Apparent paradox with amperes law (Brigning about questions with other laws)












1














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 amperes 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.






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    1














    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 amperes 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



























      1












      1








      1







      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 amperes 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 amperes 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






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      edited 4 hours ago







      Jake Rose

















      asked 5 hours ago









      Jake RoseJake Rose

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          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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            4 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
            4 hours ago



















          3














          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, $vec Bcdot dvec ell$ 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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            4 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
            4 hours ago











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

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          4














          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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            4 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
            4 hours ago
















          4














          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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            4 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
            4 hours ago














          4












          4








          4






          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 4 hours ago









          garypgaryp

          16.6k12962




          16.6k12962












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










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            4 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
            4 hours ago


















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










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
            – Jake Rose
            4 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
            4 hours ago
















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




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












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




          Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
          – Jake Rose
          4 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
          4 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
          4 hours ago











          3














          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, $vec Bcdot dvec ell$ 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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            4 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
            4 hours ago
















          3














          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, $vec Bcdot dvec ell$ 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
            4 hours ago










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            4 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
            4 hours ago














          3












          3








          3






          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, $vec Bcdot dvec ell$ 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, $vec Bcdot dvec ell$ 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










          answered 4 hours ago









          ZeroTheHeroZeroTheHero

          18.9k52956




          18.9k52956












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










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            4 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
            4 hours ago


















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










          • Ahhh, the field is perpendicular, so B (in that direction) really does = 0
            – Jake Rose
            4 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
            4 hours ago
















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




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












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




          Ahhh, the field is perpendicular, so B (in that direction) really does = 0
          – Jake Rose
          4 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
          4 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
          4 hours ago


















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