Magnetic field generated by an infinitely long current-carrying wire

  • #1
annamal
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Homework Statement:
Use Ampère’s law to calculate the magnetic field due to a steady current I in an infinitely long, thin, straight wire as shown in the image
Relevant Equations:
##\oint_{line} B\cdot dl##
Can someone explain how there can be a radial magnetic field? I thought the magnetic field was always tangent to the circle using the right hand rule where you wrap your fingers around the current and point your thumb in the direction of the current.
Screen Shot 2022-05-12 at 10.03.39 PM.png
 

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  • #2
hutchphd
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And so what is the value of ##B_r## ?
 
  • #3
annamal
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And so what is the value of ##B_r## ?
Idk. I am asking what is ##B_r##, the radial magnetic field.
 
  • #4
bob012345
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Idk. I am asking what is ##B_r##, the radial magnetic field.
Don't interpret the picture as giving the answer. The picture is testing your understanding. You have to figure out what the field looks like by using Ampère’s law.
 
  • #5
Orodruin
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I strongly suspect the idea is not to compute ##B_r## from Ampere’s law but to use symmetry arguments to deduce things about it. The real question is: how can you use Ampere’s law and symmetry arguments to find the magnetic field?
 
  • #6
annamal
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I strongly suspect the idea is not to compute ##B_r## from Ampere’s law but to use symmetry arguments to deduce things about it. The real question is: how can you use Ampere’s law and symmetry arguments to find the magnetic field?
My question is how can there be a ##B_r##?
 
  • #7
haruspex
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My question is how can there be a ##B_r##?
If there is no ##B_r## then your task is to show that using Ampères law and symmetry arguments.
 
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  • #8
Orodruin
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As an aside, but a highly relevant one:

Figures should be and are often drawn in a general fashion. There is no real need for the figure to necessarily accurately depict the result. For example, you can draw a free-body diagram with a force pointing in a certain direction and then consider the equilibrium equations based on that. If the force then turns out to be negative then it points in the opposite direction. If it turns out to be zero, then it is zero. It is the same thing here (although the problem author has cut you some slack by indicating that ##B_r## and ##B_\theta## have constant magnitude).

The right-hand rule that you are referring to is the result of Ampere’s law. You are supposed to show this here (and you may use symmetry arguments to help you along).
 
  • #9
Delta2
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This is interesting, symmetry alone can't exclude a radial field, ampere's law tell us that it can be any function of r because it gets simplified to zero in the dot product. Should we choose a proper amperian loop in order to prove that ##B_r=0##??
 
  • #10
Orodruin
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This is interesting, symmetry alone can't exclude a radial field
Yes it can.
 
  • #11
Delta2
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Yes it can.
Hmm all i can infer is that in the center will be zero , in infinity also will be zero but what about in between, it can be any function of r as i said, can't it be?
 
  • #12
Orodruin
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Hmm all i can infer is that in the center will be zero , in infinity also will be zero but what about in between, it can be any function of r as i said, can't it be?
Which symmetry are you trying to use? This system has several. Although perhaps take this conversation by DM in order to allow OP to consider this themselves?
 
  • #13
Delta2
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yes let's PM I find this something like a head cracker at least for me.
 
  • #14
alan123hk
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I just thought of a logical reasoning method based on Ampere's law to prove that ##B_r## is equal to 0 everywhere, which is convincing enough for me. :smile:
 
  • #15
Delta2
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I just thought of a logical reasoning method based on Ampere's law to prove that ##B_r## is equal to 0 everywhere, which is convincing enough for me. :smile:
Any hints? It doesn't use symmetry arguments at all ? Only Ampere's law? In integral or differential form?
 
  • #16
alan123hk
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Any hints? It doesn't use symmetry arguments at all ? Only Ampere's law? In integral or differential form?

I use Ampère's law in integral form and need to imagine a complete integral curve, starting at the origin, not necessarily extending to infinite distance, and then applying the principle of symmetry...
 
  • #17
Delta2
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I can't understand very well the symmetry arguments, for me the only clear and safe way is via the integral of the Biot Savart law, can prove that both ##B_r## and ##B_z## are zero.
 
  • #18
hutchphd
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I use Ampère's law in integral form and need to imagine a complete integral curve, starting at the origin, not necessarily extending to infinite distance, and then applying the principle of symmetry...
Yes it can.

I am left scratching my head...I don't see how this can be done using only Ampere and symmetry. Please elucidate at the proper time!

!
 
  • #19
Delta2
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I am left scratching my head...I don't see how this can be done using only Ampere and symmetry. Please elucidate at the proper time!

