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I was wondering what voltage or current, if any, would be produced if the basic magnet through a copper coil experiment had the poles rotated 90 degrees so north and south faced the top/bottom of the coil rather than the entry/exit points?

The attached image has a basic illustration of this.


enter image description here

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up vote 3 down vote accepted

In that case, by symmetry, if you take any loop of the coil, for every field line that goes through it, if it goes to the negative pole, there must be another symmetrical that comes from the positive, so the total flux will be $0$, and constant, so there will be no current induced.

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Thanks for your input. Posting this to another forum gave me an opposite response. What would your thoughts be to the below out of interest? "To give specific numbers you'd have to know the field B(x,y,z) that the magnet produces. I don't think the outcome of the experiment would be too different - you should get an induced EMF in the coil."… – Larry B Feb 4 '13 at 23:38
In that post, the guy has edited it to say it's wrong. I think my answer is right, just by symmetry, actually, you have posted my answer there and they have told you it's right. – MyUserIsThis Feb 5 '13 at 9:37
Right you are... Where can I learn in greater detail this concept of symmetry in induction by the way? – Larry B Feb 5 '13 at 9:46
@RodgersandHammertime I don't think that exists like a subject by itself. It's just analysing the system and trying to understand it by "seeing" the field lines and how the move over time. I don't think it's really complicated but the way I've learned it has been just by doing lots of problems. – MyUserIsThis Feb 6 '13 at 12:25
Be aware that actual bar magnets--such as the ones found in intro physics laboratories--typically do not have highly symmetric field shapes. The experimental result would likely be a very weak and "wibbly-wobbly" signal. – dmckee Apr 24 '14 at 14:33

Unless the magnet is kept perpendicular to the copper coil,there would no current or voltage

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