Mixing
Would a compass with unmagnetized needle work?
We know that the needle that is used in a compass is a permanently magnetized ferromagnetic material and commonly steel is used.
If we used an unmagnetized iron needle instead, would it still align with Earth's magnetic field lines? If yes, how?
magnetic-fields ferromagnetism
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
|
show 5 more comments
We know that the needle that is used in a compass is a permanently magnetized ferromagnetic material and commonly steel is used.
If we used an unmagnetized iron needle instead, would it still align with Earth's magnetic field lines? If yes, how?
magnetic-fields ferromagnetism
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
2
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
17
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
6
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
2
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
2
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49
|
show 5 more comments
We know that the needle that is used in a compass is a permanently magnetized ferromagnetic material and commonly steel is used.
If we used an unmagnetized iron needle instead, would it still align with Earth's magnetic field lines? If yes, how?
magnetic-fields ferromagnetism
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
We know that the needle that is used in a compass is a permanently magnetized ferromagnetic material and commonly steel is used.
If we used an unmagnetized iron needle instead, would it still align with Earth's magnetic field lines? If yes, how?
magnetic-fields ferromagnetism
magnetic-fields ferromagnetism
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
asked Dec 31 '18 at 15:27
physicsguy19physicsguy19
805219
805219
2
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
17
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
6
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
2
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
2
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49
|
show 5 more comments
2
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
17
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
6
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
2
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
2
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49
2
2
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
17
17
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
6
6
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
2
2
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
2
2
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49
|
show 5 more comments
4 Answers
4
active
oldest
votes
For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field.
If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the domains is zero. If you apply a strong enough external field, the domains will align to the field. The following image is from the Wikipedia page on magnetic domains (https://en.wikipedia.org/wiki/Magnetic_domain).
Then the question is whether the Earth's magnetic field is strong enough to realign the domains in an iron needle. The Landau Free Energy is used to determine this, as the domains will align in whatever way minimizes this energy. Parameters that determine this energy include things like: size and shape of the needle, material (in this case iron), and external field strength.
If the external field is strong enough to cause magnetization, the direction of the induced magnetization will be in a direction that minimizes the anisotropy energy and is pre-determined by the dimensions of the needle. The dimensions give rise to an "easy" axis, meaning the free energy is lowest when the magnetization is in a particular direction. In general, this axis could be in-plane in the x or y direction, or perpendicular to the needle in the z direction. In a graph of energy versus angle of the magnetization from the easy axis, there will be two energy minima: one along the easy axis, and another at 180 degrees (still along the easy axis, just pointed in the opposite direction).
Anyway, I haven't done the calculation, but I don't think the Earth's field is strong enough to cause realignment of the domains. I would also like to mention that once the needle has been magnetized, if you remove the external field, the
needle will keep its magnetization. It would take the addition of a lot of energy to reorient the domains/magnetization that could come from a new external field, or even thermal energy.
Edit: If your question is more about the torque that the needle would experience, it would follow the following equation assuming it was indeed magnetized:
$boldsymbol{tau}=mathbf{m}timesmathbf{B}$.
$bf{m}$ is the magnetic moment and is related to magnetization, $bf{M}$, by:
$$mathbf{m}=iiintmathbf{M} mathrm{d}V$$
For more information, here are some resources:
https://en.wikipedia.org/wiki/Magnetic_domain
Magnetism and Magnetic Materials by J.M.D. Coey. Sections on Landau Free Energy, magnetic moment, and maybe even the Stoner-Wohlfarth model would be enlightening.
https://en.wikipedia.org/wiki/Magnetic_moment
edited Jan 3 at 15:35
Chair
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
|
show 5 more comments
A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field.
However since the torque which was applied on the iron bar would be very small the chances are that there would not be an alignment even if you waited a long time.
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
add a comment |
Probably yes, in a careful experiment.
A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field.
Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker field. A needle or plate would likely align itself parallel to the field.
Shape anisotropy ensures that the magnetization is mostly parallel with long axes.
One experimental problem may be to rule out the effect of possible areas with remanence. So a careful degaussing would be necessary.
answered Jan 2 at 1:08
PieterPieter
7,66631431
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
|
show 1 more comment
There is a way of making a compass by floating the needle on a leaf in a water.
