Getting incorrect results applying Ampere's law
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
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 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
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 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
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
edited yesterday
knzhou
42.6k11117205
42.6k11117205
asked 2 days ago
Jake RoseJake Rose
10219
10219
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3 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
yesterday
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
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, $ointvec Bcdot dvec ell$ 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
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
@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
yesterday
add a comment |
You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.
Here are two examples to show that even if there is a magnetic field present the line integral is zero.
In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$
The right hand case is a little trickier.
$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$
and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$
Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.
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3 Answers
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3 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
yesterday
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
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
yesterday
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
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 2 days ago
garypgaryp
16.7k12963
16.7k12963
What about if you take a circularly symmetric path in the solenoid?
– Jake Rose
yesterday
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
add a comment |
What about if you take a circularly symmetric path in the solenoid?
– Jake Rose
yesterday
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
What about if you take a circularly symmetric path in the solenoid?
– Jake Rose
yesterday
What about if you take a circularly symmetric path in the solenoid?
– Jake Rose
yesterday
2
2
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0..
– Jake Rose
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
@JakeRose In that case $vec{B}$ is perpendicular to $dvec{l}$, so the integral is 0 (without the field being 0)
– Poon Levi
yesterday
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, $ointvec Bcdot dvec ell$ 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
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
@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
yesterday
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, $ointvec Bcdot dvec ell$ 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
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
@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
yesterday
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, $ointvec Bcdot dvec ell$ 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, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.
edited yesterday
answered 2 days ago
ZeroTheHeroZeroTheHero
19k52957
19k52957
What if you take a circularly symmetric loop in the solenoid?
– Jake Rose
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
@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
yesterday
add a comment |
What if you take a circularly symmetric loop in the solenoid?
– Jake Rose
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
@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
yesterday
What if you take a circularly symmetric loop in the solenoid?
– Jake Rose
yesterday
What if you take a circularly symmetric loop in the solenoid?
– Jake Rose
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
Ahhh, the field is perpendicular, so B (in that direction) really does = 0
– Jake Rose
yesterday
2
2
@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
yesterday
@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
yesterday
add a comment |
You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.
Here are two examples to show that even if there is a magnetic field present the line integral is zero.
In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$
The right hand case is a little trickier.
$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$
and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$
Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.
add a comment |
You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.
Here are two examples to show that even if there is a magnetic field present the line integral is zero.
In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$
The right hand case is a little trickier.
$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$
and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$
Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.
add a comment |
You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.
Here are two examples to show that even if there is a magnetic field present the line integral is zero.
In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$
The right hand case is a little trickier.
$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$
and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$
Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.
You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.
Here are two examples to show that even if there is a magnetic field present the line integral is zero.
In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$
The right hand case is a little trickier.
$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$
and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$
Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.
answered yesterday
FarcherFarcher
47.7k33796
47.7k33796
add a comment |
add a comment |
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