What is the magnetic field inside a cylinder with a hole?

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Homework Help Overview

The discussion revolves around determining the magnetic field inside a cylindrical conductor that contains a hole. The cylinder has a radius "a," the hole has a radius "b," and the distance from the center of the cylinder to the center of the hole is "d." The current density is denoted as "J," and the participants are questioning the validity of the solution presented in a textbook, which states that the magnetic field inside the hole is constant and equals 2PiJd/c.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Ampere's Law and question its validity in this scenario, suggesting that the lack of symmetry may render it ineffective. Some propose using the principle of superposition to analyze the magnetic field created by the currents in the cylinder and the hole. There are also inquiries about the area used in calculations and the meaning of "c" in the context of the problem.

Discussion Status

The conversation is ongoing, with various interpretations of the problem being explored. Some participants have offered alternative approaches, such as considering the effects of superposition and questioning the assumptions made about the current distribution. There is no explicit consensus on the correct method or interpretation at this stage.

Contextual Notes

Participants note the potential confusion regarding the area used in Ampere's Law and the implications of the hole in the cylinder. There is also mention of the lack of current passing through the hole, which raises questions about the applicability of certain laws in this context.

Planck const
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Homework Statement


First I have to say that my English isn't good:)

Ok..
There is a cylinder that at him, there is hole.
Cylinder radius is "a" and the hole radius is "b"
The distance from the cylinder center to the hole center is "d".
The current density is "J", in the hole there is nothing.
What is the field inside the hole?
Why the answer is 2PiJd/c in the book (CGS), why the field in it is constant(by heart the book) ?



Homework Equations



Amper rule?

The Attempt at a Solution


I can look in this quesion, like all the cylinder got +J and the hole - -J.
So I choose the positive direction in the direction of the J in hole.
I can use Amper rule:
B*P(hole) = mu0*J*A(hole)
So:
B*2pir=mu0*J*pir^2

B=mu0*Jr/2

Why the gave me a,d,b ? In the hole there is variable field!

Thanks in advance
 
Last edited:
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First of all, welcome to PF, the world of physics.

definitely Ampere's law I think. I am not quite sure about this question.

Is A really the area of the hole? Because current in Ampere's Law is the current passing through the area attached to the close loop. But what is c in the answer? c=confusion?
 
amperes law will not be valid in this case
revise your notes and you will see why
"fill" the hole with +current and equal -current
so the net current remains 0
now club the +current of as 1 single wire
and calculate its magnetic field at the desired point
and similarly for the -current
then add vectoricaly
i hope you get it
i find it difficult to explain
 
Last edited by a moderator:
Are you aware of the principle of superposition of fields?It follows directly from this principle that if a part of the source of the field is removed the field due to that part vanishes.So you can picture the hole in the wire as this: a part of the source of the field(the current) has been removed so the new field is actually the original filed minus the field of the current removed(of course vectorially).If you formulate the vector equation upto this point with the cross products and all you will see clearly why it comes out to be a constant.
Edit:
Ampere's Law is ALWAYS valid as long as you deal with classical physics.How do you mean that,Troglio?
 
TROGLIO said:
amperes law will not be valid in this case
revise your notes and you will see why
"fill" the hole with +current and equal -current
so the net current remains 0
now club the +current of as 1 single wire
and calculate its magnetic field at the desired point
and similarly for the -current
then add vectoricaly
i hope you get it
i find it difficult to explain

Then how to do it? Other than ampere's Law, what else can you use? Biot-Savart? I think you mean there is no current passing through the cross section of the hole. But in this case field is generated by current in the wire due to assymmetry of the wire, if they are coaxical cylinders there would not be any field. So the same question as OP's: why it's uniform? There is part where symmetry is valid but there is point where there is not.
 
Reply

Zeal - Thanks:)
Im not sure about the area.. Is it 2Pi*b*L...or 2Pib^2. I chose the secen - 2Pib^2 becuse I tought L (lenght) won't be reduced. c = velocity of light

Troglio - So which law will worked?!
about your idea... I will get +B and -B ? because its one dimension (the donations of the two currents are opposite..don't they?

Aim - I heard about superposition :)
So you say (I mean.. I say :))
I equal= A(of cylinder)J-A(of hole)J = 2PiLJ(a-b) or (As written above) PiJ(a^2-b^2)
Im not getting the solution that written in the book... if I use Amper rule.

Thanks about the replies!
 
So by c the answer implies that there is a square root of ε0μ0? But this doesn't make sense as there isn't any suggestion of using electric field here.

I can't work out the answer like that either and I don't think my answer is correct either. But at least I think that there is no current passing through the hole, so why use the cross section of the hole?
 
i can't reply in tech terms bcoz i dnt know much f them
so excuse m for that
i meant superposition
fill the hole with + and -
anperes law wouldn't work bcoz of lack of symetery
amperes law works some wat like gauss law in electro
 
TROGLIO said:
i can't reply in tech terms bcoz i dnt know much f them
so excuse m for that
i meant superposition
fill the hole with + and -
anperes law wouldn't work bcoz of lack of symetery
amperes law works some wat like gauss law in electro

Yes, they are much alike, and there is assymetry, so I think here we just use the current passint through the loop which has no symmetrical counterpart.
 
  • #10
anperes law wouldn't work bcoz of lack of symetery
amperes law works some wat like gauss law in electro
State the Ampere's Law clearly and also the reason you think makes it invalid.
[Please make a distinction between "The law works" and "The law is not valid"]
 
  • #11
Edit::Repost.
 
  • #12
aim1732 said:
Edit::Repost.

There is an option of deleting a post too :wink:
 
  • #13
There is an option of deleting a post too
And that is exercised only by moderators.They did,actually.Two posts are missing.
 

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