Weather it works is to be verified in practice. In a 'hard' water with small enough needle this could be achieved even without the leaf.
answered Jan 2 at 1:33
BartusBartus
113
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
add a comment |
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4 Answers
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4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
votes
For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field.
If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the domains is zero. If you apply a strong enough external field, the domains will align to the field. The following image is from the Wikipedia page on magnetic domains (https://en.wikipedia.org/wiki/Magnetic_domain).
Then the question is whether the Earth's magnetic field is strong enough to realign the domains in an iron needle. The Landau Free Energy is used to determine this, as the domains will align in whatever way minimizes this energy. Parameters that determine this energy include things like: size and shape of the needle, material (in this case iron), and external field strength.
If the external field is strong enough to cause magnetization, the direction of the induced magnetization will be in a direction that minimizes the anisotropy energy and is pre-determined by the dimensions of the needle. The dimensions give rise to an "easy" axis, meaning the free energy is lowest when the magnetization is in a particular direction. In general, this axis could be in-plane in the x or y direction, or perpendicular to the needle in the z direction. In a graph of energy versus angle of the magnetization from the easy axis, there will be two energy minima: one along the easy axis, and another at 180 degrees (still along the easy axis, just pointed in the opposite direction).
Anyway, I haven't done the calculation, but I don't think the Earth's field is strong enough to cause realignment of the domains. I would also like to mention that once the needle has been magnetized, if you remove the external field, the
needle will keep its magnetization. It would take the addition of a lot of energy to reorient the domains/magnetization that could come from a new external field, or even thermal energy.
Edit: If your question is more about the torque that the needle would experience, it would follow the following equation assuming it was indeed magnetized:
$boldsymbol{tau}=mathbf{m}timesmathbf{B}$.
$bf{m}$ is the magnetic moment and is related to magnetization, $bf{M}$, by:
$$mathbf{m}=iiintmathbf{M} mathrm{d}V$$
For more information, here are some resources:
https://en.wikipedia.org/wiki/Magnetic_domain
Magnetism and Magnetic Materials by J.M.D. Coey. Sections on Landau Free Energy, magnetic moment, and maybe even the Stoner-Wohlfarth model would be enlightening.
https://en.wikipedia.org/wiki/Magnetic_moment
edited Jan 3 at 15:35
Chair
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
|
show 5 more comments
For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field.
If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the domains is zero. If you apply a strong enough external field, the domains will align to the field. The following image is from the Wikipedia page on magnetic domains (https://en.wikipedia.org/wiki/Magnetic_domain).
Then the question is whether the Earth's magnetic field is strong enough to realign the domains in an iron needle. The Landau Free Energy is used to determine this, as the domains will align in whatever way minimizes this energy. Parameters that determine this energy include things like: size and shape of the needle, material (in this case iron), and external field strength.
If the external field is strong enough to cause magnetization, the direction of the induced magnetization will be in a direction that minimizes the anisotropy energy and is pre-determined by the dimensions of the needle. The dimensions give rise to an "easy" axis, meaning the free energy is lowest when the magnetization is in a particular direction. In general, this axis could be in-plane in the x or y direction, or perpendicular to the needle in the z direction. In a graph of energy versus angle of the magnetization from the easy axis, there will be two energy minima: one along the easy axis, and another at 180 degrees (still along the easy axis, just pointed in the opposite direction).
Anyway, I haven't done the calculation, but I don't think the Earth's field is strong enough to cause realignment of the domains. I would also like to mention that once the needle has been magnetized, if you remove the external field, the
needle will keep its magnetization. It would take the addition of a lot of energy to reorient the domains/magnetization that could come from a new external field, or even thermal energy.
Edit: If your question is more about the torque that the needle would experience, it would follow the following equation assuming it was indeed magnetized:
$boldsymbol{tau}=mathbf{m}timesmathbf{B}$.
$bf{m}$ is the magnetic moment and is related to magnetization, $bf{M}$, by:
$$mathbf{m}=iiintmathbf{M} mathrm{d}V$$
For more information, here are some resources:
https://en.wikipedia.org/wiki/Magnetic_domain
Magnetism and Magnetic Materials by J.M.D. Coey. Sections on Landau Free Energy, magnetic moment, and maybe even the Stoner-Wohlfarth model would be enlightening.
https://en.wikipedia.org/wiki/Magnetic_moment
edited Jan 3 at 15:35
Chair
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
|
show 5 more comments
For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field.
