Hemal Pansuriya

1 hour ago, John Cuthber said:

Yes, and, to do it, you have to feed a thousand times more power into the transformer.

No, we don't have to. Because, magnet is not going to feel resistance during delay period. 

1 hour ago, swansont said:

Transformers don't violate conservation of energy. You trade current for voltage, one direction or the other. But the power is constant. P=IV

If you attach at transformer to a coil then the load changes. You've assumed that the only impedance is resistive.

I know transformers don't Violate energy conservation law...power in primary and secondary will be the same.....but by using transformer you can draw more power by the following way :  if voltage induced in coil is varying with time than use coil as voltage source and connect it to transformer's primary coil suppose voltage produced in coil is V, then step up this voltage by factor of 1000 ( or arbitrarily large ). Suppose resistance connected in secondary is R, so current in secondary winding will be I =1000V/R. And in primary winding current will be drawn 1000*I ( because IpVp=IsVs ). So now the power output of the coil will be V*(1000*I).So now the power output of the coil will be V*(1000*I) . We have drawn 1000 times more power than previously, which was V*I.

3 hours ago, John Cuthber said:

I looked at it.

You are wrong if you think you  will violate the conservation of energy.

This will remain true, no matter what you write and who looks at it.

 

If you have looked at it. Please tell me why can't we draw large power from coil with help of transformer. 

19 minutes ago, swansont said:

You will not violate conservation of energy

please just look at what i have written.

Coil can be very large and massive such that coil will move negligible.And if voltage induced in coil is varying with time than we can use transformer to draw more power by following way :use coil as voltage source and connect it to transformer's primary coil suppose voltage produced in coil is V, then step up this voltage by factor of 1000 ( or arbitrarily large ). Suppose resistance connected in secondary is R, so current in secondary winding will be I =1000V/R. And in primary winding current will be drawn 1000*I ( because IpVp=IsVs ). So now the power output of the coil will be V*(1000*I).So now the power output of the coil will be V*(1000*I) . We have drawn 1000 times more power than previously. 

9 minutes ago, swansont said:

It's not obvious to me how a constant speed will give a constant change in flux from a source that drops off as 1/r^3

Also, you can't get from 0 current to some value without there being a changing current.

If the coil is at some speed v, it will slow down when it encounters a changing flux. It feels a force.

 

If the velocity of magnet is constant, than change in magnetic flux is also constant.

To get from 0 current to some value is due to changing magnetic flux, after that if that change in magnetic flux is constant than the EMF will also constant .

The coil is not moving in this problem . only magnet is moving.

4 minutes ago, studiot said:

You didn't answer my question.

What is the magnetic flux change in the coil?

The magnetic flux is  integral of B.da . As magnet moves towards coil with constant velocity, the B increases at the coil and so does the magnetic flux. But due to constant velocity of magnet,  d∅/dt in coil is constant .

15 minutes ago, studiot said:

What exactly does the underlined bit mean?

You realise that the change in the magnetic field means that the number of lines threading the coil must change to generate an EMF?

The coil doesn't care which lines thread, it is the total number that count.

If the magnet has constant velocity towards coil, than the magnetic flux change in coil is constant ( d∅/dt = constant ). Hence the EMF and current are constant.

20 minutes ago, swansont said:

There's self-inductance which needs to be accounted for. Your impedance will not go to zero.

If the change in magnetic field is constant , than EMF produced is constant. Than current is also constant ( dI/dt = 0). So, there is no self inductance in this case. 

The resistance value can be put as low such that we get output energy  V*I > input energy .

19 minutes ago, studiot said:

During this delay period we can produce arbitrarily large EMF in coil by keeping number of turns of coil arbitrarily large NO

the EMF still obeys Farady's Law ie it is still proportional to the rate of change of magnetice field the coild actually sees. The number of turns of the coil is a red herring because the coils takes up space so the change takes time to propagate as it covers this space. ie turns sufficiently distant from the propagating change in the magnetic field do not affect the coil until they reach it.

You can increase the current in the coil other way also, like by decreasing resistance in the circuit assuming fixed EMF during delay period. Since, the power output is = V*I , you can increase power output by decreasing resistance value in circuit. 

can u tell me the reason and what have i ignored ?

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

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. 

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. 

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. 

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 ?

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.

Please everyone look at this pdf in which i have described the problem .

magnet.pdf

9 hours ago, swansont said:

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

Than magnetic field provides resistance to whom ?

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.  

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 ) .

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 ). 

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. 

eather causality is violated or energy conservation is violated . Both can not be hold simultaneously in this situation.

since there can not be instantaneous interaction between magnet and coil. so there should be violation of energy conservation.

Refer to my question: https://physics.stackexchange.com/questions/345924/violation-of-energy-conservation-in-retarded-interaction-of-magnet-and-coil