# Electromagnet + P.Magnet interaction + C-EMF

1. Jan 8, 2013

### Miyz

Hello :)

I was working on a project for one of my classes, and it had to do with electromagnets + magnet's B field interactions(A simple demonstration of a few laws of electromagnetism).

Basically its a simple configuration like this random photo I found:
http://img577.imageshack.us/img577/353/duallayercoilforcounter.jpg [Broken]

Where you can see a permanent magnet and a coil(that we can consider as an electromagnet for the sake of illustration).

Ok, I was thinking about lenz's laws, started to think deeply about counter-EMF more and more.

Based on the configuration you see in the image:

1) In order for the C-EMF to be generated, does the magnet need to pass through the whole electromagnet? Or thats not necessary? If the magnet was rotating or moving near by the coil still C-EMF would be generated?

2) Imagine the magnets where on a mobile cart that has low friction as possibile. The B field created by the electromagnet(Em1) is weaker than of the permanent magnet(Pm1). In theory if the magnet was in motion near by the electromagnet the C-EMF would be higher than the input EMF, thus the total current flow would be zero or a negative. Due to the greater C-EMF. Now the question is the following: Although Em1's B field is weaker than Pm1's field... Can Em1 still attract/repel Pm1? Without consuming more energy from the power source.

I believe Em1 can do so... It can attract Pm1 and move the cart closer/farther depending on the type of magnetic force applied.

Hope that I made good sense :tongue:

Thank you!
Miyze,

Last edited by a moderator: May 6, 2017
2. Jan 8, 2013

### Staff: Mentor

The magnet has to move (close enough to the coil to influence it) or the current in the coil has to change. Rotating is fine, too, if it is not around its symmetry axis (=axis of the magnetic field). Moving the coil relative to the magnet works as well.
Depends on the setup.
As long as you have a field in both, they can attract or repel each other.
Might need a deeper analysis of the specific setup.

3. Jan 8, 2013

### Miyz

But wouldn't the C-EMF cause a major problem? I mean if Em1 is attracting Pm1 from a far, that force of attraction would cause the cart to move the magnet(Pm1) closer and closer, however, as the magnet gets closer! A greater C-EMF is generated and would lead the electromagnet to lose it power. I was thinking in theory that would cause the magnet to be repelled because of the C-EMF... My whole worry is that the C-EMF could cause a problem that will not allow the magnet to be fully attracted by the electromagnet.

4. Jan 8, 2013

### Miyz

What if the setup was based on the question? Weak B field from Em1 and a stronger B field from Pm1 the current can't increase the input was stable (In theory).
Generally...Would the current increase of the C-EMF was higher?

5. Jan 9, 2013

### Staff: Mentor

Problem for what?

If you power the electromagnet with the correct orientation, you will fully attract the magnet. In the wrong orientation, the magnet will turn (if it can) and be attracted afterwards. This assumes that the force is strong enough to overcome friction in the system, of course.

6. Jan 9, 2013

### Miyz

Sine the magnet's field is way stronger than of the electromagnet I worry about the C-EMF generated from the magnet being attracted closer to the electromagnet would be a problem! The C-EMF would obviously be greater than the input EMF supplied to the magnet.

Thats the problem... I feel for some reason this problem will not occur. But have to make sure!
Don't want to fail while finishing my set up!

Even if the magnet's magnetic field is much more stronger than of the electromagnet, can the electromagnet still attract it?
And when you mean by orientation... Thats the poles right? Since you said "the magnet will turn and be attracted afterword" thought it would flip for the right pole and be attracted.

7. Jan 9, 2013

### Miyz

Ah the confusion... :tongue:

8. Jan 13, 2013

### Staff: Mentor

The fact that the magnet's field is stronger is not relevant. The counter EMF is due to a change in the field. So, for sufficiently low electromagnet fields the force will be very weak so the motion will be very slow so the change in the field will be small so the counter EMF will be very low.

9. Jan 13, 2013

### Miyz

Hey Dale,
Thanks for joining in!

Why isn't it relevant?

Of both the permanent magnet and of the electromagnet?

Well wouldn't the speed of motion depend on the force of attraction we're dealing with in the set up? + The force of attraction is not solely applied by the electromagnet but of both the magnet + electromagnet acting on each other. I came up with that based on Newtons laws... Because it still will apply in electromagnetism wouldn't it?

Last edited: Jan 13, 2013
10. Jan 14, 2013

### Miyz

Ah, now I'm lost again! While reading more about Newton's third law, I found this statement here:

"Another demonstration might be to show two bar magnets. Choose one that is stronger than the other; demonstrate this by showing that one can lift a greater iron weight than the other.

Ask: If the two magnets attract one another, will one pull more strongly than the other? The answer is, no. You can feel that they pull each other equally. (This is because the force is proportional to the strength of each.)

If magnet A pulled magnet B more strongly than B pulled A, you could attach B to the front of your car and lean out, holding A in front. Your car would move effortlessly!"

You can apply this example to my set up, by replacing magnet B with my electromagnet.
Yet, this makes no sense to me... How is it that a stronger source is applying the same amount of "force" to a weaker one...

11. Jan 15, 2013

### Staff: Mentor

You get the same effect everywhere, for example with gravity: The force between two objects is mMG/r^2 - for both objects. Moon pulls on earth with the same force earth pulls on moon - but as our earth is more massive, its acceleration is smaller.

"Strong" just means "more influence between this object and another one".

12. Jan 16, 2013

### Miyz

I just realized how I CONFUSED both idea's and principles.
I'll review my statement and study my project and return with more feedback!

13. Jan 16, 2013

### Miyz

mfb and Dale, thank you for your efforts!
Much appreciated.

I've reached to a conclusion that still the electromagnet will be able to move the cart with the magnet to the desired point!
C-EMF is not an issue since, the magnet itself is not close enough to induce any current.
It has to be attracted closer and even then! It will not have much speed to even cause a problem. The electromagnet being weaker in strength will be able to fully attract the magnet.

Now the motion will depend on many factors, but its certainly not only due from the electromagnet! But from BOTH forces in the system: The electromagnet's force + The magnet!.

Even if the C-EMF were to be a problem, we can switch the power off, and due to the momentum gained, the cart can move freely to the desired point. There is low friction of course.
I hope this is right!

Last edited: Jan 16, 2013
14. Jan 17, 2013

### Miyz

I hope anyone can give me back feedback on my final post.

15. Jan 29, 2013

### Staff: Mentor

I agree with your conclusion and analysis except for the bold part. The distance only changes how much current is needed in the electromagnet, but the main point is that C-EMF is not an issue while the permanent magnet is moving slowly. Once it starts moving quickly then, as you say, it will continue due to momentum. Any C-EMF at that point can be used to generate electrical power even.

16. Jan 29, 2013

### Miyz

I'm sorry, the magnet will not be far it actually will be close, close enough to be attracted RAPIDLY by the electromagnet, since I'm limiting the input current. The electromagnet will have to draw more and more current to resist the C-EMF, however, I can manually monitor the C-EMF if it is indeed greater than the input EMF, I'll just shut the power off. Causing the magnet to be attracted and move due to the momentum. In terms of how fast? Well that will depend with the amount of force being applied more specifically the kinds of B fields Im going to work with.

But Thanks Dale!
Much appreciated!

Last edited: Jan 29, 2013