If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the domains is zero. If you apply a strong enough external field, the domains will align to the field. The following image is from the Wikipedia page on magnetic domains (https://en.wikipedia.org/wiki/Magnetic_domain).
Then the question is whether the Earth's magnetic field is strong enough to realign the domains in an iron needle. The Landau Free Energy is used to determine this, as the domains will align in whatever way minimizes this energy. Parameters that determine this energy include things like: size and shape of the needle, material (in this case iron), and external field strength.
If the external field is strong enough to cause magnetization, the direction of the induced magnetization will be in a direction that minimizes the anisotropy energy and is pre-determined by the dimensions of the needle. The dimensions give rise to an "easy" axis, meaning the free energy is lowest when the magnetization is in a particular direction. In general, this axis could be in-plane in the x or y direction, or perpendicular to the needle in the z direction. In a graph of energy versus angle of the magnetization from the easy axis, there will be two energy minima: one along the easy axis, and another at 180 degrees (still along the easy axis, just pointed in the opposite direction).
Anyway, I haven't done the calculation, but I don't think the Earth's field is strong enough to cause realignment of the domains. I would also like to mention that once the needle has been magnetized, if you remove the external field, the
needle will keep its magnetization. It would take the addition of a lot of energy to reorient the domains/magnetization that could come from a new external field, or even thermal energy.
Edit: If your question is more about the torque that the needle would experience, it would follow the following equation assuming it was indeed magnetized:
$boldsymbol{tau}=mathbf{m}timesmathbf{B}$.
$bf{m}$ is the magnetic moment and is related to magnetization, $bf{M}$, by:
$$mathbf{m}=iiintmathbf{M} mathrm{d}V$$
For more information, here are some resources:
https://en.wikipedia.org/wiki/Magnetic_domain
Magnetism and Magnetic Materials by J.M.D. Coey. Sections on Landau Free Energy, magnetic moment, and maybe even the Stoner-Wohlfarth model would be enlightening.
https://en.wikipedia.org/wiki/Magnetic_moment
edited Jan 3 at 15:35
Chair
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field.
If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the domains is zero. If you apply a strong enough external field, the domains will align to the field. The following image is from the Wikipedia page on magnetic domains (https://en.wikipedia.org/wiki/Magnetic_domain).
Then the question is whether the Earth's magnetic field is strong enough to realign the domains in an iron needle. The Landau Free Energy is used to determine this, as the domains will align in whatever way minimizes this energy. Parameters that determine this energy include things like: size and shape of the needle, material (in this case iron), and external field strength.
If the external field is strong enough to cause magnetization, the direction of the induced magnetization will be in a direction that minimizes the anisotropy energy and is pre-determined by the dimensions of the needle. The dimensions give rise to an "easy" axis, meaning the free energy is lowest when the magnetization is in a particular direction. In general, this axis could be in-plane in the x or y direction, or perpendicular to the needle in the z direction. In a graph of energy versus angle of the magnetization from the easy axis, there will be two energy minima: one along the easy axis, and another at 180 degrees (still along the easy axis, just pointed in the opposite direction).
Anyway, I haven't done the calculation, but I don't think the Earth's field is strong enough to cause realignment of the domains. I would also like to mention that once the needle has been magnetized, if you remove the external field, the
needle will keep its magnetization. It would take the addition of a lot of energy to reorient the domains/magnetization that could come from a new external field, or even thermal energy.
Edit: If your question is more about the torque that the needle would experience, it would follow the following equation assuming it was indeed magnetized:
$boldsymbol{tau}=mathbf{m}timesmathbf{B}$.
