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Electromagnetic induction and energy conservation


rajeesh

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In this whole process time delay is 2t0 . When the magnetic field change reaches to the coil , we can draw arbitrarily large amount of current from the coil by making number of turns of the coil arbitrarily large value ( EMF in the coil increases as number of turns of coils increases ). Since the signal from coil to magnet haven't reached yet,  the magnet wont feel any resistance . By that time, when the signal reaches from coil to magnet, we can produce arbitrarily large energy in the coil. 

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During the time when the signal from coil haven't reached to the magnet and the the magnet is not experiencing any resistance, at that time u can draw arbitrarily current from coil by keeping arbitrarily large numbers of turns of the coil ( EMF in the coil increases as number of turns of coils increases ). 

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42 minutes ago, Hemal Pansuriya said:

During the time when the signal from coil haven't reached to the magnet and the the magnet is not experiencing any resistance, at that time u can draw arbitrarily current from coil by keeping arbitrarily large numbers of turns of the coil ( EMF in the coil increases as number of turns of coils increases ). 

No, you can't. Any current in the coil will generate an opposite field, which increases the resistance to motion. An infinite number of turns would give you an infinite force, and you would have no motion.

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For the time during which the signal is not reached to the magnet, the magnet will not feel any resistance. so during this time we can create arbitrarily large current. Yes, i agree, When this signal will reach to magnet from coil , the magnet will stop. But we have created arbitrarily large amount of energy during this delay time which is 2t0. Where as our input energy was just the initial push to magnet ( which is a fix amount ) .

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1 hour ago, Hemal Pansuriya said:

For the time during which the signal is not reached to the magnet, the magnet will not feel any resistance. so during this time we can create arbitrarily large current. Yes, i agree, When this signal will reach to magnet from coil , the magnet will stop. But we have created arbitrarily large amount of energy during this delay time which is 2t0. Where as our input energy was just the initial push to magnet ( which is a fix amount ) .

No. There is a magnetic field in place. That provides resistance.

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No, during the delay time the magnet can not feel resistance. otherwise it would violate causality. signal from coil can not go to magnet instantaneously. Until the signal reaches from coil to magnet, the magnet will not be aware of what has happened to coil . So there can not be any resistance felt by magnet during this delay time.  

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1 hour ago, Hemal Pansuriya said:

No, during the delay time the magnet can not feel resistance. otherwise it would violate causality. signal from coil can not go to magnet instantaneously. Until the signal reaches from coil to magnet, the magnet will not be aware of what has happened to coil . So there can not be any resistance felt by magnet during this delay time.  

Did I say anything about the magnet? I said the magnetic field.

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On 7/29/2017 at 4:02 PM, Hemal Pansuriya said:

During the time when the signal from coil haven't reached to the magnet and the the magnet is not experiencing any resistance, at that time u can draw arbitrarily current from coil by keeping arbitrarily large numbers of turns of the coil ( EMF in the coil increases as number of turns of coils increases ). 

I have replied my friend.

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21 minutes ago, Hemal Pansuriya said:

I have replied my friend.

I am sorry I didn't realise that was a reply to anything I said, all of which is tailored to the OP.

I did ask you how arbitrarily large energy could appear in contravention of COE, but you simple reasserted the OP question as fact.

I have already outlined the correct way to go about this, with one exception.

Your particular misconception is based on the same fallacy that asserts that lead has infinite densite.

That is if we take the mass of a lump of lead and take the limit of the mass to volume ratio as the volume goes to zero, we get the (finite) density at a point. Similarly with force and stress or pressure at a point.

The equations of inductions are differential equations with the same sort of limiting ratio.

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12 minutes ago, Hemal Pansuriya said:

How will  energy conservation happen at the end of process ?  our input energy is, the  kinetic energy that we have given to magnet. How this input energy will be equal to output energy that we will get in the coil ?

You will have losses, like heating in the coil, but you have stated the obvious answer here: you have given the coil kinetic energy, i.e. you are doing work on it. That's the source of the energy for the current that is induced.

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  • 2 weeks later...
On 8/3/2017 at 10:14 PM, swansont said:

You will have losses, like heating in the coil, but you have stated the obvious answer here: you have given the coil kinetic energy, i.e. you are doing work on it. That's the source of the energy for the current that is induced.

The initial energy that we are giving to magnet is fixed , it will not change during the process because during the delay period magnet will not going to experience any resistance. So during this delay period whatever amount of current flows in the coil we don't have to put any extra input energy in magnet. 

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8 hours ago, Hemal Pansuriya said:

The initial energy that we are giving to magnet is fixed , it will not change during the process because during the delay period magnet will not going to experience any resistance. So during this delay period whatever amount of current flows in the coil we don't have to put any extra input energy in magnet. 

I feel like I'm talking to a wall. I said you are doing work on the coil.

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7 hours ago, Hemal Pansuriya said:

Work is only done when magnet feels any resistance, my friend. Energy in the coil produces due to change in magnetic field but for that change, magnet is not feeling any resistance during delay period. 

Aauuuuugh!

The coil sees a field, which is changing. The coil feels a resistance. You push on the coil, doing work. 

Look carefully. Was the magnet mentioned at all?

 

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2 hours ago, swansont said:

Aauuuuugh!

The coil sees a field, which is changing. The coil feels a resistance. You push on the coil, doing work. 

Look carefully. Was the magnet mentioned at all?

 

We are not moving COIL at all. We are moving magnet towards coil. So the magnet won't feel any resistance during delay period. 

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If you are moving the magnet, there will be a delay before the coil sees a change in the field. This was addressed some time ago. The coil sees no change in the field during the delay, so there is no current in it.

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4 minutes ago, swansont said:

If you are moving the magnet, there will be a delay before the coil sees a change in the field. This was addressed some time ago. The coil sees no change in the field during the delay, so there is no current in it.

I don't know where you find the patience - I would long ago have been looking for the proverbial big hammer.

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33 minutes ago, swansont said:

If you are moving the magnet, there will be a delay before the coil sees a change in the field. This was addressed some time ago. The coil sees no change in the field during the delay, so there is no current in it.

The coil will see the magnetic field change When signal reaches from magnet to coil. As soon as coil feels magnetic field change it produces current. This current will generate magnetic field that will resist the motion of magnet. But it is not instantaneous, as signal from coil to magnet will take time to travel. During this delay period we can produce arbitrarily large EMF in coil by keeping number of turns of coil arbitrarily large and hence power output of coil arbitrarily large. We will not have to give any extra input energy to magnet during this delay period.

You can refer to my PDF , in which i have described the problem more clearly. 

magnet.pdf

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30 minutes ago, Hemal Pansuriya said:

 During this delay period we can produce arbitrarily large EMF in coil by keeping number of turns of coil arbitrarily large and hence power output of coil arbitrarily large. We will not have to give any extra input energy to magnet during this delay period.

No, you can't, for reasons already described and ignored by you.

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