Apparent paradox with Ampère's law (bringing about questions with other laws)












2














Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.



I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.



Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.



To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.



This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.






electromagnetism







share|cite|improve this question





























    2














    Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.



    I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.



    Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.



    To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.



    This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.






    electromagnetism







    share|cite|improve this question



























      2












      2








      2







      Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.



      I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.



      Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.



      To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.



      This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.






      electromagnetism







      share|cite|improve this question















      Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.



      I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.



      Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.



      To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.



      This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.






      electromagnetism




      electromagnetism






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Peter Mortensen

      1,92811323




      1,92811323










      asked 9 hours ago









      Jake RoseJake Rose

      7218




      7218






















          2 Answers
          2






          active

          oldest

          votes


















          5














          You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.






          share|cite|improve this answer





















          • What about if you take a circularly symmetric path in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            8 hours ago










          • @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
            – Poon Levi
            8 hours ago



















          4














          Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.






          share|cite|improve this answer























          • What if you take a circularly symmetric loop in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            8 hours ago










          • @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
            – ZeroTheHero
            8 hours ago











          Your Answer





          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f452776%2fapparent-paradox-with-amp%25c3%25a8res-law-bringing-about-questions-with-other-laws%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5














          You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.






          share|cite|improve this answer





















          • What about if you take a circularly symmetric path in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            8 hours ago










          • @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
            – Poon Levi
            8 hours ago
















          5














          You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.






          share|cite|improve this answer





















          • What about if you take a circularly symmetric path in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            8 hours ago










          • @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
            – Poon Levi
            8 hours ago














          5












          5








          5






          You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.






          share|cite|improve this answer












          You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 9 hours ago









          garypgaryp

          16.6k12962




          16.6k12962












          • What about if you take a circularly symmetric path in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            8 hours ago










          • @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
            – Poon Levi
            8 hours ago


















          • What about if you take a circularly symmetric path in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            8 hours ago










          • @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
            – Poon Levi
            8 hours ago
















          What about if you take a circularly symmetric path in the solenoid?
          – Jake Rose
          8 hours ago




          What about if you take a circularly symmetric path in the solenoid?
          – Jake Rose
          8 hours ago












          Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
          – Jake Rose
          8 hours ago




          Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
          – Jake Rose
          8 hours ago












          @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
          – Poon Levi
          8 hours ago




          @JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
          – Poon Levi
          8 hours ago











          4














          Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.






          share|cite|improve this answer























          • What if you take a circularly symmetric loop in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            8 hours ago










          • @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
            – ZeroTheHero
            8 hours ago
















          4














          Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.






          share|cite|improve this answer























          • What if you take a circularly symmetric loop in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            8 hours ago










          • @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
            – ZeroTheHero
            8 hours ago














          4












          4








          4






          Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.






          share|cite|improve this answer














          Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 9 hours ago









          ZeroTheHeroZeroTheHero

          18.9k52956




          18.9k52956












          • What if you take a circularly symmetric loop in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            8 hours ago










          • @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
            – ZeroTheHero
            8 hours ago


















          • What if you take a circularly symmetric loop in the solenoid?
            – Jake Rose
            8 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            8 hours ago










          • @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
            – ZeroTheHero
            8 hours ago
















          What if you take a circularly symmetric loop in the solenoid?
          – Jake Rose
          8 hours ago




          What if you take a circularly symmetric loop in the solenoid?
          – Jake Rose
          8 hours ago












          Ahhh, the field is perpendicular, so B (in that direction) really does = 0
          – Jake Rose
          8 hours ago




          Ahhh, the field is perpendicular, so B (in that direction) really does = 0
          – Jake Rose
          8 hours ago












          @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
          – ZeroTheHero
          8 hours ago




          @JakeRose Yes... Ampere's law is always true, but it's now always useful in recovering $vec B$: only in some specialized symmetric situations can one do this (see physics.stackexchange.com/q/318183/36194).
          – ZeroTheHero
          8 hours ago


















          draft saved

          draft discarded




















































          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f452776%2fapparent-paradox-with-amp%25c3%25a8res-law-bringing-about-questions-with-other-laws%23new-answer', 'question_page');
          }
          );

          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







          -

          Popular posts from this blog

          Accessing regular linux commands in Huawei's Dopra Linux

          Image
          4 Backstory: I have a Huawei HG8245 router in my house and I want to change the default username/password. I've tried following this guide but the config file always gets reset. My ISP isn't helping either so I'm looking for other ways to accomplish this. The router has telnet access so this seems like a way to change the credentials. The first shell I gain access to looks like this: WAP> Pressing "?" gives me a list of available commands which I've pasted here, the two interesting ones are "su" and "shell". I can't seem to get su access, even after logging in as root (challenge password prompt, and then nothing happens), but shell gives me a proper linux shell. WAP>shell BusyBox v1.18.4 (2017-12-26 17:06:34 CST) built-in shell (ash) Enter 'hel...
          Read more

          Can't connect RFCOMM socket: Host is down

          Image
          up vote 3 down vote favorite I have bluez 5.4.3 and I am attempting to communicate with a ble device through minicom. These two deceives (my computer and a bluno nano) are alreadt paired and this has been my method so far. #bluetoothctl # pair xx:xx:xx:xx:xx:xx # exit Everything is working well so far then... #sudo rfcomm connect /dev/rfcomm0 xx:xx:xx:xx:xx:xx 1 & [1] 3665 Can't connect RFCOMM socket: Host is down Does anyone know how to resolve this issue? Any help would be very much appreciated. bluetooth bluez share | improve this question asked Feb 9 '17 at 18:46 Brandon Knape 46 2 ...
          Read more

          Kernel panic - not syncing: Fatal Exception in Interrupt

          Image
          0 Every day this error occured on my pc, kernel panic error on interrupt, what will be the problem? linux-mint share asked 6 mins ago Vinod Selvin Vinod Selvin 1 1 New contributor Vinod Selvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment ...
          Read more