Electromagnetic Induction of a permanent megnet

[markuz88]

Hello everyone,

How are you doing?

I have a doubt about electromagnetic induction, in three particular cases. I need to confirm that I have the right concepts, so I ask for your help.

The main problem:

Imagine that you have a permanent magnet, axially polarized and rotating on its axis with a constant angular speed. Surrounding this magnet, a coil (constant area section pointing in the same direction of magnet polarization). The main question is: will there be induced voltage?

This is what I think:

1) We know that, for a constant Area, flux linkage ψ = B*A*cos θ.
In this case θ = 0°, so ψ = B*A.
And the induced voltage is ε = -N*dψ/dt = -N*A*dB/dt.

In this main case, I think that there will be no variation in B, because the rotation does not change it at all. So dB/dt = 0, thus ε = 0.

2) Let's suppose the magnet is now radially polarized, but keeping the surrounding coil. In this case, can I affirm that rotation still doesn't change B at all (actually it does change B, but if we consider the whole thing it does not)? And not only because of this ε is zero, but θ = 90°, which implies ψ = 0.

3) Now suppose the coil doesn't fully surround the magnet. Let's say it covers only 270° of it (a little abstraction is needed, I know ). In this case of non-symmetry, there will be a variation in B, but ε is still zero because θ = 90°.

Am I correct? Did I miss something?

Thank you,

Marcus

 

[marcusl]

Hello everyone,

How are you doing?

I have a doubt about electromagnetic induction, in three particular cases. I need to confirm that I have the right concepts, so I ask for your help.

The main problem:

Imagine that you have a permanent magnet, axially polarized and rotating on its axis with a constant angular speed. Surrounding this magnet, a coil (constant area section pointing in the same direction of magnet polarization). The main question is: will there be induced voltage?

This is what I think:

1) We know that, for a constant Area, flux linkage ψ = B*A*cos θ.
In this case θ = 0°, so ψ = B*A.
And the induced voltage is ε = -N*dψ/dt = -N*A*dB/dt.

In this main case, I think that there will be no variation in B, because the rotation does not change it at all. So dB/dt = 0, thus ε = 0.

Correct.

2) Let's suppose the magnet is now radially polarized, but keeping the surrounding coil. In this case, can I affirm that rotation still doesn't change B at all (actually it does change B, but if we consider the whole thing it does not)? And not only because of this ε is zero, but θ = 90°, which implies ψ = 0.

Correct.

3) Now suppose the coil doesn't fully surround the magnet. Let's say it covers only 270° of it (a little abstraction is needed, I know ). In this case of non-symmetry, there will be a variation in B, but ε is still zero because θ = 90°.

I don't understand your geometry. The classic case is a bar magnet magnetized along its axis z, near a coil parallel to the x-y plane that is located a small distance away along the z axis. Now spin the magnet around the x-axis (at the magnet midline). Each time the pole swings past the coil, it introduces a large flux in the coil.

 

[markuz88]

Ah, first, I forgot to tell that these permanent magnets are round magnets.

But the third case is a bit more complicated... well, you have just described a "common" generator, right?

And thank you for your reply! If you let me, I want to ask you other geometry. This is going to help me understand a bit more. I drew it to make it easier to see the problem. The red/blue part of magnet is only north/south pole division (ie, in the picture, it is polarized along axis X).

If this magnet rotates around axis X with a constant speed ω, as the coil remain still, should I expect induced voltage? I guess not, because, again, θ = 90°. But what I can't see is: what if the magnet is polarized in Z axis? Notice that this is very similar to case (2) I described before, but the coil is in front of the magnet, not surrounding it.

Thanks again,

Marcus

 

[marcusl]

First part--you are right. Second part--what do you think? Draw your magnet as a dipole, for example, and draw a few field lines around it to see what happens.

 

[markuz88]

I think I see... dB/dt will not be zero.

Thanks for your help, marcusl.

 

[marcusl]

You are welcome.