$bf{m}$ is the magnetic moment and is related to magnetization, $bf{M}$, by:
$$mathbf{m}=iiintmathbf{M} mathrm{d}V$$
For more information, here are some resources:
https://en.wikipedia.org/wiki/Magnetic_domain
Magnetism and Magnetic Materials by J.M.D. Coey. Sections on Landau Free Energy, magnetic moment, and maybe even the Stoner-Wohlfarth model would be enlightening.
https://en.wikipedia.org/wiki/Magnetic_moment
edited Jan 3 at 15:35
Chair
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
edited Jan 3 at 15:35
Chair
3,66972034
edited Jan 3 at 15:35
Chair
3,66972034
edited Jan 3 at 15:35
Chair
3,66972034
3,66972034
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
answered Dec 31 '18 at 19:03
PhysikaPhysika
334110
334110
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
|
show 5 more comments
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
5
5
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
I learned some new things from this answer, +1. I've edited my answer to take this info into account and to point to this answer for more details.
– Ben Crowell
Dec 31 '18 at 19:26
2
2
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
Have you looked at references like this or this?
– user5713492
Jan 1 at 1:07
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
I haven't looked at these things in detail, but I have been aware of these concepts. I will be sure to give these a read! However, iron is ferromagnetic, not dia- or paramagnetic.
– Physika
Jan 1 at 1:48
1
1
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
However, if you melt down your needle and let it cool in a needle-shaped cast, it will end up weakly magnetized and want the point the same direction it was pointing when cooled. This is Geology at this point; observed on the seafloor.
– Joshua
Jan 1 at 22:34
1
1
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
There is a portion of the answer at the end that talks about the torque on a magnetized needle. This doesn't seem relevant to me, since the question asks about an unmagnetized needle. Based on the two links provided by @user5713492, it seems likely that the energy of an unmagnetized iron needle will be a non-constant function of orientation. However, symmetry considerations would seem to prevent it from being $Uproptocostheta$ as in a dipole. An unmagnetized object can never distinguish the orientations parallel and antiparallel to the field.
– Ben Crowell
Jan 2 at 15:54
|
show 5 more comments
A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field.
However since the torque which was applied on the iron bar would be very small the chances are that there would not be an alignment even if you waited a long time.
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
add a comment |
A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field.
However since the torque which was applied on the iron bar would be very small the chances are that there would not be an alignment even if you waited a long time.
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
add a comment |
A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field.
However since the torque which was applied on the iron bar would be very small the chances are that there would not be an alignment even if you waited a long time.
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field.
However since the torque which was applied on the iron bar would be very small the chances are that there would not be an alignment even if you waited a long time.
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
answered Dec 31 '18 at 16:04
FarcherFarcher
47.8k33796
47.8k33796
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
add a comment |
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
2
2
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
In the first paragraph, you seem to be assuming that the induced magnetization would be along the length of the bar, but I don't see any reason why that would be true. It could be in any random orientation relative to the bar's long dimension. The effect would then be that the bar would experience a torque that would tend to return it to whatever orientation it had when it was first able to be magnetized (e.g., at the time when the iron was first cooled below the Fermi temperature). You're then describing a bar that is magnetized, which is contrary to the question.
– Ben Crowell
Dec 31 '18 at 17:51
2
2
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
@BenCrowell I have assumed that the (soft) iron needle was initially unmagnetised and then has magnetism induced in It when placed in the Earth’s magnetic field. Where I think we disagree is that I assumed that magnetic poles will be induced near the ends of the iron needle whereas you do not? My assumption is based on my experience when an iron rod is suspended in a magnetic field much stronger than that of the Earth eg between the pole pieces of a horse shoe magnet.
– Farcher
Dec 31 '18 at 18:48
add a comment |
Probably yes, in a careful experiment.
A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field.
Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker field. A needle or plate would likely align itself parallel to the field.
Shape anisotropy ensures that the magnetization is mostly parallel with long axes.
One experimental problem may be to rule out the effect of possible areas with remanence. So a careful degaussing would be necessary.
answered Jan 2 at 1:08
PieterPieter
7,66631431
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
|
show 1 more comment
Probably yes, in a careful experiment.
A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field.
Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker field. A needle or plate would likely align itself parallel to the field.
Shape anisotropy ensures that the magnetization is mostly parallel with long axes.
One experimental problem may be to rule out the effect of possible areas with remanence. So a careful degaussing would be necessary.
answered Jan 2 at 1:08
PieterPieter
7,66631431
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
|
show 1 more comment
Probably yes, in a careful experiment.
A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field.
Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker field. A needle or plate would likely align itself parallel to the field.
