Why does a diamagnetic rod align perpendicular to a magnetic field?

  • #1
vcsharp2003
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Homework Statement:
I have the following in my textbook, but can't explain why diamagnetic rod is becoming perpendicular to the magnetic field as shown in fig 37.6 (a).
Relevant Equations:
None
I know that each material is made up of tiny magnets due to electrons orbiting the nucleus and also from electron spinning about its own axis. In ferromagnetic or paramagnetic rod these tiny magnets align with the applied field causing the net field in the rod to increase. But for diamagnetic rod, the tiny magnets align so that they oppose the applied magnetic field, which causes the net magnetic field to decrease in the rod.

From above facts, I cannot find an explanation of why rod becomes perpendicular for diamagnetic rod. Perhaps a torque acts on the rod and in perpendicular position there is no net torque and hence it comes to rest in that position.
 

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  • #2
haruspex
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If something is repelled by objects on both sides, according to an inverse square law, where is the least PE?
 
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  • #3
kuruman
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I think the argument that the excerpt you posted is this. In the ferro 0r paramagnetic case (Figure b) the magnet poles attract the end of the rod that is closest to them and repel the end that is farthest from them. In that case the rod aligns parallel to them. In the ferromagnetic case, one can label the ends of the rod with N and S regardless of whether the external field is on or off. In the paramagnetic case, one can still label the ends of the rod N and S but only when the field is on. In either case the rod's N will be closer to the magnet's S and the rod's S closer to the magnet's N.

In the case of the paramegnet diamagnet there can be no such labeling of the rod's ends. There is repulsion between each magnet pole and each rod end which get stronger when the end moves closer to the pole. Thus, if the rod turns away from being perpendicular to the field, the stronger repulsive force at each end of the rod will restore it back to being perpendicular to the field lines.
 
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  • #4
vcsharp2003
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If something is repelled by objects on both sides, according to an inverse square law, where is the least PE?
Isn't inverse square law in electrostatics and gravitation? I don't know about an inverse square law in magnetism. I know that potential energy of a magnetic dipole in a uniform magnetic field is ##U =- \vec M \cdot \vec B = -|\vec M| |\vec B| cos \theta##.

OR maybe by inverse square law you meant that when two magnetic poles come closer then the magnetic force of attraction or repulsion becomes higher?
 
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  • #5
vcsharp2003
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In the case of the paramegnet there can be no such labeling of the rod's ends. There is repulsion between each magnet pole and each rod end which get stronger when the end moves closer to the pole. Thus, if the rod turns away from being perpendicular to the field, the stronger repulsive force at each end of the rod will restore it back to being perpendicular to the field lines.
Did you mean in the case of diamagnetic rather than paramegnet?

If there's repulsion in this case, then can't we say that the rod end closer to N pole of external magnet acts like a N pole of the rod and therefore, they repel? You said we cannot label the ends of the rod as N or S, but it seems we can since repulsion happens only between like poles and so the rod end closer to N pole must be acting as a N pole.
 
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  • #6
haruspex
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when two magnetic poles come closer then the magnetic force of attraction or repulsion becomes higher?
Yes, and, roughly, according to an inverse square law. As with electric and gravitational fields, the field lines spread out in three dimensions. Since the field strength is proportional to the density of the field lines, it must fall off as the inverse square of distance from the source.
It is somewhat more complicated than in electrostatics since we do not have magnetic poles. But try treating each magnet as a dipole and see what you get for the minimum PE point. The whole system would look like
+====- -====+ +====-
Rotating the diamagnetic bar as in the book's diagram would make the central body shorter in my diagram, but the charges all stay the same.
 
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  • #7
vcsharp2003
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Rotating the diamagnetic bar as in the book's diagram would make the central body shorter in my diagram, but the charges all stay the same.
Ok, so even in the perpendicular position, if I applied the formula for PE of a magnetic dipole, I would get ## - | \vec M| |\vec B| cos 180 = -| \vec M| |\vec B| (-1)= | \vec M| |\vec B|##. Even in parallel position, the PE would be the same for a dipole.
 
  • #8
haruspex
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Ok, so even in the perpendicular position, if I applied the formula for PE of a magnetic dipole, I would get ## - | \vec M| |\vec B| cos 180 = -| \vec M| |\vec B| (-1)= | \vec M| |\vec B|##. Even in parallel position, the PE would be the same for a dipole.
The standard dipole formulae are approximations. They are derived from first principles of point charges. In the derivation, the first order terms cancel, so the dipole equation comes from the second order terms.
For the present problem the second order terms are likely to cancel too, so you would need to start from first principles again and keep the third order terms.
The diamagnetic effect is very weak.
 
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  • #9
kuruman
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Did you mean in the case of diamagnetic rather than paramegnet?
Yes, I meant diamagnet. I edited the post. Thanks for the catch.
If there's repulsion in this case, then can't we say that the rod end closer to N pole of external magnet acts like a N pole of the rod and therefore, they repel? You said we cannot label the ends of the rod as N or S, but it seems we can since repulsion happens only between like poles and so the rod end closer to N pole must be acting as a N pole.
Correct. And if the same end of the rod is closer to the S pole, it acts as a S pole. The other end would, of course, do the opposite. So each end is neither "N" nor "S". You cannot label the ends in the sense of painting letters "N" and "S" on them like you can do with bar magnets.
 
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