!
I have a private conversation with @Orodruin where he attempts to explain it in more detail, how do I invite you to that conversation?
 
  • #20
bob012345
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I have a private conversation with @Orodruin where he attempts to explain it in more detail, how do I invite you to that conversation?
Look under Conversation Info near the top right of the conversation page. There is a link called invite more.
 
  • #21
Delta2
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Look under Conversation Info near the top right of the conversation page. There is a link called invite more.
Thanks I had already found it.
 
  • #22
alan123hk
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I can't understand very well the symmetry arguments, for me the only clear and safe way is via the integral of the Biot Savart law, can prove that both Br and Bz are zero.
I am left scratching my head...I don't see how this can be done using only Ampere and symmetry. Please elucidate at the proper time
I am sorry for my carelessness and recklessness. I thought I had successfully proven this, but on second thought I realized I was wrong. Now my personal opinion is that it is impossible to prove this using just the simple integral form of Ampère's law and the principle of symmetry. 😞
 
  • #23
Orodruin
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The problem with using the integral form to get at the radial component is that any vector field on the form ##f(r)\vec e_r## is conservative and therefore contributes a net zero to any closed loop integral.

What should be used are the transformation properties of the magnetic field and current source.
 
  • #24
Orodruin
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Now that I think about it, there may be an argument using a surface and Ampere’s law on integral form, but it is not obvious.

Edit: Yes. There definitely is such an argument. It is not directly obvious and I prefer the symmetry argument though.
 
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  • #25
vela
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Can someone explain how there can be a radial magnetic field? I thought the magnetic field was always tangent to the circle using the right hand rule where you wrap your fingers around the current and point your thumb in the direction of the current.
Did you read the caption on the figure in the textbook? (bolding mine)

Figure 12.15 The possible components of the magnetic field B due to a current I, which is directed out of the page. The radial component is zero because the angle between the magnetic field and the path is at a right angle.​
 
  • #26
vela
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Now that I think about it, there may be an argument using a surface and Ampere’s law on integral form, but it is not obvious.

Edit: Yes. There definitely is such an argument. It is not directly obvious and I prefer the symmetry argument though.
The book makes a straightforward argument from symmetry and Gauss's law for magnetic fields.
 
  • #27
Orodruin
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Figure 12.15 The possible components of the magnetic field B due to a current I, which is directed out of the page. The radial component is zero because the angle between the magnetic field and the path is at a right angle.
There seems to be some context missing here. What path?
 
  • #28
Orodruin
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The book makes a straightforward argument from symmetry and Gauss's law for magnetic fields.
You can do that, yes. What I was referring to was an argument using Ampere’s law on integral form, which is also possible.
 
  • #29
vela
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There seems to be some context missing here. What path?
The circular one depicted in the figure @annamal posted in the first post. Apparently, I didn't read that sentence very carefully (I stopped at "radial component is zero"); it looks like they left a few words out.
 
  • #30
bob012345
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You can do that, yes. What I was referring to was an argument using Ampere’s law on integral form, which is also possible.
Please give us that argument? I would be interested. Thanks.
 
  • #31
Orodruin
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Please give us that argument? I would be interested. Thanks.
I have purposefully not written out those arguments since this is the homework forums and may be part of what the OP needs to do. As @vela says, Gauss’ law for magnetic fields also works and I still prefer the pure symmetry arguments.

Since there seems to be some popular demand for this though, I might summarize in a PF Insight in the future.
 
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  • #32
hutchphd
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Figure 12.15 The possible components of the magnetic field B due to a current I, which is directed out of the page. The radial component is zero because the angle between the magnetic field and the path is at a right angle.
I assume they mean the radial contribution to the line integral is zero. The fact that the dot product is zero does not directly imply that the radial field is zero. This is badly stated. What is the text?
 
  • #33
vela
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I assume they mean the radial contribution to the line integral is zero. The fact that the dot product is zero does not directly imply that the radial field is zero.
Yeah, that's what I realized when I read the sentence more carefully after first posting it. In the discussion below the figure in the OpenStax textbook, it's explained how to deduce the radial component vanishes.
 
  • #34
Orodruin
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You can do that, yes. What I was referring to was an argument using Ampere’s law on integral form, which is also possible.
Actually, scratch that. I found a hole in that argument. Pure symmetry or using Gauss law for magnetic fields is the way to go.
 
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  • #35
annamal
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I assume they mean the radial contribution to the line integral is zero. The fact that the dot product is zero does not directly imply that the radial field is zero. This is badly stated. What is the text?
I still don't understand how you can have a radial component. I thought the magnetic field was always tangent to a circle around the current.
 

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