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Magnetic Efficiency


aommaster

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Hi guys!

 

I am pretty sure I am going to regret it, but I'll ask anyway!

 

Let's say, you have two magnetics. One is an NIB (rare earth magnet, really powerful) and one is a fridge magnet (vey weak).

 

Now, you also have a fixed lenth of wire connected to an ammeter/voltmeter.

 

If you run each of the magnets down the wire, at the same speed, the rare earth magnet produces a greater current, we all know that.

 

Consider this:

Your kinetic energy is being changed by the magnetic field into kinetic energy. This means that the NIB magnet is has a greater efficiency.

 

Please correct me if I am wrong in any of the statements above, but, what exactly is going on here? Does this mean that you could get a magnet that is 99.9% efficient?

 

Thanks for your time guys.

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Hi.

 

..."Your kinetic energy is being changed by the magnetic field into electric is what you meant? energy. "...

 

I would say that the stronger magnet takes more effort to pass it over the wire; then, the kinetic energy is different in every case. More force to move the stronger magnet, less for the other.

 

The efficiency would be the ratio of the electrical energy generated and the kinetic energy requiered to move the magnet to produce such generation. In small experimenting driving magnets by hand, you cannot discern those minimal differences. You would need measuring instruments to show such.

 

Just imagine a permanent magnet toy motor, -used as generator- how much torque needs to be spinned when electrically loaded. The same motor with no magnets in it (extreme case of weak magnets), will need much less torque to spin it with the same electrical load.

 

Hope it helps,:)

Miguel

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vB=E

 

vBL=voltage

 

 

the problem becomes much easier to look at if you consider the magnet to be stationary, and the wire to be moving with some velocity v, the magnet has some field strength B.

 

 

its not really possible to get a solution without knowing the geometry of the field, but I can tell you that when you slide the magnet along the wire as you described one side, and not one end will become positive and the other negative.

 

also, no energy is used unless you make an electrical circuit with a positive and negative side. when you do this a current flows, this will inturn use some amount of power, this power will be taken out of the kinetic energy of either the wire or the magnet, (depending on which one is doing the moving), but this will then inturn reduce the velocity and thus the induced voltage in the electrical circuit, which will reduce the power being drawn out of the kinetic energy of the magnet/wire. my guess without doing the math on it is that the velocity will follow an exponential curve and thus will never be zero.

 

sorry for the ramble there, it didn't really have much to do with your question but anyway, I don't really know what you mean by efficiency in terms of the magnet so I can't be of much help there.

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Okay, let me try and explain what I meant:

 

Take it that you want to induce a current in a wire, let's say to power a very tiny bulb. You are given two magnets, a strong one and a weak one. Which magnet would you choose to make the bulb light up the most?

 

Most obviously the strong one, as for the same motion, you produce a geater current and hence a greater amount of electrical energy.

 

Does this make sense? Or am I talking a whole load of rubbish? :)

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Hmmm... didn't realise that! Does that mean that the force required for moving a magnet across a wire is dependant on the strength of the magnetic field?

 

Is there an equation that would allow you to work out the force required?

 

One more thing.... does this mean that the amount of current induced in the wire is contsant, only it is the force that makes the difference?

 

Thanks a lot for your time

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the more efficient magnet (the nib) and the fridge magnet, both of the same Mass will take X amount of energy to pass through space over the same distance.

now introduce a conductor and load, things change, Magnetic damping occurs, eddys are created and inverse magnetic fields oppose your movement.

you`ll not be able to detect this with your hand or arm as it`s only small difference, but non the less it IS there :)

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Are you referring to Lenz's Law?

 

If so, consider this: Does this mean that every magnet can only generate a certain amount of current, ragardless of the strength? If that is the case, then it makes sense that a stronger magnet generates more current because you are applying a bigger force, and hence, the conditions are not satisfied.

 

I hope I explained myself properly?

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  • 3 weeks later...
... Magnetic damping occurs' date=' eddys are created and inverse magnetic fields oppose your movement.

you`ll not be able to detect this with your hand or arm as it`s only small difference, but non the less it IS there :)[/quote']

 

Actually you can!

 

Take a fairly strong magnet and slide it quickly over a thick sqare rod or plate of copper - the braking effect of the created inverse magnetic field is clearly sensible, it feels just like sliding a magnet over iron. Its a quite funny experiment! - takes a strong magnet, though. Fridge magnets wont work.

 

Michael

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perhaps you should consider changing the wire in your thought experiment.

 

If you have a normal wire, the resistance that is created as you move the magnet goes into the current - the current does two things of interest:

1) It is a current. You could use it for something, like lighting a bulb.

2) It produces heat in the wire (entropy).

 

If you want higher efficiency, try using a superconductor, or anything with a lower resistance. By doing so, you reduce #2 and the efficiency of the energy that you expend is increased. With regards to the energy that goes into kinetic energy in the magnet ... that is, the force that you put on the magnet that does not get converted into electrical current but rather exists as kinetic energy in the magnet .. that force is irrelevant for the most part. You can reuse that by making the design circular. Even there you won't use all of it (friction with the circular track).

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