for reasons I have already explained.