Apparent paradox with amperes law (Brigning about questions with other laws)
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
add a comment |
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
add a comment |
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
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
edited 4 hours ago
Jake Rose
asked 5 hours ago
Jake RoseJake Rose
6718
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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.
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
add a comment |
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.
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
add a comment |
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2 Answers
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2 Answers
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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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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