Shape anisotropy ensures that the magnetization is mostly parallel with long axes.
One experimental problem may be to rule out the effect of possible areas with remanence. So a careful degaussing would be necessary.
answered Jan 2 at 1:08
PieterPieter
7,66631431
Probably yes, in a careful experiment.
A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field.
Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker field. A needle or plate would likely align itself parallel to the field.
Shape anisotropy ensures that the magnetization is mostly parallel with long axes.
One experimental problem may be to rule out the effect of possible areas with remanence. So a careful degaussing would be necessary.
answered Jan 2 at 1:08
PieterPieter
7,66631431
edited Jan 2 at 17:57
answered Jan 2 at 1:08
PieterPieter
7,66631431
answered Jan 2 at 1:08
PieterPieter
7,66631431
answered Jan 2 at 1:08
PieterPieter
7,66631431
7,66631431
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
|
show 1 more comment
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
This seems at least superficially to be in contradiction with Physika's answer. Are you describing an effect that occurs in a ferromagnetic material without any permanent magnetization? Could you point us to any info about the effects you're describing? A paramagnetic material like aluminum will align with a strong magnetic field Here are you talking about something needle-shaped, as opposed to plate-shaped? Clearly it's not possible to use this quite like a compass, since it can't tell north from south.
– Ben Crowell
Jan 2 at 1:23
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field.
– Pieter
Jan 2 at 8:07
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
@BenCrowell It is better to use metal plates in paramagnetic-versus-diamagnetic demonstrations because there is more magnetic damping by vortex currents. It is a classic classroom demonstrations, reference were given in a comment on Physica's post by user5713492.
– Pieter
Jan 2 at 8:23
1
1
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
I agree with Pieter. This is the best answer so far, although incomplete.
– t t t t
Jan 2 at 11:40
1
1
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
I would disagree with Physika that a small field could not move domain walls. We use a pipe of weak iron to shield the Earth's field. Doesn't that work simply because the pipe has a permeability, not because it's ferromagnetic?
– Ben Crowell
Jan 3 at 14:23
|
show 1 more comment
There is a way of making a compass by floating the needle on a leaf in a water.
Weather it works is to be verified in practice. In a 'hard' water with small enough needle this could be achieved even without the leaf.
answered Jan 2 at 1:33
BartusBartus
113
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
add a comment |
There is a way of making a compass by floating the needle on a leaf in a water.
Weather it works is to be verified in practice. In a 'hard' water with small enough needle this could be achieved even without the leaf.
answered Jan 2 at 1:33
BartusBartus
113
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
add a comment |
There is a way of making a compass by floating the needle on a leaf in a water.
Weather it works is to be verified in practice. In a 'hard' water with small enough needle this could be achieved even without the leaf.
answered Jan 2 at 1:33
BartusBartus
113
There is a way of making a compass by floating the needle on a leaf in a water.
Weather it works is to be verified in practice. In a 'hard' water with small enough needle this could be achieved even without the leaf.
answered Jan 2 at 1:33
BartusBartus
113
answered Jan 2 at 1:33
BartusBartus
113
answered Jan 2 at 1:33
BartusBartus
113
answered Jan 2 at 1:33
BartusBartus
113
113
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
add a comment |
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
This would be with a magnetized needle. The OP is asking about an unmagnetized needle.
– Ben Crowell
Jan 2 at 15:35
add a comment |
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2
Why would you think “yes”?
– ZeroTheHero
Dec 31 '18 at 15:31
17
@ZeroTheHero: Probably because they know that even an unmagnetised hunk of iron will still be attracted to a magnet, and they're applying this knowledge to a specific application of magnets.
– Sean
Dec 31 '18 at 18:03
6
@ZeroTheHero: Why would you think "no"? If you sprinkle (unmagnetized) iron fillings on a transparency on top of a magnet, you will see them align along the field lines. So you should see the same effect with a sufficiently small 'compass' needle that is free to rotate with sufficiently low friction.
– user21820
Jan 1 at 13:30
2
This is a great example of a seemingly naive question that is much deeper than it appears on the surface.
– Ben Crowell
Jan 2 at 1:24
2
related: physics.stackexchange.com/questions/451886/…
– Ben Crowell
Jan 3 at 14:49