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

[Planck const]

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

 

[ZealScience]

First of all, welcome to PF, the world of physics.

Definately 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?

 

[TROGLIO]

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

 

[aim1732]

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?

 

[ZealScience]

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.

 

[Planck const]

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!

 

[ZealScience]

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?

 

[TROGLIO]

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

 

[ZealScience]

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.

 

[aim1732]

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"]

 

[aim1732]

Edit::Repost.

 

[zorro]

Edit::Repost.

There is an option of deleting a post too

 

[aim1732]

There is an option of deleting a post too

And that is exercised only by moderators.They did,actually.Two posts are missing.