How much time does it take for a broken magnet to recover its poles?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I understand that when you cut a magnet you end up with 2 magnets but I wonder how much time does it take to the magnetic domains to rearange and form the new pole. I know the answer may vary depending on the size of the magnet, the material, and some other variable so I'm searching for an answer as general as possible and how the variables may affect the answer.
electromagnetism
$endgroup$
add a comment |
$begingroup$
I understand that when you cut a magnet you end up with 2 magnets but I wonder how much time does it take to the magnetic domains to rearange and form the new pole. I know the answer may vary depending on the size of the magnet, the material, and some other variable so I'm searching for an answer as general as possible and how the variables may affect the answer.
electromagnetism
$endgroup$
add a comment |
$begingroup$
I understand that when you cut a magnet you end up with 2 magnets but I wonder how much time does it take to the magnetic domains to rearange and form the new pole. I know the answer may vary depending on the size of the magnet, the material, and some other variable so I'm searching for an answer as general as possible and how the variables may affect the answer.
electromagnetism
$endgroup$
I understand that when you cut a magnet you end up with 2 magnets but I wonder how much time does it take to the magnetic domains to rearange and form the new pole. I know the answer may vary depending on the size of the magnet, the material, and some other variable so I'm searching for an answer as general as possible and how the variables may affect the answer.
electromagnetism
electromagnetism
asked Mar 4 at 1:08
Diego Rodríguez CidDiego Rodríguez Cid
914
914
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It takes zero time because no domains need to rearrange when a permanent magnet breaks in two. The spins in each half are still aligned and still produce a magnetic field.
The idea that magnets have “poles” is a misconception. There are no magnetic poles in nature, or at least none that we have found. (And physicists have looked hard for them.) This is the meaning of one of Maxwell’s equations,
$$nablacdotmathbfB=0.$$
The magnetic field lines of a magnet are loops than run through the interior of the magnet and then loop back around outside. The so-called “poles” are just where the field lines happen to emerge from the interior to the exterior, or return back inside. When you break a magnet, the field lines simply come out and go in in two new places, so that each half has its own loops and its own “poles”.
$endgroup$
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
add a comment |
$begingroup$
I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
$endgroup$
add a comment |
$begingroup$
The molecules that make up the magnet have a magnetic dipolar moment. You can think of them as small magnets aligned so that the total magnetic field is the sum of all the small magnets. If you cut a magnet in two, the two magnets are still made of aligned dipolar moments, so there is no rearrangement of poles. The two pieces will automatically be magnetized.
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f464256%2fhow-much-time-does-it-take-for-a-broken-magnet-to-recover-its-poles%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It takes zero time because no domains need to rearrange when a permanent magnet breaks in two. The spins in each half are still aligned and still produce a magnetic field.
The idea that magnets have “poles” is a misconception. There are no magnetic poles in nature, or at least none that we have found. (And physicists have looked hard for them.) This is the meaning of one of Maxwell’s equations,
$$nablacdotmathbfB=0.$$
The magnetic field lines of a magnet are loops than run through the interior of the magnet and then loop back around outside. The so-called “poles” are just where the field lines happen to emerge from the interior to the exterior, or return back inside. When you break a magnet, the field lines simply come out and go in in two new places, so that each half has its own loops and its own “poles”.
$endgroup$
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
add a comment |
$begingroup$
It takes zero time because no domains need to rearrange when a permanent magnet breaks in two. The spins in each half are still aligned and still produce a magnetic field.
The idea that magnets have “poles” is a misconception. There are no magnetic poles in nature, or at least none that we have found. (And physicists have looked hard for them.) This is the meaning of one of Maxwell’s equations,
$$nablacdotmathbfB=0.$$
The magnetic field lines of a magnet are loops than run through the interior of the magnet and then loop back around outside. The so-called “poles” are just where the field lines happen to emerge from the interior to the exterior, or return back inside. When you break a magnet, the field lines simply come out and go in in two new places, so that each half has its own loops and its own “poles”.
$endgroup$
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
add a comment |
$begingroup$
It takes zero time because no domains need to rearrange when a permanent magnet breaks in two. The spins in each half are still aligned and still produce a magnetic field.
The idea that magnets have “poles” is a misconception. There are no magnetic poles in nature, or at least none that we have found. (And physicists have looked hard for them.) This is the meaning of one of Maxwell’s equations,
$$nablacdotmathbfB=0.$$
The magnetic field lines of a magnet are loops than run through the interior of the magnet and then loop back around outside. The so-called “poles” are just where the field lines happen to emerge from the interior to the exterior, or return back inside. When you break a magnet, the field lines simply come out and go in in two new places, so that each half has its own loops and its own “poles”.
$endgroup$
It takes zero time because no domains need to rearrange when a permanent magnet breaks in two. The spins in each half are still aligned and still produce a magnetic field.
The idea that magnets have “poles” is a misconception. There are no magnetic poles in nature, or at least none that we have found. (And physicists have looked hard for them.) This is the meaning of one of Maxwell’s equations,
$$nablacdotmathbfB=0.$$
The magnetic field lines of a magnet are loops than run through the interior of the magnet and then loop back around outside. The so-called “poles” are just where the field lines happen to emerge from the interior to the exterior, or return back inside. When you break a magnet, the field lines simply come out and go in in two new places, so that each half has its own loops and its own “poles”.
edited Mar 4 at 1:39
answered Mar 4 at 1:20
G. SmithG. Smith
10.3k11429
10.3k11429
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
add a comment |
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
$begingroup$
Or, to use using the nomenclature "poles" for the places dense field emerges from the interior to the exterior, the field was running through the body of the magnet all along, so anywhere you break it both of the new ends will be places where dense field emerges...
$endgroup$
– dmckee♦
Mar 4 at 1:40
add a comment |
$begingroup$
I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
$endgroup$
add a comment |
$begingroup$
I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
$endgroup$
add a comment |
$begingroup$
I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
$endgroup$
I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
answered Mar 4 at 1:44
TechDroidTechDroid
70312
70312
add a comment |
add a comment |
$begingroup$
The molecules that make up the magnet have a magnetic dipolar moment. You can think of them as small magnets aligned so that the total magnetic field is the sum of all the small magnets. If you cut a magnet in two, the two magnets are still made of aligned dipolar moments, so there is no rearrangement of poles. The two pieces will automatically be magnetized.
$endgroup$
add a comment |
$begingroup$
The molecules that make up the magnet have a magnetic dipolar moment. You can think of them as small magnets aligned so that the total magnetic field is the sum of all the small magnets. If you cut a magnet in two, the two magnets are still made of aligned dipolar moments, so there is no rearrangement of poles. The two pieces will automatically be magnetized.
$endgroup$
add a comment |
$begingroup$
The molecules that make up the magnet have a magnetic dipolar moment. You can think of them as small magnets aligned so that the total magnetic field is the sum of all the small magnets. If you cut a magnet in two, the two magnets are still made of aligned dipolar moments, so there is no rearrangement of poles. The two pieces will automatically be magnetized.
$endgroup$
The molecules that make up the magnet have a magnetic dipolar moment. You can think of them as small magnets aligned so that the total magnetic field is the sum of all the small magnets. If you cut a magnet in two, the two magnets are still made of aligned dipolar moments, so there is no rearrangement of poles. The two pieces will automatically be magnetized.
answered Mar 4 at 8:40
TheAverageHijanoTheAverageHijano
4869
4869
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f464256%2fhow-much-time-does-it-take-for-a-broken-magnet-to-recover-its-poles%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown