# Electromagnetic Induction

by markuz88
Tags: coil, electromagnetic, induction, magnet
 P: 3 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
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P: 1,998
 Quote by 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.
Correct.

 Quote by markuz88 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.

 Quote by markuz88 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.
 P: 3 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
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